A Polyhedral Study on Unit Commitment with a Single Type of Binary Variables

A Polyhedral Study on Unit Commitment with a Single T ype of Binary V ariables Bin T ian a , Kai Pan a, ∗ , Chung-Lun Li b a Faculty of Business, The Hong Kong Polytechnic University , Kowloon, Hong Kong b College of Business, City University of Hong Kong, Kowloon, Hong Kong ∗ Corresponding author Contact: bin01.tian@connect.polyu.hk (BT), kai.pan@polyu.edu.hk (KP), chung-lun.li@connect.polyu.hk (CLL) Efﬁcient power production scheduling is a crucial concern for power system operators aiming to minimize operational costs. Previous mixed-integer linear programming formulations for unit commitment (UC) prob- lems have primarily used two or three types of binary variables. The investigation of str ong formulations with a single type of binary variables has been limited, as it is believed to be challenging to derive strong valid inequalities using fewer binary variables, and the reduction of the number of binary variables is often accom- panied by a compromise in tightness. T o address these issues, this paper considers a formulation for unit commitment using a single type of binary variables and develops strong valid inequality families to enhance the tightness of the formulation. Conditions under which these strong valid inequalities serve as facet-deﬁning inequalities for the single-generator UC polytope are provided. For those large-size valid inequality families, the existence of efﬁcient separation algorithms for determining the most violated inequalities is also discussed. The ef fectiveness of the proposed single-binary formulation and strong valid inequalities is demonstrated through computational experiments on network-constrained UC problems. The results indicate that the strong valid inequalities presented in this paper are effective in solving UC problems and can also be applied to UC formulations that contain more than one type of binary variables. Key words : unit commitment, polyhedral study , str ong valid inequalities, convex hull 1. Intro duction W ith the increased prevalence of extreme weather events such as droughts, wildﬁres, and ﬂood- ing ar ound the world in recent years, power consumption in many areas hit an all-time high in the summer of 2022. In addition, the continued retirements of coal-ﬁred generating plants, relatively high coal prices, and lower-than-average coal stocks at power plants have limited coal consump- tion ( US EIA 2022 ). As a r esult, the efﬁcient production and distribution of electricity have been identiﬁed as the biggest concern for power producers ar ound the globe. The unit commitment (UC) pr oblem, which involves the scheduling of power generators, has been a challenging optimization problem in the power industry for many years. It often needs to be solved multiple times per day by system operators ( Xavier et al. 2021 ). Because of such practi- cal needs, it has received considerable attention over the past decades. The UC problem involves scheduling a gr oup of generators at a possibly minimal operational cost, subject to their physi- cal and system constraints over a ﬁnite time horizon. Physical constraints specify the technical 1 2 properties of generators, and they may vary depending on the type of generation unit, such as hydro, thermal, and wind units ( V an Ackooij et al. 2018 ). Unless explicitly stated otherwise, all subsequent discussions are about thermal units. The most common physical constraints for thermal units are the ramp-up/-down rate, genera- tion lower/upper bound, and minimum up/down time constraints. System constraints typically include the load (demand) requirement, system r eserve constraint, and transmission ﬂow limit. All generation units are coupled by the system constraints to ensure the reliability of the entire system. The operational cost comprises various components, including the generation (fuel) cost and the startup and shutdown costs of generators. The generation cost is a signiﬁcant component ( Padhy 2004 ), and it is generally assumed to be an increasing quadratic convex function of the generation amount ( T akriti and Birge 2000 ). Start-up and shut-down costs are incurr ed each time the status of a generator changes ( Sen and Kothari 1998 ). Many differ ent UC problems can arise depending on the structure of the electrical power sys- tem. The ability to solve UC problems efﬁciently can have a great impact on both society and individual consumers. Even small improvements in the quality of solutions for UC pr oblems can affect the electricity price over large regions and lead to millions of U.S. dollars of savings per day ( Damcı-Kurt et al. 2016 ). Although the single-generator UC (single-UC) problem with only physical constraints and a quadratic generation cost function is proven to be polynomial-time solvable ( Frangioni and Gentile 2006b ), general UC pr oblems with system constraints have not yet been solved satisfactorily . They are often formulated as large-scale mixed-integer program- ming problems. Such problems are typically NP-hard and difﬁcult to solve when the problem sizes ar e large ( Zheng et al. 2015 , Knueven et al. 2020b , T ejada-Arango et al. 2020 ). For example, a UC problem with an arbitrary number of generators that contain only minimum up/down, gen- eration lower/upper bound, and demand constraints is classiﬁed as NP-hard even considering a single operational period ( Bendotti et al. 2019 ). In a day-ahead deregulated electricity market, an independent system operator (ISO) is expected to determine the generation schedule for a power system within a very short time. Such a generation schedule involves hundreds of thermal units, thousands of transmission lines, and 24–48 operating hours. Therefor e, over the past decades, numerous appr oaches have been devised from both formulation and algorithm perspectives in order to solve UC pr oblems efﬁciently . Priority list and heuristic algorithms are among the earliest solution approaches used to solve UC problems ( Kazarlis et al. 1996 ). The former lists all units by their operational costs and then economically dispatches the system load to generators by a pre-determined order . Heuristic algo- rithms, such as genetic algorithms and simulated annealing, are also frequently adopted as they 3 can be converted to work on parallel computers. However , these approaches usually lead to sub- optimal solutions. Dynamic programming (DP)-based approaches, in contrast, are exact solution techniques for UC problems and are widely applied in the early period ( W ang and Shahideh- pour 1993 , Baldick 1995 ). Using these approaches, UC problems are decomposed by time, and the state for each time period is repr esented by the combinations of units. However , the number of states increases dramatically as the number of units and time periods grows. Therefor e, these approaches are also integrated with heuristic methods to r educe the search space ( Ouyang and Shahidehpour 1991 ). Recently , Frangioni and Gentile ( 2006b ) and Guan et al. ( 2018 ) propose DP algorithms to solve single-UC problems and pr ovide theoretical complexity results. They show that single-UC problems with only physical constraints and suitable generation cost functions, such as quadratic cost functions, can be solved in polynomial time. In addition, given the large size of UC pr oblems, Lagrangian relaxation (LR)-based approaches are pr oposed ( Muckstadt and Koenig 1977 , Abdul-Rahman et al. 1996 , T akriti and Birge 2000 , Lu and Shahidehpour 2005 ). These approaches relax system constraints, such as load requirements, and integrate them into the objective function through Lagrangian multipliers. The resulting problem is then decomposed into subproblems either by units or by time periods, and these subproblems are solved itera- tively until the primal-dual gap is difﬁcult to shrink. However , LR-based approaches suffer from slow and unsteady convergence, and the feasibility of the ﬁnal solution cannot be guaranteed ( Ma and Shahidehpour 1999 ). T o overcome the convergence problem, quadratic terms are incor- porated into the objective function to penalize the violation of demand constraints and impr ove its convexity , resulting in augmented Lagrangian relaxation based appr oaches. A comprehensive examination of solution approaches for UC pr oblems can be found in V an Ackooij et al. ( 2018 ). The extensive advancement of mixed-integer linear programming (MILP) solvers has led to the widespread application of MILP-based approaches in formulating and solving UC problems. MILP-based approaches can guarantee convergence to the optimal solution while providing a ﬂexible and accurate modeling framework. Moreover , the optimality gap is easy to obtain. ISOs are therefor e incr easingly adopting MILP-based approaches over LR-based appr oaches to solve large-scale UC pr oblems ( Hedman et al. 2009 , W u 2011 , Ostrowski et al. 2012 , Li et al. 2021 ). T wo primary factors are generally considered in evaluations of MILP formulations for UC prob- lems: compactness and tightness . Compactness refers to the size of the problem, which can be quan- tiﬁed by the number of constraints, decision variables, or nonzero coefﬁcie nts; tightness refers to the proximity of the linear programming (LP) relaxation of the pr oblem to the convex hull of its feasible region ( Morales-Espa ˜ na et al. 2013 , Knueven et al. 2020b ). For UC problems, compact- ness can be achieved by reducing the number of integer variables in an MILP formulation, as fewer integer variables may lead to a reduction in the number of nodes of the search tree for the 4 branch-and-cut method. Note that for UC formulations, all integer variables are binary . In terms of the number of binary variables used, UC formulations can be categorized into two categories. A single-binary formulation uses a single set of binary variables to denote the on/off status of all units. A three-binary formulation uses two additional sets of binary variables to repr esent the start-up and shut-down decisions. As an important variant of the three-binary formulation, a two- binary formulation can be obtained by expr essing the shut-down variables in terms of the on/of f and start-up variables. For a UC pr oblem, a three-binary formulation can generally be tighter than a single-binary one because the addition of the start-up and shut-down binary variables in the three-binary formulation may facilitate the improvement of the physical constraints, potentially leading to a better LP bound. However , the large size of its search tree and the difﬁculty of solv- ing the subproblem at each branching node may incr ease the solution time. On the other hand, a single-binary formulation uses fewer binary variables, which reduce the size of the search tree and may decrease the solution time. Nevertheless, it usually has a weaker LP bound than the three-binary one. T ightness can be achieved by deriving strong valid inequalities for the MILP formulation to tighten its LP relaxation. Most strong valid inequalities are obtained by studying the physical constraints of a single generator (see, e.g., Lee et al. 2004 , Rajan and T akriti 2005 , Morales-Espa ˜ na et al. 2013 , Damcı-Kurt et al. 2016 , Pan and Guan 2016 , and Bendotti et al. 2018 ). Three-binary formulations for UC problems are the most widely studied. Garver ( 1962 ) is the ﬁrst to propose an MILP formulation for a UC pr oblem. In this three-binary formulation, the gen- eration cost function is assumed to be linear with respect to the generation amount. Arroyo and Conejo ( 2000 ) introduce a thr ee-binary formulation for a single-UC pr oblem. They approximate the exponential start-up cost function and the nonconvex generation cost function using stairwise and piecewise functions, respectively . Chang et al. ( 2001 ) pr esent a thr ee-binary formulation for a short-term hydro scheduling UC problem. Chang et al. ( 2004 ) put forward a new thr ee-binary formulation for UC problems, and they approximate the cubic generation cost function using a piecewise linear one with three breakpoints. Li and Shahidehpour ( 2005 ) compare the LR-based approach with the MILP-based approach in solving a price-based UC pr oblem with various types of generators based on the formulation of Chang et al. ( 2004 ). Their numerical results indicate that the MILP-based approach exhibits superior performance on small scale problems compared with the LR-based approach, and the MILP formulation must be tightened to improve its perfor- mance on large-scale pr oblems. Ostrowski et al. ( 2012 ) consider the formulation of Arroyo and Conejo ( 2000 ) and r eplace their minimum up/down constraints with those of Rajan and T akriti ( 2005 ) because the latter can reduce the computational time signiﬁcantly . By studying the physi- cal constraints of a single generator , they derive a class of strong valid inequalities to tighten the MILP formulation. Their computational results indicate that their tightened formulation is more 5 effective than Carri ´ on and Arroyo ’s ( 2006 ) single-binary formulation. Morales-Espa ˜ na et al. ( 2013 ) propose an alternative three-binary formulation based on the formulation of Ostrowski et al. ( 2012 ). The generation cost function is represented as a linear function with respect to the gen- eration amount. They introduce generation limit constraints to substitute for those in Ostrowski et al. ( 2012 ). They show that the resulting formulation has smaller size and better LP bound than that of Ostrowski et al. ( 2012 ). Morales-Espa ˜ na et al. ( 2015 ) establish a three-binary formulation based on that of Morales-Espa ˜ na et al. ( 2013 ) by considering dif ferent start-up/shut-down trajec- tories, which are ignor ed in conventional research. Damcı-Kurt et al. ( 2016 ) conduct a polyhedral study of the physical constraints based on the work of Ostrowski et al. ( 2012 ). They derive a con- vex hull for the two-period case and strong valid inequalities for the multi-period case to tighten the MILP formulation. Because the number of these strong valid inequalities can be exponential, polynomial separation algorithms are provided to apply them in the solution process. Computa- tional r esults demonstrate that this formulation outperforms the strong formulation of Ostr owski et al. ( 2012 ). Atakan et al. ( 2018 ) develop a state-transition formulation for UC problems based on the formulations of Ostrowski et al. ( 2012 ) and Morales-Espa ˜ na et al. ( 2013 ). T ransmission con- straints are not considered in their formulation. Their test results demonstrate that the proposed formulation has a shorter computational time for long-horizon problems than the formulations of Ostrowski et al. ( 2012 ) and Morales-Espa ˜ na et al. ( 2013 ). Gentile et al. ( 2017 ) pr ovide the convex hull description for the single-UC polytope considering the minimum up/down time constraints, start-up/shut-down capabilities constraints, and generation limit constraints. Other variants of three-binary formulations have also received considerable attention. Rajan and T akriti ( 2005 ) study a single-UC problem with minimum up/down times constraints using three types of binary variables. They provide a complete description of the convex hull of the poly- tope. Based on Rajan and T akriti ’s three-binary formulation, Pan et al. ( 2016 ) derive several fam- ilies of strong valid inequalities for UC problems with gas turbine generators. Their strong valid inequalities are facet-deﬁning for the polytope of physical constraints under speciﬁc conditions. Pan and Guan ( 2016 ) conduct a polyhedral study of physical constraints based on the formulation of Pan et al. ( 2016 ). They derive the complete convex hull descriptions for the two- and three- period polytopes under different parameter settings. They also develop strong valid inequalities for the multi-period case and provide polynomial-time separation algorithms for exponentially large valid inequality families. Bendotti et al. ( 2018 ) analyze the minimum up/down polytope of multiple generators based on the formulation of Pan and Guan ( 2016 ); their generation cost func- tion is linear in the generation amount. They obtain up-set and interval up-set valid inequalities to accelerate the branch-and-cut algorithm. However , given a fractional solution, the problems of separating these two types of inequalities are NP-complete and NP-hard, respectively . Pan et al. 6 ( 2022 ) perform a polyhedral study of a single generator by incorporating fuel constraints. They prove that the single-UC problem with a fuel constraint is NP-hard, and they derive strong valid inequalities to impr ove the computational performance. Dupin ( 2017 ) present two main formula- tions for a UC problem with min-stop ramping constraints based on the deﬁnitions of the so-called state and level variables. The two formulations are compar ed and exhibit an isomorphism. Studies on single-binary formulations are limited. Lee et al. ( 2004 ) investigate the minimum up/down polytope using a single type of binary variables. They give a complete convex hull description of the polytope, obtain valid inequalities, and design an efﬁcient separation proce- dure for using these valid inequalities. Carri ´ on and Arroyo ( 2006 ) propose a single-binary MILP formulation for UC pr oblems. They approximate the generation cost function and the exponential start-up cost function using linear functions as in Arroyo and Conejo ( 2000 ). They also estab- lish new minimum up/down constraints. They then compare the pr oposed formulation with the three-binary formulation of Arr oyo and Conejo ( 2000 ), as well as with its variant in which one type of binary variables in Arroyo and Conejo ( 2000 ) is relaxed. They computationally demon- strate that their single-binary formulation outperforms the other two formulations signiﬁcantly . Frangioni and Gentile ( 2006a ) derive perspective cuts for the mixed-integer quadratic pr ogram- ming problem with semi-continuous variables. They test these cuts by solving a single-binary UC formulation with a quadratic generation cost function. These cuts can substantially improve the performance of the branch-and-cut method. However , their formulation does not consider the ramp-up/-down, system reserve, and transmission constraints. Frangioni et al. ( 2009 ) apply the perspective cuts of Frangioni and Gentile ( 2006a ) to pr ovide a new piecewise linear appr ox- imation of the generation cost function for a short-term UC problem with hydro and thermal generators. Brandenber g et al. ( 2017 ) consider the summed start-up cost across all time periods in a single-binary UC formulation by introducing a single continuous variable for each unit. They derive the H-repr esentation of its epigraph and provide an exact linear separation algorithm. Generally , enhancing the tightness of a formulation for a UC problem necessitates the inclusion of additional variables or constraints, which may lead to an increase in the solution time since the solver needs to solve larger LP subproblems repeatedly; conversely , improving the compactness of a formulation often comes at the expense of weakening tightness, resulting in a weak lower bound ( Morales-Espa ˜ na et al. 2013 , Knueven et al. 2020b ). Therefore, in practice, tightness and compactness must be balanced. In most cases, improvement in tightness for UC formulations is preferr ed over that in compactness because of the potential reduction in solution time, despite the increased complexity r esulting fr om additional binary variables ( Hedman et al. 2009 , Ostrowski et al. 2012 ). Moreover , a lack of binary variables for start-up/shut-down decisions makes it difﬁ- cult to generate strong valid inequalities ( Ostrowski et al. 2012 ). Thus, few studies examine for - mulations with a single type of binary variables and derive str ong valid inequalities to improve 7 their tightness. T o bridge this gap, this paper studies a single-binary formulation for a UC pr ob- lem with thermal units and derives strong valid inequalities to speed up the solution process. The studied single-binary formulation also offers an alternative solution approach for UC pr ob- lems, which can be leveraged in both theoretical resear ch and practical applications. The main contributions of this study are summarized as follows: • Through an investigation of the physical constraints, we provide a complete convex hull description of the two-period single-UC polytope of the single-binary formulation. • W e develop str ong valid inequality families for the multi-period single-UC polytope, and we derive the conditions under which the strong valid inequalities ar e facet-deﬁning. W e also develop efﬁcient separation algorithms for determining a most violated inequality in each valid inequality family . • W e demonstrate the effectiveness of our strong valid inequalities in tightening our single- binary formulation through computational experiments. The r esults indicate that our strong valid inequalities are ef fective in solving UC pr oblems and can also be applied to UC formu- lations that contain more than one type of binary variables. The rest of the paper is organized as follows. Section 2 describes the UC problem under study, presents a single-binary MILP formulation for it, and intr oduces the single-UC polytope. Sec- tion 3 pr ovides the complete description of the convex hull for the two-period case and discusses its importance for solving our UC problem. Section 4 presents various str ong valid inequalities to tighten the single-UC polytope and discusses the existence of ef ﬁcient separation algorithms. Section 5 reports the results of a computational study conducted to assess the effectiveness of our strong valid inequalities in tightening the single-binary MILP formulation and speeding up the solution pr ocess. Section 6 concludes the paper and offers suggestions for future resear ch. All mathematical proofs are pr ovided in Online Appendix A . Some additional computational r esults are pr ovided in Online Appendix D . 2. MILP Mo del fo r Unit Commitment In this section, we ﬁrst present an MILP model for the UC problem with thermal units, and then present some important properties of the single-generator case. In the UC problem being studied, a system operator plans the generation schedule of a set of generators G for a number of time periods at minimal operating costs while satisfying physical and system constraints. The system includes a set of buses B and a set of transmission lines E that link the buses, allowing surplus power to be distributed. Each bus can be equipped with multiple generators and is responsible for the load requir ement of a geographical region. Surplus power at one bus can be transferr ed to neighboring buses thr ough transmission lines to satisfy the load requir ements of other regions. 8 The power ﬂow on each transmission line should not exceed the line’s capacity . T o ensure the reliability of the power supply , some generation capacity should be reserved for outages. All of the generators should operate without violating their physical conﬁgurations. Every time a generator starts up or shuts down, a ﬁxed cost is incurr ed. For each time period, an operational cost is incurred depending on the generation amount and the online/of ﬂine status of the generators. W e let R n denote the n -dimensional real vector space, R n + denote the n -dimensional nonneg- ative real vector space, and B n denote the n -dimensional binary vector space. Given any non- negative integers a and b , we let [ a , b ] Z denote the set of all integers between a and b ; that is, [ a , b ] Z = { a , a + 1, . . . , b } if a ≤ b , and [ a , b ] Z = ∅ if a > b . Let T be the number of time periods in the operation horizon. For each generator g ∈ G , let L g > 0 and ℓ g > 0 be the minimum up and minimum down time requir ements, respectively . That is, once the generator starts up, it must stay online for at least L g time periods, and once it shuts down, it must stay ofﬂine for at least ℓ g time periods. For each g ∈ G , let C g and C g be the gener- ation upper and lower bounds, where C g > C g > 0. For each g ∈ G , let V g > 0 be the maximum change in the generation amount between two consecutive online time periods, i.e., the ramp-up rate is assumed to be equal to the ramp-down rate. For each g ∈ G , let V g > 0 be the start-up/shut- down ramp limit, i.e., the start-up ramp limit is assumed to be equal to the shut-down ramp limit. Thus, when a generator g is online, its generation amount should be within the range [ C g , C g ] . When the generator starts up, its generation amount in the start-up period should be within the range [ C g , V g ] . When the generator shuts down, its generation amount in the pr evious time period should also be within the range [ C g , V g ] . W e assume that V g + V g ≤ C g for all g ∈ G . This condition guarantees that a generator can ramp up at its full rate V g for at least one period after it starts up. W e also assume that C g < V g < C g + V g for all g ∈ G , which holds in most industrial settings as indicated by Morales-Espa ˜ na et al. ( 2015 ), Damcı-Kurt et al. ( 2016 ), Pan et al. ( 2016 ), and Gentile et al. ( 2017 ). For each bus b ∈ B , let G b be the set of generators at bus b (note: S b ∈ B G b = G and G b ∩ G b ′ = ∅ for all b , b ′ ∈ B such that b  = b ′ ). The other parameters are deﬁned as follows: • f g ( · ) : The generation cost function for generator g (for each g ∈ G , f g ( · ) is a non-decreasing convex piecewise linear function with a ﬁxed number of linear segments). • c g : The ﬁxed cost incurred if generator g is online ( c g ≥ 0 for all g ∈ G ). • ϕ g : The ﬁxed start-up cost of generator g ( ϕ g ≥ 0 for all g ∈ G ). • ψ g : The ﬁxed shut-down cost of generator g ( ψ g ≥ 0 for all g ∈ G ). • d b t : The load (demand) at bus b in time period t ( d b t ≥ 0 for all t ∈ [ 1, T ] Z and b ∈ B ). • C e : The capacity limit of transmission line e ( C e ≥ 0 for all e ∈ E ). • K b e : Line ﬂow distribution factor for the ﬂow on transmission line e contributed by the net injection at bus b ( K b e ≥ 0 for all e ∈ E and b ∈ B ). 9 • r t : The system reserve factor of time period t ( r t ≥ 0 for all t ∈ [ 1, T ] Z ). Here, the non-decreasing convex piecewise linear generation cost function f g ( x ) is used to approximate the convex quadratic cost function a g x 2 + b g x ; see Carri ´ on and Arroyo ( 2006 ) and Pan et al. ( 2022 ) for similar approximations. Our model has the following decision variables: • x g t : The generation amount of generator g ∈ G in period t ∈ [ 1, T ] Z . • y g t : The online/ofﬂine status of generator g ∈ G in period t ∈ [ 1, T ] Z , where y g t = 1 if g is online in period t , and y g t = 0 otherwise. • u g t : The start-up cost of generator g ∈ G in period t ∈ [ 1, T ] Z . • v g t : The shut-down cost of generator g ∈ G in period t ∈ [ 1, T ] Z . V ariables x g t , u g t , and v g t are continuous, whereas variable y g t is binary . W e assume that the values of y g − max { L g , ℓ g } + 1 , y g − max { L g , ℓ g } + 2 , . . . , y g − 1 , y g 0 , and x g 0 (for all g ∈ G ) are given as initial conditions. The UC problem is formulated as follows: Problem 1: min ∑ g ∈ G ∑ T t = 1 ( u g t + v g t + c g y g t + f g ( x g t )) (1a) s.t. − y g t − 1 + y g t − y g k ≤ 0, ∀ t ∈ [ − L g + 2, T ] Z , ∀ k ∈ [ t , min { T , t + L g − 1 } ] Z , ∀ g ∈ G , (1b) y g t − 1 − y g t + y g k ≤ 1, ∀ t ∈ [ − ℓ g + 2, T ] Z , ∀ k ∈ [ t , min { T , t + ℓ g − 1 } ] Z , ∀ g ∈ G , (1c) − x g t + C g y g t ≤ 0, ∀ t ∈ [ 1, T ] Z , ∀ g ∈ G , (1d) x g t − C g y g t ≤ 0, ∀ t ∈ [ 1, T ] Z , ∀ g ∈ G , (1e) x g t − x g t − 1 ≤ V g y g t − 1 + V g ( 1 − y g t − 1 ) , ∀ t ∈ [ 1, T ] Z , ∀ g ∈ G , (1f) x g t − 1 − x g t ≤ V g y g t + V g ( 1 − y g t ) , ∀ t ∈ [ 1, T ] Z , ∀ g ∈ G , (1g) u g t ≥ ϕ g ( y g t − y g t − 1 ) , ∀ t ∈ [ 1, T ] Z , ∀ g ∈ G , (1h) v g t ≥ ψ g ( y g t − 1 − y g t ) , ∀ t ∈ [ 1, T ] Z , ∀ g ∈ G , (1i) ∑ g ∈ G x g t = ∑ b ∈ B d b t , ∀ t ∈ [ 1, T ] Z , (1j) ∑ g ∈ G C g y g t ≥ ( 1 + r t ) ∑ b ∈ B d b t , ∀ t ∈ [ 1, T ] Z , (1k) − C e ≤ ∑ b ∈ B K b e  ∑ g ∈ G b x g t − d b t  ≤ C e , ∀ t ∈ [ 1, T ] Z , ∀ e ∈ E , (1l) y g t ∈ { 0, 1 } , x g t ≥ 0, u g t ≥ 0, v g t ≥ 0, ∀ t ∈ [ 1, T ] Z , ∀ g ∈ G . (1m) Objective function ( 1a ) minimizes the total cost, which includes the start-up costs, shut-down costs, and ﬁxed and variable generation costs. Constraint ( 1b ) states the minimum up requir e- ment for generator g . It requires generator g to stay online in periods [ t , min { T , t + L g − 1 } ] Z if it starts up in period t . Constraint ( 1c ) states the minimum down requir ement for generator g . It requir es generator g to stay of ﬂine in periods [ t , min { T , t + ℓ g − 1 } ] Z if it shuts down in period t . Constraints ( 1d ) and ( 1e ) ensure that the generation amount of generator g in period t is 0 if the generator is ofﬂine and is within the range [ C g , C g ] if the generator is online. Constraints ( 1f ) and ( 1g ) guarantee that generator g ramps up/down within its limit V g between two consecutive online time periods. They also guarantee that generator g ramps up by no more than V g units 10 when it starts up and ramps down by no more than V g units when it shuts down. Constraint ( 1h ) and objective function ( 1a ), together with the nonnegativity constraint of u g t , imply that the start- up cost for generator g in period t is ϕ g if the generator starts up in period t , and 0 otherwise. Con- straint ( 1i ) and the objective function ( 1a ), together with the nonnegativity constraint of v g t , imply that the shut-down cost for generator g in period t is ψ g if the generator shuts down in period t , and 0 otherwise. Constraint ( 1j ) is the load balance constraint in period t , which requir es the total generation amount to satisfy the total demand in the period. Constraint ( 1k ) is the system reserve requir ement, which requires the total generation capacity of all online generators to exceed the load requirement by a system reserve factor to deal with demand variations. Constraint ( 1l ) states the transmission ﬂow limit. In the distribution process, a bus b contributes a factor K b e of its net injection ∑ g ∈ G b x g t − d b t to each transmission line e , and constraint ( 1l ) requires the absolute value of the total net injection contributed by all buses to each transmission line to stay below its capacity limit to prevent it from being overloaded; see Ma and Shahidehpour ( 1999 ), Shahidehpour et al. ( 2002 , p. 290), and Xavier et al. ( 2021 ) for similar settings. Constraint ( 1m ) states the nonnegativity and binary requir ements of the decision variables. Note that Problem 1 uses only a single type of binary variables, y g t , and thus is a single-binary formulation. However , in the optimal solution, the continuous variables u g t and v g t have only two possible values. Hence, Problem 1 can also be formulated using two additional vectors of discrete variables. Because the polytope conv ( P ) that we are analyzing in this paper is independent of u g t and v g t (see the deﬁnition of set P below), the polytope is a single-binary polytope regardless of whether u g t and v g t are declared as continuous or discr ete. Note also that the objective function of Problem 1 is piecewise linear . Following the literature (see, e.g., Arr oyo and Conejo 2000 ), Problem 1 can be converted into an MILP . Bendotti et al. ( 2019 ) consider a UC problem in which there is a linear generation cost, a mini- mum demand requirement, no ramp-up, ramp-down, start-up, and shut-down limits, no system reservation requirement, no transmission ﬂow limit, and some initial conditions. They prove that the problem is str ongly NP-har d. It is easy to verify that their NP-har dness pr oof remains valid when applied to our UC problem. Thus, Pr oblem 1 is also NP-hard in the strong sense. In Problem 1, constraints ( 1b )–( 1g ) specify the physical properties of the generators. Constraints ( 1h ) and ( 1i ) determine the start-up and shut-down costs, respectively . Once the y g t values are determined for all g ∈ G and t ∈ [ 1, T ] Z , the u g t and v g t values can be easily obtained by these con- straints. Constraints ( 1j )–( 1l ) are the coupling constraints, or system constraints, that link all of the generators. Because of the scale and complexity of the UC problem, one way of reducing the solving time is to decompose the problem into smaller subproblems with one subproblem corre- sponding to each generator (see Knueven et al. 2020a ). For example, in the Lagrangian r elaxation method, the coupling constraints can be integrated into the objective function through Lagrangian 11 multipliers, and the resulting problem is decomposed into subproblems that contain only the physical constraints ( Baldick 1995 , T akriti and Birge 2000 ). Thus, most improvements in UC mod- els result from studying the properties of an individual generator ’s feasible region ( Knueven et al. 2018 ). Mor eover , strong valid inequalities for the physical constraints are valid for Problem 1 and can be used to tighten its linear relaxation. A tighter linear relaxation can often improve compu- tational efﬁciency by r educing the amount of enumeration required to ﬁnd and pr ove an optimal solution ( Knueven et al. 2020b ). Hence, in the mathematical analysis presented in Sections 3 and 4 , we focus on deriving strong valid inequalities for the physical constraints for the generators in Problem 1. Because all of the generators have the same set of physical constraints, it sufﬁces to concentrate on the physical constraints of a single generator , and the results obtained can be applied to all of other generators. Therefore, in the following analysis, the superscript g in the parameters and decision variables is dropped. Denote x = ( x 1 , . . . , x T ) and y = ( y 1 , . . . , y T ) . Thus, the vector ( x , y ) ∈ R T + × B T contains the generation amount and on/off status of the generator in the T time periods. The set of ( x , y ) values that satisfy the physical constraints of Problem 1 is given as P =  ( x , y ) ∈ R T + × B T : − y t − 1 + y t − y k ≤ 0, ∀ t ∈ [ 2, T ] Z , ∀ k ∈ [ t , min { T , t + L − 1 } ] Z , (2a) y t − 1 − y t + y k ≤ 1, ∀ t ∈ [ 2, T ] Z , ∀ k ∈ [ t , min { T , t + ℓ − 1 } ] Z , (2b) − x t + C y t ≤ 0, ∀ t ∈ [ 1, T ] Z , (2c) x t − C y t ≤ 0, ∀ t ∈ [ 1, T ] Z , (2d) x t − x t − 1 ≤ Vy t − 1 + V ( 1 − y t − 1 ) , ∀ t ∈ [ 2, T ] Z , (2e) x t − 1 − x t ≤ Vy t + V ( 1 − y t ) , ∀ t ∈ [ 2, T ] Z  . (2f) Here, the assumptions C > C > 0, V > 0, V + V ≤ C , and C < V < C + V remain valid. Note that inequalities ( 2a )–( 2f ) in P are the same as inequalities ( 1b )–( 1g ) in Problem 1 for a speciﬁc generator g , except that t is restricted to the range [ 2, T ] Z in ( 2a ), ( 2b ), ( 2e ), and ( 2f ) (i.e., constraints dependent on the initial conditions are not included in P ). Let conv ( P ) denote the convex hull of P , and we refer to conv ( P ) as the single-UC poly- tope. Note that conv ( P ) is full dimensional, which is shown in Appendix C.1 . Obviously , a valid inequality for P is also valid for Problem 1 for any generator g . Hence, the strong valid inequal- ities developed for conv ( P ) can be used to tighten the formulation of Problem 1. The following two lemmas provide some important pr operties of P . L E M M A 1 . Consider any point ( x , y ) ∈ P and t ∈ [ 2, T ] Z . (i) If y t = 0 , then y t − j − y t − j − 1 ≤ 0 for all j ∈ [ 0, min { t − 2, L − 1 } ] Z . (ii) If y t = 1 , then there exists at most one j ∈ [ 0, min { t − 2, L } ] Z such that y t − j − y t − j − 1 = 1 . 12 L E M M A 2 . Denote x ′ = ( x ′ 1 , . . . , x ′ T ) and y ′ = ( y ′ 1 , . . . , y ′ T ) . Let P ′ =  ( x ′ , y ′ ) ∈ R T + × B T : − y ′ T − t + 2 + y ′ T − t + 1 − y ′ T − k + 1 ≤ 0, ∀ t ∈ [ 2, T ] Z , ∀ k ∈ [ t , min { T , t + L − 1 } ] Z , y ′ T − t + 2 − y ′ T − t + 1 + y ′ T − k + 1 ≤ 1, ∀ t ∈ [ 2, T ] Z , ∀ k ∈ [ t , min { T , t + ℓ − 1 } ] Z , − x ′ T − t + 1 + C y ′ T − t + 1 ≤ 0, ∀ t ∈ [ 1, T ] Z , x ′ T − t + 1 − C y ′ T − t + 1 ≤ 0, ∀ t ∈ [ 1, T ] Z , x ′ T − t + 1 − x ′ T − t + 2 ≤ Vy ′ T − t + 2 + V ( 1 − y ′ T − t + 2 ) , ∀ t ∈ [ 2, T ] Z , x ′ T − t + 2 − x ′ T − t + 1 ≤ Vy ′ T − t + 1 + V ( 1 − y ′ T − t + 1 ) , ∀ t ∈ [ 2, T ] Z  . Then, P = P ′ . Lemma 1 states a relationship among the y t variables. This relationship, derived from the min- imum up time requir ement, allows us to simplify the validity proofs of the inequality families presented in Section 4 . Lemma 2 states that if variables x t and y t are replaced by x T − t + 1 and y T − t + 1 , respectively , then the set P remains unchanged. This property enables us to show that if a given inequality family is known to be valid and facet-deﬁning, then the corr esponding “mirr or image” of that inequality family is also valid and facet-deﬁning. 3. The Tw o-p erio d Convex Hull In this section, we investigate the properties of the set P when there are only two periods. Then, we demonstrate that the strong valid inequalities resulting from our investigation can be used not only to tighten the single-UC polytope conv ( P ) but also to derive other forms of str ong valid inequalities for conv ( P ) . Consider any two consecutive periods t − 1 and t , where t ∈ [ 2, T ] Z . Denote P 2 =  ( x t − 1 , x t , y t − 1 , y t ) ∈ R 2 + × B 2 : − x i + C y i ≤ 0, ∀ i ∈ { t − 1, t } , (3a) x i − C y i ≤ 0, ∀ i ∈ { t − 1, t } , (3b) x t − x t − 1 ≤ Vy t − 1 + V ( 1 − y t − 1 ) , (3c) x t − 1 − x t ≤ Vy t + V ( 1 − y t )  . (3d) Note that when t = 2, the set P 2 is the same as the set P with T = 2. Note also that when T = 2, inequalities ( 2a ) and ( 2b ) become redundant, which signiﬁcantly simpliﬁes the set P 2 . Let conv ( P 2 ) denote the convex hull of P 2 . The following theor em provides a complete description of conv ( P 2 ) . 13 T H E O R E M 1 . Denote Q 2 =  ( x t − 1 , x t , y t − 1 , y t ) ∈ R 4 : y i ≤ 1, ∀ i ∈ { t − 1, t } , (4a) C y i ≤ x i ≤ C y i , ∀ i ∈ { t − 1, t } , (4b) x t − 1 ≤ V y t − 1 + ( C − V ) y t , (4c) x t ≤ ( C − V ) y t − 1 + V y t , (4d) x t − x t − 1 ≤ ( C + V ) y t − C y t − 1 , (4e) x t − x t − 1 ≤ V y t − ( V − V ) y t − 1 , (4f) x t − 1 − x t ≤ ( C + V ) y t − 1 − C y t , (4g) x t − 1 − x t ≤ V y t − 1 − ( V − V ) y t  . (4h) Then, Q 2 = conv ( P 2 ) . Note that for every t ∈ [ 2, T ] Z , any inequality in ( 4a )–( 4h ) is valid for conv ( P ) . Note also that inequalities ( 4c )–( 4h ) do not exist in the description of P . Hence, they can be added to the con- straint set of P to tighten the linear relaxation of P . In particular , because V y t − ( V − V ) y t − 1 ≤ Vy t − 1 + V ( 1 − y t − 1 ) for any y t ≤ 1, the right-hand side of ( 4f ) is no gr eater than the right-hand side of ( 2e ), and thus inequality ( 4f ) dominates inequality ( 2e ) and can effectively tighten the linear relaxation of P . Similarly , inequality ( 4h ) dominates inequality ( 2f ) and can effectively tighten the linear relaxation of P . Therefor e, the following inequality families can be used as valid inequali- ties for conv ( P ) : x t ≤ V y t + ( C − V ) y t + 1 , ∀ t ∈ [ 1, T − 1 ] Z ; (5) x t ≤ ( C − V ) y t − 1 + V y t , ∀ t ∈ [ 2, T ] Z ; (6) x t − x t − 1 ≤ ( C + V ) y t − C y t − 1 , ∀ t ∈ [ 2, T ] Z ; (7) x t − x t − 1 ≤ V y t − ( V − V ) y t − 1 , ∀ t ∈ [ 2, T ] Z ; (8) x t − x t + 1 ≤ ( C + V ) y t − C y t + 1 , ∀ t ∈ [ 1, T − 1 ] Z ; (9) x t − x t + 1 ≤ V y t − ( V − V ) y t + 1 , ∀ t ∈ [ 1, T − 1 ] Z . (10) These valid inequalities provide upper bounds on the generation amount x t for each time period t , upper bounds on x t − x t − 1 for each pair of consecutive time periods t and t − 1, and upper bounds on x t − x t + 1 for each pair of consecutive time periods t and t + 1. Inequalities ( 5 )–( 10 ) also enable us to develop other strong valid inequalities for conv ( P ) . W e demonstrate this by presenting a strong valid inequality derived from ( 7 ). Consider any point ( x , y ) ∈ P . For any k ∈ [ 1, T − 1 ] Z and any t ∈ [ k + 1, T ] Z , because inequality ( 7 ) is valid for conv ( P ) , we have t ∑ τ = t − k + 1 ( x τ − x τ − 1 ) ≤ t ∑ τ = t − k + 1  ( C + V ) y τ − C y τ − 1  = V t ∑ τ = t − k + 1 y τ + C t ∑ τ = t − k + 1 ( y τ − y τ − 1 ) , 14 which implies that x t − x t − k ≤ V t ∑ τ = t − k + 1 y τ + C y t − C y t − k . (11) If y t = 1, then ∑ t τ = t − k + 1 y τ ≤ k y t , and by ( 11 ), x t − x t − k ≤ ( C + kV ) y t − C y t − k . If y t = 0, then by ( 2c ) and ( 2d ), − x t − k ≤ − C y t − k and x t = 0, which also imply that x t − x t − k ≤ ( C + k V ) y t − C y t − k . Thus, in both cases, x t − x t − k ≤ ( C + kV ) y t − C y t − k . (12) Hence, ( 12 ) is a valid inequality for conv ( P ) for any k ∈ [ 1, T − 1 ] Z and t ∈ [ k + 1, T ] Z . It is worth noting that inequality ( 19 ) presented in Pr oposition 9 in Section 4.2 is reduced to inequality ( 12 ) when m = 0 and S = ∅ . Therefore, inequality ( 12 ) is a special case of the facet-deﬁning valid inequality ( 19 ). 4. Multi-p erio d Strong V alid Inequalities When there are mor e than two periods, the set P is signiﬁcantly more complex than the set P 2 because it involves not only more time periods but also minimum up/down constraints. In this section, we present a collection of strong valid inequalities that can effectively enhance the tight- ness of Problem 1. W e provide the validity proofs for these inequalities, and we identify the con- ditions under which these inequalities are facet-deﬁning for conv ( P ) . For each family of valid inequalities, we also show that for any given point ( x , y ) ∈ R 2 T + , an efﬁcient separation algorithm exists for determining a most violated inequality . 4.1. Valid Inequalities with a Single Continuous Va riable In this subsection, we present strong valid inequalities that provide upper bounds on the gen- eration amount x t for each time period t . Families of such inequalities appear in pairs. The ﬁrst family consists of inequalities for which the upper bound on x t depends mainly on the values of y t − s − y t − s − 1 for some s ≥ 0, and the second family consists of inequalities for which the upper bound on x t depends mainly on the values of y t + s − y t + s + 1 for some s ≥ 0. The following proposi- tion presents a pair of such inequality families. P R O P O S I T I O N 1 . Consider any S ⊆ [ 0, min { L − 1, T − 2, ⌊ ( C − V ) / V ⌋ } ] Z . For any t ∈ [ 1, T ] Z such that t ≥ s + 2 for all s ∈ S , the inequality x t ≤ C y t − ∑ s ∈ S ( C − V − s V ) ( y t − s − y t − s − 1 ) (13) is valid and facet-deﬁning for conv ( P ) . For any t ∈ [ 1, T ] Z such that t ≤ T − s − 1 for all s ∈ S , the inequality x t ≤ C y t − ∑ s ∈ S ( C − V − s V ) ( y t + s − y t + s + 1 ) (14) is valid and facet-deﬁning for conv ( P ) . 15 In Proposition 1 , inequalities ( 13 ) and ( 14 ) provide upper bounds on the generation amount x t . These upper bounds can be explained as follows. Let s max denote the largest element of S . The condition “ S ⊆ [ 0, min { L − 1, T − 2, ⌊ ( C − V ) / V ⌋ } ] Z ” implies that s max ≤ L − 1, which in turn implies that there is at most one startup and at most one shutdown during the time interval [ t − s max , t ] , and that there is at most one shutdown and at most one startup during the time interval [ t + 1, t + s max + 1 ] . Consider the situation in which a generator starts up in period t − s 1 , stays online until period t + s 2 , and shuts down in period t + s 2 + 1, where s 1 , s 2 ∈ [ 0, s max ] Z , t − s 1 ≥ 2, and t + s 2 + 1 ≤ T . Then, y t − s 1 − 1 = 0, y t − s 1 = y t − s 1 + 1 = · · · = y t + s 2 = 1, and y t + s 2 + 1 = 0. If s 1 ∈ S and none of the time periods in { t − s ≥ 2 : s ∈ S } is a shut-down period, then the right-hand side of inequality ( 13 ) becomes C − ( C − V − s 1 V ) . This upper bound limits the value of x t to be no more than V + s 1 V , which is smaller than the generation upper bound C . Similarly , if s 2 ∈ S and none of the periods in { t + s + 1 ≤ T : s ∈ S } is a start-up period, then the right-hand side of inequality ( 14 ) becomes C − ( C − V − s 2 V ) . This upper bound limits the value of x t to be no more than V + s 2 V , which is smaller than the generation upper bound C . In Proposition 1 , the set S only contains elements that are less than L . The following proposition states that under certain conditions, inequalities ( 13 ) and ( 14 ) remain valid and facet-deﬁning when S contains some elements that are greater than or equal to L . P R O P O S I T I O N 2 . Consider any integers α , β , and s max such that (a) L ≤ s max ≤ min { T − 2, ⌊ ( C − V ) / V ⌋ } , (b) 0 ≤ α < β ≤ s max , and (c) β = α + 1 or s max ≤ L + α . Let S = [ 0, α ] Z ∪ [ β , s max ] Z . For any t ∈ [ s max + 2, T ] Z , inequality ( 13 ) is valid and facet-deﬁning for conv ( P ) . For any t ∈ [ 1, T − s max − 1 ] Z , inequality ( 14 ) is valid and facet-deﬁning for conv ( P ) . E X A M P L E 1 . Let T = 16, C = 80, C = 8, L = ℓ = 5, V = 15, and V = 10. Then, ⌊ ( C − V ) / V ⌋ = 6. By Proposition 1 , we obtain the following pair of valid inequalities if we set S = { 0, 2, 4 } and t = 8:  x 8 ≤ 25 y 3 − 25 y 4 + 45 y 5 − 45 y 6 + 65 y 7 + 15 y 8 ; x 8 ≤ 15 y 8 + 65 y 9 − 45 y 10 + 45 y 11 − 25 y 12 + 25 y 13 . By Proposition 2 , we obtain the following pair of valid inequalities if we set S = { 0, 1, 2, 5, 6 } (i.e., α = 2, β = 5, and s max = 6) and t = 8:  x 8 ≤ 5 y 1 + 10 y 2 − 15 y 3 + 45 y 5 + 10 y 6 + 10 y 7 + 15 y 8 ; x 8 ≤ 15 y 8 + 10 y 9 + 10 y 10 + 45 y 11 − 15 y 13 + 10 y 14 + 5 y 15 . The next proposition extends Pr oposition 1 and pr esents another two families of str ong valid inequalities. P R O P O S I T I O N 3 . Consider any set S ⊆ [ 0, min { L − 1, T − 3, ⌊ ( C − V ) / V ⌋ } ] Z and any real number η such that 0 ≤ η ≤ min { L − 1, ( C − V ) / V } . For any t ∈ [ 1, T − 1 ] Z such that t ≥ s + 2 for all s ∈ S , the inequality x t ≤ ( C − η V ) y t + η Vy t + 1 − ∑ s ∈ S ( C − V − s V ) ( y t − s − y t − s − 1 ) (15) 16 is valid for conv ( P ) . For any t ∈ [ 2, T ] Z such that t ≤ T − s − 1 for all s ∈ S , the inequality x t ≤ ( C − η V ) y t + η Vy t − 1 − ∑ s ∈ S ( C − V − s V ) ( y t + s − y t + s + 1 ) (16) is valid for conv ( P ) . Furthermore, inequalities ( 15 ) and ( 16 ) are facet-deﬁning for conv ( P ) when η ∈ { 0, ( C − V ) / V } or η = L − 1 ∈ S . When t  = T , inequality ( 15 ) is a generalization of inequality ( 13 ). Speciﬁcally , the right-hand side of ( 15 ) dif fers from the right-hand side of ( 13 ) by η Vy t + 1 − η Vy t , and this dif ference is zero if η = 0. Similarly , when t  = 1, inequality ( 16 ) is a generalization of inequality ( 14 ), and the right- hand side of ( 16 ) differs from the right-hand side of ( 14 ) by η Vy t − 1 − η Vy t . In Pr oposition 3 , the set S only contains elements that are less than L . The following proposition, which extends Proposition 2 , states that under certain conditions, inequalities ( 15 ) and ( 16 ) remain valid and facet-deﬁning when S contains some elements that are greater than or equal to L . P R O P O S I T I O N 4 . Consider any real number η such that 0 ≤ η ≤ min { L − 1, ( C − V ) / V } and any integers α , β , and s max such that (a) L ≤ s max ≤ min { T − 3, ⌊ ( C − V ) / V ⌋ } , (b) 0 ≤ α < β ≤ s max , and (c) β = α + 1 or s max ≤ L + α . Let S = [ 0, α ] Z ∪ [ β , s max ] Z . For any t ∈ [ s max + 2, T − 1 ] Z , inequality ( 15 ) is valid for conv ( P ) . For any t ∈ [ 2, T − s max − 1 ] Z , inequality ( 16 ) is valid for conv ( P ) . Furthermore, ( 15 ) and ( 16 ) are facet-deﬁning for conv ( P ) when η ∈ { 0, ( C − V ) / V } or η = L − 1 ∈ S . E X A M P L E 2 ( C O N T I N U AT I O N O F E X A M P L E 1 ) . In Example 1 , if we set η = 2.5, S = { 0, 2, 4 } , and t = 8, then by Proposition 3 , we obtain the following pair of valid inequalities:  x 8 ≤ 25 y 3 − 25 y 4 + 45 y 5 − 45 y 6 + 65 y 7 − 10 y 8 + 25 y 9 ; x 8 ≤ 25 y 7 − 10 y 8 + 65 y 9 − 45 y 10 + 45 y 11 − 25 y 12 + 25 y 13 . Note that the right-hand sides of the ﬁrst and second inequalities differ from those in the ﬁrst pair of inequalities in Example 1 by η Vy t + 1 − η Vy t (i.e., 25 y 9 − 25 y 8 ) and η Vy t − 1 − η Vy t (i.e., 25 y 7 − 25 y 8 ), r espectively . If we set η = 2.5, S = { 0, 1, 2, 5, 6 } (i.e., α = 2, β = 5, and s max = 6), and t = 8, then by Proposition 4 , we obtain the following pair of valid inequalities:  x 8 ≤ 5 y 1 + 10 y 2 − 15 y 3 + 45 y 5 + 10 y 6 + 10 y 7 − 10 y 8 + 25 y 9 ; x 8 ≤ 25 y 7 − 10 y 8 + 10 y 9 + 10 y 10 + 45 y 11 − 15 y 13 + 10 y 14 + 5 y 15 . Similarly , the right-hand sides of the ﬁrst and second inequalities differ from those in the second pair of inequalities in Example 1 by η Vy t + 1 − η Vy t (i.e., 25 y 9 − 25 y 8 ) and η Vy t − 1 − η Vy t (i.e., 25 y 7 − 25 y 8 ), respectively . The next proposition also extends Proposition 1 and presents another two families of strong valid inequalities. P R O P O S I T I O N 5 . Consider any S ⊆ [ 1, min { L , T − 2, ⌊ ( C − V ) / V ⌋ } ] Z and any real number η such that 0 ≤ η ≤ min { L , ( C − V ) / V } . For any t ∈ [ 2, T ] Z such that t ≥ s + 2 for all s ∈ S , the inequality x t ≤ ( V + η V ) y t + ( C − V − η V ) y t − 1 − ∑ s ∈ S ( C − V − s V ) ( y t − s − y t − s − 1 ) (17) 17 is valid for conv ( P ) . For any t ∈ [ 1, T − 1 ] Z such that t ≤ T − s − 1 for all s ∈ S , the inequality x t ≤ ( V + η V ) y t + ( C − V − η V ) y t + 1 − ∑ s ∈ S ( C − V − s V ) ( y t + s − y t + s + 1 ) (18) is valid for conv ( P ) . Furthermore, inequalities ( 17 ) and ( 18 ) are facet-deﬁning for conv ( P ) when η ∈ { 0, ( C − V ) / V } or η = L ∈ S . In Pr oposition 5 , the set S only contains elements that ar e less than or equal to L . The following proposition states that under certain conditions, inequalities ( 17 ) and ( 18 ) remain valid and facet- deﬁning when S contains some elements that are greater than L . P R O P O S I T I O N 6 . Consider any integers α , β , and s max such that (a) L + 1 ≤ s max ≤ min { T − 2, ⌊ ( C − V ) / V ⌋ } , (b) 1 ≤ α < β ≤ s max , and (c) β = α + 1 or s max ≤ L + α . Let S = [ 1, α ] Z ∪ [ β , s max ] Z . For any t ∈ [ s max + 2, T ] Z , inequality ( 17 ) is valid for conv ( P ) . For any t ∈ [ 1, T − s max − 1 ] Z , inequality ( 18 ) is valid for conv ( P ) . Furthermore, ( 17 ) and ( 18 ) are facet-deﬁning for conv ( P ) when η ∈ { 0, ( C − V ) / V } or η = L ∈ S . E X A M P L E 3 ( C O N T I N U AT I O N O F E X A M P L E 1 ) . In Example 1 , if we set η = 2.5, S = { 1, 3, 5 } , and t = 8, then by Proposition 5 , we obtain the following pair of valid inequalities:  x 8 ≤ 15 y 2 − 15 y 3 + 35 y 4 − 35 y 5 + 55 y 6 − 15 y 7 + 40 y 8 ; x 8 ≤ 40 y 8 − 15 y 9 + 55 y 10 − 35 y 11 + 35 y 12 − 15 y 13 + 15 y 14 . If we set η = 2.5, S = { 1, 2, 5, 6 } (i.e., α = 2, β = 5, and s max = 6), and t = 8, then by Proposition 6 , we obtain the following pair of valid inequalities:  x 8 ≤ 5 y 1 + 10 y 2 − 15 y 3 + 45 y 5 + 10 y 6 − 15 y 7 + 40 y 8 ; x 8 ≤ 40 y 8 − 15 y 9 + 10 y 10 + 45 y 11 − 15 y 13 + 10 y 14 + 5 y 15 . Propositions 1 – 6 present dif ferent families of valid inequalities. For each family of valid inequalities and any given point ( x , y ) with non-binary y values, it is important to have an ef ﬁ- cient separation algorithm that can identify a most violated inequality in the family , if such a violated inequality exists. In Propositions 1 , 3 , and 5 , the number of combinations of S is expo- nential in T . Furthermor e, in Pr opositions 3 and 5 , η is a r eal value. However , the next pr oposition states that given any point ( x , y ) with non-binary y values, a most violated inequality in each of the inequality families stated in Propositions 1 , 3 , and 5 can be determined in linear time. P R O P O S I T I O N 7 . For any given point ( x , y ) ∈ R 2 T + , the most violated inequalities ( 13 ) – ( 14 ) , ( 15 ) – ( 16 ) , and ( 17 ) – ( 18 ) in Propositions 1 , 3 , and 5 , respectively , can be determined in O ( T ) time if such violated inequalities exist. In Propositions 2 , 4 , and 6 , the number of combinations of α , β , s max , and t is O ( T 4 ) . Furthermore, in Propositions 4 and 6 , η is a real value. However , the next proposition states that given any point ( x , y ) with non-binary y values, a most violated inequality in each of the inequality families stated in Propositions 2 , 4 , and 6 can be determined in O ( T 3 ) time. 18 P R O P O S I T I O N 8 . For any given point ( x , y ) ∈ R 2 T + , the most violated inequalities ( 13 ) – ( 14 ) , ( 15 ) – ( 16 ) , and ( 17 ) – ( 18 ) in Propositions 2 , 4 , and 6 , respectively , can be determined in O ( T 3 ) time if such violated inequalities exist. 4.2. Valid Inequalities with Two Continuous Va riables In this subsection, we present strong valid inequalities that provide upper bounds on x t − x t − k (respectively x t − x t + k ) for each pair of time periods t and t − k (respectively t and t + k ). The following proposition pr esents a pair of such inequality families. P R O P O S I T I O N 9 . Consider any k ∈ [ 1, T − 1 ] Z such that C − C − kV > 0 , any m ∈ [ 0, k − 1 ] Z , and any S ⊆ [ 0, min { k − 1, L − m − 1 } ] Z . For any t ∈ [ k + 1, T − m ] Z , the inequality x t − x t − k ≤ ( C + ( k − m ) V ) y t + V m ∑ i = 1 y t + i − C y t − k − ∑ s ∈ S ( C + ( k − s ) V − V ) ( y t − s − y t − s − 1 ) (19) is valid for conv ( P ) . For any t ∈ [ m + 1, T − k ] Z , the inequality x t − x t + k ≤ ( C + ( k − m ) V ) y t + V m ∑ i = 1 y t − i − C y t + k − ∑ s ∈ S ( C + ( k − s ) V − V ) ( y t + s − y t + s + 1 ) (20) is valid for conv ( P ) . Furthermore, ( 19 ) and ( 20 ) are facet-deﬁning for conv ( P ) when m = 0 and s ≥ min { k − 1, 1 } for all s ∈ S . In Proposition 9 , the number of combinations of S , t , k , and m is exponential in T . Thus, the sizes of the inequality families ( 19 ) and ( 20 ) are exponential in T . However , the next proposition states that given any point ( x , y ) with non-binary y values, the most violated inequalities ( 19 ) and ( 20 ) can be determined in polynomial time. P R O P O S I T I O N 1 0 . For any given point ( x , y ) ∈ R 2 T + , the most violated inequalities ( 19 ) and ( 20 ) can be determined in O ( T 3 ) time if such violated inequalities exist. P R O P O S I T I O N 1 1 . Consider any k ∈ [ 1, T − 1 ] Z such that C − C − k V > 0 , any m ∈ [ 0, k − 1 ] Z , and any S ⊆ [ 0, min { k − 1, L − m − 2 } ] Z . For any t ∈ [ k + 1, T − m − 1 ] Z , the inequality x t − x t − k ≤ ( C + ( k − m ) V − V ) y t + m + 1 + V m ∑ i = 1 y t + i + V y t − C y t − k − ∑ s ∈ S ( C + ( k − s ) V − V ) ( y t − s − y t − s − 1 ) (21) is valid and facet-deﬁning for conv ( P ) . For any t ∈ [ m + 2, T − k ] Z , the inequality x t − x t + k ≤ ( C + ( k − m ) V − V ) y t − m − 1 + V m ∑ i = 1 y t − i + V y t − C y t + k − ∑ s ∈ S ( C + ( k − s ) V − V ) ( y t + s − y t + s + 1 ) (22) is valid and facet-deﬁning for conv ( P ) . 19 In Proposition 11 , the number of combinations of S , t , k , and m is exponential in T . Thus, the sizes of the inequality families ( 21 ) and ( 22 ) are exponential in T . However , the next proposition states that given any point ( x , y ) with non-binary y values, the most violated inequalities ( 21 ) and ( 22 ) can be determined in polynomial time. P R O P O S I T I O N 1 2 . For any given point ( x , y ) ∈ R 2 T + , the most violated inequalities ( 21 ) and ( 22 ) can be determined in O ( T 3 ) time if such violated inequalities exist. 5. Computational Exp eriments W e conduct a computational study to evaluate the ef fectiveness of our strong valid inequalities in tightening the pr oposed single-binary MILP formulation for the UC pr oblem. In Section 5.1 , we describe the problem instances that we use in this computational study . In Section 5.2 , we present the computational results. All of the computational experiments ar e performed on a computer node with Intel(R) Xeon(R) CPU E5-2699 v3 at 2.30GHz and 16 cor es. The addressable memory is 32GB. IBM ILOG CPLEX 22.1 is used as the MILP solver to run all of the experiments. The MILP solver is called thr ough its Python application programming interface under the default settings. Note that the performance of a formulation is affected by the inher ent random component of the heuristic pr ocess used in solvers ( T ejada-Arango et al. 2020 ). Thus, to accurately evaluate the effectiveness of our strong valid inequalities, “traditional branch-and-cut” is chosen to be the search strategy . 5.1. T est Instances W e conduct three computational experiments. These experiments are based on a network- constrained UC problem. Recall that in Sections 3 and 4 , the superscript g was omitted when we focused on deriving strong valid inequalities for the polytope conv ( P ) that consists of a sin- gle generator . In the test instances of these three experiments, we reinstate the superscript g in the str ong valid inequalities that are used to tighten the UC formulations. Thus, when a family of strong valid inequalities is added to a UC formulation in these three experiments, it will be added to each generator with its corresponding parameters at the same time. In all three experi- ments, the non-decreasing convex piecewise cost function is obtained by approximating the given quadratic cost function a g x 2 + b g x . W e apply the method developed by Frangioni et al. ( 2009 ) to perform this piecewise linear approximation, using nine line segments with the x -coordinates of the breakpoints spr ead evenly between the lower bound C and the upper bound C . In the ﬁrst experiment, we use the data obtained fr om Ostr owski et al. ( 2012 ) and Pan and Guan ( 2016 ). Because of the absence of transmission ﬂow data in this data set, the transmission con- straint ( 1l ) is not considered in this experiment. The removal of the transmission constraint does 20 not have a major impact on our computational study because we focus primarily on evaluating the effectiveness of the str ong valid inequalities in tightening the single-binary formulation. The system contains eight types of generators. T able 1 contains the data of these eight generator types. The generation cost function for generator g is a g x 2 + b g x , where the values of a g and b g are provided in the 10th and 11th columns, respectively , of the table. The data set comprises 20 test instances, as shown in T able 2 . For each instance, the operation horizon is set equal to 24 hours, i.e., T = 24, and the system r eserve factor is set equal to 3% for all periods, i.e., r t = 0.03 for all t ∈ [ 1, T ] Z . The system load ∑ b ∈ B d b t in each period t is shown in T able 3 , and it is expressed as a percentage of the total generation capacity ∑ g ∈ G C g . T able 1 Generator Data ( Ostrowski et al. 2012 , Pan and Guan 2016 ) Generator T ype C g C g L g ℓ g V g V g ϕ g ψ g a g b g c g (MW) (MW) (h) (h) (MW/h) (MW/h) ( $ /h) ( $ /h) ( $ /M W 2 h) ( $ /MWh) ( $ /h) 1 150 455 8 8 91 180 2000 2000 0.00048 16.19 1000 2 150 455 8 8 91 180 2000 2000 0.00031 17.26 970 3 20 130 5 5 26 35 500 500 0.00200 16.6 700 4 20 130 5 5 26 35 500 500 0.00211 16.5 680 5 25 162 6 6 32.4 40 700 700 0.00398 19.7 450 6 20 80 3 3 16 28 150 150 0.00712 22.26 370 7 25 85 3 3 17 33 200 200 0.00079 27.74 480 8 10 55 1 1 11 15 60 60 0.00413 25.92 660 T able 2 Problem Instances ( Ostr owski et al. 2012 , Pan and Guan 2016 ) Instance Number of generators T otal no. of generators T ype 1 T ype 2 T ype 3 T ype 4 T ype 5 T ype 6 T ype 7 T ype 8 1 12 11 0 0 1 4 0 0 28 2 13 15 2 0 4 0 0 1 35 3 15 13 2 6 3 1 1 3 44 4 15 11 0 1 4 5 6 3 45 5 15 13 3 7 5 3 2 1 49 6 10 10 2 5 7 5 6 5 50 7 17 16 1 3 1 7 2 4 51 8 17 10 6 5 2 1 3 7 51 9 12 17 4 7 5 2 0 5 52 10 13 12 5 7 2 5 4 6 54 11 46 45 8 0 5 0 12 16 132 12 40 54 14 8 3 15 9 13 156 13 50 41 19 11 4 4 12 15 156 14 51 58 17 19 16 1 2 1 165 15 43 46 17 15 13 15 6 12 167 16 50 59 8 15 1 18 4 17 172 17 53 50 17 15 16 5 14 12 182 18 45 57 19 7 19 19 5 11 182 19 58 50 15 7 16 18 7 12 183 20 55 48 18 5 18 17 15 11 187 T able 3 System Load—Percentage of T otal Generation Capacity ( Ostrowski et al. 2012 , Pan and Guan 2016 ) Period 1 2 3 4 5 6 7 8 9 10 11 12 System Load 71% 65% 62% 60% 58% 58% 60% 64% 73% 80% 82% 83% Period 13 14 15 16 17 18 19 20 21 22 23 24 System Load 82% 80% 79% 79% 83% 91% 90% 88% 85% 84% 79% 74% 21 In this experiment, we compare the following two formulations: F1: minimize objective function ( 1a ) subject to constraints ( 1b )–( 1k ), ( 1m ). F1-X: minimize objective function ( 1a ) subject to constraints ( 1b )–( 1k ), ( 1m ); constraints ( 5 )–( 10 ); cutting planes ( 13 )–( 22 ). Formulation F1 is the original formulation of Problem 1 with the transmission constraint ( 1l ) removed. In F1-X, the strong valid inequality families ( 5 )–( 10 ) obtained from the two-period single-UC polytope are added to the formulation as constraints, and the multi-period strong valid inequality families ( 13 )–( 22 ) derived in Section 4 are added to the user cut pool of the CPLEX optimizer and are applied at any stage of the optimization. Note that each of the inequality fam- ilies ( 13 )–( 22 ) contains a large number of inequalities. Thus, for each of these inequality families, only some of the inequalities are added to F1-X as user cuts , because the use of separation algo- rithms through the callbacks of CPLEX will slow down the solution pr ocess. This is attributable to the fr equent invocation of the separation algorithms during the solution process. Speciﬁcally , for each of the inequality families ( 13 )–( 22 ), S is restricted to the empty set and the set that con- tains all of the elements in its range, and the other parameters such as t , k , and m are allowed to take any values in their respective ranges such that the inequality obtained is facet-deﬁning for conv ( P ) . For example, for inequality family ( 19 ), we consider each k ∈ [ 1, T − 1 ] Z , m = 0, S = { ∅ , [ 0, min { k − 1, L − 1 } ] Z } , and t ∈ [ k + 1, T ] Z such that s ≥ min { k − 1, 1 } for all s ∈ S . In the second experiment, we use the same data as in the ﬁrst experiment. W e compare the effectiveness of our str ong valid inequalities with that of the valid inequalities in Pan and Guan ( 2016 ) in tightening Pan and Guan ’s formulation. W e also test the effectiveness of our strong valid inequalities when they are combined with the valid inequalities in Pan and Guan ( 2016 ). T o do so, we solve the following four formulations: F2: minimize objective function (38a) in Pan and Guan ( 2016 ) subject to constraints (38b)–(38i), (38k) in Pan and Guan ( 2016 ). F2-X: minimize objective function (38a) in Pan and Guan ( 2016 ) subject to constraints (38b)–(38i), (38k) in Pan and Guan ( 2016 ); constraints ( 5 )–( 10 ) in this paper; cutting planes ( 13 )–( 22 ) in this paper . F2-Y : minimize objective function (38a) in Pan and Guan ( 2016 ) subject to constraints (38b)–(38i), (38k) in Pan and Guan ( 2016 ); constraints (2d)–(2g) in Pan and Guan ( 2016 ); cutting planes (4)–(7), (10)–(13), (24d), (24h)–(24i), (24o)–(24r), (28)–(36) in Pan and Guan ( 2016 ). 22 F2-Z: minimize objective function (38a) in Pan and Guan ( 2016 ) subject to constraints (38b)–(38i), (38k) in Pan and Guan ( 2016 ); constraints (2d)–(2g) in Pan and Guan ( 2016 ); constraints ( 5 )–( 10 ) in this paper; cutting planes (4)–(7), (10)–(13), (24d), (24h)–(24i), (24o)–(24r), (28)–(36) in Pan and Guan ( 2016 ). cutting planes ( 13 )–( 22 ) in this paper . Formulation F2 is the two-binary UC formulation in Pan and Guan ( 2016 ), except that the trans- mission constraint (38j) has been excluded. In formulation F2-X, the valid inequalities ( 5 )–( 10 ) ar e added as constraints, and the valid inequalities ( 13 )–( 22 ) are added as user cuts in the same way as in the ﬁrst experiment. In formulation F2-Y , the strong valid inequalities in Pan and Guan ( 2016 ) are used the same way as in Pan and Guan’s computational study to tighten formulation F2. Speciﬁcally , valid inequalities in the two-period convex hull, (2d)–(2g), are added as constraints, and other valid inequalities are added as user cuts. For inequality families that have an exponen- tial size, the S set is restricted to the empty set and the set that contains all of the elements in its range. The other parameters, such as t , m , and n , are allowed to take any values in their respective ranges such that the inequality obtained is facet-deﬁning for conv ( P ) . Formulation F2-Z contains all the valid inequalities in formulations F2-X and F2-Y . The third experiment examines a network-constrained UC problem based on the modiﬁed IEEE 118-bus system. The system comprises 54 thermal generators, 118 buses, and 186 trans- mission lines. System data such as C g , C g , L g , ℓ g , a g , b g , c g , etc., as well as the load amount of each load bus, are obtained from http://motor.ece.iit.edu/data/SCUC 118/ . Each instance has a 24-hour operation horizon, i.e., T = 24. The system reserve factor of each time period is set equal to 3%, as in the ﬁrst experiment. The maximum hourly load of the system is ran- domly generated from a uniform distribution on [ 0.5 ∑ g ∈ G C g , ∑ g ∈ G C g ] . The maximum hourly load of each load bus is then obtained by allocating the maximum hourly load of the sys- tem to each load bus in proportion to their load amounts. For each load bus, the loads in dif- ferent time periods ar e then obtained by following a daily load proﬁle such that the maxi- mum load of the day is equal to the maximum hourly load. This daily load proﬁle is obtained from https://www.pjm.com/markets- and- operations/data- dictionary , which was generated based on the average values of the actual hourly electricity demand over 30 days in the western market. T wenty instances with randomly generated loads are created using this method. Each instance is solved using the following formulations: F1 + : minimize objective function ( 1a ) subject to constraints ( 1b )–( 1m ). 23 F1 + -X: minimize objective function ( 1a ) subject to constraints ( 1b )–( 1m ); constraints ( 5 )–( 10 ); cutting planes ( 13 )–( 22 ). Formulations F1 + and F1 + -X resemble formulations F1 and F1-X, respectively , with the transmis- sion constraint ( 1l ) reinstated. In F1 + -X, the strong valid inequalities ( 5 )–( 10 ) and ( 13 )–( 22 ) are added as constraints and user cuts, respectively , in the same way as in the ﬁrst experiment. 5.2. Computational Results In this subsection, we report the computational results of the three experiments described in Sec- tion 5.1 . In these experiments, each test instance is executed once using each of the formulations in the experiment, and the time limit for each execution is set to one hour . T ables 4 – 8 summarize the computational results. In these tables, the “# var ,” “# bin var ,” and “# cstr” columns report the total number of decision variables, the number of binary decision variables, and the number of constraints (excluding nonnegativity and binary constraints), respectively , in the formulation. The “IGap” columns report the root node integrality gaps of the differ ent formulations, where IGap is given as | Z ∗ − Z LP | / Z ∗ × 100%, where Z ∗ is the best objective function value obtained by solving the formulations in the experiment and Z LP is the optimal objective function value of the LP relaxation of the formulation concerned. Note that for each strong formulation, the objective function value of its LP relaxation (i.e., Z LP ) is obtained such that there is no violation of all valid inequality families. For example, to obtain Z LP for F1-X, we solve its LP relaxation with a subset of valid inequalities ( 13 )–( 22 ) added as constraints ﬁrst. When an optimal solution is obtained, the separation algorithms are invoked to check for any violation of valid inequalities ( 13 )–( 22 ). If there exists a violation, the most violated inequality will be added to the LP pr oblem as constraints and the new pr oblem is solved again to obtain an optimal solution. This process is repeated until ther e is no violation of valid inequalities ( 13 )–( 22 ). Thus, the obtained solution satisﬁes valid inequal- ities ( 13 )–( 22 ), and its objective function value Z LP is used to calculate the integrality gap. This integrality gap measures the tightness of the formulation. T o evaluate the effectiveness of the strong valid inequalities in tightening the formulation, we r eport the per centage r eduction in inte- grality gap in the “Pct. reduction” columns, where Pct. reduction = IGap without cuts − IGap with cuts IGap without cuts × 100%, IGap without cuts is the integrality gap of the formulation without our derived strong valid inequal- ities, and IGap with cuts is the integrality gap of the current formulation. The “CPU time [TGap]” columns report the computational time (in seconds) requir ed to solve the instance to optimality (with a default optimality gap of 0.01%). Instances that could not be solved to optimality within one hour are marked with “**,” and the terminating gaps of those instances are reported (enclosed 24 in squar e brackets). The “# nodes” columns report the number of branch-and-cut nodes explored. The “# user cuts” columns report the number of user cuts added to each formulation. T able 4 Performance of MIP Formulations in the First Experiment Instance # var # bin var # cstr CPU time [TGap] # nodes # user cuts F1 F1-X F1 F1-X F1 F1-X F1-X 1 3360 672 19422 23286 666.8 38.6 289339 11563 211 2 4200 840 24514 29344 ** [0.06%] 301.8 1009424 137851 367 3 5280 1056 29556 35628 ** [0.03%] 231.2 596629 53267 554 4 5400 1080 29304 35514 ** [0.01%] 1335.1 1584979 494770 299 5 5880 1176 32812 39574 ** [0.06%] 703.7 481348 149614 525 6 6000 1200 31562 38462 ** [0.07%] 190.5 604379 94790 520 7 6120 1224 33610 40648 ** [0.05%] 3194.2 463539 903454 445 8 6120 1224 32932 39970 ** [0.09%] 662.4 504831 171291 518 9 6240 1248 34346 41522 ** [0.09%] 1630.6 456636 425148 556 10 6480 1296 34356 41808 ** [0.09%] 2119.2 410349 433936 697 11 15840 3168 87526 105742 ** [0.12%] ** [0.04%] 274195 255185 1006 12 18720 3744 102148 123676 ** [0.09%] ** [0.02%] 212690 279909 1923 13 18720 3744 102174 123702 ** [0.10%] 927.5 172698 59414 1457 14 19800 3960 113314 136084 ** [0.12%] ** [0.01%] 88150 123673 2899 15 20040 4008 109194 132240 ** [0.17%] ** [0.01%] 114417 150664 1566 16 20640 4128 112994 136730 ** [0.11%] ** [0.01%] 240391 234442 2214 17 21840 4368 120156 145272 ** [0.13%] 660.0 177862 15758 1208 18 21840 4368 119936 145052 ** [0.13%] 2880.9 207181 130559 1533 19 21960 4392 120816 146070 ** [0.09%] 1958.5 208972 54786 2312 20 22440 4488 122468 148274 ** [0.12%] 920.3 257848 19072 1198 T able 5 The Strength of LP Relaxations of MIP Formulations in the First Experiment IGap Instance 1 2 3 4 5 6 7 8 9 10 F1 0.45% 0.39% 0.41% 0.36% 0.51% 0.61% 0.34% 0.55% 0.51% 0.64% F1-X 0.20% 0.14% 0.08% 0.06% 0.05% 0.04% 0.07% 0.06% 0.06% 0.04% Instance 11 12 13 14 15 16 17 18 19 20 F1 0.32% 0.32% 0.40% 0.38% 0.50% 0.29% 0.45% 0.42% 0.37% 0.43% F1-X 0.06% 0.02% 0.02% 0.03% 0.02% 0.02% 0.02% 0.02% 0.02% 0.02% Pct. reduction Instance 1 2 3 4 5 6 7 8 9 10 F1-X 55.7% 63.8% 80.4% 82.3% 90.0% 93.3% 78.7% 89.4% 89.1% 93.1% Instance 11 12 13 14 15 16 17 18 19 20 F1-X 82.6% 94.0% 95.2% 90.9% 96.4% 94.1% 96.1% 95.9% 94.1% 96.1% The computational results of the ﬁrst experiment are presented in T ables 4 and 5 . The integrality gaps generated by formulation F1-X are considerably smaller than those generated by formula- tion F1, particularly for large instances (i.e., instances 11–20, where the total number of generators exceeds 100 and the number of binary decision variables exceeds 2000). This suggests that for- mulation F1-X is tighter than formulation F1. Using formulation F1, CPLEX is able to solve only one of the 20 test instances to optimality within one hour . In contrast, using formulation F1-X, CPLEX is able to solve 15 instances to optimality within the same time limit. For instances that cannot be solved to optimality using formulation F1-X, the terminating gaps are all within 0.05% and are much smaller than those using formulation F1. Formulation F1-X tends to explore fewer nodes than formulation F1, and the number of user cuts added by F1-X in the solution process is 25 small compared with the total number of constraints in formulations F1 and F1-X. These results demonstrate that our proposed strong valid inequalities can signiﬁcantly tighten the single-binary formulation of the network-constrained UC problem and thus speed up the solution pr ocess. T ables 6 and 7 present the computational results of the second experiment, in which four formu- lations, namely F2, F2-X, F2-Y, and F2-Z, are used to solve the network-constrained UC problem. Using formulation F2, CPLEX solves only two of the 20 instances within the one-hour time limit. Using formulation F2-X, which includes our proposed valid inequalities, CPLEX is able to solve 16 instances to optimality . Using formulation F2-Y, which includes the valid inequalities developed by Pan and Guan ( 2016 ), CPLEX is also able to solve 16 instances to optimality . The integrality gaps and CPU time of formulations F2-X and F2-Y are signiﬁcantly smaller than those of formu- lation F2, wher eas the integrality gaps and CPU time of formulation F2-X are comparable to those of formulation F2-Y . T able 6 Performance of MIP Formulations in the Second Experiment Instance # var # bin var # cstr CPU time [TGap] # nodes # user cuts F2 F2-X F2-Y F2-Z F2 F2-X F2-Y F2-Z F2 F2-X F2-Y F2-Z F2-X F2-Y F2-Z 1 3304 1960 12300 16164 14876 18096 710.8 26.0 33.8 17.3 365652 12926 25641 11508 168 98 203 2 4130 2450 15350 20180 18570 22595 ** [0.07%] 819.0 316.2 292.0 900272 577704 321040 213721 294 126 307 3 5192 3080 19354 25426 13402 28462 ** [0.02%] 134.2 104.5 234.3 666918 57024 44507 86655 494 252 439 4 5310 3150 19842 26052 23982 29157 ** [0.01%] 222.7 1675.7 497.4 760760 260255 1732340 505512 203 130 234 5 5782 3430 21556 28318 26064 31699 ** [0.05%] 127.1 458.9 406.6 516797 65305 309318 244711 448 243 463 6 5900 3500 22098 28998 26698 32448 1908.6 219.9 147.1 151.2 375616 268065 213464 151584 515 202 475 7 6018 3570 22458 29496 27150 33015 ** [0.04%] ** [0.01%] 417.0 564.4 618447 3089892 357940 456912 291 132 317 8 6018 3570 22496 29534 27188 33053 ** [0.05%] 419.0 238.9 72.5 526007 240221 147181 27169 407 208 344 9 6136 3640 22896 30072 27680 33660 ** [0.05%] 930.8 1034.6 985.1 434524 440662 495795 397759 535 261 517 10 6372 3780 23846 31298 28814 35024 ** [0.08%] 1252.6 ** [0.02%] 2206.9 310975 455369 1538372 787578 449 233 507 11 15576 9240 58012 76228 70156 85336 ** [0.12%] ** [0.03%] ** [0.03%] 1036.6 180807 256996 570824 184432 806 353 836 12 18408 10920 68630 90158 82982 100922 ** [0.08%] ** [0.01%] 1739.3 **[0.01%] 108631 285196 276194 382696 1484 815 1763 13 18408 10920 68630 90158 82982 100922 ** [0.09%] 733.4 887.2 1205.6 84351 44990 92311 125922 1420 810 1603 14 19470 11550 72312 95082 87492 106467 ** [0.11%] ** [0.01%] ** [0.01%] 1187.2 62170 179443 412241 69854 2285 1570 2619 15 19706 11690 73482 96528 88846 108051 ** [0.13%] 539.7 180.6 438.4 50839 15365 6643 26615 1410 919 1717 16 20296 12040 75640 99376 91464 111244 ** [0.10%] 2790.1 ** [0.01%] 1498.7 142997 191482 249999 157638 1690 762 1630 17 21476 12740 80014 105130 96758 117688 ** [0.10%] 444.3 180.4 244.3 70436 14938 7444 7710 1294 757 1753 18 21476 12740 80026 105142 96770 117700 ** [0.10%] 1686.3 239.2 225.3 41514 148584 24456 7093 1515 1067 1923 19 21594 12810 80450 105704 97286 118331 ** [0.09%] 1685.8 242.5 632.2 93893 136577 19084 51614 1703 1050 1977 20 22066 13090 82264 108070 99468 120973 ** [0.12%] 293.2 245.7 187.7 62970 19072 10870 4069 1299 970 1746 This demonstrates that strong valid inequalities developed for a single-binary formulation can be used for a formulation with more than a single type of binary variables and can achieve compa- rable effectiveness. Comparing the results presented in T ables 6 and 7 with the results presented in T ables 4 and 5 , we observe that formulations F1-X, F2-X, and F2-Y have similar performance. This shows that a single-binary formulation has a similar performance as a formulation that uses more than a single type of binary variables when strong valid inequalities are added to these formula- tions. It also shows that the strong valid inequalities obtained from our single-binary formulation have an effectiveness similar to those strong valid inequalities obtained fr om a formulation that 26 uses more than a single type of binary variables. However , formulation F2-Z outperforms for- mulations F1-X, F2-X, and F2-Y , as CPLEX can solve 19 instances to optimality with F2-Z, and the instance that is not solved optimally has a TGap of only 0.01%. This indicates that our strong valid inequalities derived fr om a single-binary formulation can be used in conjunction with those valid inequalities obtained fr om a formulation with mor e than a single type of binary variables to achieve better performance. T able 7 The Strength of LP Relaxations of MIP Formulations in the Second Experiment IGap Instance 1 2 3 4 5 6 7 8 9 10 F2 0.45% 0.39% 0.38% 0.35% 0.47% 0.60% 0.34% 0.52% 0.48% 0.63% F2-X 0.20% 0.14% 0.07% 0.06% 0.05% 0.04% 0.07% 0.06% 0.06% 0.04% F2-Y 0.20% 0.14% 0.08% 0.05% 0.05% 0.05% 0.08% 0.06% 0.05% 0.05% F2-Z 0.20% 0.14% 0.07% 0.05% 0.05% 0.04% 0.07% 0.06% 0.05% 0.04% Instance 11 12 13 14 15 16 17 18 19 20 F2 0.32% 0.31% 0.36% 0.34% 0.47% 0.29% 0.41% 0.40% 0.34% 0.41% F2-X 0.05% 0.02% 0.02% 0.02% 0.02% 0.02% 0.02% 0.02% 0.02% 0.02% F2-Y 0.06% 0.02% 0.02% 0.02% 0.02% 0.02% 0.02% 0.01% 0.02% 0.01% F2-Z 0.05% 0.02% 0.02% 0.02% 0.02% 0.02% 0.02% 0.01% 0.02% 0.01% Pct. reduction Instance 1 2 3 4 5 6 7 8 9 10 F2-X 55.7% 64.0% 81.2% 82.2% 89.3% 93.2% 79.4% 88.8% 88.3% 93.0% F2-Y 55.7% 63.9% 80.3% 84.3% 90.3% 92.0% 77.5% 88.1% 89.0% 92.4% F2-Z 55.7% 64.0% 81.2% 84.3% 90.3% 93.5% 79.4% 89.0% 89.0% 93.0% Instance 11 12 13 14 15 16 17 18 19 20 F2-X 84.0% 93.7% 94.7% 94.6% 96.1% 94.5% 95.6% 95.7% 93.2% 95.7% F2-Y 82.5% 92.1% 94.5% 94.1% 96.4% 92.0% 96.2% 96.7% 94.4% 96.3% F2-Z 84.0% 93.9% 94.8% 94.7% 96.5% 94.5% 96.2% 96.7% 94.4% 96.4% T ables 8 and 9 present the computational results of the third experiment, in which formulations F1 + and F1 + -X are used to solve the network-constrained UC pr oblem based on the modiﬁed IEEE 118-bus system. The integrality gaps generated by formulation F1 + -X are 44.1% to 66.3% smaller than those generated by formulation F1 + . Using formulation F1 + , CPLEX is able to solve only one of the 20 instances to optimality within one hour . In contrast, using formulation F1 + -X, CPLEX is able to solve 17 instances to optimality within the same time limit. Formulation F1 + -X explores fewer nodes than formulation F1 + , and the number of user cuts added by F1 + -X in the solution process is quite small. These results demonstrate the effectiveness of the strong formulation F1 + -X. Some additional computational results on formulations F1 + and F1 + -X using a more congested demand setting and a less congested demand setting are pr esented in Online Appendix B, which show that formulation F1 + -X is more ef fective when the demand is more congested. 6. Conclusions This paper considers a UC formulation with a single type of binary variables. By analyzing the physical constraints of a single generator , we obtain the convex hull description of the two-period single-UC polytope, which can be used to tighten the original MILP formulation and derive other strong valid inequalities. For the multi-period single-UC polytope, we derive strong valid 27 T able 8 Performance of MIP Formulations in the Third Experiment Instance # var # bin var # cstr CPU time [TGap] # nodes # user cuts F1 + F1 + -X F1 + F1 + -X F1 + F1 + -X F1 + -X 1 6372 1296 36124 43576 ** [0.15%] 571.7 109080 14069 316 2 ** [0.06%] 2324.7 97424 84034 474 3 ** [0.02%] 2251.6 158482 31631 445 4 ** [0.11%] 2803.9 131942 43108 544 5 ** [0.01%] 288.5 175810 7220 394 6 ** [0.17%] 1653.5 116598 24813 441 7 ** [0.27%] 1800.0 122958 43624 545 8 ** [0.11%] 1799.3 180499 22310 591 9 ** [0.13%] 851.4 250053 19961 464 10 2938.4 850.7 181549 18889 449 11 ** [0.08%] 2425.3 196002 59158 463 12 ** [0.15%] ** [0.05%] 287079 44675 619 13 ** [0.14%] 3351.3 164549 82738 563 14 ** [0.17%] 1969.8 235344 37421 486 15 ** [0.30%] ** [0.03%] 266093 111169 696 16 ** [0.10%] 1666.2 112045 42146 438 17 ** [0.13%] 461.9 139242 13578 448 18 ** [0.14%] 2433.3 139375 61670 531 19 ** [0.15%] ** [0.02%] 169572 133348 742 20 ** [0.21%] 3298.6 158536 50471 615 T able 9 The Strength of LP Relaxations of MIP Formulations in the Thir d Experiment IGap Instance 1 2 3 4 5 6 7 8 9 10 F1 + 0.85% 1.03% 0.76% 0.93% 0.86% 0.88% 1.04% 0.92% 0.91% 0.80% F1 + -X 0.36% 0.39% 0.39% 0.52% 0.33% 0.36% 0.35% 0.42% 0.38% 0.40% Instance 11 12 13 14 15 16 17 18 19 20 F1 + 0.90% 1.17% 0.87% 0.91% 1.02% 0.89% 0.84% 0.85% 0.95% 0.89% F1 + -X 0.40% 0.47% 0.37% 0.43% 0.44% 0.40% 0.40% 0.44% 0.39% 0.40% Pct. reduction Instance 1 2 3 4 5 6 7 8 9 10 F1 + -X 57.4% 62.2% 49.0% 44.1% 61.9% 59.7% 66.3% 54.8% 58.8% 52.2% Instance 11 12 13 14 15 16 17 18 19 20 F1 + -X 55.9% 60.0% 57.5% 52.2% 57.4% 55.6% 53.0% 48.4% 59.0% 55.6% inequalities with one and two continuous variables. Conditions under which these valid inequal- ities are facet-deﬁning for the multi-period single-UC polytope are provided. Because the number of inequalities in each valid inequality family is very large, efﬁcient separation algorithms are provided to identify the most violated inequalities. The effectiveness of the proposed strong valid inequalities is demonstrated in solving network-constrained UC problems. Computational results show that our valid inequalities can speed up the solution process signiﬁcantly . Moreover , these strong valid inequalities exhibit effectiveness comparable to two-binary valid inequalities and thus can be used to tighten two/three-binary formulations. V arious intriguing resear ch directions can be pursued following this line of work. First, it would be interesting to investigate the complete convex hull descriptions of the multi-period single-UC polytopes, such as the three-period polytope, and derive str ong valid inequalities with mor e than two continuous variables to further tighten Pr oblem 1. In addition, the discussion of the single- 28 UC polytope can be extended to different parameter settings, such as the case wher e V ≥ C + V or the case where V + V > C . Second, it would be appealing to incorporate differ ent start-up/shut- down trajectories of generators into the physical constraints to accurately r epresent the operation of units and to conduct a polyhedral study on the obtained single-UC polytope to derive str ong valid inequalities. Third, considering the demand and electricity price ﬂuctuations that often occur in practice when dealing with UC problems, it would be interesting to formulate the corr e- sponding stochastic UC problems to better reﬂect real-world scenarios. Fourth, given that differ- ent types of electrical generators (e.g., pumped storage hydro units) may have differ ent physical constraints in addition to those considered in this paper , it would be interesting to derive strong valid inequalities for the UC problems with these speciﬁc generators. For example, consider the hydro UC problem described in Cheng et al. ( 2016 ) in which there are constraints that impose restrictions on the power output in the safe operating zones. By employing a similar approach to the derivation of inequality ( 12 ), we can obtain a valid inequality by deriving an upper bound for “ p i , t − p i , t − k ,” where p i , t and p i , t − k are the power output of unit i in periods t and t − k , r espectively . Fifth, as mentioned in Section 1 , some studies have utilized LR-based methods to tackle complex UC problems by decomposing the multiple-generator pr oblem into independent single-generator subproblems. The strong valid inequalities derived in this paper can be applied to these subprob- lems, thus enhancing their effectiveness. It would be useful to develop mathematical techniques that can integrate them into the decomposition procedur e to improve its ef ﬁciency . Sixth, it would be intriguing to make a theoretical comparison between our strong valid inequalities and those derived in prior studies, such as Pan and Guan ( 2016 ) and Damcı-Kurt et al. ( 2016 ), to reveal the relationships and distinctions between them. W e leave these issues for future resear ch. References Abdul-Rahman, K., Shahidehpour , S., Aganagic, M., and Mokhtari, S. (1996). A practical resour ce schedul- ing with OPF constraints. IEEE T ransactions on Power Systems , 11(1):254–259. Arroyo, J. M. and Conejo, A. J. (2000). Optimal response of a thermal unit to an electricity spot market. IEEE T ransactions on Power Systems , 15(3):1098–1104. Atakan, S., Lulli, G., and Sen, S. (2018). A state transition MIP formulation for the unit commitment problem. IEEE T ransactions on Power Systems , 33(1):736–748. Baldick, R. (1995). The generalized unit commitment problem. IEEE T ransactions on Power Systems , 10(1):465–475. Bendotti, P ., Fouilhoux, P ., and Rottner , C. (2018). The min-up/min-down unit commitment polytope. Jour- nal of Combinatorial Optimization , 36(3):1024–1058. Bendotti, P ., Fouilhoux, P ., and Rottner , C. (2019). On the complexity of the unit commitment problem. Annals of Operations Research , 274(1–2):119–130. Brandenberg, R., Huber , M., and Silbernagl, M. (2017). The summed start-up costs in a unit commitment problem. EURO Journal on Computational Optimization , 5(1):203–238. Carri ´ on, M. and Arroyo, J. M. (2006). A computationally efﬁcient mixed-integer linear formulation for the thermal unit commitment problem. IEEE T ransactions on Power Systems , 21(3):1371–1378. 29 Chang, G. W ., Aganagic, M., W aight, J. G., Medina, J., Burton, T ., Reeves, S., and Christoforidis, M. (2001). Experiences with mixed integer linear programming based approaches on short-term hydr o schedul- ing. IEEE T ransactions on Power Systems , 16(4):743–749. Chang, G. W ., T sai, Y . D., Lai, C. Y ., and Chung, J. S. (2004). A practical mixed integer linear programming based approach for unit commitment. In Proceedings of the IEEE Power Engineering Society General Meeting, 2004 , pages 221–225. IEEE. Cheng, C., W ang, J., and W u, X. (2016). Hydr o unit commitment with a head-sensitive r eservoir and multi- ple vibration zones using MILP. IEEE T ransactions on Power Systems , 31(6):4842–4852. Damcı-Kurt, P ., K ¨ uc ¸ ¨ ukyavuz, S., Rajan, D., and Atamt ¨ urk, A. (2016). A polyhedral study of production ramping. Mathematical Programming , 158(1–2):175–205. Dupin, N. (2017). T ighter MIP formulations for the discr etised unit commitment pr oblem with min-stop ramping constraints. EURO Journal on Computational Optimization , 5(1-2):149–176. Frangioni, A. and Gentile, C. (2006a). Perspective cuts for a class of convex 0–1 mixed integer programs. Mathematical Programming , 106(2):225–236. Frangioni, A. and Gentile, C. (2006b). Solving nonlinear single-unit commitment problems with ramping constraints. Operations Research , 54(4):767–775. Frangioni, A., Gentile, C., and Lacalandra, F . (2009). T ighter approximated MILP formulations for unit commitment problems. IEEE T ransactions on Power Systems , 24(1):105–113. Garver , L. L. (1962). Power generation scheduling by integer programming—Development of theory . T rans. of the American Institute of Electrical Engineers. Part III: Power Apparatus and Systems , 81(3):730–734. Gentile, C., Morales-Espana, G., and Ramos, A. (2017). A tight MIP formulation of the unit commitment problem with start-up and shut-down constraints. EURO J. Comput. Optim. , 5(1):177–201. Guan, Y ., Pan, K., and Zhou, K. (2018). Polynomial time algorithms and extended formulations for unit commitment problems. IISE transactions , 50(8):735–751. Hedman, K. W ., O’Neill, R. P ., and Oren, S. S. (2009). Analyzing valid inequalities of the generation unit commitment problem. In 2009 IEEE/PES Power Systems Confer ence and Exposition , pages 1–6. IEEE. Kazarlis, S. A., Bakirtzis, A. G., and Petridis, V . (1996). A genetic algorithm solution to the unit commitment problem. IEEE T ransactions on Power Systems , 11(1):83–92. Knueven, B., Ostrowski, J., and W ang, J. (2018). The ramping polytope and cut generation for the unit commitment problem. INFORMS Journal on Computing , 30(4):739–749. Knueven, B., Ostrowski, J., and W atson, J.-P . (2020a). A novel matching formulation for startup costs in unit commitment. Mathematical Programming Computation , 12(2):225–248. Knueven, B., Ostrowski, J., and W atson, J.-P . (2020b). On mixed-integer programming formulations for the unit commitment problem. INFORMS Journal on Computing , 32(4):857–876. Lee, J., Leung, J., and Margot, F . (2004). Min-up/min-down polytopes. Discrete Optimization , 1(1):77–85. Li, C., Zhang, M., and Hedman, K. (2021). Extr eme ray feasibility cuts for unit commitment with uncertainty . INFORMS Journal on Computing , 33(3):1037–1055. Li, T . and Shahidehpour , M. (2005). Price-based unit commitment: A case of Lagrangian relaxation versus mixed integer programming. IEEE T ransactions on Power Systems , 20(4):2015–2025. Lu, B. and Shahidehpour , M. (2005). Unit commitment with ﬂexible generating units. IEEE T ransactions on Power Systems , 20(2):1022–1034. Ma, H. and Shahidehpour , S. M. (1999). Unit commitment with transmission security and voltage con- straints. IEEE T ransactions on Power Systems , 14(2):757–764. Morales-Espa ˜ na, G., Gentile, C., and Ramos, A. (2015). T ight MIP formulations of the power-based unit commitment problem. OR Spectrum , 37(4):929–950. Morales-Espa ˜ na, G., Latorre, J. M., and Ramos, A. (2013). T ight and compact MILP formulation for the thermal unit commitment problem. IEEE T ransactions on Power Systems , 28(4):4897–4908. 30 Muckstadt, J. A. and Koenig, S. A. (1977). An application of Lagrangian relaxation to scheduling in power - generation systems. Operations Research , 25(3):387–403. Ostrowski, J., Anjos, M. F ., and V annelli, A. (2012). T ight mixed integer linear programming formulations for the unit commitment problem. IEEE T ransactions on Power Systems , 27(1):39–46. Ouyang, Z. and Shahidehpour , S. M. (1991). An intelligent dynamic programming for unit commitment application. IEEE T ransactions on Power Systems , 6(3):1203–1209. Padhy , N. P . (2004). Unit commitment—A bibliographical survey . IEEE T ransactions on Power Systems , 19(2):1196–1205. Pan, K. and Guan, Y . (2016). A polyhedral study of the integrated minimum-up/-down time and ramping polytope. arXiv preprint . Pan, K., Guan, Y ., W atson, J.-P ., and W ang, J. (2016). Str engthened MILP formulation for certain gas turbine unit commitment problems. IEEE T ransactions on Power Systems , 31(2):1440–1448. Pan, K., Zhao, M., Li, C.-L., and Qiu, F . (2022). A polyhedral study on fuel-constrained unit commitment. INFORMS Journal on Computing , 34(6):3309–3324. Rajan, D. and T akriti, S. (2005). Minimum up/down polytopes of the unit commitment problem with start- up costs. T echnical Report, IBM, Y orktown Heights, NY . Sen, S. and Kothari, D. P . (1998). Optimal thermal generating unit commitment: a review . International Journal of Electrical Power & Energy Systems , 20(7):443–451. Shahidehpour , M., Y amin, H., and Li, Z. (2002). Market Operations in Electric Power Systems: Forecasting, Scheduling, and Risk Management . John W iley & Sons, New Y ork. T akriti, S. and Birge, J. R. (2000). Using integer programming to reﬁne Lagrangian-based unit commitment solutions. IEEE T ransactions on Power Systems , 15(1):151–156. T ejada-Arango, D. A., Lumbreras, S., S ´ anchez-Mart ´ ın, P ., and Ramos, A. (2020). Which unit-commitment formulation is best? A comparison framework. IEEE T ransactions on Power Systems , 35(4):2926–2936. US EIA (2022). Daily U.S. electricity generation from natural gas hit a record in mid-July . Accessed April 23, 2023, https://www.eia.gov/todayinenergy/detail.php ? id = 53559 . V an Ackooij, W ., Danti Lopez, I., Frangioni, A., Lacalandra, F ., and T ahanan, M. (2018). Large-scale unit commitment under uncertainty: an updated literature survey . Annals of Oper . Res. , 271(1):11–85. W ang, C. and Shahidehpour , S. M. (1993). Effects of ramp-rate limits on unit commitment and economic dispatch. IEEE T ransactions on Power Systems , 8(3):1341–1350. W u, L. (2011). A tighter piecewise linear approximation of quadratic cost curves for unit commitment problems. IEEE T ransactions on Power Systems , 26(4):2581–2583. Xavier , ´ A. S., Qiu, F ., and Ahmed, S. (2021). Learning to solve large-scale security-constrained unit commit- ment problems. INFORMS Journal on Computing , 33(2):739–756. Zheng, Q. P ., W ang, J., and Liu, A. L. (2015). Stochastic optimization for unit commitment—A review . IEEE T ransactions on Power Systems , 30(4):1913–1924. ec.1 Online Supplement for “A Polyhedral Study on Unit Commitment with a Single T ype of Binary V ariables” App endix A: Supplement to Section 2 A.1. Pro of of Lemma 1 Lemma 1 . Consider any point ( x , y ) ∈ P and t ∈ [ 2, T ] Z . (i) If y t = 0 , then y t − j − y t − j − 1 ≤ 0 for all j ∈ [ 0, min { t − 2, L − 1 } ] Z . (ii) If y t = 1 , then there exists at most one j ∈ [ 0, min { t − 2, L } ] Z such that y t − j − y t − j − 1 = 1 . Proof. (i) Consider any ( x , y ) ∈ P and any t ∈ [ 2, T ] Z such that y t = 0. Suppose, to the contrary , that there exists j ∈ [ 0, min { t − 2, L − 1 } ] Z such that y t − j − y t − j − 1 = 1. Then, t − j ∈ [ 2, T ] Z . Thus, by ( 2a ), y k = 1 for all k ∈ [ t − j , min { T , t − j + L − 1 } ] Z . It is easy to check that t ∈ [ t − j , min { T , t − j + L − 1 } ] Z . Hence, y t = 1, which is a contradiction. Therefor e, y t − j − y t − j − 1 ≤ 0 for all j ∈ [ 0, min { t − 2, L − 1 } ] Z . (ii) Consider any ( x , y ) ∈ P and any t ∈ [ 2, T ] Z such that y t = 1. Suppose, to the contrary , that there exist j 1 , j 2 ∈ [ 0, min { t − 2, L } ] Z such that j 1 < j 2 and y t − j 1 − y t − j 1 − 1 = y t − j 2 − y t − j 2 − 1 = 1. Because t − j 2 ∈ [ 2, T ] Z , and y t − j 2 − y t − j 2 − 1 = 1, by ( 2a ), y k = 1 for all k ∈ [ t − j 2 , min { T , t − j 2 + L − 1 } ] Z . Note that t − j 1 − 1 ≥ t − j 2 , t − j 1 − 1 ≤ T , and t − j 1 − 1 ≤ t − 1 ≤ t − j 2 + L − 1. Thus, t − j 1 − 1 ∈ [ t − j 2 , min { T , t − j 2 + L − 1 } ] Z . Hence, y t − j 1 − 1 = 1, which contradicts that y t − j 1 − y t − j 1 − 1 = 1. Ther efore, there exists at most one j ∈ [ 0, min { t − 2, L } ] Z such that y t − j − y t − j − 1 = 1. □ A.2. Pro of of Lemma 2 Lemma 2 . Denote x ′ = ( x ′ 1 , . . . , x ′ T ) and y ′ = ( y ′ 1 , . . . , y ′ T ) . Let P ′ =  ( x ′ , y ′ ) ∈ R T + × B T : − y ′ T − t + 2 + y ′ T − t + 1 − y ′ T − k + 1 ≤ 0, ∀ t ∈ [ 2, T ] Z , ∀ k ∈ [ t , min { T , t + L − 1 } ] Z , (EC.1a) y ′ T − t + 2 − y ′ T − t + 1 + y ′ T − k + 1 ≤ 1, ∀ t ∈ [ 2, T ] Z , ∀ k ∈ [ t , min { T , t + ℓ − 1 } ] Z , (EC.1b) − x ′ T − t + 1 + C y ′ T − t + 1 ≤ 0, ∀ t ∈ [ 1, T ] Z , (EC.1c) x ′ T − t + 1 − C y ′ T − t + 1 ≤ 0, ∀ t ∈ [ 1, T ] Z , (EC.1d) x ′ T − t + 1 − x ′ T − t + 2 ≤ Vy ′ T − t + 2 + V ( 1 − y ′ T − t + 2 ) , ∀ t ∈ [ 2, T ] Z , (EC.1e) x ′ T − t + 2 − x ′ T − t + 1 ≤ Vy ′ T − t + 1 + V ( 1 − y ′ T − t + 1 ) , ∀ t ∈ [ 2, T ] Z  . (EC.1f) Then, P = P ′ . Proof. First, consider any element ( x , y ) of P , and we show that ( x , y ) ∈ P ′ . Consider inequality ( EC.1a ) and any t ∈ [ 2, T ] Z . Obviously , ( x , y ) satisﬁes ( EC.1a ) if − y T − t + 2 + y T − t + 1 ≤ 0. Consider the case where − y T − t + 2 + y T − t + 1 > 0 (i.e., y T − t + 2 = 0 and y T − t + 1 = 1). Suppose, to the contrary , that y T − k + 1 = 0 for some k ∈ [ t , min { T , t + L − 1 } ] Z . Then, because y T − k + 1 = 0 and y T − t + 1 = 1, there exists p ∈ [ t , k − 1 ] Z such that y T − p = 0 and y T − p + 1 = 1. This implies that − y T − p + y T − p + 1 − y T − t + 2 = 1. Note that T − p + 1 ∈ [ 2, T ] Z and T − t + 2 ∈ [ T − p + 1, min { T , ( T − p + 1 ) + L − 1 } ] Z . Thus, ( x , y ) violates inequality ( 2a ), which contradicts that ( x , y ) ∈ P . Hence, y T − k + 1 = 1 for all k ∈ [ t , min { T , t + L − 1 } ] Z . Thus, ( x , y ) satisﬁes inequality ( EC.1a ). Consider inequality ( EC.1b ) and any t ∈ [ 2, T ] Z . Obviously , ( x , y ) satisﬁes ( EC.1b ) if y T − t + 2 − y T − t + 1 ≤ 0. Consider the case where y T − t + 2 − y T − t + 1 > 0 (i.e., y T − t + 2 = 1 and y T − t + 1 = 0). Suppose, to the contrary , that y T − k + 1 = 1 for some k ∈ [ t , min { T , t + ℓ − 1 } ] Z . Then, because y T − k + 1 = 1 and y T − t + 1 = 0, there exists p ∈ [ t , k − 1 ] Z such that y T − p = 1 and y T − p + 1 = 0. This implies that y T − p − y T − p + 1 + y T − t + 2 = 2. Note that T − p + 1 ∈ [ 2, T ] Z and T − t + 2 ∈ [ T − p + 1, min { T , ( T − p + 1 ) + ℓ − 1 } ] Z . Thus, ( x , y ) violates inequality ( 2b ), which contradicts that ( x , y ) ∈ P . Hence, y T − k + 1 = 0 for all k ∈ [ t , min { T , t + ℓ − 1 } ] Z . Thus, ( x , y ) satisﬁes inequality ( EC.1b ). Consider inequalities ( EC.1c ) and ( EC.1d ). For any t ∈ [ 1, T ] Z , because ( x , y ) satisﬁes inequalities ( 2c ) and ( 2d ), ( x , y ) also satisﬁes inequalities ( EC.1c ) and ( EC.1d ). Consider inequality ( EC.1e ) and any t ∈ [ 2, T ] Z . Because T − t + 2 ∈ [ 2, T ] Z , by ( 2f ), x T − t + 1 − x T − t + 2 ≤ Vy T − t + 2 + V ( 1 − y T − t + 2 ) . Thus, ( x , y ) satisﬁes inequality ( EC.1e ). Consider inequality ( EC.1f ) and any t ∈ [ 2, T ] Z . Because T − t + 2 ∈ [ 2, T ] Z , by ( 2e ), x T − t + 2 − x T − t + 1 ≤ Vy T − t + 1 + V ( 1 − y T − t + 1 ) . Thus, ( x , y ) satisﬁes inequality ( EC.1f ). Summarizing the above analysis, we conclude that ( x , y ) ∈ P ′ . Next, consider any element ( x ′ , y ′ ) of P ′ , and we show that ( x ′ , y ′ ) ∈ P . ec.2 Consider inequality ( 2a ) and any t ∈ [ 2, T ] Z . Obviously , ( x ′ , y ′ ) satisﬁes ( 2a ) if − y ′ t − 1 + y ′ t ≤ 0. Consider the case wher e − y ′ t − 1 + y ′ t > 0 (i.e., y ′ t − 1 = 0 and y ′ t = 1). Suppose, to the contrary , that y ′ k = 0 for some k ∈ [ t , min { T , t + L − 1 } ] Z . Then, because y ′ t = 1 and y ′ k = 0, there exists p ∈ [ t , k − 1 ] Z such that y ′ p = 1 and y ′ p + 1 = 0. This implies that − y ′ T − ( T − p + 1 )+ 2 + y ′ T − ( T − p + 1 )+ 1 − y ′ T − ( T − t + 2 ) + 1 = 1. Note that T − p + 1 ∈ [ 2, T ] Z and T − t + 2 ∈ [ T − p + 1, min { T , ( T − p + 1 ) + L − 1 } ] Z . Thus, ( x ′ , y ′ ) violates inequality ( EC.1a ), which contradicts that ( x ′ , y ′ ) ∈ P ′ . Hence, y ′ k = 1 for all k ∈ [ t , min { T , t + L − 1 } ] Z . Thus, ( x ′ , y ′ ) satisﬁes inequality ( 2a ). Consider inequality ( 2b ) and any t ∈ [ 2, T ] Z . Obviously , ( x ′ , y ′ ) satisﬁes ( 2b ) if y ′ t − 1 − y ′ t ≤ 0. Consider the case where y ′ t − 1 − y ′ t > 0 (i.e., y ′ t − 1 = 1 and y ′ t = 0). Suppose, to the contrary , that y ′ k = 1 for some k ∈ [ t , min { T , t + ℓ − 1 } ] Z . Then, because y ′ t = 0 and y ′ k = 1, there exists p ∈ [ t , k − 1 ] Z such that y ′ p = 0 and y ′ p + 1 = 1. This implies that y ′ T − ( T − p + 1 )+ 2 − y ′ T − ( T − p + 1 )+ 1 + y ′ T − ( T − t + 2 ) + 1 = 2. Note that T − p + 1 ∈ [ 2, T ] Z and T − t + 2 ∈ [ T − p + 1, min { T , ( T − p + 1 ) + ℓ − 1 } ] Z . Thus, ( x ′ , y ′ ) violates inequality ( EC.1b ), which contradicts that ( x ′ , y ′ ) ∈ P ′ . Hence, y ′ k = 0 for all k ∈ [ t , min { T , t + ℓ − 1 } ] Z . Thus, ( x ′ , y ′ ) satisﬁes inequality ( 2b ). Consider inequalities ( 2c ) and ( 2d ). For any t ∈ [ 1, T ] Z , because ( x ′ , y ′ ) satisﬁes inequalities ( EC.1c ) and ( EC.1d ), ( x ′ , y ′ ) also satisﬁes inequalities ( 2c ) and ( 2d ). Consider inequality ( 2e ) and any t ∈ [ 2, T ] Z . Because T − t + 2 ∈ [ 2, T ] Z , by ( EC.1f ), x ′ T − ( T − t + 2 ) + 2 − x ′ T − ( T − t + 2 ) + 1 ≤ Vy ′ T − ( T − t + 2 ) + 1 + V ( 1 − y ′ T − ( T − t + 2 ) + 1 ) . Hence, x ′ t − x ′ t − 1 ≤ Vy ′ t − 1 + V ( 1 − y ′ t − 1 ) . Thus, ( x ′ , y ′ ) satisﬁes inequality ( 2e ). Consider inequality ( 2f ) and any t ∈ [ 2, T ] Z . Because T − t + 2 ∈ [ 2, T ] Z , by ( EC.1e ), x ′ T − ( T − t + 2 ) + 1 − x ′ T − ( T − t + 2 ) + 2 ≤ Vy ′ T − ( T − t + 2 ) + 2 + V ( 1 − y ′ T − ( T − t + 2 ) + 2 ) . Hence, x ′ t − 1 − x ′ t ≤ Vy ′ t + V ( 1 − y ′ t ) . Thus, ( x ′ , y ′ ) satisﬁes inequality ( 2f ). Summarizing the above analysis, we conclude that ( x ′ , y ′ ) ∈ P . Therefor e, P = P ′ . □ App endix B: Supplement to Section 3 B.1. Pro of of Theorem 1 Theorem 1 . Denote Q 2 =  ( x t − 1 , x t , y t − 1 , y t ) ∈ R 4 : y i ≤ 1, ∀ i ∈ { t − 1, t } , ( 4a ) C y i ≤ x i ≤ C y i , ∀ i ∈ { t − 1, t } , ( 4b ) x t − 1 ≤ V y t − 1 + ( C − V ) y t , ( 4c ) x t ≤ ( C − V ) y t − 1 + V y t , ( 4d ) x t − x t − 1 ≤ ( C + V ) y t − C y t − 1 , ( 4e ) x t − x t − 1 ≤ V y t − ( V − V ) y t − 1 , ( 4f ) x t − 1 − x t ≤ ( C + V ) y t − 1 − C y t , ( 4g ) x t − 1 − x t ≤ V y t − 1 − ( V − V ) y t  . ( 4h ) Then, Q 2 = conv ( P 2 ) . Proof. W e divide the proof into two parts. For the sake of simplicity , we let t = 2 in P 2 and Q 2 . Part I: conv ( P 2 ) ⊆ Q 2 . W e prove that the linear inequalities ( 4a )–( 4h ) are valid for conv ( P 2 ) . T o do so, it sufﬁces to show that ( 4a )–( 4h ) ar e valid for P 2 . Clearly , inequalities ( 4a ) and ( 4b ) hold for any element of P 2 . In the following, we show that ( 4c )–( 4h ) hold for any element of P 2 . For inequality ( 4c ), consider any element ( x 1 , x 2 , y 1 , y 2 ) of P 2 . W e consider three differ ent cases. Case 1: y 1 = 0. In this case, by ( 3b ), x 1 = 0. Thus, ( x 1 , x 2 , y 1 , y 2 ) satisﬁes inequality ( 4c ). Case 2: y 1 = 1 and y 2 = 0. In this case, by ( 3b ), x 2 = 0. Then, by ( 3d ), x 1 ≤ V . Thus, ( x 1 , x 2 , y 1 , y 2 ) satisﬁes inequality ( 4c ). Case 3: y 1 = 1 and y 2 = 1. In this case, by ( 3b ), x 1 ≤ C . Thus, ( x 1 , x 2 , y 1 , y 2 ) satisﬁes inequality ( 4c ). For inequality ( 4d ), consider any element ( x 1 , x 2 , y 1 , y 2 ) of P 2 . W e consider three different cases. Case 1: y 2 = 0. In this case, by ( 3b ), x 2 = 0. Thus, ( x 1 , x 2 , y 1 , y 2 ) satisﬁes inequality ( 4d ). Case 2: y 2 = 1 and y 1 = 0. In ec.3 this case, by ( 3b ), x 1 = 0. Then, by ( 3c ), x 2 ≤ V . Thus, ( x 1 , x 2 , y 1 , y 2 ) satisﬁes inequality ( 4d ). Case 3: y 2 = 1 and y 1 = 1. In this case, by ( 3b ), x 2 ≤ C . Thus, ( x 1 , x 2 , y 1 , y 2 ) satisﬁes inequality ( 4d ). For inequality ( 4e ), consider any element ( x 1 , x 2 , y 1 , y 2 ) of P 2 . W e consider four different cases. Case 1: y 1 = y 2 = 0. In this case, by ( 3b ), x 1 = x 2 = 0. Thus, ( x 1 , x 2 , y 1 , y 2 ) satisﬁes inequality ( 4e ). Case 2: y 1 = y 2 = 1. In this case, by ( 3c ), x 2 − x 1 ≤ V . Thus, ( x 1 , x 2 , y 1 , y 2 ) satisﬁes inequality ( 4e ). Case 3: y 1 = 1 and y 2 = 0. In this case, by ( 3b ), x 2 = 0. By ( 3a ), C ≤ x 1 . Thus, ( x 1 , x 2 , y 1 , y 2 ) satisﬁes inequality ( 4e ). Case 4: y 1 = 0 and y 2 = 1. In this case, by ( 3b ), x 1 = 0. By ( 3c ), x 2 ≤ V < C + V . Thus, ( x 1 , x 2 , y 1 , y 2 ) satisﬁes inequality ( 4e ). For inequality ( 4f ), consider any element ( x 1 , x 2 , y 1 , y 2 ) of P 2 . W e consider four dif ferent cases. Case 1: y 1 = y 2 = 0. In this case, by ( 3b ), x 1 = x 2 = 0. Thus, ( x 1 , x 2 , y 1 , y 2 ) satisﬁes inequality ( 4f ). Case 2: y 1 = y 2 = 1. In this case, by ( 3c ), x 2 − x 1 ≤ V . Thus, ( x 1 , x 2 , y 1 , y 2 ) satisﬁes inequality ( 4f ). Case 3: y 1 = 1 and y 2 = 0. In this case, by ( 3b ), x 2 = 0. By ( 3a ), C ≤ x 1 , which implies that − x 1 < V − V (because V < C + V ). Thus, ( x 1 , x 2 , y 1 , y 2 ) satisﬁes inequality ( 4f ). Case 4: y 1 = 0 and y 2 = 1. In this case, by ( 3b ), x 1 = 0. Then, by ( 3c ), x 2 ≤ V . Thus, ( x 1 , x 2 , y 1 , y 2 ) satisﬁes inequality ( 4f ). For inequality ( 4g ), consider any element ( x 1 , x 2 , y 1 , y 2 ) of P 2 . W e consider four differ ent cases. Case 1: y 1 = y 2 = 0. In this case, by ( 3b ), x 1 = x 2 = 0. Thus, ( x 1 , x 2 , y 1 , y 2 ) satisﬁes inequality ( 4g ). Case 2: y 1 = y 2 = 1. In this case, by ( 3d ), x 1 − x 2 ≤ V . Thus, ( x 1 , x 2 , y 1 , y 2 ) satisﬁes inequality ( 4g ). Case 3: y 1 = 1 and y 2 = 0. In this case, by ( 3b ), x 2 = 0. Then, by ( 3d ), x 1 ≤ V < C + V . Thus, ( x 1 , x 2 , y 1 , y 2 ) satisﬁes inequality ( 4g ). Case 4: y 1 = 0 and y 2 = 1. In this case, by ( 3b ), x 1 = 0. By ( 3a ), C ≤ x 2 . Thus, ( x 1 , x 2 , y 1 , y 2 ) satisﬁes inequality ( 4g ). For inequality ( 4h ), consider any element ( x 1 , x 2 , y 1 , y 2 ) of P 2 . W e consider four differ ent cases. Case 1: y 1 = y 2 = 0. In this case, by ( 3b ), x 1 = x 2 = 0. Thus, ( x 1 , x 2 , y 1 , y 2 ) satisﬁes inequality ( 4h ). Case 2: y 1 = y 2 = 1. In this case, by ( 3d ), x 1 − x 2 ≤ V . Thus, ( x 1 , x 2 , y 1 , y 2 ) satisﬁes inequality ( 4h ). Case 3: y 1 = 1 and y 2 = 0. In this case, by ( 3b ), x 2 = 0. Then, by ( 3d ), x 1 ≤ V . Thus, ( x 1 , x 2 , y 1 , y 2 ) satisﬁes inequality ( 4h ). Case 4: y 1 = 0 and y 2 = 1. In this case, by ( 3b ), x 1 = 0. By ( 3a ), C ≤ x 2 , which implies that − x 2 < V − V (because V < C + V ). Thus, ( x 1 , x 2 , y 1 , y 2 ) satisﬁes inequality ( 4h ). Part II: Q 2 ⊆ conv ( P 2 ) . Consider any given ( ¯ x 1 , ¯ x 2 , ¯ y 1 , ¯ y 2 ) ∈ Q 2 . W e have ¯ y 1 ≤ 1; (EC.3) ¯ y 2 ≤ 1; (EC.4) C ¯ y 1 ≤ ¯ x 1 ≤ C ¯ y 1 ; (EC.5) C ¯ y 2 ≤ ¯ x 2 ≤ C ¯ y 2 ; (EC.6) ¯ x 1 ≤ V ¯ y 1 + ( C − V ) ¯ y 2 ; (EC.7) ¯ x 2 ≤ ( C − V ) ¯ y 1 + V ¯ y 2 ; (EC.8) ¯ x 2 − ¯ x 1 ≤ ( C + V ) ¯ y 2 − C ¯ y 1 ; (EC.9) ¯ x 2 − ¯ x 1 ≤ V ¯ y 2 − ( V − V ) ¯ y 1 ; (EC.10) ¯ x 1 − ¯ x 2 ≤ ( C + V ) ¯ y 1 − C ¯ y 2 ; (EC.11) ¯ x 1 − ¯ x 2 ≤ V ¯ y 1 − ( V − V ) ¯ y 2 . (EC.12) W e prove that ( ¯ x 1 , ¯ x 2 , ¯ y 1 , ¯ y 2 ) can be expr essed as a convex combination of some elements of P 2 . Speciﬁcally , we prove that ther e exist real numbers ρ 1 , ρ 2 , ρ 3 , ρ 4 , λ 1 , λ 2 , λ 3 , λ 4 ≥ 0 such that ( ¯ x 1 , ¯ x 2 , ¯ y 1 , ¯ y 2 ) = λ 1 ( ρ 1 , ρ 2 , 1, 1 ) + λ 2 ( ρ 3 , 0, 1, 0 ) + λ 3 ( 0, ρ 4 , 0, 1 ) + λ 4 ( 0, 0, 0, 0 ) , (EC.13) ( ρ 1 , ρ 2 , 1, 1 ) , ( ρ 3 , 0, 1, 0 ) , ( 0, ρ 4 , 0, 1 ) , ( 0, 0, 0, 0 ) ∈ P 2 , and λ 1 + λ 2 + λ 3 + λ 4 = 1. W e set λ 1 = min { ¯ y 1 , ¯ y 2 } ; λ 2 = ¯ y 1 − λ 1 ; λ 3 = ¯ y 2 − λ 1 ; λ 4 = 1 − ¯ y 1 − ¯ y 2 + λ 1 . Clearly , λ 1 + λ 2 + λ 3 + λ 4 = 1. By ( EC.3 )–( EC.6 ), we have 0 ≤ ¯ y 1 ≤ 1 and 0 ≤ ¯ y 2 ≤ 1. It is easy to verify that λ 1 , λ 2 , λ 3 , λ 4 ≥ 0. W e consider ﬁve differ ent cases. ec.4 Case 1: ¯ y 1 = 0. In this case, λ 1 = 0, λ 2 = 0, λ 3 = ¯ y 2 , and λ 4 = 1 − ¯ y 2 . W e set ρ 1 = ρ 2 = ρ 3 = C and ρ 4 =  ¯ x 2 / ¯ y 2 , if ¯ y 2 > 0; C , if ¯ y 2 = 0. By ( EC.5 ), ¯ x 1 = 0. It is easy to verify that equation ( EC.13 ) holds and that ( ρ 1 , ρ 2 , 1, 1 ) , ( ρ 3 , 0, 1, 0 ) , ( 0, 0, 0, 0 ) ∈ P 2 . Therefor e, it sufﬁces to show that ( 0, ρ 4 , 0, 1 ) ∈ P 2 . Clearly , ( 0, ρ 4 , 0, 1 ) satisﬁes ( 3d ). By ( EC.6 ), ¯ x 2 ≥ C ¯ y 2 , which implies that ρ 4 ≥ C . Thus, ( 0, ρ 4 , 0, 1 ) satisﬁes ( 3a ). By ( EC.6 ), ¯ x 2 ≤ C ¯ y 2 , which implies that ρ 4 ≤ C . Thus, ( 0, ρ 4 , 0, 1 ) satisﬁes ( 3b ). By ( EC.10 ), ¯ x 2 ≤ V ¯ y 2 , which implies that ρ 4 ≤ V . Thus, ( 0, ρ 4 , 0, 1 ) satisﬁes ( 3c ). Therefor e, ( 0, ρ 4 , 0, 1 ) ∈ P 2 . Case 2: ¯ y 2 = 0. In this case, λ 1 = 0, λ 2 = ¯ y 1 , λ 3 = 0, and λ 4 = 1 − ¯ y 1 . W e set ρ 1 = ρ 2 = ρ 4 = C and ρ 3 =  ¯ x 1 / ¯ y 1 , if ¯ y 1 > 0; C , if ¯ y 1 = 0. By ( EC.6 ), ¯ x 2 = 0. It is easy to verify that equation ( EC.13 ) holds and that ( ρ 1 , ρ 2 , 1, 1 ) , ( 0, ρ 4 , 0, 1 ) , ( 0, 0, 0, 0 ) ∈ P 2 . Therefore, it sufﬁces to show that ( ρ 3 , 0, 1, 0 ) ∈ P 2 . Clearly , ( ρ 3 , 0, 1, 0 ) satisﬁes ( 3c ). By ( EC.5 ), ¯ x 1 ≥ C ¯ y 1 , which implies that ρ 3 ≥ C . Thus, ( ρ 3 , 0, 1, 0 ) satisﬁes ( 3a ). By ( EC.5 ), ¯ x 1 ≤ C ¯ y 1 , which implies that ρ 3 ≤ C . Thus, ( ρ 3 , 0, 1, 0 ) satisﬁes ( 3b ). By ( EC.12 ), ¯ x 1 ≤ V ¯ y 1 , which implies that ρ 3 ≤ V . Thus, ( ρ 3 , 0, 1, 0 ) satisﬁes ( 3d ). Therefor e, ( ρ 3 , 0, 1, 0 ) ∈ P 2 . Case 3: 0 < ¯ y 1 < ¯ y 2 . In this case, λ 1 = ¯ y 1 , λ 2 = 0, λ 3 = ¯ y 2 − ¯ y 1 , and λ 4 = 1 − ¯ y 2 . W e set ρ 1 = ¯ x 1 ¯ y 1 ; ρ 2 = 1 ¯ y 1 [ ¯ x 2 − ( ¯ y 2 − ¯ y 1 ) ρ 4 ] ; ρ 3 = C ; ρ 4 = min  ¯ x 2 − C ¯ y 1 ¯ y 2 − ¯ y 1 , ( ¯ x 2 − ¯ x 1 ) + V ¯ y 1 ¯ y 2 − ¯ y 1 , V  . By ( EC.5 ), ρ 1 ≥ C . Note that ρ 4 ≤ ¯ x 2 − C ¯ y 1 ¯ y 2 − ¯ y 1 , which implies that ρ 2 ≥ 1 ¯ y 1 [ ¯ x 2 − ( ¯ y 2 − ¯ y 1 ) ¯ x 2 − C ¯ y 1 ¯ y 2 − ¯ y 1 ] = C . By ( EC.6 ), ¯ x 2 ≥ C ¯ y 2 , which implies that ¯ x 2 − C ¯ y 1 ¯ y 2 − ¯ y 1 ≥ C . By ( EC.11 ), ( ¯ x 2 − ¯ x 1 )+ V ¯ y 1 ¯ y 2 − ¯ y 1 ≥ C . Thus, min { ¯ x 2 − C ¯ y 1 ¯ y 2 − ¯ y 1 , ( ¯ x 2 − ¯ x 1 )+ V ¯ y 1 ¯ y 2 − ¯ y 1 , V } ≥ C ; that is, ρ 4 ≥ C . Hence, ρ 1 , ρ 2 , ρ 3 , ρ 4 ≥ C . It is easy to verify that equation ( EC.13 ) holds and that ( ρ 3 , 0, 1, 0 ) , ( 0, 0, 0, 0 ) ∈ P 2 . Therefor e, it sufﬁces to show that ( ρ 1 , ρ 2 , 1, 1 ) , ( 0, ρ 4 , 0, 1 ) ∈ P 2 . T o show that ( ρ 1 , ρ 2 , 1, 1 ) ∈ P 2 , we ﬁrst note that ( ρ 1 , ρ 2 , 1, 1 ) satisﬁes ( 3a ) (because ρ 1 , ρ 2 ≥ C ). By ( EC.5 ), ρ 1 ≤ C . Note that ρ 2 = 1 ¯ y 1  ¯ x 2 − ( ¯ y 2 − ¯ y 1 ) min  ¯ x 2 − C ¯ y 1 ¯ y 2 − ¯ y 1 , ( ¯ x 2 − ¯ x 1 ) + V ¯ y 1 ¯ y 2 − ¯ y 1 , V  = 1 ¯ y 1  ¯ x 2 − min  ¯ x 2 − C ¯ y 1 , ( ¯ x 2 − ¯ x 1 ) + V ¯ y 1 , ( ¯ y 2 − ¯ y 1 ) V  = 1 ¯ y 1 max  C ¯ y 1 , ¯ x 1 − V ¯ y 1 , ¯ x 2 − ( ¯ y 2 − ¯ y 1 ) V  ≤ 1 ¯ y 1 max  C ¯ y 1 , ¯ x 1 , ¯ x 2 − ( ¯ y 2 − ¯ y 1 ) V  = max  C , ¯ x 1 ¯ y 1 , ¯ x 2 − ( ¯ y 2 − ¯ y 1 ) V ¯ y 1  ≤ C , where the last inequality follows fr om ( EC.5 ) and ( EC.8 ). Hence, ( ρ 1 , ρ 2 , 1, 1 ) satisﬁes ( 3b ). Note that ρ 2 − ρ 1 = 1 ¯ y 1  ( ¯ x 2 − ¯ x 1 ) − ( ¯ y 2 − ¯ y 1 ) min  ¯ x 2 − C ¯ y 1 ¯ y 2 − ¯ y 1 , ( ¯ x 2 − ¯ x 1 ) + V ¯ y 1 ¯ y 2 − ¯ y 1 , V  = 1 ¯ y 1  ( ¯ x 2 − ¯ x 1 ) − min  ¯ x 2 − C ¯ y 1 , ( ¯ x 2 − ¯ x 1 ) + V ¯ y 1 , ( ¯ y 2 − ¯ y 1 ) V  ec.5 = 1 ¯ y 1 max  C ¯ y 1 − ¯ x 1 , − V ¯ y 1 , ( ¯ x 2 − ¯ x 1 ) − ( ¯ y 2 − ¯ y 1 ) V  = max  C − ¯ x 1 ¯ y 1 , − V , ( ¯ x 2 − ¯ x 1 ) − ( ¯ y 2 − ¯ y 1 ) V ¯ y 1  . By ( EC.5 ), C − ¯ x 1 ¯ y 1 ≤ 0. By ( EC.10 ), ( ¯ x 2 − ¯ x 1 ) − ( ¯ y 2 − ¯ y 1 ) V ¯ y 1 ≤ V . Thus, ρ 2 − ρ 1 ≤ V . Hence, ( ρ 1 , ρ 2 , 1, 1 ) satisﬁes ( 3c ). Because ρ 1 − ρ 2 = 1 ¯ y 1  − ( ¯ x 2 − ¯ x 1 ) + ( ¯ y 2 − ¯ y 1 ) ρ 4  ≤ 1 ¯ y 1  − ( ¯ x 2 − ¯ x 1 ) + ( ¯ y 2 − ¯ y 1 ) · ( ¯ x 2 − ¯ x 1 ) + V ¯ y 1 ¯ y 2 − ¯ y 1  = V , ( ρ 1 , ρ 2 , 1, 1 ) satisﬁes ( 3d ). Therefore, ( ρ 1 , ρ 2 , 1, 1 ) ∈ P 2 . T o show that ( 0, ρ 4 , 0, 1 ) ∈ P 2 , we note that ( 0, ρ 4 , 0, 1 ) satisﬁes ( 3a ) and ( 3d ) (because ρ 4 ≥ C ). Because ρ 4 ≤ V ≤ C , ( 0, ρ 4 , 0, 1 ) satisﬁes ( 3b ) and ( 3c ). Therefore, ( 0, ρ 4 , 0, 1 ) ∈ P 2 . Case 4: 0 < ¯ y 2 < ¯ y 1 . In this case, λ 1 = ¯ y 2 , λ 2 = ¯ y 1 − ¯ y 2 , λ 3 = 0, and λ 4 = 1 − ¯ y 1 . W e set ρ 1 = 1 ¯ y 2 [ ¯ x 1 − ( ¯ y 1 − ¯ y 2 ) ρ 3 ] ; ρ 2 = ¯ x 2 ¯ y 2 ; ρ 3 = min  ¯ x 1 − C ¯ y 2 ¯ y 1 − ¯ y 2 , ( ¯ x 1 − ¯ x 2 ) + V ¯ y 2 ¯ y 1 − ¯ y 2 , V  ; ρ 4 = C . Note that ρ 3 ≤ ¯ x 1 − C ¯ y 2 ¯ y 1 − ¯ y 2 , which implies that ρ 1 ≥ 1 ¯ y 2 [ ¯ x 1 − ( ¯ y 1 − ¯ y 2 ) ¯ x 1 − C ¯ y 2 ¯ y 1 − ¯ y 2 ] = C . By ( EC.6 ), ρ 2 ≥ C . By ( EC.5 ), ¯ x 1 ≥ C ¯ y 1 , which implies that ¯ x 1 − C ¯ y 2 ¯ y 1 − ¯ y 2 ≥ C . By ( EC.9 ), ( ¯ x 1 − ¯ x 2 )+ V ¯ y 2 ¯ y 1 − ¯ y 2 ≥ C . Thus, min { ¯ x 1 − C ¯ y 2 ¯ y 1 − ¯ y 2 , ( ¯ x 1 − ¯ x 2 )+ V ¯ y 2 ¯ y 1 − ¯ y 2 , V } ≥ C ; that is, ρ 3 ≥ C . Hence, ρ 1 , ρ 2 , ρ 3 , ρ 4 ≥ C . It is easy to verify that equation ( EC.13 ) holds and that ( 0, ρ 4 , 0, 1 ) , ( 0, 0, 0, 0 ) ∈ P 2 . Therefor e, it sufﬁces to show that ( ρ 1 , ρ 2 , 1, 1 ) , ( ρ 3 , 0, 1, 0 ) ∈ P 2 . T o show that ( ρ 1 , ρ 2 , 1, 1 ) ∈ P 2 , we ﬁrst note that ( ρ 1 , ρ 2 , 1, 1 ) satisﬁes ( 3a ) (because ρ 1 , ρ 2 ≥ C ). Note that ρ 1 = 1 ¯ y 2  ¯ x 1 − ( ¯ y 1 − ¯ y 2 ) min  ¯ x 1 − C ¯ y 2 ¯ y 1 − ¯ y 2 , ( ¯ x 1 − ¯ x 2 ) + V ¯ y 2 ¯ y 1 − ¯ y 2 , V  = 1 ¯ y 2  ¯ x 1 − min  ¯ x 1 − C ¯ y 2 , ( ¯ x 1 − ¯ x 2 ) + V ¯ y 2 , ( ¯ y 1 − ¯ y 2 ) V  = 1 ¯ y 2 max  C ¯ y 2 , ¯ x 2 − V ¯ y 2 , ¯ x 1 − ( ¯ y 1 − ¯ y 2 ) V  ≤ 1 ¯ y 2 max  C ¯ y 2 , ¯ x 2 , ¯ x 1 − ( ¯ y 1 − ¯ y 2 ) V  = max  C , ¯ x 2 ¯ y 2 , ¯ x 1 − ( ¯ y 1 − ¯ y 2 ) V ¯ y 2  ≤ C , where the last inequality follows fr om ( EC.6 ) and ( EC.7 ). By ( EC.6 ), ρ 2 ≤ C . Hence, ( ρ 1 , ρ 2 , 1, 1 ) satisﬁes ( 3b ). Because ρ 2 − ρ 1 = 1 ¯ y 2  − ( ¯ x 1 − ¯ x 2 ) + ( ¯ y 1 − ¯ y 2 ) ρ 3  ≤ 1 ¯ y 2  − ( ¯ x 1 − ¯ x 2 ) + ( ¯ y 1 − ¯ y 2 ) · ( ¯ x 1 − ¯ x 2 ) + V ¯ y 2 ¯ y 1 − ¯ y 2  = V , ( ρ 1 , ρ 2 , 1, 1 ) satisﬁes ( 3c ). Note that ρ 1 − ρ 2 = 1 ¯ y 2  ( ¯ x 1 − ¯ x 2 ) − ( ¯ y 1 − ¯ y 2 ) min  ¯ x 1 − C ¯ y 2 ¯ y 1 − ¯ y 2 , ( ¯ x 1 − ¯ x 2 ) + V ¯ y 2 ¯ y 1 − ¯ y 2 , V  = 1 ¯ y 2  ( ¯ x 1 − ¯ x 2 ) − min  ¯ x 1 − C ¯ y 2 , ( ¯ x 1 − ¯ x 2 ) + V ¯ y 2 , ( ¯ y 1 − ¯ y 2 ) V  ec.6 = 1 ¯ y 2 max  C ¯ y 2 − ¯ x 2 , − V ¯ y 2 , ( ¯ x 1 − ¯ x 2 ) − ( ¯ y 1 − ¯ y 2 ) V  = max  C − ¯ x 2 ¯ y 2 , − V , ( ¯ x 1 − ¯ x 2 ) − ( ¯ y 1 − ¯ y 2 ) V ¯ y 2  . By ( EC.6 ), C − ¯ x 2 ¯ y 2 ≤ 0. By ( EC.12 ), ( ¯ x 1 − ¯ x 2 ) − ( ¯ y 1 − ¯ y 2 ) V ¯ y 2 ≤ V . Thus, ρ 1 − ρ 2 ≤ V . Hence, ( ρ 1 , ρ 2 , 1, 1 ) satisﬁes ( 3d ). Therefor e, ( ρ 1 , ρ 2 , 1, 1 ) ∈ P 2 . T o show that ( ρ 3 , 0, 1, 0 ) ∈ P 2 , we note that ( ρ 3 , 0, 1, 0 ) satisﬁes ( 3a ) and ( 3c ) (because ρ 3 ≥ C ). Because ρ 3 ≤ V ≤ C , ( ρ 3 , 0, 1, 0 ) satisﬁes ( 3b ) and ( 3d ). Therefore, ( ρ 3 , 0, 1, 0 ) ∈ P 2 . Case 5: 0 < ¯ y 1 = ¯ y 2 . In this case, λ 1 = ¯ y 1 = ¯ y 2 , λ 2 = 0, λ 3 = 0, and λ 4 = 1 − ¯ y 1 = 1 − ¯ y 2 . W e set ρ 1 = ¯ x 1 ¯ y 1 ; ρ 2 = ¯ x 2 ¯ y 2 ; ρ 3 = C ; ρ 4 = C . Clearly , ρ 1 , ρ 2 , ρ 3 , ρ 4 ≥ 0. It is easy to verify that equation ( EC.13 ) holds and that ( ρ 3 , 0, 1, 0 ) , ( 0, ρ 4 , 0, 1 ) , ( 0, 0, 0, 0 ) ∈ P 2 . Therefore, it sufﬁces to show that ( ρ 1 , ρ 2 , 1, 1 ) ∈ P 2 . By ( EC.5 ) and ( EC.6 ), ρ 1 , ρ 2 ≥ C and ρ 1 , ρ 2 ≤ C . Thus, ( ρ 1 , ρ 2 , 1, 1 ) satisﬁes ( 3a ) and ( 3b ). By ( EC.9 ), ¯ x 2 − ¯ x 1 ≤ V ¯ y 1 , which implies that ρ 2 − ρ 1 ≤ V . Thus, ( ρ 1 , ρ 2 , 1, 1 ) satisﬁes ( 3c ). By ( EC.11 ), ¯ x 1 − ¯ x 2 ≤ V ¯ y 1 , which implies that ρ 1 − ρ 2 ≤ V . Thus, ( ρ 1 , ρ 2 , 1, 1 ) satisﬁes ( 3d ). Therefore, ( ρ 1 , ρ 2 , 1, 1 ) ∈ P 2 . Combining Cases 1–5, we conclude that there exist ρ 1 , ρ 2 , ρ 3 , ρ 4 , λ 1 , λ 2 , λ 3 , λ 4 ≥ 0 that satisfy equation ( EC.13 ). Hence, Q 2 ⊆ conv ( P 2 ) . Note that Damcı-Kurt et al. ( 2016 ) have also derived the two-period convex hulls for the two-period ramp-up and ramp-down polytopes separately . In contrast, our Theorem 1 provides the convex hull for the two-period UC polytope containing both the ramp-up and ramp-down polytopes. □ App endix C: Supplement to Section 4 C.1. Pro of of Prop osition 1 Proposition 1 . Consider any S ⊆ [ 0, min { L − 1, T − 2, ⌊ ( C − V ) / V ⌋ } ] Z . For any t ∈ [ 1, T ] Z such that t ≥ s + 2 for all s ∈ S , the inequality x t ≤ C y t − ∑ s ∈ S ( C − V − s V ) ( y t − s − y t − s − 1 ) ( 13 ) is valid and facet-deﬁning for conv ( P ) . For any t ∈ [ 1, T ] Z such that t ≤ T − s − 1 for all s ∈ S , the inequality x t ≤ C y t − ∑ s ∈ S ( C − V − s V ) ( y t + s − y t + s + 1 ) ( 14 ) is valid and facet-deﬁning for conv ( P ) . Proof. W e ﬁrst pr ove that inequality ( 13 ) is valid and facet-deﬁning for conv ( P ) . Note that the pr oof of facet-deﬁning of ( 13 ) here can also be used to prove the facet-deﬁning of ( 13 ) in Pr oposition 2 . For notational convenience, we deﬁne s max = max { s : s ∈ S } if S  = ∅ , and s max = − 1 if S = ∅ . Consider any t ∈ [ s max + 2, T ] Z (i.e., t ∈ [ 1, T ] Z such that t ≥ s + 2 for all s ∈ S ). T o prove that linear inequalities ( 13 ) is valid for conv ( P ) , it sufﬁces to show that they are valid for P . Consider any element ( x , y ) of P . W e show that ( x , y ) satisﬁes ( 13 ). Case 1: y t = 0. By Lemma 1 (i), y t − j − y t − j − 1 ≤ 0 for all j ∈ [ 0, min { t − 2, L − 1 } ] Z . Because S ⊆ [ 0, L − 1 ] Z and s max ≤ t − 2, we have S ⊆ [ 0, min { t − 2, L − 1 } ] Z . Thus, y t − s − y t − s − 1 ≤ 0 for all s ∈ S . Because S ⊆ [ 0, ⌊ ( C − V ) / V ⌋ ] Z , we have C − V − sV ≥ 0 for all s ∈ S . Thus, the right-hand side of inequality ( 13 ) is nonnegative. Because y t = 0, by ( 2d ), x t = 0. Therefor e, in this case, ( x , y ) satisﬁes ( 13 ). Case 2: y t = 1 and y t − s ′ − y t − s ′ − 1 = 1 for some s ′ ∈ S . By Lemma 1 (ii), there exists at most one j ∈ [ 0, min { t − 2, L } ] Z such that y t − j − y t − j − 1 = 1. Because S ⊆ [ 0, L ] Z and s max ≤ t − 2, we have S ⊆ [ 0, min { t − 2, L } ] Z . This implies that y t − s − y t − s − 1 ≤ 0, for all for all s ∈ S \ { s ′ } . For any s ∈ S , because s ≤ ⌊ ( C − V ) / V ⌋ , we have C − V − sV ≥ 0. Thus, for any s ∈ S \ { s ′ } , ( C − V − sV ) ( y t − s − y t − s − 1 ) is either zero or negative. Hence, ∑ s ∈ S ( C − V − s V )( y t − s − y t − s − 1 ) ≤ C − V − s ′ V . Thus, the right-hand side of inequality ( 13 ) is at least s ′ V + V . By ( 2e ), ∑ t τ = t − s ′ ( x τ − x τ − 1 ) ≤ ∑ t τ = t − s ′ Vy τ − 1 + ∑ t τ = t − s j V ( 1 − y τ − 1 ) , which implies ec.7 that x t − x t − s ′ − 1 ≤ s ′ V + V . Because y t − s ′ − y t − s ′ − 1 = 1, we have y t − s ′ − 1 = 0. By ( 2d ), x t − s ′ − 1 = 0. Hence, x t ≤ s ′ V + V . Therefor e, in this case, ( x , y ) satisﬁes ( 13 ). Case 3: y t = 1 and y t − s − y t − s − 1  = 1 for all s ∈ S . In this case, y t − s − y t − s − 1 ≤ 0 for all s ∈ S . For any s ∈ S , because s ≤ ⌊ ( C − V ) / V ⌋ , we have C − V − sV ≥ 0. Thus, the right-hand side of inequality ( 13 ) is at least C . By ( 2d ), x t ≤ C . Therefor e, in this case, ( x , y ) satisﬁes ( 13 ). Summarizing Cases 1–3, we conclude that ( x , y ) satisﬁes ( 13 ). Hence, ( 13 ) is valid for conv ( P ) . Consider any t ∈ [ s max + 2, T ] Z . T o prove that inequality ( 13 ) is facet-deﬁning for conv ( P ) , it suf ﬁces to show that there exist 2 T afﬁnely independent points in conv ( P ) that satisfy ( 13 ) at equality . Because 0 ∈ conv ( P ) and 0 satisﬁes ( 13 ) at equality , it sufﬁces to create the remaining 2 T − 1 nonzero linearly independent points. W e denote these 2 T − 1 points as ( ¯ x r , ¯ y r ) for r ∈ [ 1, T ] Z \ { t } and ( ˆ x r , ˆ y r ) for r ∈ [ 1, T ] Z , and denote the q th component of ¯ x r , ¯ y r , ˆ x r , and ˆ y r as ¯ x r q , ¯ y r q , ˆ x r q , and ˆ y r q , respectively . Let ϵ = V − C > 0. W e divide these 2 T − 1 points into the following four groups: (A1) For each r ∈ [ 1, T ] Z \ { t } , we create a point ( ¯ x r , ¯ y r ) as follows: ¯ x r q =  C , for q ∈ [ 1, T ] Z \ { r } ; C − ϵ , for q = r ; and ¯ y r q = 1 for all q ∈ [ 1, T ] Z . It is easy to verify that ( ¯ x r , ¯ y r ) satisﬁes ( 2a )–( 2f ). Thus, ( ¯ x r , ¯ y r ) ∈ conv ( P ) . It is also easy to verify that ( ¯ x r , ¯ y r ) satisﬁes ( 13 ) at equality . (A2) For each r ∈ [ 1, t − 1 ] Z , we create a point ( ˆ x r , ˆ y r ) as follows: If t − r − 1 / ∈ S , then ˆ x r q =  C , for q ∈ [ 1, r ] Z ; 0, for q ∈ [ r + 1, T ] Z ; and ˆ y r q =  1, for q ∈ [ 1, r ] Z ; 0, for q ∈ [ r + 1, T ] Z . If t − r − 1 ∈ S , then ˆ x r q =    0, for q ∈ [ 1, r ] Z ; V + ( q − r − 1 ) V , for q ∈ [ r + 1, t ] Z ; V + ( t − r − 1 ) V , for q ∈ [ t + 1, T ] Z ; and ˆ y r q =  0, for q ∈ [ 1, r ] Z ; 1, for q ∈ [ r + 1, T ] Z . W e ﬁrst consider the case where t − r − 1 / ∈ S . In this case, it is easy to verify that ( ˆ x r , ˆ y r ) satisﬁes ( 2a )– ( 2f ) and is therefor e in conv ( P ) . Note that in this case ˆ x r t = ˆ y r t = 0, and t − s − 1  = r for all s ∈ S , which implies that ˆ y r t − s − ˆ y r t − s − 1 = 0 for all s ∈ S . Thus, ( ˆ x r , ˆ y r ) satisﬁes ( 13 ) at equality . Next, we consider the case where t − r − 1 ∈ S . In this case, it is easy to verify that ( ˆ x r , ˆ y r ) satisﬁes ( 2a ) and ( 2b ). For each q ∈ [ 1, r ] Z , we have ˆ x r q = ˆ y r q = 0. For each q ∈ [ r + 1, T ] Z , because t − r − 1 ∈ S , we have t − r − 1 ≤ ⌊ ( C − V ) / V ⌋ , which implies that V + ( t − r − 1 ) V ≤ C , which in turn implies that C ≤ ˆ x r q ≤ C . Hence, ( ˆ x r , ˆ y r ) satisﬁes ( 2c ) and ( 2d ). Note that ˆ x r q − ˆ x r q − 1 = 0 when q ∈ [ 2, r ] Z , ˆ x r q − ˆ x r q − 1 = V when q = r + 1, and 0 ≤ ˆ x r q − ˆ x r q − 1 ≤ V when q ∈ [ r + 2, T ] Z . Thus, − V ˆ y r q − V ( 1 − ˆ y r q ) ≤ ˆ x r q − ˆ x r q − 1 ≤ V ˆ y r q − 1 + V ( 1 − ˆ y r q − 1 ) for all q ∈ [ 2, T ] Z . Hence, ( ˆ x r , ˆ y r ) satisﬁes ( 2e ) and ( 2f ). Therefor e, ( ˆ x r , ˆ y r ) ∈ conv ( P ) . Note that in this case ˆ x r t = V + ( t − r − 1 ) V , ˆ y r t = 1, ˆ y r t − s − ˆ y r t − s − 1 = 1 when s = t − r − 1, and ˆ y r t − s − ˆ y r t − s − 1 = 0 when s  = t − r − 1. Thus, ( ˆ x r , ˆ y r ) satisﬁes ( 13 ) at equality . (A3) W e create a point ( ˆ x t , ˆ y t ) by setting ˆ x t q = C and ˆ y t q = 1 for q ∈ [ 1, T ] Z . It is easy to verify that ( ˆ x t , ˆ y t ) satisﬁes ( 2a )–( 2f ). Thus, ( ˆ x t , ˆ y t ) ∈ conv ( P ) . It is also easy to verify that ( ˆ x t , ˆ y t ) satisﬁes ( 13 ) at equality . (A4) For each r ∈ [ t + 1, T ] Z , we create a point ( ˆ x r , ˆ y r ) as follows: ˆ x r q =  0, for q ∈ [ 1, r − 1 ] Z ; C , for q ∈ [ r , T ] Z ; and ˆ y r q =  0, for q ∈ [ 1, r − 1 ] Z ; 1, for q ∈ [ r , T ] Z . It is easy to verify that ( ˆ x r , ˆ y r ) satisﬁes ( 2a )–( 2f ). Thus, ( ˆ x r , ˆ y r ) ∈ conv ( P ) . It is also easy to verify that ( ˆ x r , ˆ y r ) satisﬁes ( 13 ) at equality . ec.8 T able EC.1 shows a matrix with 2 T − 1 rows, where each row represents a point created by this process. This matrix can be transformed into the matrix in T able EC.2 via the following Gaussian elimination process: (i) For each r ∈ [ 1, T ] Z \ { t } , the point with index r in group (B1), denoted ( ¯ x r , ¯ y r ) , is obtained by setting ( ¯ x r , ¯ y r ) = ( ¯ x r , ¯ y r ) − ( ˆ x t , ˆ y t ) . Here, ( ¯ x r , ¯ y r ) is the point with index r in group (A1) , and ( ˆ x t , ˆ y t ) is the point in group (A3) . (ii) For each r ∈ [ 1, t − 1 ] Z , the point with index r in group (B2), denoted ( ˆ x r , ˆ y r ) , is obtained by setting ( ˆ x r , ˆ y r ) = ( ˆ x r , ˆ y r ) if t − r − 1 / ∈ S , and setting ( ˆ x r , ˆ y r ) = ( ˆ x t , ˆ y t ) − ( ˆ x r , ˆ y r ) if t − r − 1 ∈ S . Here, ( ˆ x r , ˆ y r ) is the point with index r in group (A2) , and ( ˆ x t , ˆ y t ) is the point in group (A3) . (iii) The point in group (B3), denoted ( ˆ x t , ˆ y t ) , is obtained by setting ( ˆ x t , ˆ y t ) = ( ˆ x t , ˆ y t ) − ( ˆ x t + 1 , ˆ y t + 1 ) . Here, ( ˆ x t , ˆ y t ) is the point in group (A3) , and ( ˆ x t + 1 , ˆ y t + 1 ) is the point with index t + 1 in group (A4) . (iv) For each r ∈ [ t + 1, T ] Z , the point with index r in group (B4), denoted ( ˆ x r , ˆ y r ) , is obtained by setting ( ˆ x r , ˆ y r ) = ( ˆ x r , ˆ y r ) − ( ˆ x r + 1 , ˆ y r + 1 ) if r  = T , and setting ( ˆ x r , ˆ y r ) = ( ˆ x r , ˆ y r ) if r = T . Here, ( ˆ x r , ˆ y r ) and ( ˆ x r + 1 , ˆ y r + 1 ) are the points with indices r and r + 1, respectively , in group (A4) . The matrix shown in T able EC.2 is lower triangular; that is, the position of the last nonzero component of a row of the matrix is greater than the position of the last nonzero component of the previous row . This implies that the 2 T − 1 points in gr oups (A1) – (A4) are linearly independent. Ther efore, inequality ( 13 ) is facet-deﬁning for conv ( P ) . Next, we show that inequality ( 14 ) is valid and facet-deﬁning for conv ( P ) . Note that this proof can also be used to prove the validity and facet-deﬁning of inequality ( 14 ) in Proposition 2 . Denote x ′ t = x T − t + 1 and y ′ t = y T − t + 1 for t ∈ [ 1, T ] Z . Because inequality ( 13 ) is valid and facet-deﬁning for conv ( P ) for any t ∈ [ s max + 2, T ] Z , the inequality x ′ T − t + 1 ≤ C y ′ T − t + 1 − ∑ s ∈ S ( C − V − s V ) ( y ′ T − t + s + 1 − y ′ T − t + s + 2 ) is valid and facet-deﬁning for conv ( P ′ ) for any t ∈ [ s max + 2, T ] Z . Let t ′ = T − t + 1. Then, the inequality x ′ t ′ ≤ C y ′ t ′ − ∑ s ∈ S ( C − V − s V ) ( y ′ t ′ + s − y ′ t ′ + s + 1 ) is valid and facet-deﬁning for conv ( P ′ ) for any t ′ ∈ [ 1, T − s max − 1 ] Z . Hence, by Lemma 2 , inequality ( 14 ) is valid and facet-deﬁning for conv ( P ) for any t ∈ [ 1, T − s max − 1 ] Z . □ C.2. Pro of of Prop osition 2 Proposition 2 . Consider any integers α , β , and s max such that (a) L ≤ s max ≤ min { T − 2, ⌊ ( C − V ) / V ⌋ } , (b) 0 ≤ α < β ≤ s max , and (c) β = α + 1 or s max ≤ L + α . Let S = [ 0, α ] Z ∪ [ β , s max ] Z . For any t ∈ [ s max + 2, T ] Z , inequality ( 13 ) is valid and facet-deﬁning for conv ( P ) . For any t ∈ [ 1, T − s max − 1 ] Z , inequality ( 14 ) is valid and facet-deﬁning for conv ( P ) . Proof. Consider any t ∈ [ s max + 2, T ] Z . T o prove that the linear inequality ( 13 ) is valid for conv ( P ) when S = [ 0, α ] Z ∪ [ β , s max ] Z , it sufﬁces to show that ( 13 ) is valid for P when S = [ 0, α ] Z ∪ [ β , s max ] Z . Consider any element ( x , y ) of P . W e show that ( x , y ) satisﬁes ( 13 ) when S = [ 0, α ] Z ∪ [ β , s max ] Z . W e divide the analysis into four cases. Case 1: y t = 0 and y t − s − y t − s − 1 ≤ 0 for all s ∈ S . Because S ⊆ [ 0, ⌊ ( C − V ) / V ⌋ ] Z , we have C − V − s V ≥ 0 for all s ∈ S . Thus, in this case, the right-hand side of inequality ( 13 ) is nonnegative. Because y t = 0, by ( 2d ), x t = 0. Therefor e, in this case, ( x , y ) satisﬁes ( 13 ). Case 2: y t = 0 and y t − s − y t − s − 1 > 0 for some s ∈ S . Let ˜ S = { σ ∈ S : y t − σ − y t − σ − 1 > 0 } and v = | ˜ S | . Then, v ≥ 1. Denote ˜ S = { σ 1 , σ 2 , . . . , σ v } , where σ 1 < σ 2 < · · · < σ v . Note that y t − σ j − 1 = 0 and y t − σ j = 1 for j = 1, . . . , v . Denote σ 0 = − 1. Then for each j = 1, . . . , v , there exists σ ′ j ∈ [ σ j − 1 + 1, σ j − 1 ] Z such that y t − σ ′ j − 1 = 1 and y t − σ ′ j = 0. Thus, 0 ≤ σ ′ 1 < σ 1 < σ ′ 2 < σ 2 · · · < σ ′ v < σ v ≤ s max . Because y t − σ v − y t − σ v − 1 = 1 and t − σ v ∈ [ 2, T ] Z , by ( 2a ), y k = 1 for all k ∈ [ t − σ v , min { T , t − σ v + L − 1 } ] Z , which implies that t − σ ′ j ≥ t − σ v + L for j = 1, . . . , v . Hence, for j = 1, . . . , v , we have σ ′ j ≤ σ v − L , which implies that σ ′ j ≤ s max − L . (EC.14) ec.9 T able EC.1 A matrix with the rows r epresenting 2 T − 1 points in conv ( P ) that satisfy inequality ( 13 ) at equality Group Point Index r x y 1 · · · t − 1 t t + 1 · · · T 1 · · · t − 1 t t + 1 · · · T (A1) ( ¯ x r , ¯ y r ) 1 C − ϵ · · · C C C · · · C 1 · · · 1 1 1 · · · 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . t − 1 C · · · C − ϵ C C · · · C 1 · · · 1 1 1 · · · 1 t + 1 C · · · C C C − ϵ · · · C 1 · · · 1 1 1 · · · 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . T C · · · C C C · · · C − ϵ 1 · · · 1 1 1 · · · 1 (A2) ( ˆ x r , ˆ y r ) 1 (See Note EC.1 -1) (See Note EC.1 -1) . . . t − 1 (A3) t C · · · C C C · · · C 1 · · · 1 1 1 · · · 1 (A4) t + 1 0 · · · 0 0 C · · · C 0 · · · 0 0 1 · · · 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . T 0 · · · 0 0 0 · · · C 0 · · · 0 0 0 · · · 1 Note EC.1 -1: For r ∈ [ 1, t − 1 ] Z , the x and y vectors in group (A2) are given as follows: ˆ x r = ( C , . . . , C | {z } r terms , 0, . . . , 0 | { z } T − r terms ) and ˆ y r = ( 1, . . . , 1 | { z } r terms , 0, . . . , 0 | { z } T − r terms ) if t − r − 1 / ∈ S ; ˆ x r = ( 0, . . . , 0 | { z } r terms , V , V + V , V + 2 V , . . . , V + ( t − r − 1 ) V | { z } t − r terms , V + ( t − r − 1 ) V , . . . , V + ( t − r − 1 ) V | {z } T − t terms ) and ˆ y r = ( 0, . . . , 0 | { z } r terms , 1, . . . , 1 | { z } T − r terms ) if t − r − 1 ∈ S . ec.10 T able EC.2 Lower triangular matrix obtained from T able EC.1 via Gaussian elimination Group Point Index r x y 1 · · · t − 1 t t + 1 · · · T 1 · · · t − 1 t t + 1 · · · T (B1) ( ¯ x r , ¯ y r ) 1 − ϵ · · · 0 0 0 · · · 0 0 · · · 0 0 0 · · · 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . t − 1 0 · · · − ϵ 0 0 · · · 0 0 · · · 0 0 0 · · · 0 t + 1 0 · · · 0 0 − ϵ · · · 0 0 · · · 0 0 0 · · · 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . T 0 · · · 0 0 0 · · · − ϵ 0 · · · 0 0 0 · · · 0 (B2) ( ˆ x r , ˆ y r ) 1 1 · · · 0 0 0 · · · 0 . . . (Omitted) . . . . . . . . . . . . . . . . . . t − 1 1 · · · 1 0 0 · · · 0 (B3) t (Omitted) 1 · · · 1 1 0 · · · 0 (B4) t + 1 0 · · · 0 0 1 · · · 0 . . . (Omitted) . . . . . . . . . . . . . . . . . . T 0 · · · 0 0 0 · · · 1 ec.11 If β = α + 1, then S = [ 0, s max ] Z , which implies that σ ′ j ∈ S for j = 1, . . . , v . If β  = α + 1, then condition (c) of Proposition 2 implies that s max ≤ L + α , which, by ( EC.14 ), implies that σ ′ j ≤ α for j = 1, . . . , v . Thus, in both cases, σ ′ j ∈ S for j = 1, . . . , v . Because y t = 0, by ( 2d ), x t = 0. Hence, the left-hand side of inequality ( 13 ) is 0. Because S ⊆ [ 0, ⌊ ( C − V ) / V ⌋ ] Z , we have C − V − sV ≥ 0 for all s ∈ S . Note that { σ ′ 1 , . . . , σ ′ v } ⊆ S \ ˜ S and y t − s − y t − s − 1 ≤ 0 for all s ∈ S \ ˜ S . Thus, ∑ s ∈ S \ ˜ S ( C − V − s V ) ( y t − s − y t − s − 1 ) ≤ ∑ v j = 1 ( C − V − σ ′ j V )( y t − σ ′ j − y t − σ ′ j − 1 ) . Hence, the right-hand side of inequality ( 13 ) is C y t − ∑ s ∈ S ( C − V − s V ) ( y t − s − y t − s − 1 ) = − ∑ s ∈ ˜ S ( C − V − s V ) ( y t − s − y t − s − 1 ) − ∑ s ∈ S \ ˜ S ( C − V − s V ) ( y t − s − y t − s − 1 ) ≥ − v ∑ j = 1 ( C − V − σ j V )( y t − σ j − y t − σ j − 1 ) − v ∑ j = 1 ( C − V − σ ′ j V )( y t − σ ′ j − y t − σ ′ j − 1 ) = − v ∑ j = 1 ( C − V − σ j V ) + v ∑ j = 1 ( C − V − σ ′ j V ) = v ∑ j = 1 ( σ j − σ ′ j ) V > 0. Therefor e, in this case, ( x , y ) satisﬁes ( 13 ). Case 3: y t = 1 and y t − s − y t − s − 1 ≤ 0 for all s ∈ S . Because S ⊆ [ 0, ⌊ ( C − V ) / V ⌋ ] Z , we have C − V − s V ≥ 0 for all s ∈ S . Thus, in this case, the right-hand side of inequality ( 13 ) is at least C . By ( 2d ), x t ≤ C . Ther efore, in this case, ( x , y ) satisﬁes ( 13 ). Case 4: y t = 1 and y t − s − y t − s − 1 > 0 for some s ∈ S . Let ˜ S = { σ ∈ S : y t − σ − y t − σ − 1 > 0 } and v = | ˜ S | . Then, v ≥ 1. Denote ˜ S = { σ 1 , σ 2 , . . . , σ v } , where σ 1 < σ 2 < · · · < σ v . Note that y t − σ j − 1 = 0 and y t − σ j = 1 for j = 1, . . . , v . Then, for each j = 2, . . . , v , ther e exists σ ′ j ∈ [ σ j − 1 + 1, σ j − 1 ] Z such that y t − σ ′ j − 1 = 1 and y t − σ ′ j = 0. Thus, 0 ≤ σ 1 < σ ′ 2 < σ 2 < · · · < σ ′ v < σ v ≤ s max . In addition, y k = 1 for all k ∈ [ t − σ 1 , t ] Z . Because y t − σ v − y t − σ v − 1 = 1 and t − σ v ∈ [ 2, T ] Z , by ( 2a ), y k = 1 for all k ∈ [ t − σ v , min { T , t − σ v + L − 1 } ] Z , which implies that t − σ ′ j ≥ t − σ v + L for j = 2, . . . , v . Hence, for j = 2, . . . , v , we have σ ′ j ≤ σ v − L , which implies that σ ′ j ≤ s max − L . (EC.15) If β = α + 1, then S = [ 0, s max ] Z , which implies that σ ′ j ∈ S for j = 2, . . . , v . If β  = α + 1, then condition (c) of Proposition 2 implies that s max ≤ L + α , which, by ( EC.15 ) implies that σ ′ j ≤ α for all j = 2, . . . , v . Thus, in both cases, σ ′ j ∈ S for j = 2, . . . , v . By ( 2e ), t ∑ τ = t − σ 1 ( x τ − x τ − 1 ) ≤ t ∑ τ = t − σ 1 Vy τ − 1 + t ∑ τ = t − σ 1 V ( 1 − y τ − 1 ) , which implies that x t − x t − σ 1 − 1 ≤ t ∑ τ = t − σ 1 Vy τ − 1 + t ∑ τ = t − σ 1 V ( 1 − y τ − 1 ) = σ 1 V + V . Because y t − σ 1 − 1 = 0, by ( 2d ), x t − σ 1 − 1 = 0. Hence, x t ≤ σ 1 V + V ; that is, the left-hand side of inequality ( 13 ) is at most σ 1 V + V . Because S ⊆ [ 0, ⌊ ( C − V ) / V ⌋ ] Z , we have C − V − sV ≥ 0 for all s ∈ S . Note that { σ ′ 2 , . . . , σ ′ v } ⊆ S \ ˜ S and y t − s − y t − s − 1 ≤ 0 for all s ∈ S \ ˜ S . Thus, ∑ s ∈ S \ ˜ S ( C − V − s V )( y t − s − y t − s − 1 ) ≤ ∑ v j = 2 ( C − V − σ ′ j V )( y t − σ ′ j − y t − σ ′ j − 1 ) . Hence, the right-hand side of inequality ( 13 ) is C y t − ∑ s ∈ S ( C − V − s V ) ( y t − s − y t − s − 1 ) ec.12 = C − ∑ s ∈ ˜ S ( C − V − s V ) ( y t − s − y t − s − 1 ) − ∑ s ∈ S \ ˜ S ( C − V − s V ) ( y t − s − y t − s − 1 ) ≥ C − v ∑ j = 1 ( C − V − σ j V )( y t − σ j − y t − σ j − 1 ) − v ∑ j = 2 ( C − V − σ ′ j V )( y t − σ ′ j − y t − σ ′ j − 1 ) = C − v ∑ j = 1 ( C − V − σ j V ) + v ∑ j = 2 ( C − V − σ ′ j V ) = σ 1 V + V + v ∑ j = 2 ( σ j − σ ′ j ) V ≥ σ 1 V + V . Therefor e, in this case, ( x , y ) satisﬁes ( 13 ). Summarizing Cases 1–4, we conclude that ( x , y ) satisﬁes ( 13 ). Hence, ( 13 ) is valid for conv ( P ) when S = [ 0, α ] Z ∪ [ β , s max ] Z . It is easy to verify that the proof of facet-deﬁning of inequality ( 13 ) in the proof of Proposition 1 r emains valid when S = [ 0, α ] Z ∪ [ β , s max ] Z . Therefor e, inequality ( 13 ) is facet-deﬁning for conv ( P ) under the con- ditions stated in Proposition 2 . It is also easy to verify that the proof of validity and facet-deﬁning of inequality ( 14 ) in the proof of Proposition 1 remains valid when S = [ 0, α ] Z ∪ [ β , s max ] Z . Therefor e, inequality ( 14 ) is valid and facet- deﬁning for conv ( P ) under the conditions stated in Proposition 2 . □ C.3. Pro of of Prop osition 3 Proposition 3 . Consider any set S ⊆ [ 0, min { L − 1, T − 3, ⌊ ( C − V ) / V ⌋} ] Z and any real number η such that 0 ≤ η ≤ min { L − 1, ( C − V ) / V } . For any t ∈ [ 1, T − 1 ] Z such that t ≥ s + 2 for all s ∈ S , the inequality x t ≤ ( C − η V ) y t + η Vy t + 1 − ∑ s ∈ S ( C − V − s V ) ( y t − s − y t − s − 1 ) ( 15 ) is valid for conv ( P ) . For any t ∈ [ 2, T ] Z such that t ≤ T − s − 1 for all s ∈ S , the inequality x t ≤ ( C − η V ) y t + η Vy t − 1 − ∑ s ∈ S ( C − V − s V ) ( y t + s − y t + s + 1 ) ( 16 ) is valid for conv ( P ) . Furthermore, inequalities ( 15 ) and ( 16 ) ar e facet-deﬁning for conv ( P ) when η ∈ { 0, ( C − V ) / V } or η = L − 1 ∈ S . Proof. W e ﬁrst pr ove that inequality ( 15 ) is valid and facet-deﬁning for conv ( P ) . Note that the pr oof of facet-deﬁning of ( 15 ) here can also be used to prove the facet-deﬁning of ( 15 ) in Pr oposition 4 . For notational convenience, we deﬁne s max = max { s : s ∈ S } if S  = ∅ , and s max = − 1 if S = ∅ . Consider any t ∈ [ s max + 2, T − 1 ] Z (i.e., t ∈ [ 1, T − 1 ] Z such that t ≥ s + 2 for all s ∈ S ). T o prove that the linear inequality ( 15 ) is valid for conv ( P ) , it sufﬁces to show that it is valid for P . Consider any element ( x , y ) of P . W e show that ( x , y ) satisﬁes ( 15 ). W e divide the analysis into three cases. Case 1: y t = 0. By Lemma 1 (i), y t − j − y t − j − 1 ≤ 0 for all j ∈ [ 0, min { t − 2, L − 1 } ] Z . Because S ⊆ [ 0, L − 1 ] Z and s max ≤ t − 2, we have S ⊆ [ 0, min { t − 2, L − 1 } ] Z . Thus, y t − s − y t − s − 1 ≤ 0 for all s ∈ S . Because S ⊆ [ 0, ⌊ ( C − V ) / V ⌋ ] Z , we have C − V − s V ≥ 0 for all s ∈ S . Thus, the right-hand side of ( 15 ) is at least η Vy t + 1 ≥ 0. Because y t = 0, by ( 2d ), x t = 0. Therefor e, in this case, ( x , y ) satisﬁes ( 15 ). Case 2: y t = 1 and y t − s ′ − y t − s ′ − 1 = 1 for some s ′ ∈ S . In this case, y t − s ′ = 1 and y t − s ′ − 1 = 0. Because s ′ ≤ s max ≤ t − 2, we have t − s ′ ∈ [ 2, T ] Z . By ( 2a ), y k = 1 for all k ∈ [ t − s ′ , min { T , t − s ′ + L − 1 } ] Z . Because S ⊆ [ 0, L − 1 ] Z , we have s ′ ≤ L − 1, or equivalently t − s ′ + L − 1 ≥ t , and thus y t − s = 1 for all s ∈ [ 0, s ′ ] Z , which implies that y t − s − y t − s − 1 = 0 for all s ∈ [ 0, s ′ − 1 ] Z . Because s ′ ≤ t − 2, either s ′ = t − 2 or s ′ ≤ t − 3. If s ′ = t − 2, then it does not exist any s ∈ S such that s > s ′ . If s ′ ≤ t − 3, then t − s ′ − 1 ∈ [ 2, T ] Z , and by Lemma 1 (i), y t − s ′ − j − 1 − y t − s ′ − j − 2 ≤ 0 for all j ∈ [ 0, min { t − s ′ − 3, L − 1 } ] Z , which implies that y t − s − y t − s − 1 ≤ 0 for all s ∈ [ s ′ + 1, min { t − 2, L + s ′ } ] Z , which in turn implies that y t − s − y t − s − 1 ≤ 0 for all s ∈ S such that s > s ′ . Hence, y t − s − y t − s − 1 ≤ 0 for all s ∈ S \ { s ′ } . Because S ⊆ [ 0, ⌊ ( C − V ) / V ⌋ ] Z , we have C − V − s V ≥ 0 for all s ∈ S . Thus, − ∑ s ∈ S ( C − V − s V ) ( y t − s − y t − s − 1 ) ≥ − ( C − V − s ′ V ) . (EC.16) ec.13 Because y t = 1, by ( EC.16 ), the right-hand side of ( 15 ) is at least s ′ V + V when y t + 1 = 1 and is at least V + ( s ′ − η ) V when y t + 1 = 0. By ( 2d ), x t − s ′ − 1 = 0. By ( 2e ), ∑ t τ = t − s ′ ( x τ − x τ − 1 ) ≤ ∑ t τ = t − s ′ Vy τ − 1 + ∑ t τ = t − s ′ V ( 1 − y τ − 1 ) , which implies that x t ≤ s ′ V + V . If y t + 1 = 0, then by ( 2d ) and ( 2f ), x t + 1 = 0 and x t − x t + 1 ≤ Vy t + 1 + V ( 1 − y t + 1 ) , implying that x t ≤ V . In addition, if y t + 1 = 0, then because y k = 1 for all k ∈ [ t − s ′ , min { T , t − s ′ + L − 1 } ] Z , we have t + 1 ≥ t − s ′ + L , which implies that s ′ ≥ L − 1 ≥ η . Thus, if y t + 1 = 0, then x t ≤ V + ( s ′ − η ) V . Hence, x t is at most s ′ V + V , and it is at most V + ( s ′ − η ) V when y t + 1 = 0. Ther efore, in this case, ( x , y ) satisﬁes ( 15 ). Case 3: y t = 1 and y t − s − y t − s − 1  = 1 for all s ∈ S . In this case, y t − s − y t − s − 1 ≤ 0 for all s ∈ S . Because S ⊆ [ 0, ⌊ ( C − V ) / V ⌋ ] Z , we have C − V − sV ≥ 0 for all s ∈ S . Thus, ∑ s ∈ S ( C − V − sV ) ( y t − s − y t − s − 1 ) ≤ 0. Because y t = 1, the right-hand side of ( 15 ) is at least C when y t + 1 = 1 and is at least C − η V ≥ V when y t + 1 = 0 (as η ≤ ( C − V ) / V ). By ( 2d ), x t ≤ C . If y t + 1 = 0, then by ( 2d ) and ( 2f ), x t + 1 = 0 and x t − x t + 1 ≤ Vy t + 1 + V ( 1 − y t + 1 ) , which imply that x t ≤ V . Hence, x t is at most C , and it is at most V when y t + 1 = 0. Therefor e, in this case, ( x , y ) satisﬁes ( 15 ). Summarizing Cases 1–3, we conclude that ( x , y ) satisﬁes ( 15 ). Hence, ( 15 ) is valid for conv ( P ) . W e ﬁrst show that conv ( P ) is full dimensional. As the conv ( P ) contains 2 T decision variables, to show dim ( conv ( P )) = 2 T , we need to ﬁnd 2 T + 1 afﬁnely independent points in conv ( P ) . Because 0 ∈ conv ( P ) , it sufﬁces to cr eate the r emaining 2 T nonzero linearly independent points. W e denote these 2 T points as ( ¯ x r , ¯ y r ) and ( ˆ x r , ˆ y r ) , and denote the q th component of ¯ x r , ¯ y r , ˆ x r , and ˆ y r as ¯ x r q , ¯ y r q , ˆ x r q and ˆ y r q , r espectively . Let ϵ = V − C > 0. (A1) For each r ∈ [ 1, T ] Z , we create a point ( ¯ x r , ¯ y r ) as follows: ¯ x r q =  C + ϵ , q ∈ [ 1, r ] Z ; 0, q ∈ [ r + 1, T ] Z ; and ¯ y r q =  1, q ∈ [ 1, r ] Z ; 0, q ∈ [ r + 1, T ] Z . It is easy to observe that ( ¯ x r , ¯ y r ) satisﬁes ( 2a )–( 2f ) and thus ( ¯ x r , ¯ y r ) ∈ conv ( P ) for r ∈ [ 1, T ] Z . (A2) For each r ∈ [ 1, T ] Z , we create a point ( ˆ x r , ˆ y r ) as follows: ˆ x r q =  C , q ∈ [ 1, r ] Z ; 0, q ∈ [ r + 1, T ] Z ; and ˆ y r q =  1, q ∈ [ 1, r ] Z ; 0, q ∈ [ r + 1, T ] Z . It is easy to observe that ( ˆ x r , ˆ y r ) satisﬁes ( 2a )–( 2f ) and thus ( ˆ x r , ˆ y r ) ∈ conv ( P ) for r ∈ [ 1, T ] Z . It is also easy to observe that the above 2 T points ar e linearly independent. Therefor e, the dimension of conv ( P ) is 2 T , i.e., conv ( P ) is full dimensional. Consider any t ∈ [ s max + 2, T − 1 ] Z . T o prove that inequality ( 15 ) is facet-deﬁning for conv ( P ) when η ∈ { 0, ( C − V ) / V } or η = L − 1 ∈ S , it sufﬁces to show that there exist 2 T af ﬁnely independent points in conv ( P ) that satisfy ( 15 ) at equality when η ∈ { 0, ( C − V ) / V } or η = L − 1 ∈ S . When η = 0, inequalities ( 15 ) become inequality ( 13 ), and by Pr oposition 1 , it is facet-deﬁning for conv ( P ) . Hence, in the following, we only consider the case where η = ( C − V ) / V or η = L − 1 ∈ S . Because 0 ∈ conv ( P ) and 0 satisﬁes ( 15 ) at equality , it sufﬁces to create the remaining 2 T − 1 nonzero linearly independent points. W e denote these 2 T − 1 points as ( ¯ x r , ¯ y r ) for r ∈ [ 1, T ] Z \ { t } and ( ˆ x r , ˆ y r ) for r ∈ [ 1, T ] Z . W e divide these 2 T − 1 points into the following ﬁve groups: (A1) For each r ∈ [ 1, T ] Z \ { t } , we create the same point ( ¯ x r , ¯ y r ) as in group (A1) in the proof of Proposi- tion 1 . Thus, ( ¯ x r , ¯ y r ) ∈ conv ( P ) . It is easy to verify that ( ¯ x r , ¯ y r ) satisﬁes ( 15 ) at equality . (A2) For each r ∈ [ 1, t − 1 ] Z , we create the same point ( ˆ x r , ˆ y r ) as in gr oup (A2) in the proof of Pr oposition 1 . Thus, ( ˆ x r , ˆ y r ) ∈ conv ( P ) . Consider the case wher e t − r − 1 / ∈ S . In this case, ˆ x r t = ˆ y r t = ˆ y r t + 1 = 0. In addition, t − s − 1  = r for all s ∈ S , which implies that ˆ y r t − s − ˆ y r t − s − 1 = 0 for all s ∈ S . Hence, ( ˆ x r , ˆ y r ) satisﬁes ( 15 ) at equality . Next, consider the case where t − r − 1 ∈ S . In this case, ˆ x r t = V + ( t − r − 1 ) V and ˆ y r t = ˆ y r t + 1 = 1. In addition, ˆ y r t − s − ˆ y r t − s − 1 = 1 when s = t − r − 1, and ˆ y r t − s − ˆ y r t − s − 1 = 0 when s  = t − r − 1. Hence, ( ˆ x r , ˆ y r ) satisﬁes ( 15 ) at equality . ec.14 (A3) W e create a point ( ˆ x t , ˆ y t ) as follows: If η = ( C − V ) / V , then ˆ x t q =  V , for q ∈ [ 1, t ] Z ; 0, for q ∈ [ t + 1, T ] Z ; and ˆ y t q =  1, for q ∈ [ 1, t ] Z ; 0, for q ∈ [ t + 1, T ] Z . If η = L − 1 ∈ S , then ˆ x t q =  V , for q ∈ [ t − L + 1, t ] Z ; 0, for q ∈ [ 1, t − L ] Z ∪ [ t + 1, T ] Z ; and ˆ y t q =  1, q ∈ [ t − L + 1, t ] Z ; 0, q ∈ [ 1, t − L ] Z ∪ [ t + 1, T ] Z . W e ﬁrst consider the case where η = ( C − V ) / V . It is easy to verify that ( ˆ x t , ˆ y t ) satisﬁes ( 2a )–( 2f ). Thus, ( ˆ x t , ˆ y t ) ∈ conv ( P ) . In this case, ˆ x t t = V , ˆ y t t = 1, and ˆ y t t + 1 = 0, and ˆ y t t − s − ˆ y t t − s − 1 = 0 for all s ∈ S . Hence, ( ˆ x t , ˆ y t ) satisﬁes ( 15 ) at equality . Next, we consider the case where η = L − 1 ∈ S . In this case, for any q ∈ [ 2, T ] Z , ˆ y t q − ˆ y t q − 1 ≤ 0 if q  = t − L + 1, while ˆ y t q − ˆ y t q − 1 = 1 and ˆ y t k = 1 for all k ∈ [ q , min { T , q + L − 1 } ] Z if q = t − L + 1. Thus, ( ˆ x t , ˆ y t ) satisﬁes ( 2a ). For any q ∈ [ 2, T ] Z , ˆ y t q − 1 − ˆ y t q ≤ 0 if q  = t + 1, while ˆ y t q − 1 − ˆ y t q = 1 and ˆ y t k = 0 for all k ∈ [ q , min { T , q + ℓ − 1 } ] Z if q = t + 1. Thus, ( ˆ x t , ˆ y t ) satisﬁes ( 2b ). It is easy to verify that ( ˆ x t , ˆ y t ) satisﬁes ( 2c )–( 2f ). Thus, ( ˆ x t , ˆ y t ) ∈ conv ( P ) . Note that ˆ x t t = V , ˆ y t t = 1, ˆ y t t + 1 = 0, ˆ y t t − s − ˆ y t t − s − 1 = 0 for all s ∈ S \ { L − 1 } , ˆ y t t − L + 1 − ˆ y t t − L = 1, and ( C − η V ) − ( C − V − ( L − 1 ) V ) = V . Thus, ( ˆ x t , ˆ y t ) satisﬁes ( 15 ) at equality . (A4) W e create a point ( ˆ x t + 1 , ˆ y t + 1 ) by setting ˆ x t + 1 q = C and ˆ y t + 1 q = 1 for q ∈ [ 1, T ] Z . It is easy to verify that ( ˆ x t + 1 , ˆ y t + 1 ) satisﬁes ( 2a )–( 2f ). Thus, ( ˆ x t + 1 , ˆ y t + 1 ) ∈ conv ( P ) . It is also easy to verify that ( ˆ x t + 1 , ˆ y t + 1 ) satisﬁes ( 15 ) at equality . (A5) For each r ∈ [ t + 2, T ] Z , we create the same point ( ˆ x r , ˆ y r ) as in group (A4) in the proof of Proposition 1 . Thus, ( ˆ x r , ˆ y r ) ∈ conv ( P ) . It is easy to verify that ( ˆ x r , ˆ y r ) satisﬁes ( 15 ) at equality . T able EC.3 shows a matrix with 2 T − 1 rows, where each row represents a point created by this process. This matrix can be transformed into the matrix in T able EC.4 via the following Gaussian elimination process: (i) For each r ∈ [ 1, T ] Z \ { t } , the point with index r in group (B1), denoted ( ¯ x r , ¯ y r ) , is obtained by setting ( ¯ x r , ¯ y r ) = ( ¯ x r , ¯ y r ) − ( ˆ x t + 1 , ˆ y t + 1 ) . Her e, ( ¯ x r , ¯ y r ) is the point with index r in gr oup (A1) , and ( ˆ x t + 1 , ˆ y t + 1 ) is the point in group (A5) . (ii) For each r ∈ [ 1, t − 1 ] Z , the point with index r in group (B2), denoted ( ˆ x r , ˆ y r ) , is obtained by setting ( ˆ x r , ˆ y r ) = ( ˆ x r , ˆ y r ) if t − r − 1 / ∈ S , and setting ( ˆ x r , ˆ y r ) = ( ˆ x t + 1 , ˆ y t + 1 ) − ( ˆ x r , ˆ y r ) if t − r − 1 ∈ S . Here, ( ˆ x r , ˆ y r ) is the point with index r in group (A2) , and ( ˆ x t + 1 , ˆ y t + 1 ) is the point in group (A4) . (iii) The point in group (B3), denoted ( ˆ x t , ˆ y t ) , is obtained by setting ( ˆ x t , ˆ y t ) = ( ˆ x t , ˆ y t ) . Here, ( ˆ x t , ˆ y t ) is the point in group (A3) . (iv) The point in group (B4), denoted ( ˆ x t + 1 , ˆ y t + 1 ) , is obtained by setting ( ˆ x t + 1 , ˆ y t + 1 ) = ( ˆ x t + 1 , ˆ y t + 1 ) − ( ˆ x t + 2 , ˆ y t + 2 ) . Her e, ( ˆ x t + 1 , ˆ y t + 1 ) is the point in gr oup (A4) , and ( ˆ x t + 2 , ˆ y t + 2 ) is the point with index t + 2 in group (A5) . (v) For each r ∈ [ t + 2, T ] Z , the point with index r in group (B5), denoted ( ˆ x r , ˆ y r ) , is obtained by setting ( ˆ x r , ˆ y r ) = ( ˆ x r , ˆ y r ) − ( ˆ x r + 1 , ˆ y r + 1 ) if r  = T , and setting ( ˆ x r , ˆ y r ) = ( ˆ x r , ˆ y r ) if r = T . Here, ( ˆ x r , ˆ y r ) and ( ˆ x r + 1 , ˆ y r + 1 ) are the points with indices r and r + 1, respectively , in group (A5) . The matrix shown in T able EC.4 is lower triangular; that is, the position of the last nonzero component of a row of the matrix is greater than the position of the last nonzero component of the previous row . This implies that the 2 T − 1 points in gr oups (A1) – (A5) are linearly independent. Ther efore, inequality ( 15 ) is facet-deﬁning for conv ( P ) . Next, we show that inequality ( 16 ) is valid for conv ( P ) and is facet-deﬁning for conv ( P ) when η ∈ { 0, ( C − V ) / V } or η = L − 1 ∈ S . Note that this proof can also be used to prove the validity and facet- deﬁning of inequality ( 16 ) in Proposition 4 . Denote x ′ t = x T − t + 1 and y ′ t = y T − t + 1 for t ∈ [ 1, T ] Z . Because ec.15 inequality ( 15 ) is valid for conv ( P ) and is facet-deﬁning for conv ( P ) when η ∈ { 0, ( C − V ) / V } or η = L − 1 ∈ S for any t ∈ [ s max + 2, T − 1 ] Z , the inequality x ′ T − t + 1 ≤ ( C − η V ) y ′ T − t + 1 + η Vy ′ T − t − ∑ s ∈ S ( C − V − s V ) ( y ′ T − t + s + 1 − y ′ T − t + s + 2 ) is valid for conv ( P ′ ) and is facet-deﬁning for conv ( P ′ ) when η ∈ { 0, ( C − V ) / V } or η = L − 1 ∈ S for any t ∈ [ s max + 2, T − 1 ] Z . Let t ′ = T − t + 1. Then, the inequality x ′ t ′ ≤ ( C − η V ) y ′ t ′ + η Vy ′ t ′ − 1 − ∑ s ∈ S ( C − V − s V ) ( y ′ t ′ + s − y ′ t ′ + s + 1 ) is valid for conv ( P ′ ) and is facet-deﬁning for conv ( P ′ ) when η ∈ { 0, ( C − V ) / V } or η = L − 1 ∈ S for any t ′ ∈ [ 2, T − s max − 1 ] Z . Hence, by Lemma 2 , inequality ( 16 ) is valid for conv ( P ) and is facet-deﬁning for conv ( P ) when η ∈ { 0, ( C − V ) / V } or η = L − 1 ∈ S for any t ∈ [ 2, T − s max − 1 ] Z . □ C.4. Pro of of Prop osition 4 Proposition 4 . Consider any real number η such that 0 ≤ η ≤ min { L − 1, ( C − V ) / V } and any integers α , β , and s max such that (a) L ≤ s max ≤ min { T − 3, ⌊ ( C − V ) / V ⌋} , (b) 0 ≤ α < β ≤ s max , and (c) β = α + 1 or s max ≤ L + α . Let S = [ 0, α ] Z ∪ [ β , s max ] Z . For any t ∈ [ s max + 2, T − 1 ] Z , inequality ( 15 ) is valid for conv ( P ) . For any t ∈ [ 2, T − s max − 1 ] Z , inequality ( 16 ) is valid for conv ( P ) . Furthermore, ( 15 ) and ( 16 ) are facet-deﬁning for conv ( P ) when η ∈ { 0, ( C − V ) / V } or η = L − 1 ∈ S . Proof. Consider any t ∈ [ s max + 2, T − 1 ] Z . T o pr ove that the linear inequality ( 15 ) is valid for conv ( P ) when S = [ 0, α ] Z ∪ [ β , s max ] Z , it sufﬁces to show that ( 15 ) is valid for P when S = [ 0, α ] Z ∪ [ β , s max ] Z . Consider any element ( x , y ) of P . W e show that ( x , y ) satisﬁes ( 15 ) when S = [ 0, α ] Z ∪ [ β , s max ] Z . W e divide the analysis into four cases. Case 1: y t = 0 and y t − s − y t − s − 1 ≤ 0 for all s ∈ S . Because S ⊆ [ 0, ⌊ ( C − V ) / V ⌋ ] Z , we have C − V − s V ≥ 0 for all s ∈ S . Thus, in this case, the right-hand side of inequality ( 15 ) is nonnegative. Because y t = 0, by ( 2d ), x t = 0. Therefor e, ( x , y ) satisﬁes ( 15 ). Case 2: y t = 0 and y t − s − y t − s − 1 > 0 for some s ∈ S . Let ˜ S = { σ ∈ S : y t − σ − y t − σ − 1 > 0 } and v = | ˜ S | . Then, v ≥ 1. Denote ˜ S = { σ 1 , σ 2 , . . . , σ v } , where σ 1 < σ 2 < · · · < σ v . Note that y t − σ j − 1 = 0 and y t − σ j = 1 for j = 1, . . . , v . Denote σ 0 = − 1. Then, for each j = 1, . . . , v , ther e exists σ ′ j ∈ [ σ j − 1 + 1, σ j − 1 ] Z such that y t − σ ′ j − 1 = 1 and y t − σ ′ j = 0. Thus, 0 ≤ σ ′ 1 < σ 1 < σ ′ 2 < σ 2 < · · · < σ ′ v < σ v ≤ s max . Because y t − σ v − y t − σ v − 1 = 1 and t − σ v ∈ [ 2, T ] Z , by ( 2a ), y k = 1 for all k ∈ [ t − σ v , min { T , t − σ v + L − 1 } ] Z , which implies that t − σ ′ j ≥ t − σ v + L for j = 1, . . . , v . Hence, for j = 1, . . . , v , we have σ ′ j ≤ σ v − L , which implies that σ ′ j ≤ s max − L . (EC.17) If β = α + 1, then S = [ 0, s max ] Z , which implies that σ ′ j ∈ S for j = 1, . . . , v . If β  = α + 1, then condition (c) of Proposition 4 implies that s max ≤ L + α , which, by ( EC.17 ), implies that σ ′ j ≤ α for j = 1, . . . , v . Thus, in both cases, σ ′ j ∈ S for j = 1, . . . , v . Because y t = 0, by ( 2d ), x t = 0. Hence, the left-hand side of inequality ( 15 ) is 0. Because S ⊆ [ 0, ⌊ ( C − V ) / V ⌋ ] Z , we have C − V − sV ≥ 0 for all s ∈ S . Note that { σ ′ 1 , . . . , σ ′ v } ⊆ S \ ˜ S and y t − s − y t − s − 1 ≤ 0 for all s ∈ S \ ˜ S . Thus, ∑ s ∈ S \ ˜ S ( C − V − s V ) ( y t − s − y t − s − 1 ) ≤ ∑ v j = 1 ( C − V − σ ′ j V )( y t − σ ′ j − y t − σ ′ j − 1 ) . Hence, the right-hand side of inequality ( 15 ) is ( C − η V ) y t + η Vy t + 1 − ∑ s ∈ S ( C − V − s V ) ( y t − s − y t − s − 1 ) = η Vy t + 1 − ∑ s ∈ ˜ S ( C − V − s V ) ( y t − s − y t − s − 1 ) − ∑ s ∈ S \ ˜ S ( C − V − s V ) ( y t − s − y t − s − 1 ) ≥ η Vy t + 1 − v ∑ j = 1 ( C − V − σ j V )( y t − σ j − y t − σ j − 1 ) − v ∑ j = 1 ( C − V − σ ′ j V )( y t − σ ′ j − y t − σ ′ j − 1 ) ec.16 T able EC.3 A matrix with the rows r epresenting 2 T − 1 points in conv ( P ) that satisfy inequality ( 15 ) at equality Group Point Index r x y 1 · · · t − 1 t t + 1 t + 2 · · · T 1 · · · t − 1 t t + 1 t + 2 · · · T (A1) ( ¯ x r , ¯ y r ) 1 C − ϵ · · · C C C C · · · C 1 · · · 1 1 1 1 · · · 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . t − 1 C · · · C − ϵ C C C · · · C 1 · · · 1 1 1 1 · · · 1 t + 1 C · · · C C C − ϵ C · · · C 1 · · · 1 1 1 1 · · · 1 t + 2 C · · · C C C C − ϵ · · · C 1 · · · 1 1 1 1 · · · 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . T C · · · C C C C · · · C − ϵ 1 · · · 1 1 1 1 · · · 1 (A2) ( ˆ x r , ˆ y r ) 1 (See Note EC.3 -1) (See Note EC.3 -1) . . . t − 1 (A3) t (See Note EC.3 -2) (See Note EC.3 -2) (A4) t + 1 C · · · C C C C · · · C 1 · · · 1 1 1 1 · · · 1 (A5) t + 2 0 · · · 0 0 0 C · · · C 0 · · · 0 0 0 1 · · · 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . T 0 · · · 0 0 0 0 · · · C 0 · · · 0 0 0 0 · · · 1 Note EC.3 -1: For r ∈ [ 1, t − 1 ] Z , the x and y vectors in group (A2) are given as follows: ˆ x r = ( C , . . . , C | {z } r terms , 0, . . . , 0 | { z } T − r terms ) and ˆ y r = ( 1, . . . , 1 | { z } r terms , 0, . . . , 0 | { z } T − r terms ) if t − r − 1 / ∈ S ; ˆ x r = ( 0, . . . , 0 | { z } r terms , V , V + V , V + 2 V , . . . , V + ( t − r − 1 ) V | { z } t − r terms , V + ( t − r − 1 ) V , . . . , V + ( t − r − 1 ) V | {z } T − t terms ) and ˆ y r = ( 0, . . . , 0 | { z } r terms , 1, . . . , 1 | { z } T − r terms ) if t − r − 1 ∈ S . Note EC.3 -2: The x and y vectors in group (A3) are given as follows: ˆ x t = ( V , . . . , V | { z } t terms , 0, . . . , 0 | { z } T − t terms ) and ˆ y t = ( 1, . . . , 1 | { z } t terms , 0, . . . , 0 | { z } T − t terms ) if η = ( C − V ) / V ; ˆ x t = ( 0, . . . , 0 | { z } t − L terms , V , . . . , V | { z } L terms , 0, . . . , 0 | { z } T − t terms ) and ˆ y t = ( 0, . . . , 0 | { z } t − L terms , 1, . . . , 1 | { z } L terms , 0, . . . , 0 | { z } T − t terms ) if η = L − 1 ∈ S . ec.17 T able EC.4 Lower triangular matrix obtained from T able EC.3 via Gaussian elimination Group Point Index r x y 1 · · · t − 1 t t + 1 t + 2 · · · T 1 · · · t − 1 t t + 1 t + 2 · · · T (B1) ( ¯ x r , ¯ y r ) 1 − ϵ · · · 0 0 0 0 · · · 0 0 · · · 0 0 0 0 · · · 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . t − 1 0 · · · − ϵ 0 0 0 · · · 0 0 · · · 0 0 0 0 · · · 0 t + 1 0 · · · 0 0 − ϵ 0 · · · 0 0 · · · 0 0 0 0 · · · 0 t + 2 0 · · · 0 0 0 − ϵ · · · 0 0 · · · 0 0 0 0 · · · 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . T 0 · · · 0 0 0 0 · · · − ϵ 0 · · · 0 0 0 0 · · · 0 (B2) ( ˆ x r , ˆ y r ) 1 1 · · · 0 0 0 0 · · · 0 . . . (Omitted) . . . . . . . . . . . . . . . . . . . . . t − 1 1 · · · 1 0 0 0 · · · 0 (B3) t (Omitted) (See Note EC.4 -1) (B4) t + 1 (Omitted) 1 · · · 1 1 1 0 · · · 0 (B5) t + 2 0 · · · 0 0 0 1 · · · 0 . . . (Omitted) . . . . . . . . . . . . . . . . . . . . . T 0 · · · 0 0 0 0 · · · 1 Note EC.4 -1: The y vector in group (B3) is given as follows: ˆ y t = ( 1, . . . , 1 | { z } t terms , 0, . . . , 0 | { z } T − t terms ) if η = ( C − V ) / V ; ˆ y t = ( 0, . . . , 0 | { z } t − L terms , 1, . . . , 1 | { z } L terms , 0, . . . , 0 | { z } T − t terms ) if η = L − 1 ∈ S . ec.18 = η Vy t + 1 − v ∑ j = 1 ( C − V − σ j V ) + v ∑ j = 1 ( C − V − σ ′ j V ) = η Vy t + 1 + v ∑ j = 1 ( σ j − σ ′ j ) V > 0. Therefor e, in this case, ( x , y ) satisﬁes ( 15 ). Case 3: y t = 1 and y t − s − y t − s − 1 ≤ 0 for all s ∈ S . Because S ⊆ [ 0, ⌊ ( C − V ) / V ⌋ ] Z , we have C − V − s V ≥ 0 for all s ∈ S . Thus, ∑ s ∈ S ( C − V − s V ) ( y t − s − y t − s − 1 ) ≤ 0. Because y t = 1, the right-hand side of ( 15 ) is at least C when y t + 1 = 1 and is at least C − η V ≥ V when y t + 1 = 0 (as η ≤ ( C − V ) / V ). By ( 2d ), x t ≤ C . If y t + 1 = 0, then by ( 2d ) and ( 2f ), x t + 1 = 0 and x t − x t + 1 ≤ Vy t + 1 + V ( 1 − y t + 1 ) , which imply that x t ≤ V . Hence, x t is at most C , and it is at most V when y t + 1 = 0. Therefor e, in this case, ( x , y ) satisﬁes ( 15 ). Case 4: y t = 1 and y t − s − y t − s − 1 > 0 for some s ∈ S . If y t + 1 = 1, then inequality ( 15 ) becomes inequality ( 13 ), and by Proposition 1 , ( x , y ) satisﬁes the inequality . In the following, we consider the case where y t + 1 = 0. Let ˜ S = { σ ∈ S : y t − σ − y t − σ − 1 > 0 } and v = | ˜ S | . Then, v ≥ 1. Denote ˜ S = { σ 1 , σ 2 , . . . , σ v } , wher e σ 1 < σ 2 < · · · < σ v . Note that y t − σ j − 1 = 0 and y t − σ j = 1 for j = 1, . . . , v . Then, for each j = 2, . . . , v , there exists σ ′ j ∈ [ σ j − 1 + 1, σ j − 1 ] Z such that y t − σ ′ j − 1 = 1 and y t − σ ′ j = 0. Thus, 0 ≤ σ 1 < σ ′ 2 < σ 2 < · · · < σ ′ v < σ v ≤ s max . In addition, y k = 1 for all k ∈ [ t − σ 1 , t ] Z . Because y t − σ v − y t − σ v − 1 = 1 and t − σ v ∈ [ 2, T ] Z , by ( 2a ), y k = 1 for all k ∈ [ t − σ v , min { T , t − σ v + L − 1 } ] Z , which implies that t − σ ′ j ≥ t − σ v + L for j = 2, . . . , v . Hence, for j = 2, . . . , v , we have σ ′ j ≤ σ v − L , which implies that σ ′ j ≤ s max − L . (EC.18) If β = α + 1, then S = [ 0, s max ] Z , which implies that σ ′ j ∈ S for j = 2, . . . , v . If β  = α + 1, then condition (c) of Proposition 4 implies that s max ≤ L + α , which, by ( EC.18 ), implies that σ ′ j ≤ α for j = 2, . . . , v . Thus, in both cases, σ ′ j ∈ S for j = 2, . . . , v . Because y t + 1 = 0, by ( 2d ) and ( 2f ), x t + 1 = 0 and x t − x t + 1 ≤ Vy t + 1 + V ( 1 − y t + 1 ) , which imply that x t ≤ V ; that is, the left-hand side of inequality ( 15 ) is at most V . Because y t − σ 1 − y t − σ 1 − 1 = 1 and t − σ 1 ∈ [ 2, T ] Z , by ( 2a ), y k = 1 for all k ∈ [ t − σ 1 , min { T , t − σ 1 + L − 1 } ] Z . Because y t + 1 = 0, this implies that t + 1 ≥ t − σ 1 + L , or equivalently , L − 1 ≤ σ 1 . Because η ≤ L − 1, we have η ≤ σ 1 . (EC.19) Because S ⊆ [ 0, ⌊ ( C − V ) / V ⌋ ] Z , we have C − V − sV ≥ 0 for all s ∈ S . Note that { σ ′ 2 , . . . , σ ′ v } ⊆ S \ ˜ S and y t − s − y t − s − 1 ≤ 0 for all s ∈ S \ ˜ S . Thus, ∑ s ∈ S \ ˜ S ( C − V − s V ) ( y t − s − y t − s − 1 ) ≤ ∑ v j = 2 ( C − V − σ ′ j V )( y t − σ ′ j − y t − σ ′ j − 1 ) . Hence, the right-hand side of inequality ( 15 ) is ( C − η V ) y t + η Vy t + 1 − ∑ s ∈ S ( C − V − s V ) ( y t − s − y t − s − 1 ) = ( C − η V ) − ∑ s ∈ ˜ S ( C − V − s V ) ( y t − s − y t − s − 1 ) − ∑ s ∈ S \ ˜ S ( C − V − s V ) ( y t − s − y t − s − 1 ) ≥ ( C − η V ) − v ∑ j = 1 ( C − V − σ j V )( y t − σ j − y t − σ j − 1 ) − v ∑ j = 2 ( C − V − σ ′ j V )( y t − σ ′ j − y t − σ ′ j − 1 ) = ( C − η V ) − v ∑ j = 1 ( C − V − σ j V ) + v ∑ j = 2 ( C − V − σ ′ j V ) = V + ( σ 1 − η ) V + v ∑ j = 2 ( σ j − σ ′ j ) V ≥ V + ( σ 1 − η ) V ≥ V , where the last inequality follows fr om ( EC.19 ). Therefore, in this case, ( x , y ) satisﬁes ( 15 ). ec.19 Summarizing Cases 1–4, we conclude that ( x , y ) satisﬁes ( 15 ). Hence, ( 15 ) is valid for conv ( P ) when S = [ 0, α ] Z ∪ [ β , s max ] Z . It is easy to verify that the proof of facet-deﬁning of inequality ( 15 ) in the proof of Proposition 3 r emains valid when S = [ 0, α ] Z ∪ [ β , s max ] Z . Therefor e, inequality ( 15 ) is facet-deﬁning for conv ( P ) under the con- ditions stated in Proposition 4 . It is also easy to verify that the proof of validity and facet-deﬁning of inequality ( 16 ) in the proof of Proposition 3 remains valid when S = [ 0, α ] Z ∪ [ β , s max ] Z . Therefor e, inequality ( 16 ) is valid and facet- deﬁning for conv ( P ) under the conditions stated in Proposition 4 . □ C.5. Pro of of Prop osition 5 Proposition 5 . Consider any S ⊆ [ 1, min { L , T − 2, ⌊ ( C − V ) / V ⌋} ] Z and any real number η such that 0 ≤ η ≤ min { L , ( C − V ) / V } . For any t ∈ [ 2, T ] Z such that t ≥ s + 2 for all s ∈ S , the inequality x t ≤ ( V + η V ) y t + ( C − V − η V ) y t − 1 − ∑ s ∈ S ( C − V − s V ) ( y t − s − y t − s − 1 ) ( 17 ) is valid for conv ( P ) . For any t ∈ [ 1, T − 1 ] Z such that t ≤ T − s − 1 for all s ∈ S , the inequality x t ≤ ( V + η V ) y t + ( C − V − η V ) y t + 1 − ∑ s ∈ S ( C − V − s V ) ( y t + s − y t + s + 1 ) ( 18 ) is valid for conv ( P ) . Furthermore, inequalities ( 17 ) and ( 18 ) ar e facet-deﬁning for conv ( P ) when η ∈ { 0, ( C − V ) / V } or η = L ∈ S . Proof. W e ﬁrst pr ove that inequality ( 17 ) is valid and facet-deﬁning for conv ( P ) . Note that the pr oof of facet-deﬁning of ( 17 ) here can also be used to prove the facet-deﬁning of ( 17 ) in Pr oposition 6 . For notational convenience, we deﬁne s max = max { s : s ∈ S } if S  = ∅ , and s max = 0 if S = ∅ . Consider any t ∈ [ s max + 2, T ] Z (i.e., t ∈ [ 2, T ] Z such that t ≥ s + 2 for all s ∈ S ). T o prove that the linear inequality ( 17 ) is valid for conv ( P ) , it sufﬁces to show that it is valid for P . Consider any element ( x , y ) of P . W e show that ( x , y ) satisﬁes ( 17 ). W e divide the analysis into three cases. Case 1: y t = 0. In this case, by ( 2d ), x t = 0. Thus, the left-hand side of ( 17 ) and the ﬁrst term on the right-hand side of ( 17 ) are 0. Because y t = 0, by Lemma 1 (i), y t − j − y t − j − 1 ≤ 0 for all j ∈ [ 0, min { t − 2, L − 1 } ] Z . Because s max ≤ t − 2, we have S ⊆ [ 0, min { t − 2, L } ] Z . Thus, y t − s − y t − s − 1 ≤ 0 for all s ∈ S \ { L } . Because η ≤ ( C − V ) / V , the coef ﬁcient “ C − V − η V ” on the right-hand side of ( 17 ) is nonnegative. Because S ⊆ [ 1, ⌊ ( C − V ) / V ⌋ ] Z , for any s ∈ S , the coefﬁcient “ C − V − s V ” on the right-hand side of ( 17 ) is also nonnegative. Hence, if s max ≤ L − 1 or y t − L − y t − L − 1 ≤ 0, then the right-hand side of ( 17 ) is nonnegative. Now , consider the situation where s max = L and y t − L − y t − L − 1 > 0. Then, y t − L = 1 and y t − L − 1 = 0. By ( 2a ), y t − 1 = 1. Thus, the right-hand side of ( 17 ) is at least ( C − V − η V ) y t − 1 − ( C − V − L V )( y t − L − y t − L − 1 ) = ( C − V − η V ) − ( C − V − L V ) = ( L − η ) V ≥ 0. Therefore, in this case, ( x , y ) satisﬁes ( 17 ). Case 2: y t = 1 and y t − s ′ − y t − s ′ − 1 = 1 for some s ′ ∈ S . In this case, y t − s ′ = 1 and y t − s ′ − 1 = 0. Because s max ≤ t − 2, we have s ′ ≤ t − 2. If s ′ = t − 2, then it does not exist any s ∈ S such that s > s ′ . If s ′ ≤ t − 3, then t − s ′ − 1 ∈ [ 2, T ] Z , and by Lemma 1 (i), y t − s ′ − j − 1 − y t − s ′ − j − 2 ≤ 0 for all j ∈ [ 0, min { t − s ′ − 3, L − 1 } ] Z , which implies that y t − s − y t − s − 1 ≤ 0 for all s ∈ [ s ′ + 1, min { t − 2, L + s ′ } ] Z . Thus, y t − s − y t − s − 1 ≤ 0 for all s ∈ S such that s > s ′ . Because y t − s ′ − y t − s ′ − 1 = 1 and t − s ′ ∈ [ 2, T ] Z , by ( 2a ), y k = 1 for all k ∈ [ t − s ′ , min { T , t − s ′ + L − 1 } ] Z . This implies that y t − s − y t − s − 1 = 0 for all s ∈ [ 1, s ′ − 1 ] Z (as s ′ ≤ L ). This in turn implies that y t − s − y t − s − 1 = 0 for all s ∈ S such that s < s ′ . Hence, y t − s − y t − s − 1 ≤ 0 for all s ∈ S \ { s ′ } . Because S ⊆ [ 1, ⌊ ( C − V ) / V ⌋ ] Z , we have C − V − sV ≥ 0 for all s ∈ S . Thus, − ∑ s ∈ S ( C − V − s V ) ( y t − s − y t − s − 1 ) ≥ − ( C − V − s ′ V ) . (EC.20) Note that t − 1 ∈ [ t − s ′ , min { T , t − s ′ + L − 1 } ] Z . Hence, y t − 1 = 1. Because y t = 1 and y t − 1 = 1, by ( EC.20 ), the right-hand side of inequality ( 17 ) is at least s ′ V + V . By ( 2e ), ∑ t τ = t − s ′ ( x τ − x τ − 1 ) ≤ ∑ t τ = t − s ′ Vy τ − 1 + ∑ t τ = t − s ′ V ( 1 − y τ − 1 ) , which implies that x t − x t − s ′ − 1 ≤ s ′ V + V . Because y t − s ′ − 1 = 0, we have x t − s ′ − 1 = 0. Thus, x t ≤ s ′ V + V . Therefor e, in this case, ( x , y ) satisﬁes ( 17 ). Case 3: y t = 1 and y t − s − y t − s − 1  = 1 for all s ∈ S . In this case, y t − s − y t − s − 1 ≤ 0 for all s ∈ S . Because S ⊆ [ 1, ⌊ ( C − V ) / V ⌋ ] Z , we have C − V − sV ≥ 0 for all s ∈ S . Thus, ∑ s ∈ S ( C − V − s V ) ( y t − s − y t − s − 1 ) ≤ 0. The right-hand side of ( 17 ) is at least V + η V when y t − 1 = 0, and is at least C when y t − 1 = 1. If y t − 1 = 0, ec.20 then by ( 2d ) and ( 2e ), x t − 1 = 0 and x t − x t − 1 ≤ V , which imply that x t ≤ V , and hence, x t is less than or equal to the right-hand side of ( 17 ). If y t − 1 = 1, then by ( 2d ), x t ≤ C , and hence, x t is less than or equal to the right-hand side of ( 17 ). Therefor e, in this case, ( x , y ) satisﬁes ( 17 ). Summarizing Cases 1–3, we conclude that ( x , y ) satisﬁes ( 17 ). Hence, ( 17 ) is valid for conv ( P ) . Consider any t ∈ [ s max + 2, T ] Z . T o prove that inequality ( 17 ) is facet-deﬁning for conv ( P ) when η ∈ { 0, ( C − V ) / V } or η = L ∈ S , it sufﬁces to show that there exist 2 T afﬁnely independent points in conv ( P ) that satisfy ( 17 ) at equality when η ∈ { 0, ( C − V ) / V } or η = L ∈ S . Because 0 ∈ conv ( P ) and 0 satisﬁes ( 17 ) at equality , it sufﬁces to create the remaining 2 T − 1 nonzero linearly independent points. W e denote these 2 T − 1 points as ( ¯ x r , ¯ y r ) for r ∈ [ 1, T ] Z \ { t } and ( ˆ x r , ˆ y r ) for r ∈ [ 1, T ] Z , and denote the q th component of ¯ x r , ¯ y r , ˆ x r , and ˆ y r as ¯ x r q , ¯ y r q , ˆ x r q , and ˆ y r q , respectively . Let ϵ = V − C > 0. W e divide these 2 T − 1 points into the following ﬁve groups: (A1) For each r ∈ [ 1, T ] Z \ { t } , we create the same point ( ¯ x r , ¯ y r ) as in gr oup (A1) in the proof of Pr oposition 1 . Thus, ( ¯ x r , ¯ y r ) ∈ conv ( P ) . It is easy to verify that ( ¯ x r , ¯ y r ) satisﬁes ( 17 ) at equality . (A2) For each r ∈ [ 1, t − 2 ] Z , we create the same point ( ˆ x r , ˆ y r ) as in group (A2) in the proof of Proposition 1 . Thus, ( ˆ x r , ˆ y r ) ∈ conv ( P ) . Consider the case where t − r − 1 / ∈ S . In this case, ˆ x r t = ˆ y r t = ˆ y r t − 1 = 0. In addition, t − s − 1  = r for all s ∈ S , which implies that ˆ y r t − s − ˆ y r t − s − 1 = 0 for all s ∈ S . Hence, ( ˆ x r , ˆ y r ) satisﬁes ( 17 ) at equality . Next, consider the case where t − r − 1 ∈ S . In this case, ˆ x r t = V + ( t − r − 1 ) V and ˆ y r t = ˆ y r t − 1 = 1. In addition, ˆ y r t − s − ˆ y r t − s − 1 = 1 when s = t − r − 1, and ˆ y r t − s − ˆ y r t − s − 1 = 0 when s  = t − r − 1. Hence, ( ˆ x r , ˆ y r ) satisﬁes ( 17 ) at equality . (A3) W e create a point ( ˆ x t − 1 , ˆ y t − 1 ) as follows: If η = 0, then ˆ x t − 1 q =  0, for q ∈ [ 0, t − 1 ] Z ; V , for q ∈ [ t , T ] Z ; and ˆ y t − 1 q =  0, for q ∈ [ 0, t − 1 ] Z ; 1, for q ∈ [ t , T ] Z . If η = ( C − V ) / V , then ˆ x t − 1 q =  C , for q ∈ [ 1, t − 1 ] Z ; 0, for q ∈ [ t , T ] Z ; and ˆ y t − 1 q =  1, for q ∈ [ 1, t − 1 ] Z ; 0, for q ∈ [ t , T ] Z . If η = L ∈ S , then ˆ x t − 1 q =  0, for q ∈ [ 1, t − L − 1 ] Z ∪ [ t , T ] Z ; V , for q ∈ [ t − L , t − 1 ] Z ; and ˆ y t − 1 q =  0, for q ∈ [ 1, t − L − 1 ] Z ∪ [ t , T ] Z ; 1, for q ∈ [ t − L , t − 1 ] Z . W e ﬁrst consider the case wher e η = 0. It is easy to verify that ( ˆ x t − 1 , ˆ y t − 1 ) satisﬁes ( 2a )–( 2f ). Thus, ( ˆ x t − 1 , ˆ y t − 1 ) ∈ conv ( P ) . In this case, ˆ x t − 1 t = V , ˆ y t − 1 t = 1, ˆ y t − 1 t − 1 = 0, and ˆ y t − 1 t − s − ˆ y t − 1 t − s − 1 = 0 for all s ∈ S . Hence, ( ˆ x t − 1 , ˆ y t − 1 ) satisﬁes ( 17 ) at equality . Next, we consider the case where η = ( C − V ) / V . It is easy to verify that ( ˆ x t − 1 , ˆ y t − 1 ) satisﬁes ( 2a )–( 2f ). Thus, ( ˆ x t − 1 , ˆ y t − 1 ) ∈ conv ( P ) . In this case, ˆ x t − 1 t = ˆ y t − 1 t = 0, C − V − η V = 0, and ˆ y t − 1 t − s = ˆ y t − 1 t − s − 1 = 0 for all s ∈ S . Hence, ( ˆ x t − 1 , ˆ y t − 1 ) satisﬁes ( 17 ) at equality . Next, we consider the case wher e η = L ∈ S . In this case, for any q ∈ [ 2, T ] Z , ˆ y t − 1 q − ˆ y t − 1 q − 1 ≤ 0 if q  = t − L , while ˆ y t − 1 q − ˆ y t − 1 q − 1 = 1 and ˆ y t − 1 k = 1 for all k ∈ [ q , min { T , q + L − 1 } ] Z if q = t − L . Thus, ( ˆ x t − 1 , ˆ y t − 1 ) satisﬁes ( 2a ). For any q ∈ [ 2, T ] Z , ˆ y t − 1 q − 1 − ˆ y t − 1 q ≤ 0 if q  = t , while ˆ y t − 1 q − 1 − ˆ y t − 1 q = 1 and ˆ y t − 1 q = 0 for all k ∈ [ q , min { T , q + ℓ − 1 } ] Z if q = t . Thus, ( ˆ x t − 1 , ˆ y t − 1 ) satisﬁes ( 2b ). It is easy to verify that ( ˆ x t − 1 , ˆ y t − 1 ) satisﬁes ( 2c )–( 2f ). Hence, ( ˆ x t − 1 , ˆ y t − 1 ) ∈ conv ( P ) . Note that ˆ x t − 1 t = ˆ y t − 1 t = 0, ˆ y t − 1 t − 1 = 1, ˆ y t − 1 t − s − ˆ y t − 1 t − s − 1 = 0 for all s ∈ S \ { L } , ˆ y t − 1 t − L − ˆ y t − 1 t − L − 1 = 1, and C − V − η V = C − V − L V . Thus, ( ˆ x t − 1 , ˆ y t − 1 ) satisﬁes ( 17 ) at equality . (A4) W e create the same point ( ˆ x t , ˆ y t ) as in group (A3) in the proof of Proposition 1 . Thus, ( ˆ x t , ˆ y t ) ∈ conv ( P ) . It is easy to verify that ( ˆ x t , ˆ y t ) satisﬁes ( 17 ) at equality . ec.21 (A5) For each r ∈ [ t + 1, T ] Z , we create the same point ( ˆ x r , ˆ y r ) as in group (A4) in the proof of Proposition 1 . Thus, ( ˆ x r , ˆ y r ) ∈ conv ( P ) . It is easy to verify that ( ˆ x r , ˆ y r ) satisﬁes ( 17 ) at equality . T able EC.5 shows a matrix with 2 T − 1 rows, where each row represents a point created by this process. This matrix can be transformed into the matrix in T able EC.6 via the following Gaussian elimination process: (i) For each r ∈ [ 1, T ] Z \ { t } , the point with index r in group (B1), denoted ( ¯ x r , ¯ y r ) , is obtained by setting ( ¯ x r , ¯ y r ) = ( ¯ x r , ¯ y r ) − ( ˆ x t , ˆ y t ) . Here, ( ¯ x r , ¯ y r ) is the point with index r in group (A1) , and ( ˆ x t , ˆ y t ) is the point in group (A4) . (ii) For each r ∈ [ 1, t − 2 ] Z , the point with index r in group (B2), denoted ( ˆ x r , ˆ y r ) , is obtained by setting ( ˆ x r , ˆ y r ) = ( ˆ x r , ˆ y r ) if t − r − 1 / ∈ S , and setting ( ˆ x r , ˆ y r ) = ( ˆ x t , ˆ y t ) − ( ˆ x r , ˆ y r ) if t − r − 1 ∈ S . Here, ( ˆ x r , ˆ y r ) is the point with index r in group (A2) , and ( ˆ x t , ˆ y t ) is the point in group (A4) . (iii) The point in group (B3), denoted ( ˆ x t − 1 , ˆ y t − 1 ) , is obtained by setting ( ˆ x t − 1 , ˆ y t − 1 ) = ( ˆ x t − 1 , ˆ y t − 1 ) − ( ˆ x t , ˆ y t ) if η = 0, and setting ( ˆ x t − 1 , ˆ y t − 1 ) = ( ˆ x t − 1 , ˆ y t − 1 ) if η = ( C − V ) / V or η = L ∈ S . Her e, ( ˆ x t − 1 , ˆ y t − 1 ) is the point in group (A3) , and ( ˆ x t , ˆ y t ) is the point in group (A4) . (iv) The point in group (B4), denoted ( ˆ x t , ˆ y t ) , is obtained by setting ( ˆ x t , ˆ y t ) = ( ˆ x t , ˆ y t ) − ( ˆ x t + 1 , ˆ y t + 1 ) . Here, ( ˆ x t , ˆ y t ) is the point in group (A4) , and ( ˆ x t + 1 , ˆ y t + 1 ) is the point with index t + 1 in group (A5) . (v) For each r ∈ [ t + 1, T ] Z , the point with index r in group (B5), denoted ( ˆ x r , ˆ y r ) , is obtained by setting ( ˆ x r , ˆ y r ) = ( ˆ x r , ˆ y r ) − ( ˆ x r + 1 , ˆ y r + 1 ) if r  = T , and setting ( ˆ x r , ˆ y r ) = ( ˆ x r , ˆ y r ) if r = T . Here, ( ˆ x r , ˆ y r ) and ( ˆ x r + 1 , ˆ y r + 1 ) are the points with indices r and r + 1, respectively , in group (A5) . The matrix shown in T able EC.6 is lower triangular; that is, the position of the last nonzero component of a row of the matrix is greater than the position of the last nonzero component of the previous row . This implies that the 2 T − 1 points in gr oups (A1) – (A5) are linearly independent. Ther efore, inequality ( 17 ) is facet-deﬁning for conv ( P ) . Next, we show that inequality ( 18 ) is valid for conv ( P ) and is facet-deﬁning for conv ( P ) when η ∈ { 0, ( C − V ) / V } or η = L ∈ S . Note that this proof can also be used to prove the validity and facet-deﬁning of inequality ( 18 ) in Proposition 6 . Denote x ′ t = x T − t + 1 and y ′ t = y T − t + 1 for t ∈ [ 1, T ] Z . Because inequality ( 17 ) is valid for conv ( P ) and is facet-deﬁning for conv ( P ) when η ∈ { 0, ( C − V ) / V } or η = L ∈ S for any t ∈ [ s max + 2, T ] Z , the inequality x ′ T − t + 1 ≤ ( V + η V ) y ′ T − t + 1 + ( C − V − η V ) y ′ T − t + 2 − ∑ s ∈ S ( C − V − s V ) ( y ′ T − t + s + 1 − y ′ T − t + s + 2 ) is valid for conv ( P ′ ) and is facet-deﬁning for conv ( P ′ ) when η ∈ { 0, ( C − V ) / V } or η = L ∈ S for any t ∈ [ s max + 2, T ] Z . Let t ′ = T − t + 1. Then, the inequality x ′ t ′ ≤ ( V + η V ) y ′ t ′ + ( C − V − η V ) y ′ t ′ + 1 − ∑ s ∈ S ( C − V − s V ) ( y ′ t ′ + s − y ′ t ′ + s + 1 ) is valid for conv ( P ′ ) and is facet-deﬁning for conv ( P ′ ) when η ∈ { 0, ( C − V ) / V } or η = L ∈ S for any t ′ ∈ [ 1, T − s max − 1 ] Z . Hence, by Lemma 2 , inequality ( 18 ) is valid for conv ( P ) and is facet-deﬁning for conv ( P ) when η ∈ { 0, ( C − V ) / V } or η = L ∈ S for any t ∈ [ 1, T − s max − 1 ] Z . □ C.6. Pro of of Prop osition 6 Proposition 6 . Consider any integers α , β , and s max such that (a) L + 1 ≤ s max ≤ min { T − 2, ⌊ ( C − V ) / V ⌋ } , (b) 1 ≤ α < β ≤ s max , and (c) β = α + 1 or s max ≤ L + α . Let S = [ 1, α ] Z ∪ [ β , s max ] Z . For any t ∈ [ s max + 2, T ] Z , inequality ( 17 ) is valid for conv ( P ) . For any t ∈ [ 1, T − s max − 1 ] Z , inequality ( 18 ) is valid for conv ( P ) . Furthermore, ( 17 ) and ( 18 ) ar e facet-deﬁning for conv ( P ) when η ∈ { 0, ( C − V ) / V } or η = L ∈ S . Proof. Consider any t ∈ [ s max + 2, T ] Z . T o prove that the linear inequality ( 17 ) is valid for conv ( P ) when S = [ 1, α ] Z ∪ [ β , s max ] Z , it sufﬁces to show that ( 17 ) is valid for P when S = [ 1, α ] Z ∪ [ β , s max ] Z . Consider any element ( x , y ) of P . W e show that ( x , y ) satisﬁes ( 17 ) when S = [ 1, α ] Z ∪ [ β , s max ] Z . W e divide the analysis into four cases. Case 1: y t = 0 and y t − s − y t − s − 1 ≤ 0 for all s ∈ S . Because S ⊆ [ 1, ⌊ ( C − V ) / V ⌋ ] Z , we have C − V − s V ≥ 0 for all s ∈ S . Because η ≤ ( C − V ) / V , we have C − V − η V ≥ 0. Thus, in this case, the right-hand side of inequality ( 17 ) is nonnegative. Because y t = 0, by ( 2d ), x t = 0. Therefor e, in this case, ( x , y ) satisﬁes ( 17 ). Case 2: y t = 0 and y t − s − y t − s − 1 > 0 for some s ∈ S . Let ˜ S = { σ ∈ S : y t − σ − y t − σ − 1 > 0 } and v = | ˜ S | . Then, v ≥ 1. Denote ˜ S = { σ 1 , σ 2 , . . . , σ v } , where σ 1 < σ 2 < · · · < σ v . Note that y t − σ j − 1 = 0 and y t − σ j = 1 ec.22 T able EC.5 A matrix with the rows r epresenting 2 T − 1 points in conv ( P ) that satisfy inequality ( 17 ) at equality Group Point Index r x y 1 · · · t − 2 t − 1 t t + 1 · · · T 1 · · · t − 2 t − 1 t t + 1 · · · T (A1) ( ¯ x r , ¯ y r ) 1 C − ϵ · · · C C C C · · · C 1 · · · 1 1 1 1 · · · 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . t − 2 C · · · C − ϵ C C C · · · C 1 · · · 1 1 1 1 · · · 1 t − 1 C · · · C C − ϵ C C · · · C 1 · · · 1 1 1 1 · · · 1 t + 1 C · · · C C C C − ϵ · · · C 1 · · · 1 1 1 1 · · · 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . T C · · · C C C C · · · C − ϵ 1 · · · 1 1 1 1 · · · 1 (A2) ( ˆ x r , ˆ y r ) 1 (See Note EC.5 -1) (See Note EC.5 -1) . . . t − 2 (A3) t − 1 (See Note EC.5 -2) (See Note EC.5 -2) (A4) t C · · · C C C C · · · C 1 · · · 1 1 1 1 · · · 1 (A5) t + 1 0 · · · 0 0 0 C · · · C 0 · · · 0 0 0 1 · · · 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . T 0 · · · 0 0 0 0 · · · C 0 · · · 0 0 0 0 · · · 1 Note EC.5 -1: For r ∈ [ 1, t − 2 ] Z , the x and y vectors in group (A2) are given as follows: ˆ x r = ( C , . . . , C | {z } r terms , 0, . . . , 0 | { z } T − r terms ) and ˆ y r = ( 1, . . . , 1 | { z } r terms , 0, . . . , 0 | { z } T − r terms ) if t − r − 1 / ∈ S ; ˆ x r = ( 0, . . . , 0 | { z } r terms , V , V + V , V + 2 V , . . . , V + ( t − r − 1 ) V | { z } t − r terms , V + ( t − r − 1 ) V , . . . , V + ( t − r − 1 ) V | {z } T − t terms ) and ˆ y r = ( 0, . . . , 0 | { z } r terms , 1, . . . , 1 | { z } T − r terms ) if t − r − 1 ∈ S . Note EC.5 -2: The x and y vectors in group (A3) are given as follows: ˆ x t − 1 = ( 0, . . . , 0 | { z } t − 1 terms , V , . . . , V | { z } T − t + 1 terms ) and ˆ y t − 1 = ( 0, . . . , 0 | { z } t − 1 terms , 1, . . . , 1 | { z } T − t + 1 terms ) if η = 0; ˆ x t − 1 = ( C , . . . , C | {z } t − 1 terms , 0, . . . , 0 | { z } T − t + 1 terms ) and ˆ y t − 1 = ( 1, . . . , 1 | { z } t − 1 terms , 0, . . . , 0 | { z } T − t + 1 terms ) if η = ( C − V ) / V ; ˆ x t − 1 = ( 0, . . . , 0 | { z } t − L − 1 terms , V , . . . , V | { z } L terms , 0, . . . , 0 | { z } T − t + 1 terms ) and ˆ y t − 1 = ( 0, . . . , 0 | { z } t − L − 1 terms , 1, . . . , 1 | { z } L terms , 0, . . . , 0 | { z } T − t + 1 terms ) if η = L ∈ S . ec.23 T able EC.6 Lower triangular matrix obtained from T able EC.5 via Gaussian elimination Group Point Index r x y 1 · · · t − 2 t − 1 t t + 1 · · · T 1 · · · t − 2 t − 1 t t + 1 · · · T (B1) ( ¯ x r , ¯ y r ) 1 − ϵ · · · 0 0 0 0 · · · 0 0 · · · 0 0 0 0 · · · 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . t − 2 0 · · · − ϵ 0 0 0 · · · 0 0 · · · 0 0 0 0 · · · 0 t − 1 0 · · · 0 − ϵ 0 0 · · · 0 0 · · · 0 0 0 0 · · · 0 t + 1 0 · · · 0 0 0 − ϵ · · · 0 0 · · · 0 0 0 0 · · · 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . T 0 · · · 0 0 0 0 · · · − ϵ 0 · · · 0 0 0 0 · · · 0 (B2) ( ˆ x r , ˆ y r ) 1 1 · · · 0 0 0 0 · · · 0 . . . (Omitted) . . . . . . . . . . . . . . . . . . . . . t − 2 1 · · · 1 0 0 0 · · · 0 (B3) t − 1 (Omitted) (See Note EC.6 -1) (B4) t (Omitted) 1 · · · 1 1 1 0 · · · 0 (B5) t + 1 0 · · · 0 0 0 1 · · · 0 . . . (Omitted) . . . . . . . . . . . . . . . . . . . . . T 0 · · · 0 0 0 0 · · · 1 Note EC.6 -1: The y vector in group (B3) is given as follows: ˆ y t − 1 = ( − 1, . . . , − 1 | {z } t − 1 terms , 0, . . . , 0 | { z } T − t + 1 terms ) if η = 0; ˆ y t − 1 = ( 1, . . . , 1 | { z } t − 1 terms , 0, . . . , 0 | { z } T − t + 1 terms ) if η = ( C − V ) / V ; ˆ y t − 1 = ( 0, . . . , 0 | { z } t − L − 1 terms , 1, . . . , 1 | { z } L terms , 0, . . . , 0 | { z } T − t + 1 terms ) if η = L ∈ S . ec.24 for j = 1, . . . , v . Denote σ 0 = − 1. Then, for each j = 1, . . . , v , ther e exists σ ′ j ∈ [ σ j − 1 + 1, σ j − 1 ] Z such that y t − σ ′ j − 1 = 1 and y t − σ ′ j = 0. Thus, 0 ≤ σ ′ 1 < σ 1 < σ ′ 2 < σ 2 < · · · < σ ′ v < σ v ≤ s max . Because y t − σ v − y t − σ v − 1 = 1 and t − σ v ∈ [ 2, T ] Z , by ( 2a ), y k = 1 for all k ∈ [ t − σ v , min { T , t − σ v + L − 1 } ] Z , which implies that t − σ ′ j ≥ t − σ v + L for j = 1, . . . , v . Hence, for j = 1, . . . , v , we have σ ′ j ≤ σ v − L , which implies that σ ′ j ≤ s max − L . (EC.21) If β = α + 1, then S = [ 1, s max ] Z , which implies that σ ′ j ∈ S for j = 2, . . . , v . If β  = α + 1, then condition (c) of Proposition 6 implies that s max ≤ L + α , which, by ( EC.21 ), implies that σ ′ j ≤ α for j = 1, . . . , v . Because σ ′ 2 > σ 1 ≥ 1, we have σ ′ j ∈ S for j = 2, . . . , v . Thus, in both cases, σ ′ j ∈ S for j = 2, . . . , v . Because y t − σ 1 − y t − σ 1 − 1 = 1 and t − σ 1 ∈ [ 2, T ] Z , by ( 2a ), y k = 1 for all k ∈ [ t − σ 1 , min { T , t − σ 1 + L − 1 } ] Z . Because y t = 0, , this implies that t ≥ t − σ 1 + L , or equivalent, σ 1 ≥ L . Because η ≤ L , we have η ≤ σ 1 . (EC.22) Because y t = 0, by ( 2d ), x t = 0. Hence, the left-hand side of inequality ( 17 ) is 0. Because s max ≤ ⌊ ( C − V ) / V ⌋ , we have C − V − sV ≥ 0 for all s ∈ S . Note that if y t − 1 = 0, then σ ′ 1 ≥ 1 and σ ′ 1 ∈ S ; if y t − 1 = 1, then σ ′ 1 = 0 and σ ′ 1 / ∈ S . Note that { σ ′ 2 , . . . , σ ′ v } ⊆ S \ ˜ S and y t − s − y t − s − 1 ≤ 0 for all s ∈ S \ ˜ S . Thus, ∑ s ∈ S \ ˜ S ( C − V − s V ) ( y t − s − y t − s − 1 ) ≤ ∑ v j = 2 ( C − V − σ ′ j V )( y t − σ ′ j − y t − σ ′ j − 1 ) + ( C − V − σ ′ 1 V )( y t − σ ′ 1 − y t − σ ′ 1 − 1 )( 1 − y t − 1 ) . Hence, the right-hand side of inequality ( 17 ) is ( V + η V ) y t + ( C − V − η V ) y t − 1 − ∑ s ∈ S ( C − V − s V ) ( y t − s − y t − s − 1 ) = ( C − V − η V ) y t − 1 − ∑ s ∈ ˜ S ( C − V − s V ) ( y t − s − y t − s − 1 ) − ∑ s ∈ S \ ˜ S ( C − V − s V ) ( y t − s − y t − s − 1 ) ≥ ( C − V − η V ) y t − 1 − v ∑ j = 1 ( C − V − σ j V )( y t − σ j − y t − σ j − 1 ) − v ∑ j = 2 ( C − V − σ ′ j V )( y t − σ ′ j − y t − σ ′ j − 1 ) − ( C − V − σ ′ 1 V )( y t − σ ′ 1 − y t − σ ′ 1 − 1 )( 1 − y t − 1 ) = ( C − V − η V ) y t − 1 − ( C − V − σ ′ 1 V ) y t − 1 − ( C − V − σ 1 V ) + ( C − V − σ ′ 1 V ) − v ∑ j = 2 ( C − V − σ j V ) + v ∑ j = 2 ( C − V − σ ′ j V ) = ( σ ′ 1 − η ) Vy t − 1 + ( σ 1 − σ ′ 1 ) V + v ∑ j = 2 ( σ j − σ ′ j ) V ≥ ( σ ′ 1 − η ) Vy t − 1 + ( σ 1 − σ ′ 1 ) V ≥ 0. where the last inequality follows from y t − 1 ∈ { 0, 1 } , σ 1 > σ ′ 1 , and ( EC.22 ). Therefore, in this case, ( x , y ) satisﬁes ( 17 ). Case 3: y t = 1 and y t − s − y t − s − 1 ≤ 0 for all s ∈ S . Because S ⊆ [ 0, ⌊ ( C − V ) / V ⌋ ] Z , we have C − V − s V ≥ 0 for all s ∈ S . Thus, ∑ s ∈ S ( C − V − s V ) ( y t − s − y t − s − 1 ) ≤ 0. Because y t = 1, the right-hand side of inequality ( 17 ) is at least C when y t − 1 = 1 and is at least V + η V ≥ V when y t − 1 = 0 (as η ≥ 0). By ( 2d ), x t ≤ C . If y t − 1 = 0, then by ( 2d ) and ( 2e ), x t − 1 = 0 and x t − x t − 1 ≤ Vy t − 1 + V ( 1 − y t − 1 ) , which imply that x t ≤ V . Hence, x t is at most C , and is at most V when y t − 1 = 0. Therefor e, in this case, ( x , y ) satisﬁes ( 17 ). Case 4: y t = 1 and y t − s − y t − s − 1 > 0 for some s ∈ S . Let ˜ S = { σ ∈ S : y t − σ − y t − σ − 1 > 0 } and v = | ˜ S | . Then, v ≥ 1. Denote ˜ S = { σ 1 , σ 2 , . . . , σ v } , where σ 1 < σ 2 < · · · < σ v . Note that y t − σ j − 1 = 0 and y t − σ j = 1 for j = 1, . . . , v . Then, for each j = 2, . . . , v , ther e exists σ ′ j ∈ [ σ j − 1 + 1, σ j − 1 ] Z such that y t − σ ′ j − 1 = 1 and y t − σ ′ j = 0. Thus, 1 ≤ σ 1 < σ ′ 2 < σ 2 < · · · < σ ′ v < σ v ≤ s max . ec.25 Because y t − σ v − y t − σ v − 1 = 1 and t − σ v ∈ [ 2, T ] Z , by ( 2a ), y k = 1 for all k ∈ [ t − σ v , min { T , t − σ v + L − 1 } ] Z , which implies that t − σ ′ j ≥ t − σ v + L for j = 2, . . . , v . Hence, for j = 2, . . . , v , we have σ ′ j ≤ σ v − L , which implies that σ ′ j ≤ s max − L . (EC.23) If β = α + 1, then S = [ 1, s max ] Z , which implies that σ ′ j ∈ S for j = 2, . . . , v . If β  = α + 1, then condition (c) of Proposition 6 implies that s max ≤ L + α , which, by ( EC.23 ), implies that 1 < σ ′ j ≤ α for j = 2, . . . , v . Thus, in both cases, σ ′ j ∈ S for j = 2, . . . , v . If y t − 1 = 0, by ( 2d ) and ( 2e ), then x t − 1 = 0 and x t − x t − 1 ≤ Vy t − 1 + V ( 1 − y t − 1 ) = V , which implies that x t ≤ V . In addition, there exists σ ′ 1 ∈ [ 1, σ 1 − 1 ] Z such that y t − σ ′ 1 = 0, y t − σ ′ 1 − 1 = 1, and σ ′ 1 ∈ S . If y t − 1 = 1, then we have y k = 1 for all k ∈ [ t − σ 1 , t ] Z . Because y t − σ 1 − 1 = 0, by ( 2d ) and ( 2e ), we have x t − σ 1 − 1 = 0 and t ∑ τ = t − σ 1 ( x τ − x τ − 1 ) ≤ t ∑ τ = t − σ 1 Vy τ − 1 + t ∑ τ = t − σ 1 V ( 1 − y τ − 1 ) , which implies that x t − x t − σ 1 − 1 ≤ t ∑ τ = t − σ 1 Vy τ − 1 + t ∑ τ = t − σ 1 V ( 1 − y τ − 1 ) = σ 1 V + V . Thus, the left-hand side of inequality ( 17 ) is at most σ 1 V + V . Because S ⊆ [ 0, ⌊ ( C − V ) / V ⌋ ] Z , we have C − V − sV ≥ 0 for all s ∈ S . Note that { σ ′ 2 , . . . , σ ′ v } ⊆ S \ ˜ S . Thus, ∑ s ∈ S \ ˜ S ( C − V − s V ) ( y t − s − y t − s − 1 ) ≤ ∑ v j = 2 ( C − V − σ ′ j V )( y t − σ ′ j − y t − σ ′ j − 1 ) . Hence, when y t − 1 = 0, the right-hand side of inequality ( 17 ) is ( V + η V ) y t + ( C − V − η V ) y t − 1 − ∑ s ∈ S ( C − V − s V ) ( y t − s − y t − s − 1 ) = V + η V − ∑ s ∈ ˜ S ( C − V − s V ) ( y t − s − y t − s − 1 ) − ∑ s ∈ S \ ˜ S ( C − V − s V ) ( y t − s − y t − s − 1 ) ≥ V + η V − v ∑ j = 1 ( C − V − σ j V )( y t − σ j − y t − σ j − 1 ) − v ∑ j = 1 ( C − V − σ ′ j V )( y t − σ ′ j − y t − σ ′ j − 1 ) = V + η V + v ∑ j = 1 ( σ j − σ ′ j ) V ≥ V + η V ≥ V . When y t − 1 = 1, the right-hand side of inequality ( 17 ) is ( V + η V ) y t + ( C − V − η V ) y t − 1 − ∑ s ∈ S ( C − V − s V ) ( y t − s − y t − s − 1 ) = C − ∑ s ∈ ˜ S ( C − V − s V ) ( y t − s − y t − s − 1 ) − ∑ s ∈ S \ ˜ S ( C − V − s V ) ( y t − s − y t − s − 1 ) ≥ C − v ∑ j = 1 ( C − V − σ j V )( y t − σ j − y t − σ j − 1 ) − v ∑ j = 2 ( C − V − σ ′ j V )( y t − σ ′ j − y t − σ ′ j − 1 ) = C − ( C − V − σ 1 V ) − v ∑ j = 2 ( C − V − σ j V ) + v ∑ j = 2 ( C − V − σ ′ j V ) = σ 1 V + V + v ∑ j = 2 ( σ j − σ ′ j ) V ≥ σ 1 V + V . Therefor e, in this case, ( x , y ) satisﬁes ( 17 ). Summarizing Cases 1–4, we conclude that ( x , y ) satisﬁes ( 17 ). Hence, ( 17 ) is valid for conv ( P ) . ec.26 It is easy to verify that the proof of facet-deﬁning of inequality ( 17 ) in the proof of Proposition 5 r emains valid when S = [ 1, α ] Z ∪ [ β , s max ] Z . Therefor e, inequality ( 17 ) is facet-deﬁning under the conditions stated in Proposition 6 . It is also easy to verify that the proof of validity and facet-deﬁning of inequality ( 18 ) in the proof of Proposition 5 remains valid when S = [ 1, α ] Z ∪ [ β , s max ] Z . Therefor e, inequality ( 18 ) is valid and facet- deﬁning for conv ( P ) under the conditions stated in Proposition 6 . □ C.7. Pro of of Prop osition 7 Proposition 7 . For any given point ( x , y ) ∈ R 2 T + , the most violated inequalities ( 13 ) – ( 14 ) , ( 15 ) – ( 16 ) , and ( 17 ) – ( 18 ) in Propositions 1 , 3 , and 5 , r espectively, can be determined in O ( T ) time if such violated inequalities exist. Proof. Let ˆ η and a 1 , . . . , a 6 be any r eal numbers such that ˆ η ≥ 0. Let ˇ s and ˆ s be any integers such that 0 ≤ ˇ s ≤ ˆ s ≤ min { T − 2, ⌊ ( C − V ) / V ⌋ } . Let ˇ t = 1 if a 1 = a 2 = 0, and let ˇ t = 2 otherwise. Let ˆ t = T if a 5 = a 6 = 0, and let ˆ t = T − 1 otherwise. (i) Consider the following family of inequalities: x t ≤ ( a 1 + a 2 η ) y t − 1 + ( a 3 + a 4 η ) y t + ( a 5 + a 6 η ) y t + 1 − ∑ s ∈ S ( C − V − s V ) ( y t − s − y t − s − 1 ) , (EC.24) where η ∈ [ 0, ˆ η ] , S ⊆ [ ˇ s , ˆ s ] Z , t ∈ [ ˇ t , ˆ t ] Z , and t ≥ s + 2 for all s ∈ S . Note that inequality family ( 13 ) in Propo- sition 1 , inequality family ( 15 ) in Proposition 3 , and inequality family ( 17 ) in Proposition 5 ar e special cases of this inequality family . Consider any given point ( x , y ) ∈ R 2 T + . W e show that the set S , the real number η , and the integer t corresponding to a most violated inequality ( EC.24 ) can be determined in O ( T ) time. For any integer t ≤ T , let θ ( t ) = t ∑ τ = 2 max { y τ − y τ − 1 , 0 } . Then, for any t ∈ [ 2, T ] Z , ∑ s ∈ [ ˇ s , ˆ s ] Z t − s ≥ 2 max { y t − s − y t − s − 1 , 0 } = θ ( t − ˇ s ) − θ ( t − ˆ s − 1 ) (EC.25) and ∑ s ∈ [ ˇ s + 1, ˆ s + 1 ] Z t − s ≥ 2 max { y t − s − y t − s − 1 , 0 } = θ ( t − ˇ s − 1 ) − θ ( t − ˆ s − 2 ) . (EC.26) Furthermore, for any t ∈ [ 2, T ] Z , ∑ s ∈ [ ˇ s , ˆ s ] Z t − s ≥ 2 s max { y t − s − y t − s − 1 , 0 } − ∑ s ∈ [ ˇ s + 1, ˆ s + 1 ] Z t − s ≥ 2 s max { y t − s − y t − s − 1 , 0 } =    ˇ s max { y t − ˇ s − y t − ˇ s − 1 , 0 } − ( ˆ s + 1 ) max { y t − ˆ s − 1 − y t − ˆ s − 2 , 0 } , if 2 ≤ t − ˆ s − 1; ˇ s max { y t − ˇ s − y t − ˇ s − 1 , 0 } , if t − ˆ s − 1 < 2 ≤ t − ˇ s ; 0, if t − ˇ s < 2. Note that θ ( t − ˇ s ) − θ ( t − ˇ s − 1 ) =  max { y t − ˇ s − y t − ˇ s − 1 , 0 } , if t − ˇ s ≥ 2; 0, if t − ˇ s < 2. and θ ( t − ˆ s − 1 ) − θ ( t − ˆ s − 2 ) =  max { y t − ˆ s − 1 − y t − ˆ s − 2 , 0 } , if t − ˆ s − 1 ≥ 2; 0, if t − ˆ s − 1 < 2. Hence, for any t ∈ [ 2, T ] Z , ∑ s ∈ [ ˇ s , ˆ s ] Z t − s ≥ 2 s max { y t − s − y t − s − 1 , 0 } − ∑ s ∈ [ ˇ s + 1, ˆ s + 1 ] Z t − s ≥ 2 s max { y t − s − y t − s − 1 , 0 } = ˇ s [ θ ( t − ˇ s ) − θ ( t − ˇ s − 1 ) ] − ( ˆ s + 1 ) [ θ ( t − ˆ s − 1 ) − θ ( t − ˆ s − 2 ) ] . (EC.27) ec.27 For any η ∈ [ 0, ˆ η ] , S ⊆ [ ˇ s , ˆ s ] Z , and t ∈ [ ˇ t , ˆ t ] Z such that t ≥ s + 2 ∀ s ∈ S , let ˜ v ( η , S , t ) = x t − ( a 1 + a 2 η ) y t − 1 − ( a 3 + a 4 η ) y t − ( a 5 + a 6 η ) y t + 1 + ∑ s ∈ S ( C − V − s V ) ( y t − s − y t − s − 1 ) . If ˜ v ( η , S , t ) > 0, then ˜ v ( η , S , t ) is the amount of violation of inequality ( EC.24 ). If ˜ v ( η , S , t ) ≤ 0, then there is no violation of inequality ( EC.24 ). For any η ∈ [ 0, ˆ η ] and t ∈ [ ˇ t , ˆ t ] Z , let v ( η , t ) = max S ⊆ [ ˇ s ,min { ˆ s , t − 2 } ] Z { ˜ v ( η , S , t ) } . If v ( η , t ) > 0, then v ( η , t ) is the lar gest possible violation of inequality ( EC.24 ) for this combination of η and t . If v ( η , t ) ≤ 0, then the largest possible violation of inequality ( EC.24 ) is zero for this combination of η and t . Note that C − V − sV ≥ 0 for any s ∈ [ ˇ s , ˆ s ] Z . Thus, for any given η ∈ [ 0, ˆ η ] and t ∈ [ ˇ t , ˆ t ] Z , ˜ v ( η , S , t ) is maximized when S contains all s ∈ [ ˇ s , min { ˆ s , t − 2 } ] Z such that y t − s − y t − s − 1 > 0 (if any). If it does not exist any s ∈ [ ˇ s , min { ˆ s , t − 2 } ] Z such that y t − s − y t − s − 1 > 0, then ˜ v ( η , S , t ) is maximized when S = ∅ , and v ( η , t ) = x t − ( a 1 + a 2 η ) y t − 1 − ( a 3 + a 4 η ) y t − ( a 5 + a 6 η ) y t + 1 . Hence, for any η ∈ [ 0, ˆ η ] and t ∈ [ ˇ t , ˆ t ] Z , v ( η , t ) = x t − ( a 1 + a 2 η ) y t − 1 − ( a 3 + a 4 η ) y t − ( a 5 + a 6 η ) y t + 1 + ( C − V ) ∑ s ∈ [ ˇ s , ˆ s ] Z t − s ≥ 2 max { y t − s − y t − s − 1 , 0 } − V ∑ s ∈ [ ˇ s , ˆ s ] Z t − s ≥ 2 s max { y t − s − y t − s − 1 , 0 } . When t = ˇ t , we have v ( η , ˇ t ) =          x 1 − ( a 3 + a 4 η ) y 1 − ( a 5 + a 6 η ) y 2 , if ˇ t = 1; x 2 − ( a 1 + a 2 η ) y 1 − ( a 3 + a 4 η ) y 2 − ( a 5 + a 6 η ) y 3 +( C − V ) max { y 2 − y 1 , 0 } , if ˇ t = 2 and ˇ s = 0; x 2 − ( a 1 + a 2 η ) y 1 − ( a 3 + a 4 η ) y 2 − ( a 5 + a 6 η ) y 3 , otherwise. For any η ∈ [ 0, ˆ η ] and t ∈ [ ˇ t + 1, ˆ t ] Z , v ( η , t ) − v ( η , t − 1 ) = ( x t − x t − 1 ) − ( a 1 + a 2 η ) ( y t − 1 − y t − 2 ) − ( a 3 + a 4 η ) ( y t − y t − 1 ) − ( a 5 + a 6 η ) ( y t + 1 − y t ) + ( C − V )     ∑ s ∈ [ ˇ s , ˆ s ] Z t − s ≥ 2 max { y t − s − y t − s − 1 , 0 } − ∑ s ∈ [ ˇ s + 1, ˆ s + 1 ] Z t − s ≥ 2 max { y t − s − y t − s − 1 , 0 }     − V     ∑ s ∈ [ ˇ s , ˆ s ] Z t − s ≥ 2 s max { y t − s − y t − s − 1 , 0 } − ∑ s ∈ [ ˇ s + 1, ˆ s + 1 ] Z t − s ≥ 2 ( s − 1 ) max { y t − s − y t − s − 1 , 0 }     , which implies that v ( η , t ) = v ( η , t − 1 ) + ( x t − x t − 1 ) − ( a 1 + a 2 η ) ( y t − 1 − y t − 2 ) − ( a 3 + a 4 η ) ( y t − y t − 1 ) − ( a 5 + a 6 η ) ( y t + 1 − y t ) + ( C − V )     ∑ s ∈ [ ˇ s , ˆ s ] Z t − s ≥ 2 max { y t − s − y t − s − 1 , 0 } − ∑ s ∈ [ ˇ s + 1, ˆ s + 1 ] Z t − s ≥ 2 max { y t − s − y t − s − 1 , 0 }     − V ∑ s ∈ [ ˇ s + 1, ˆ s + 1 ] Z t − s ≥ 2 max { y t − s − y t − s − 1 , 0 } − V     ∑ s ∈ [ ˇ s , ˆ s ] Z t − s ≥ 2 s max { y t − s − y t − s − 1 , 0 } − ∑ s ∈ [ ˇ s + 1, ˆ s + 1 ] Z t − s ≥ 2 s max { y t − s − y t − s − 1 , 0 }     . ec.28 Algorithm 1 Determination of a most violated inequality ( EC.24 ) for any given ( x , y ) ∈ R 2 T + 1: θ ( t ) ← 0 ∀ t ∈ [ − ˆ s , 1 ] Z 2: for t = 2, . . . , ˆ t do 3: θ ( t ) ← θ ( t − 1 ) + max { y t − y t − 1 , 0 } 4: end for 5: for η = 0, ˆ η do 6: if ˇ t = 1 then 7: v ( η , ˇ t ) ← x 1 − ( a 3 + a 4 η ) y 1 − ( a 5 + a 6 η ) y 2 8: else if ˇ s = 0 then 9: v ( η , ˇ t ) ← x 2 − ( a 1 + a 2 η ) y 1 − ( a 3 + a 4 η ) y 2 − ( a 5 + a 6 η ) y 3 + ( C − V ) max { y 2 − y 1 , 0 } 10: else 11: v ( η , ˇ t ) ← x 2 − ( a 1 + a 2 η ) y 1 − ( a 3 + a 4 η ) y 2 − ( a 5 + a 6 η ) y 3 12: end if 13: for t = ˇ t + 1, . . . , ˆ t do 14: v ( η , t ) ← v ( η , t − 1 ) + ( x t − x t − 1 ) − ( a 1 + a 2 η ) ( y t − 1 − y t − 2 ) − ( a 3 + a 4 η ) ( y t − y t − 1 ) − ( a 5 + a 6 η ) ( y t + 1 − y t ) + ( C − V ) [ θ ( t − ˇ s ) − θ ( t − ˆ s − 1 ) − θ ( t − ˇ s − 1 ) + θ ( t − ˆ s − 2 ) ] − V [ ˇ s θ ( t − ˇ s ) − ( ˇ s − 1 ) θ ( t − ˇ s − 1 ) − ( ˆ s + 1 ) θ ( t − ˆ s − 1 ) + ˆ s θ ( t − ˆ s − 2 ) ] 15: end for 16: end for 17: ( η ∗ , t ∗ ) ← argmax ( η , t ) ∈ { 0, ˆ η }× [ ˇ t , ˆ t ] Z { v ( η , t ) } 18: S ∗ ← ∅ 19: for s = ˇ s , . . . , min { ˆ s , t ∗ − 2 } do 20: if y t ∗ − s − y t ∗ − s − 1 > 0 then S ∗ ← S ∗ ∪ { s } 21: end for Thus, by ( EC.25 ), ( EC.26 ), and ( EC.27 ), v ( η , t ) = v ( η , t − 1 ) + ( x t − x t − 1 ) − ( a 1 + a 2 η ) ( y t − 1 − y t − 2 ) − ( a 3 + a 4 η ) ( y t − y t − 1 ) − ( a 5 + a 6 η ) ( y t + 1 − y t ) + ( C − V ) [ θ ( t − ˇ s ) − θ ( t − ˆ s − 1 ) − θ ( t − ˇ s − 1 ) + θ ( t − ˆ s − 2 ) ] − V [ ˇ s θ ( t − ˇ s ) − ( ˇ s − 1 ) θ ( t − ˇ s − 1 ) − ( ˆ s + 1 ) θ ( t − ˆ s − 1 ) + ˆ s θ ( t − ˆ s − 2 ) ] (EC.28) for any η ∈ [ 0, ˆ η ] and t ∈ [ ˇ t + 1, ˆ t ] Z . Note that ˜ v ( η , S , t ) is linear in η . Thus, for any given t , v ( η , t ) is maxi- mized when η = 0 or η = ˆ η . That is, the largest possible value of v ( η , t ) is equal to v ( 0, t ) if a 2 y t − 1 + a 4 y t + a 6 y t + 1 ≥ 0, and the lar gest possible value of v ( η , t ) is equal to v ( ˆ η , t ) if a 2 y t − 1 + a 4 y t + a 6 y t + 1 < 0. Hence, to determine the η and t values corresponding to the largest violation of inequality ( EC.24 ), it sufﬁces to deter - mine v ( 0, ˇ t ) , v ( 0, ˇ t + 1 ) , . . . , v ( 0, ˆ t ) and v ( ˆ η , ˇ t ) , v ( ˆ η , ˇ t + 1 ) , . . . , v ( ˆ η , ˆ t ) . Algorithm 1 performs this computation. In Algorithm 1 , step 1 sets θ ( t ) to zero when t ≤ 1. Steps 2–4 determine the θ ( t ) values recursively for t = 2, 3, . . . , ˆ t . These steps r equire O ( T ) time. Steps 5–16 consider the case η = 0 and the case η = ˆ η . For each of these two η values, these steps ﬁrst determine v ( η , ˇ t ) , and then determine v ( η , ˇ t + 1 ) , v ( η , ˇ t + 2 ) , . . . , v ( η , ˆ t ) recursively using equation ( EC.28 ). These steps requir e O ( T ) time. Steps 17–21 identify a most violated inequality ( EC.24 ) by setting the η and t values to ( η ∗ , t ∗ ) = argmax ( η , t ) ∈ { 0, ˆ η }× [ ˇ t , ˆ t ] Z { v ( η , t ) } and setting S equal to the set of s values such that s ∈ [ ˇ s , min { ˆ s , t ∗ − 2 } ] Z and y t ∗ − s − y t ∗ − s − 1 > 0. These steps also requir e O ( T ) time. Therefore, the total computational time of Algorithm 1 is O ( T ) . (ii) Consider the following family of inequalities: x t ≤ ( a 1 + a 2 η ) y t − 1 + ( a 3 + a 4 η ) y t + ( a 5 + a 6 η ) y t + 1 − ∑ s ∈ S ( C − V − s V ) ( y t + s − y t + s + 1 ) , (EC.29) where η ∈ [ 0, ˆ η ] , S ⊆ [ ˇ s , ˆ s ] Z , t ∈ [ ˇ t , ˆ t ] Z , and t ≥ s + 2 for all s ∈ S . Note that inequality family ( 14 ) in Proposi- tion 1 , inequality family ( 16 ) in Pr oposition 3 , and inequality family ( 18 ) in Proposition 5 ar e special cases of this inequality family . Consider any given point ( x , y ) ∈ R 2 T + . Let x ′ t = x T − t + 1 and y ′ t = y T − t + 1 for t ∈ [ 1, T ] Z . Inequality ( EC.29 ) becomes x ′ T − t + 1 ≤ ( a 1 + a 2 η ) y ′ T − t + 2 + ( a 3 + a 4 η ) y ′ T − t + 1 + ( a 5 + a 6 η ) y ′ T − t − ∑ s ∈ S ( C − V − s V ) ( y ′ T − t − s + 1 − y ′ T − t − s ) . ec.29 Letting t ′ = T − t + 1, this inequality becomes x ′ t ′ ≤ ( a 5 + a 6 η ) y ′ t ′ − 1 + ( a 3 + a 4 η ) y ′ t ′ + ( a 1 + a 2 η ) y ′ t ′ + 1 − ∑ s ∈ S ( C − V − s V ) ( y ′ t ′ − s − y ′ t ′ − s − 1 ) . (EC.30) Let ˇ t ′ = 1 if a 5 = a 6 = 0, and let ˇ t ′ = 2 otherwise. Let ˆ t ′ = T if a 1 = a 2 = 0, and let ˆ t ′ = T − 1 otherwise. From the analysis in part (i), the set S ⊆ [ ˇ s , ˆ s ] , the r eal number η ∈ [ 0, ˆ η ] Z , and the integer t ′ ∈ [ ˇ t ′ , ˆ t ′ ] Z with t ′ ≥ s + 2 ∀ s ∈ S corresponding to a most violated inequality ( EC.30 ) can be obtained in O ( T ) time using Algorithm 1 . Hence, the set S ⊆ [ ˇ s , ˆ s ] , the real number η ∈ [ 0, ˆ η ] Z , and the integer t ∈ [ ˇ t , ˆ t ] Z with t ≤ T − s − 1 ∀ s ∈ S corr esponding to a most violated inequality ( EC.29 ) can be obtained in O ( T ) time. Summarizing (i) and (ii), we conclude that for any given point ( x , y ) ∈ R 2 T + , the most violated inequalities ( 13 )–( 14 ), ( 15 )–( 16 ), and ( 17 )–( 18 ) in Propositions 1 , 3 , and 5 , r espectively , can be determined in O ( T ) time if such violated inequalities exist. □ C.8. Pro of of Prop osition 8 Proposition 8 . For any given point ( x , y ) ∈ R 2 T + , the most violated inequalities ( 13 ) – ( 14 ) , ( 15 ) – ( 16 ) , and ( 17 ) – ( 18 ) in Propositions 2 , 4 , and 6 , r espectively, can be determined in O ( T 3 ) time if such violated inequalities exist. Proof. Let ˆ η and a 1 , . . . , a 6 be any real numbers such that ˆ η ≥ 0. Let ˇ s , ˇ s max , and ˆ s max be any integers such that 0 ≤ ˇ s ≤ ˇ s max ≤ ˆ s max ≤ min { T − 2, ⌊ ( C − V ) / V ⌋} . Let ˇ t = 1 if a 1 = a 2 = 0, and let ˇ t = 2 otherwise. Let ˆ t = T if a 5 = a 6 = 0, and let ˆ t = T − 1 otherwise. (i) Consider the following family of inequalities: x t ≤ ( a 1 + a 2 η ) y t − 1 + ( a 3 + a 4 η ) y t + ( a 5 + a 6 η ) y t + 1 − ∑ s ∈ S ( C − V − s V ) ( y t − s − y t − s − 1 ) , (EC.31) where η ∈ [ 0, ˆ η ] , S = [ ˇ s , α ] Z ∪ [ β , s max ] Z , t ∈ [ s max + 2, ˆ t ] Z , and α , β , and s max are integers such that (a) ˇ s max ≤ s max ≤ ˆ s max , (b) ˇ s ≤ α < β ≤ s max , and (c) β = α + 1 or s max ≤ L + α . Note that inequality family ( 13 ) in Proposition 2 , inequality family ( 15 ) in Prop osition 4 , and inequality family ( 17 ) in Proposition 6 are special cases of this inequality family . Consider any given point ( x , y ) ∈ R 2 T + . W e show that the integers α , β , s max , t and the real number η corresponding to a most violated inequality can be determined in O ( T 3 ) time. For any η ∈ [ 0, ˆ η ] Z , s max ∈ [ ˇ s max , ˆ s max ] Z , S ⊆ [ ˇ s , s max ] Z , t ∈ [ s max + 2, ˆ t ] Z , let ˜ v ( η , S , t ) = x t − ( a 1 + a 2 η ) y t − 1 − ( a 3 + a 4 η ) y t − ( a 5 + a 6 η ) y t + 1 + ∑ s ∈ S ( C − V − s V ) ( y t − s − y t − s − 1 ) . If ˜ v ( η , S , t ) > 0, then ˜ v ( η , S , t ) is the amount of violation of inequality ( EC.31 ). If ˜ v ( η , S , t ) ≤ 0, then there is no violation of inequality ( EC.31 ). Note that ˜ v ( η , S , t ) is linear in η . Thus, for any given S and t , the function ˜ v ( η , S , t ) is maximized at η = 0 if a 2 y t − 1 + a 4 y t + a 6 y t + 1 ≥ 0, and is maximized at η = ˆ η if a 2 y t − 1 + a 4 y t + a 6 y t + 1 < 0. For any s max ∈ [ ˇ s max , ˆ s max ] Z , t ∈ [ s max + 2, ˆ t ] Z , and i ∈ [ ˇ s , s max ] Z , let v 1 ( s max , t , i ) =  ˜ v ( 0, [ ˇ s , i ] Z , t ) , if a 2 y t − 1 + a 4 y t + a 6 y t + 1 ≥ 0; ˜ v ( ˆ η , [ ˇ s , i ] Z , t ) , if a 2 y t − 1 + a 4 y t + a 6 y t + 1 < 0; that is, v 1 ( s max , t , i ) =          x t − a 1 y t − 1 − a 3 y t − a 5 y t + 1 + ∑ i s = ˇ s ( C − V − s V ) ( y t − s − y t − s − 1 ) , if a 2 y t − 1 + a 4 y t + a 6 y t + 1 ≥ 0; x t − ( a 1 + a 2 ˆ η ) y t − 1 − ( a 3 + a 4 ˆ η ) y t − ( a 5 + a 6 ˆ η ) y t + 1 + ∑ i s = ˇ s ( C − V − s V ) ( y t − s − y t − s − 1 ) , if a 2 y t − 1 + a 4 y t + a 6 y t + 1 < 0. For any s max ∈ [ ˇ s max , ˆ s max ] Z , t ∈ [ s max + 2, ˆ t ] Z , and j ∈ [ ˇ s + 1, s max ] Z , let v 2 ( s max , t , j ) = s max ∑ s = j ( C − V − s V ) ( y t − s − y t − s − 1 ) . Note that v 1 ( s max , t , i ) =              v 1 ( s max , t , i − 1 ) + ( C − V − i V ) ( y t − i − y t − i − 1 ) , if i ≥ ˇ s + 1; x t − a 1 y t − 1 − a 3 y t − a 5 y t + 1 +( C − V − ˇ sV ) ( y t − ˇ s − y t − ˇ s − 1 ) , if i = ˇ s and a 2 y t − 1 + a 4 y t + a 6 y t + 1 ≥ 0; x t − ( a 1 + a 2 ˆ η ) y t − 1 − ( a 3 + a 4 ˆ η ) y t − ( a 5 + a 6 ˆ η ) y t + 1 + ( C − V − ˇ sV ) ( y t − ˇ s − y t − ˇ s − 1 ) , if i = ˇ s and a 2 y t − 1 + a 4 y t + a 6 y t + 1 < 0. ec.30 Thus, for each s max and t , the values of v 1 ( s max , t , ˇ s ) , v 1 ( s max , t , ˇ s + 1 ) , . . . , v 1 ( s max , t , s max ) can be determined recursively in O ( T ) time. This implies that the v 1 ( s max , t , i ) values (for all s max , t , and i ) can be determined in O ( T 3 ) time. Similarly , the v 2 ( s max , t , j ) values (for all s max , t , and j ) can be determined in O ( T 3 ) time. For each s max , t , and j , let ˆ v 2 ( s max , t , j ) = max β ∈ [ j , s max ] Z { v 2 ( s max , t , β ) } and ˆ β ( s max , t , j ) = ar gmax β ∈ [ j , s max ] Z { v 2 ( s max , t , β ) } . Note that for each s max and t , the values of ˆ v 2 ( s max , t , ˇ s + 1 ) , ˆ v 2 ( s max , t , ˇ s + 2 ) , . . . , ˆ v 2 ( s max , t , s max ) and ˆ β 2 ( s max , t , ˇ s + 1 ) , ˆ β 2 ( s max , t , ˇ s + 2 ) , . . . , ˆ β 2 ( s max , t , s max ) can be determined in O ( T ) time by setting ˆ v 2 ( s max , t , j ) =  max { ˆ v 2 ( s max , t , j + 1 ) , v 2 ( s max , t , j ) } , if j ≤ s max − 1; v 2 ( s max , t , s max ) , if j = s max ; and ˆ β ( s max , t , j ) =  ˆ β ( s max , t , j + 1 ) , if ˆ v 2 ( s max , t , j + 1 ) ≥ v 2 ( s max , t , j ) ; j , if ˆ v 2 ( s max , t , j + 1 ) < v 2 ( s max , t , j ) . This implies that the ˆ v 2 ( s max , t , j ) and ˆ β ( s max , t , j ) values (for all s max , t , and j ) can be determined in O ( T 3 ) time. Note that the condition “ β = α + 1 or s max ≤ L + α ” in Pr oposition 8 implies that “ S = [ ˇ s , s max ] Z ” or “ s max ≤ L + α .” For any s max ∈ [ ˇ s max , ˆ s max ] Z and t ∈ [ s max + 2, ˆ t ] Z , if S = [ ˇ s , s max ] Z , then the largest possible amount of violation of inequality ( EC.31 ) is equal to v 1 ( s max , t , s max ) . For any s max ∈ [ ˇ s max , ˆ s max ] Z , t ∈ [ s max + 2, ˆ t ] Z , and α ∈ [ ˇ s , s max − 1 ] Z , if s max ≤ L + α , then the lar gest possible amount of violation of inequality ( EC.31 ) is equal to v 1 ( s max , t , α ) + v 2 ( s max , t , ˆ β ( s max , t , α + 1 ) ) = v 1 ( s max , t , α ) + ˆ v 2 ( s max , t , α + 1 ) . T o determine the most violated inequality ( EC.31 ) that satisﬁes the conditions in Proposition 8 , we ﬁrst determine all v 1 ( s max , t , i ) , v 2 ( s max , t , j ) , ˆ v 2 ( s max , t , j ) , and ˆ β ( s max , t , j ) values, which requir es O ( T 3 ) time. Next, we sear ch for the s max and t values such that v 1 ( s max , t , s max ) is the lar gest possible. This r equires O ( T 2 ) time. Let s ∗ max and t ∗ be the s max and t values obtained, and let S ∗ = [ ˇ s , s ∗ max ] Z . Let η ∗ = 0 if a 2 y t ∗ − 1 + a 4 y t ∗ + a 6 y t ∗ + 1 ≥ 0, and η ∗ = ˆ η otherwise. Next, we search for the s max , t , and α values, where α ∈ [ s max − L , s max − 1 ] Z , such that v 1 ( s max , t , α ) + ˆ v 2 ( s max , t , α + 1 ) is the largest possible. This requir es O ( T 3 ) time. Let s ∗∗ max , t ∗∗ , and α ∗∗ be the s max , t , and α values obtained, and let S ∗∗ = [ ˇ s , α ∗∗ ] Z ∪ [ β ∗∗ , s ∗∗ max ] Z , where β ∗∗ = ˆ β ( s ∗∗ max , t ∗∗ , α ∗∗ + 1 ) . Let η ∗∗ = 0 if a 2 y t ∗∗ − 1 + a 4 y t ∗∗ + a 6 y t ∗∗ + 1 ≥ 0, and η ∗∗ = ˆ η otherwise. If v 1 ( s ∗ max , t ∗ , s ∗ max ) > v 1 ( s ∗∗ max , t ∗∗ , α ∗∗ ) + ˆ v 2 ( s ∗∗ max , t ∗∗ , α ∗∗ + 1 ) , then a most violated inequality ( EC.31 ) is obtained by setting S = S ∗ , η = η ∗ , and t = t ∗ . Otherwise, it is obtained by setting S = S ∗∗ , η = η ∗∗ , and t = t ∗∗ . The overall computational time of this process is O ( T 3 ) . (ii) Consider the following family of inequalities: x t ≤ ( a 1 + a 2 η ) y t − 1 + ( a 3 + a 4 η ) y t + ( a 5 + a 6 η ) y t + 1 − ∑ s ∈ S ( C − V − s V ) ( y t + s − y t + s + 1 ) , (EC.32) where η ∈ [ 0, ˆ η ] , S = [ ˇ s , α ] Z ∪ [ β , s max ] Z , t ∈ [ ˇ t , T − s max − 1 ] Z , and α , β , and s max are integers such that (a) ˇ s max ≤ s max ≤ ˆ s max , (b) ˇ s ≤ α < β ≤ s max , and (c) β = α + 1 or s max ≤ L + α . Note that inequality family ( 14 ) in Pr oposition 2 , inequality family ( 16 ) in Proposition 4 , and inequality family ( 18 ) in Proposition 6 are special cases of this inequality family . Consider any given point ( x , y ) ∈ R 2 T + . Let x ′ t = x T − t + 1 and y ′ t = y T − t + 1 for t ∈ [ 1, T ] Z . Inequality ( EC.32 ) becomes x ′ T − t + 1 ≤ ( a 1 + a 2 η ) y ′ T − t + 2 + ( a 3 + a 4 η ) y ′ T − t + 1 + ( a 5 + a 6 η ) y ′ T − t − ∑ s ∈ S ( C − V − s V ) ( y ′ T − t − s + 1 − y ′ T − t − s ) . Letting t ′ = T − t + 1, this inequality becomes x ′ t ′ ≤ ( a 5 + a 6 η ) y ′ t ′ − 1 + ( a 3 + a 4 η ) y ′ t ′ + ( a 1 + a 2 η ) y ′ t ′ + 1 − ∑ s ∈ S ( C − V − s V ) ( y ′ t ′ − s − y ′ t ′ − s − 1 ) . (EC.33) Let ˆ t ′ = T if a 1 = a 2 = 0, and let ˆ t ′ = T − 1 otherwise. From the analysis in part (i), the integers α , β , s max , and t ′ ∈ [ s max + 2, ˆ t ′ ] Z corresponding to a most violated inequality ( EC.33 ) can be obtained in O ( T 3 ) time. Hence, the integers α , β , s max , and t ∈ [ ˇ t , T − s max − 1 ] Z corresponding to a most violated inequality ( EC.32 ) can be obtained in O ( T 3 ) time. Summarizing (i) and (ii), we conclude that for any given point ( x , y ) ∈ R 2 T + , the most violated inequalities ( 13 )–( 14 ), ( 15 )–( 16 ), and ( 17 )–( 18 ) in Propositions 2 , 4 , and 6 , respectively , can be determined in O ( T 3 ) time if such violated inequalities exist. □ ec.31 C.9. Pro of of Prop osition 9 Proposition 9 . Consider any k ∈ [ 1, T − 1 ] Z such that C − C − kV > 0 , any m ∈ [ 0, k − 1 ] Z , and any S ⊆ [ 0, min { k − 1, L − m − 1 } ] Z . For any t ∈ [ k + 1, T − m ] Z , the inequality x t − x t − k ≤ ( C + ( k − m ) V ) y t + V m ∑ i = 1 y t + i − C y t − k − ∑ s ∈ S ( C + ( k − s ) V − V ) ( y t − s − y t − s − 1 ) ( 19 ) is valid for conv ( P ) . For any t ∈ [ m + 1, T − k ] Z , the inequality x t − x t + k ≤ ( C + ( k − m ) V ) y t + V m ∑ i = 1 y t − i − C y t + k − ∑ s ∈ S ( C + ( k − s ) V − V ) ( y t + s − y t + s + 1 ) ( 20 ) is valid for conv ( P ) . Furthermor e, ( 19 ) and ( 20 ) are facet-deﬁning for conv ( P ) when m = 0 and s ≥ min { k − 1, 1 } for all s ∈ S . Proof. W e ﬁrst prove that inequality ( 19 ) is valid and facet-deﬁning for conv ( P ) . For notational conve- nience, we deﬁne s max = max { s : s ∈ S } if S  = ∅ , and s max = − 1 if S = ∅ . Consider any t ∈ [ k + 1, T − m ] Z . T o pr ove that the linear inequality ( 19 ) is valid for conv ( P ) , it suf ﬁces to show that it is valid for P . Consider any element ( x , y ) of P . W e show that ( x , y ) satisﬁes ( 19 ). W e divide the analysis into three cases: Case 1: y t = 0. In this case, by ( 2c ) and ( 2d ), − x t − k ≤ − C y t − k and x t = 0. Thus, the left-hand side of ( 19 ) is at most − C y t − k and the ﬁrst term on the right-hand side of ( 19 ) is 0. Because y t = 0 and t ∈ [ 2, T ] Z , by Lemma 1 (i), we have y t − j − y t − j − 1 ≤ 0 for all j ∈ [ 0, min { t − 2, L − 1 } ] Z . Because S ⊆ [ 0, min { k − 1, L − m − 1 } ] Z , m ≥ 0 and, t ≥ k + 1, we have S ⊆ [ 0, min { t − 2, L − 1 } ] Z . Thus, y t − s − y t − s − 1 ≤ 0 for all s ∈ S . Because S ⊆ [ 0, k − 1 ] Z and C + V > V , for any s ∈ S , the coefﬁcient “ C + ( k − s ) V − V ” on the right-hand side of ( 19 ) is positive. Hence, the right-hand side of ( 19 ) is at least − C y t − k . Therefor e, in this case, ( x , y ) satisﬁes ( 19 ). Case 2: y t = 1 and y t − s ′ − y t − s ′ − 1 = 1 for some s ′ ∈ S . In this case, y t − s ′ = 1 and y t − s ′ − 1 = 0. Because y t = 1 and t ∈ [ 2, T ] Z , by Lemma 1 (ii), there exists at most one j ∈ [ 0, min { t − 2, L } ] Z such that y t − j − y t − j − 1 = 1. Because S ⊆ [ 0, min { k − 1, L − m − 1 } ] Z , m ≥ 0 and, t ≥ k + 1, we have S ⊆ [ 0, min { t − 2, L } ] Z . Thus, y t − s − y t − s − 1 ≤ 0 for all s ∈ S \ { s ′ } . Because y t − s ′ − y t − s ′ − 1 = 1 and t − s ′ ∈ [ 2, T ] Z , by ( 2a ), we have y τ = 1 for all τ ∈ [ t − s ′ , min { T , t − s ′ + L − 1 } ] Z . Because S ⊆ [ 0, L − m − 1 ] Z , we have t − s ′ + L − 1 ≥ t + m . Thus, y τ = 1 for all τ ∈ [ t − s ′ , t + m ] Z , which implies that ( C + ( k − m ) V ) y t + V ∑ m i = 1 y t + i = C + k V . Because S ⊆ [ 0, k − 1 ] Z and C + V > V , the coefﬁcient “ C + ( k − s ) V − V ” on the right-hand side of ( 19 ) is positive for all s ∈ S . Hence, the right-hand side of ( 19 ) is at least C + k V − C y t − k − ( C + ( k − s ′ ) V − V ) = s ′ V + V − C y t − k . By ( 2e ), ∑ t τ = t − s ′ ( x τ − x τ − 1 ) ≤ ∑ t τ = t − s ′ Vy τ − 1 + ∑ t τ = t − s ′ V ( 1 − y τ − 1 ) , which implies that x t − x t − s ′ − 1 ≤ s ′ V + V . Because y t − s ′ − 1 = 0, by ( 2d ), x t − s ′ − 1 = 0. By ( 2c ), − x t − k ≥ − C y t − k . Thus, x t − x t − k ≤ s ′ V + V − C y t − k . Therefor e, in this case, ( x , y ) satisﬁes ( 19 ). Case 3: y t = 1 and y t − s − y t − s − 1 ≤ 0 for all s ∈ S . Because S ⊆ [ 0, k − 1 ] Z and C + V > V , we have C + ( k − s ) V − V > 0 for each s ∈ S . Hence, the term “ − ∑ s ∈ S ( C + ( k − s ) V − V ) ( y t − s − y t − s − 1 ) ” on the right-hand side of inequality ( 19 ) is nonnegative. W e divide our analysis into three subcases. Case 3.1: y τ = 0 for some τ ∈ [ t , t + m ] Z . Let t ′ = min { τ ∈ [ t , t + m ] Z : y τ = 0 } . Then, y τ = 1 for all τ ∈ [ t , t ′ − 1 ] Z . Thus, the right-hand side of ( 19 ) is at least C + ( k − m ) V + ( t ′ − t − 1 ) V − C y t − k . Because k − m ≥ 1 and C + V > V , the right-hand side of ( 19 ) is at least V + ( t ′ − t − 1 ) V − C y t − k . By ( 2f ), ∑ t ′ τ = t + 1 ( x τ − 1 − x τ ) ≤ ∑ t ′ τ = t + 1 Vy τ + ∑ t ′ τ = t + 1 V ( 1 − y τ ) , which implies that x t − x t ′ ≤ V + ( t ′ − t − 1 ) V . Because y t ′ = 0, by ( 2d ), x t ′ = 0. By ( 2c ), − x t − k ≤ − C y t − k . Hence, x t − x t − k ≤ V + ( t ′ − t − 1 ) V − C y t − k . Thus, the left-hand side of ( 19 ) is less than or equal to the right-hand side. Case 3.2: y τ = 1 for all τ ∈ [ t , t + m ] Z and y τ = 0 for some τ ∈ [ t − k , t − 1 ] Z . In this case, the right-hand side of ( 19 ) is at least C + k V − C y t − k . Let t ′ = max { τ ∈ [ t − k , t − 1 ] Z : y τ = 0 } . Because t ′ ≥ t − k and C + V > V , the right-hand side of ( 19 ) is greater than V + ( t − t ′ − 1 ) V − C y t − k . By ( 2e ), ∑ t τ = t ′ + 1 ( x τ − x τ − 1 ) ≤ ∑ t τ = t ′ + 1 Vy τ − 1 + ∑ t τ = t ′ + 1 V ( 1 − y τ − 1 ) , which implies that x t − x t ′ ≤ V + ( t − t ′ − 1 ) V . Because y t ′ = 0, by ( 2d ), x t ′ = 0. By ( 2c ), − x t − k ≤ − C y t − k . Hence, x t − x t − k ≤ V + ( t − t ′ − 1 ) V − C y t − k . Thus, the left-hand side of ( 19 ) is less than the right-hand side. Case 3.3: y τ = 1 for all τ ∈ [ t − k , t + m ] Z . In this case, the right-hand side of ( 19 ) is at least kV . By ( 2e ), ∑ t τ = t − k + 1 ( x τ − x τ − 1 ) ≤ ∑ t τ = t − k + 1 Vy τ − 1 + ∑ t τ = t − k + 1 V ( 1 − y τ − 1 ) , which implies that x t − x t − k ≤ k V . Thus, the left-hand side of ( 19 ) is less than or equal to the right-hand side. In Cases 3.1–3.3, ( x , y ) satisﬁes ( 19 ). Summarizing Cases 1–3, we conclude that ( 19 ) is valid for conv ( P ) . ec.32 Consider any t ∈ [ k + 1, T − m ] Z . T o prove that inequality ( 19 ) is facet-deﬁning for conv ( P ) when m = 0 and s ≥ min { k − 1, 1 } for all s ∈ S , it sufﬁces to show that ther e exist 2 T afﬁnely independent points in conv ( P ) that satisfy ( 19 ) at equality when m = 0 and s ≥ min { k − 1, 1 } for all s ∈ S . Because 0 ∈ conv ( P ) and 0 satisﬁes ( 19 ) at equality , it sufﬁces to create the remaining 2 T − 1 nonzero linearly independent points. W e denote these 2 T − 1 points as ( ¯ x r , ¯ y r ) for r ∈ [ 1, T ] Z \ { t − k } and ( ˆ x r , ˆ y r ) for r ∈ [ 1, T ] Z , and denote the q th component of ¯ x r , ¯ y r , ˆ x r , and ˆ y r as ¯ x r q , ¯ y r q , ˆ x r q , and ˆ y r q , respectively . Let ϵ = min { V − C , C − C − k V } > 0. W e divide these 2 T − 1 points into the following eight groups: (A1) For each r ∈ [ 1, t − k − 1 ] Z , we create a point ( ¯ x r , ¯ y r ) as follows: ¯ x r q =    C , for q ∈ [ 1, r − 1 ] Z ; C + ϵ , for q = r ; 0, for q ∈ [ r + 1, T ] ; and ¯ y r q =  1, for q ∈ [ 1, r ] Z ; 0, for q ∈ [ r + 1, T ] Z . It is easy to verify that ( ¯ x r , ¯ y r ) satisﬁes ( 2a )–( 2f ). Thus, ( ¯ x r , ¯ y r ) ∈ conv ( P ) . Note that ¯ x r t = ¯ x r t − k = ¯ y r t = ¯ y r t − k = 0 and m = 0. Because t − s − 1  = r for all s ∈ S , we have ¯ y r t − s − ¯ y r t − s − 1 = 0 for all s ∈ S . Hence, ( ¯ x r , ¯ y r ) satisﬁes ( 20 ) at equality . (A2) For each r ∈ [ t − k + 1, t − 1 ] Z , we create a point ( ¯ x r , ¯ y r ) as follows: ¯ x r q =    C , for q ∈ [ 1, t − 1 ] Z \ { r } ; C + ϵ , for q = r ; 0, for q ∈ [ t , T ] Z ; and ¯ y r q =  1, for q ∈ [ 1, t − 1 ] Z ; 0, for q ∈ [ t , T ] Z . It is easy to verify that ( ¯ x r , ¯ y r ) satisﬁes ( 2a )–( 2f ). Thus, ( ¯ x r , ¯ y r ) ∈ conv ( P ) . Note that ¯ x r t = ¯ y r t = 0, ¯ x r t − k = C , ¯ y r t − k = 1, and m = 0. The existence of r ∈ [ t − k + 1, t − 1 ] Z implies that k ≥ 2, which in turn implies that s ≥ 1 for all s ∈ S . Hence, ¯ y r t − s − ¯ y r t − s − 1 = 0 for all s ∈ S . Thus, ( ¯ x r , ¯ y r ) satisﬁes ( 20 ) at equality . (A3) W e create a point ( ¯ x t , ¯ y t ) as follows: ¯ x t q =    C , for q ∈ [ 1, t − k − 1 ] Z ; C + ( q − t + k ) V + ϵ , for q ∈ [ t − k , t ] Z ; C + kV , for q ∈ [ t + 1, T ] Z ; and ¯ y t q = 1 for all q ∈ [ 1, T ] Z . It is easy to verify that ( ¯ x t , ¯ y t ) satisﬁes ( 2a )–( 2d ). Note that ¯ x t q − ¯ x t q − 1 = 0 when q ∈ [ 2, t − k − 1 ] Z , 0 < ¯ x t q − ¯ x t q − 1 ≤ V when q ∈ [ t − k , t ] Z , and − ϵ ≤ ¯ x t q − ¯ x t q − 1 ≤ 0 when q ∈ [ t + 1, T ] Z . Thus, − V ¯ y t q − V ( 1 − ¯ y t q ) ≤ ¯ x t q − ¯ x t q − 1 ≤ V ¯ y t q − 1 + V ( 1 − ¯ y t q − 1 ) for all q ∈ [ 2, T ] Z . Hence, ( ¯ x t , ¯ y t ) satisﬁes ( 2e ) and ( 2f ). Therefore, ( ¯ x t , ¯ y t ) ∈ conv ( P ) . Note that ¯ x t t = C + k V + ϵ , ¯ x t t − k = C + ϵ , ¯ y t t = ¯ y t t − k = 1, m = 0, and ¯ y t t − s − ¯ y t t − s − 1 = 0 for all s ∈ S . Thus, ( ¯ x t , ¯ y t ) satisﬁes ( 19 ) at equality . (A4) For each r ∈ [ t + 1, T ] Z , we create a point ( ¯ x r , ¯ y r ) as follows: ¯ x r q =    0, for q ∈ [ 1, r − 1 ] Z ; C + ϵ , for q = r ; C , for q ∈ [ r + 1, T ] Z ; and ¯ y r q =  0, for q ∈ [ 1, r − 1 ] Z ; 1, for q ∈ [ r , T ] Z . It is easy to verify that ( ¯ x r , ¯ y r ) satisﬁes ( 2a )–( 2f ). Thus, ( ¯ x r , ¯ y r ) ∈ conv ( P ) . Note that ¯ x r t = ¯ x r t − k = ¯ y r t = ¯ y r t − k = 0, m = 0, and ¯ y r t − s − ¯ y r t − s − 1 = 0 for all s ∈ S . Hence, ( ¯ x t , ¯ y t ) satisﬁes ( 19 ) at equality . ec.33 (A5) For each r ∈ [ 1, t − 1 ] Z , we create the same point ( ˆ x r , ˆ y r ) as in gr oup (A2) in the proof of Pr oposition 1 . Thus, ( ˆ x r , ˆ y r ) ∈ conv ( P ) . T o show that ( ˆ x r , ˆ y r ) satisﬁes ( 19 ) at equality , we ﬁrst consider the case where t − r − 1 / ∈ S . In this case, ˆ x r t = ˆ y r t = 0 and m = 0. Because t − k ≤ t − s max − 1 ≤ r , we have ˆ x r t − k = C and ˆ y r t − k = 1. Because t − s − 1  = r for all s ∈ S , we have ˆ y r t − s − ˆ y r t − s − 1 = 0 for all s ∈ S . Hence, ( ˆ x r , ˆ y r ) satisﬁes ( 19 ) at equality . Next, we consider the case where t − r − 1 ∈ S . In this case, ˆ x r t = V + ( t − r − 1 ) V , ˆ y r t = 1, ˆ x r t − k = ˆ y r t − k = 0, and m = 0. In addition, ˆ y r t − s − ˆ y r t − s − 1 = 1 when s = t − r − 1, and ˆ y r t − s − ˆ y r t − s − 1 = 0 when s  = t − r − 1. Hence, ( ˆ x r , ˆ y r ) satisﬁes ( 19 ) at equality . (A6) W e create a point ( ˆ x t , ˆ y t ) as follows: ˆ x t q =    C , for q ∈ [ 1, t − k − 1 ] Z ; C + ( q − t + k ) V , for q ∈ [ t − k , t ] Z ; C + kV , for q ∈ [ t + 1, T ] Z ; and ˆ y t q = 1 for all q ∈ [ 1, T ] Z . It is easy to verify that ( ˆ x t , ˆ y t ) satisﬁes ( 2a )–( 2d ). Note that ˆ x t q − ˆ x t q − 1 = 0 when q ∈ [ 2, t − k ] Z , ˆ x t q − ˆ x t q − 1 = V when q ∈ [ t − k + 1, t ] Z , and ˆ x t q − ˆ x t q − 1 = 0 when q ∈ [ t + 1, T ] Z . Thus, − V ˆ y t q − V ( 1 − ˆ y t q ) ≤ ˆ x t q − ˆ x t q − 1 ≤ V ˆ y t q − 1 + V ( 1 − ˆ y t q − 1 ) for all q ∈ [ 2, T ] Z . Hence, ( ˆ x t , ˆ y t ) satisﬁes ( 2e ) and ( 2f ). Therefore, ( ˆ x t , ˆ y t ) ∈ conv ( P ) . Note that ˆ x t t = C + kV , ˆ x t t − k = C , ˆ y t t = ˆ y t t − k = 1, m = 0, and ˆ y t t − s − ˆ y t t − s − 1 = 0 for all s ∈ S , Thus, ( ˆ x t , ˆ y t ) satisﬁes ( 19 ) at equality . (A7) For each r ∈ [ t + 1, T ] Z , we create the same point ( ˆ x r , ˆ y r ) as in group (A4) in the pr oof of Proposition 1 . Thus, ( ˆ x r , ˆ y r ) ∈ conv ( P ) . Note that ˆ x r t = ˆ x r t − k = ˆ y r t = ˆ y r t − k = 0, m = 0, and ˆ y r t − s − ˆ y r t − s − 1 = 0 for all s ∈ S . Hence, ( ˆ x r , ˆ y r ) satisﬁes ( 19 ) at equality . T able EC.7 shows a matrix with 2 T − 1 rows, where each row represents a point created by this process. This matrix can be transformed into the matrix in T able EC.8 via the following Gaussian elimination process: (i) For each r ∈ [ 1, t − k − 1 ] Z , the point with index r in group (B1), denoted ( ¯ x r , ¯ y r ) , is obtained by setting ( ¯ x r , ¯ y r ) = ( ¯ x r , ¯ y r ) − ( ˆ x r , ˆ y r ) . Here, ( ¯ x r , ¯ y r ) is the point with index r in group (A1) , and ( ˆ x r , ˆ y r ) is the point with index r in group (A5) . Note that t − r − 1 / ∈ S for all r ∈ [ 1, t − k − 1 ] Z . Thus, when r ≤ t − k − 1, the point with index r in group (A5) is given by ˆ x r q = C and ˆ y r q = 1 for q ∈ [ 1, r ] Z , and ˆ x r q = ˆ y r q = 0 for q ∈ [ r + 1, T ] Z . (ii) For each r ∈ [ t − k + 1, t − 1 ] Z , the point with index r in group (B2), denoted ( ¯ x r , ¯ y r ) , is obtained by setting ( ¯ x r , ¯ y r ) = ( ¯ x r , ¯ y r ) − ( ˆ x t − 1 , ˆ y t − 1 ) . Here, ( ¯ x r , ¯ y r ) is the point in group (A2) , and ( ˆ x t − 1 , ˆ y t − 1 ) is the point with index t − 1 in group (A5) . Note that because s ≥ 1 for all s ∈ S , the point with index t − 1 in group (A5) is given by ˆ x t − 1 q = C and ˆ y t − 1 q = 1 for q ∈ [ 1, t − 1 ] Z , and ˆ x t − 1 q = ˆ y t − 1 q = 0 for q ∈ [ t , T ] Z . (iii) The point in group (B3), denoted ( ¯ x t , ¯ y t ) , is obtained by setting ( ¯ x t , ¯ y t ) = ( ¯ x t , ¯ y t ) − ( ˆ x t , ˆ y t ) . Here, ( ¯ x t , ¯ y t ) is the point in group (A3) , and ( ˆ x t , ˆ y t ) is the point in group (A6) . (iv) For each r ∈ [ t + 1, T ] Z , the point with index r in group (B4), denoted ( ¯ x r , ¯ y r ) , is obtained by setting ( ¯ x r , ¯ y r ) = ( ¯ x r , ¯ y r ) − ( ˆ x r , ˆ y r ) . Here, ( ¯ x r , ¯ y r ) is the point with index r in group (A4) , and ( ˆ x r , ˆ y r ) is the point in group (A7) . (v) For each r ∈ [ 1, t − 1 ] Z , the point with index r in group (B5), denoted ( ˆ x r , ˆ y r ) , is obtained by setting ( ˆ x r , ˆ y r ) = ( ˆ x r , ˆ y r ) if t − r − 1 / ∈ S , and setting ( ˆ x r , ˆ y r ) = ( ˆ x t , ˆ y t ) − ( ˆ x r , ˆ y r ) if t − r − 1 ∈ S . Here, ( ˆ x r , ˆ y r ) is the point with index r in group (A5) , and ( ˆ x t , ˆ y t ) is the point in group (A6) . (vi) The point in group (B6), denoted ( ˆ x t , ˆ y t ) , is obtained by setting ( ˆ x t , ˆ y t ) = ( ˆ x t , ˆ y t ) − ( ˆ x t + 1 , ˆ y t + 1 ) . Here, ( ˆ x t , ˆ y t ) is the point in group (A6) , and ( ˆ x t + 1 , ˆ y t + 1 ) is the point with index t + 1 in group (A7) . (vii) For each r ∈ [ t + 1, T ] Z , the point with index r in group (B7), denoted ( ˆ x r , ˆ y r ) , is obtained by setting ( ˆ x r , ˆ y r ) = ( ˆ x r , ˆ y r ) − ( ˆ x r + 1 , ˆ y r + 1 ) if r  = T , and setting ( ˆ x r , ˆ y r ) = ( ˆ x r , ˆ y r ) if r = T . Here, ( ˆ x r , ˆ y r ) and ( ˆ x r + 1 , ˆ y r + 1 ) are the points with indices r and r + 1, respectively , in group (A7) . The matrix shown in T able EC.8 is lower triangular; that is, the position of the last nonzero component of a row of the matrix is greater than the position of the last nonzero component of the previous row . This implies that these 2 T − 1 points in groups (A1) – (A7) are linearly independent. Therefore, inequality ( 19 ) is facet-deﬁning for conv ( P ) . Next, we show that inequality ( 20 ) is valid for conv ( P ) and is facet-deﬁning for conv ( P ) when m = 0 and s ≥ min { k − 1, 1 } ∀ s ∈ S . Denote x ′ t = x T − t + 1 and y ′ t = y T − t + 1 for t ∈ [ 1, T ] Z . Because inequality ( 19 ) ec.34 T able EC.7 A matrix with the rows r epresenting 2 T − 1 linearly independent points in conv ( P ) satisfying inequality ( 19 ) at equality Group Point Index r x y 1 · · · t − k − 1 t − k t − k + 1 · · · t − 1 t t + 1 · · · T 1 · · · t − k − 1 t − k t − k + 1 · · · t − 1 t t + 1 · · · T (A1) ( ¯ x r , ¯ y r ) 1 C + ϵ · · · 0 0 0 · · · 0 0 0 · · · 0 1 · · · 0 0 0 · · · 0 0 0 · · · 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . t − k − 1 C · · · C + ϵ 0 0 · · · 0 0 0 · · · 0 1 · · · 1 0 0 · · · 0 0 0 · · · 0 (A2) t − k + 1 C · · · C C C + ϵ · · · C 0 0 · · · 0 1 · · · 1 1 1 · · · 1 0 0 · · · 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . t − 1 C · · · C C C · · · C + ϵ 0 0 · · · 0 1 · · · 1 1 1 · · · 1 0 0 · · · 0 (A3) t C · · · C C + ϵ C + V + ϵ · · · C + ( k − 1 ) V + ϵ C + k V + ϵ C + kV · · · C + k V 1 · · · 1 1 1 · · · 1 1 1 · · · 1 (A4) t + 1 0 · · · 0 0 0 · · · 0 0 C + ϵ · · · C 0 · · · 0 0 0 · · · 0 0 1 · · · 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . T 0 · · · 0 0 0 · · · 0 0 0 · · · C + ϵ 0 · · · 0 0 0 · · · 0 0 0 · · · 1 (A5) ( ˆ x r , ˆ y r ) 1 (See Note EC.7 -1) (See Note EC.7 -1) . . . t − 1 (A6) t C · · · C C C + V · · · C + ( k − 1 ) V C + k V C + k V · · · C + k V 1 · · · 1 1 1 · · · 1 1 1 · · · 1 (A7) t + 1 0 · · · 0 0 0 · · · 0 0 C · · · C 0 · · · 0 0 0 · · · 0 0 1 · · · 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . T 0 · · · 0 0 0 · · · 0 0 0 · · · C 0 · · · 0 0 0 · · · 0 0 0 · · · 1 Note EC.7 -1: For r ∈ [ 1, t − 1 ] Z , the x and y vectors in group (A5) are given as follows: ˆ x r = ( C , . . . , C | {z } r terms , 0, . . . , 0 | { z } T − r terms ) and ˆ y r = ( 1, . . . , 1 | { z } r terms , 0, . . . , 0 | { z } T − r terms ) if t − r − 1 / ∈ S ; ˆ x r = ( 0, . . . , 0 | { z } r terms , V , V + V , V + 2 V , . . . , V + ( t − r − 1 ) V | { z } t − r terms , V + ( t − r − 1 ) V , . . . , V + ( t − r − 1 ) V | {z } T − t terms ) and ˆ y r = ( 0, . . . , 0 | { z } r terms , 1, . . . , 1 | { z } T − r terms ) if t − r − 1 ∈ S . ec.35 T able EC.8 Lower triangular matrix obtained from T able EC.7 via Gaussian elimination Group Point Index r x y 1 · · · t − k − 1 t − k t − k + 1 · · · t − 1 t t + 1 · · · T 1 · · · t − 1 t t + 1 · · · T (B1) ( ¯ x r , ¯ y r ) 1 ϵ · · · 0 0 0 · · · 0 0 0 · · · 0 0 · · · 0 0 0 · · · 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . t − k − 1 0 · · · ϵ 0 0 · · · 0 0 0 · · · 0 0 · · · 0 0 0 · · · 0 (B2) t − k + 1 0 · · · 0 0 ϵ · · · 0 0 0 · · · 0 0 · · · 0 0 0 · · · 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . t − 2 0 · · · 0 0 0 · · · ϵ 0 0 · · · 0 0 · · · 0 0 0 · · · 0 (B3) t 0 · · · 0 ϵ ϵ · · · ϵ ϵ 0 · · · 0 0 · · · 0 0 0 · · · 0 (B4) t + 1 0 · · · 0 0 0 · · · 0 0 ϵ · · · 0 0 · · · 0 0 0 · · · 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . T 0 · · · 0 0 0 · · · 0 0 0 · · · ϵ 0 · · · 0 0 0 · · · 0 (B5) ( ˆ x r , ˆ y r ) 1 1 · · · 0 0 0 · · · 0 . . . (Omitted) . . . . . . . . . . . . . . . . . . t − 1 1 · · · 1 0 0 · · · 0 (B6) t (Omitted) 1 · · · 1 1 0 · · · 0 (B7) t + 1 0 · · · 0 0 1 · · · 0 . . . (Omitted) . . . . . . . . . . . . . . . . . . T 0 · · · 0 0 0 · · · 1 ec.36 is valid for conv ( P ) and is facet-deﬁning for conv ( P ) when m = 0 and s ≥ min { k − 1, 1 } ∀ s ∈ S for any t ∈ [ k + 1, T − m ] Z , the inequality x ′ T − t + 1 − x ′ T − t + k + 1 ≤ ( C + ( k − m ) V ) y ′ T − t + 1 + V m ∑ i = 1 y ′ T − t − i + 1 − C y ′ T − t + k + 1 − ∑ s ∈ S ( C + ( k − s ) V − V ) ( y ′ T − t + s + 1 − y ′ T − t + s + 2 ) is valid for conv ( P ′ ) and is facet-deﬁning for conv ( P ′ ) when m = 0 and s ≥ min { k − 1, 1 } ∀ s ∈ S for any t ∈ [ k + 1, T − m ] Z . Let t ′ = T − t + 1. Then, the inequality x ′ t ′ − x ′ t ′ + k ≤ ( C + ( k − m ) V ) y ′ t ′ + V m ∑ i = 1 y ′ t ′ − i − C y ′ t ′ + k − ∑ s ∈ S ( C + ( k − s ) V − V ) ( y ′ t ′ + s − y ′ t ′ + s + 1 ) is valid for conv ( P ′ ) and is facet-deﬁning for conv ( P ′ ) when m = 0 and s ≥ min { k − 1, 1 } ∀ s ∈ S for any t ′ ∈ [ m + 1, T − k ] Z . Hence, by Lemma 2 , inequality ( 20 ) is valid for conv ( P ) and is facet-deﬁning for conv ( P ) when m = 0 and s ≥ min { k − 1, 1 } ∀ s ∈ S for any t ∈ [ m + 1, T − k ] Z . □ C.10. Pro of of Prop osition 10 Proposition 10 . For any given point ( x , y ) ∈ R 2 T + , the most violated inequalities ( 19 ) and ( 20 ) can be determined in O ( T 3 ) time if such violated inequalities exist. Proof. W e ﬁrst consider inequality ( 19 ). Consider any given ( x , y ) ∈ R 2 T + . For notational convenience, denote ˆ k = max { k ∈ [ 1, T − 1 ] Z : C − C − kV > 0 } , and denote ˆ s km = min { k − 1, L − m − 1 } for any k ∈ [ 1, ˆ k ] Z and m ∈ [ 0, k − 1 ] Z . For any t ∈ [ 1, T ] Z , let θ ( t ) = t ∑ τ = 2 max { y τ − y τ − 1 , 0 } . Then, for any k ∈ [ 1, ˆ k ] Z , m ∈ [ 0, k − 1 ] Z , and t ∈ [ k + 1, T − m ] Z , ˆ s km ∑ s = 1 max { y t − s − y t − s − 1 , 0 } = t − 1 ∑ τ = t − ˆ s km max { y τ − y τ − 1 , 0 } = θ ( t − 1 ) − θ ( t − ˆ s km − 1 ) . (EC.34) For any k ∈ [ 1, ˆ k ] Z , m ∈ [ 0, k − 1 ] Z , t ∈ [ k + 1, T − m ] Z , and S ⊆ [ 0, ˆ s km ] Z , let ˜ v km ( S , t ) = x t − x t − k − ( C + ( k − m ) V ) y t − V m ∑ i = 1 y t + i + C y t − k + ∑ s ∈ S ( C + ( k − s ) V − V ) ( y t − s − y t − s − 1 ) . If ˜ v km ( S , t ) > 0, then ˜ v km ( S , t ) is the amount of violation of inequality ( 19 ). If ˜ v km ( S , t ) ≤ 0, there is no violation of inequality ( 19 ). For any k ∈ [ 1, ˆ k ] Z , m ∈ [ 0, k − 1 ] Z , and t ∈ [ k + 1, T − m ] Z , let v km ( t ) = max S ⊆ [ 0, ˆ s km ] Z { ˜ v km ( S , t ) } . If v km ( t ) > 0, then v km ( t ) is the lar gest possible violation of inequality ( 19 ) for this combination of k , m , and t . If v km ( t ) ≤ 0, the largest possible violation of inequality ( 19 ) is zero for this combination of k , m , and t . Because C + V > V , we have C + ( k − s ) V − V > 0 for all k ∈ [ 1, ˆ k ] Z , s ∈ [ 0, ˆ s km ] Z , and m ∈ [ 0, k − 1 ] Z . Thus, for any k ∈ [ 1, ˆ k ] Z , m ∈ [ 0, k − 1 ] Z , and t ∈ [ k + 1, T − m ] Z , ˜ v km ( S , t ) is maximized when S contains all s ∈ [ 0, ˆ s km ] Z such that y t − s − y t − s − 1 > 0 (if any). If it does not exist any s ∈ [ 0, ˆ s ] Z such that y t − s − y t − s − 1 > 0, then ˜ v km ( S , t ) is maximized when S = ∅ , and v km ( t ) = x t − x t − k − ( C + ( k − m ) V ) y t − V ∑ m i = 1 y t + i + C y t − k . Hence, for any k ∈ [ 1, ˆ k ] Z , m ∈ [ 0, k − 1 ] Z , and t ∈ [ k + 1, T − m ] Z , v km ( t ) = x t − x t − k − ( C + ( k − m ) V ) y t − V m ∑ i = 1 y t + i + C y t − k + ˆ s km ∑ s = 0 ( C + ( k − s ) V − V ) max { y t − s − y t − s − 1 , 0 } . Determining θ ( t ) for all t ∈ [ 1, T ] Z can be done recursively in O ( T ) time by setting θ ( 1 ) = 0 and setting θ ( t ) = θ ( t − 1 ) + max { y t − y t − 1 , 0 } for t = 2, . . . , T . Clearly , for each k ∈ [ 1, ˆ k ] Z and each m ∈ [ 0, k − 1 ] Z , the ec.37 value of v km ( k + 1 ) can be determined in O ( T ) time. For any k ∈ [ 1, ˆ k ] Z , m ∈ [ 0, k − 1 ] Z , and t ∈ [ k + 2, T − m ] Z , v km ( t ) − v km ( t − 1 ) = ( x t − x t − 1 ) − ( x t − k − x t − k − 1 ) − ( C + ( k − m ) V )( y t − y t − 1 ) − V " m ∑ i = 1 y t + i − m ∑ i = 1 y t + i − 1 # + C ( y t − k − y t − k − 1 ) + ( C + kV − V ) " ˆ s km ∑ s = 0 max { y t − s − y t − s − 1 , 0 } − ˆ s km ∑ s = 0 max { y t − s − 1 − y t − s − 2 , 0 } # − V " ˆ s km ∑ s = 0 s max { y t − s − y t − s − 1 , 0 } − ˆ s km ∑ s = 0 s max { y t − s − 1 − y t − s − 2 , 0 } # = ( x t − x t − 1 ) − ( x t − k − x t − k − 1 ) − ( C + ( k − m ) V )( y t − y t − 1 ) − V ( y t + m − y t ) + C ( y t − k − y t − k − 1 ) + ( C + kV − V )  max { y t − y t − 1 , 0 } − max { y t − ˆ s km − 1 − y t − ˆ s km − 2 , 0 }  − V " ˆ s km ∑ s = 1 max { y t − s − y t − s − 1 , 0 } − ˆ s km max { y t − ˆ s km − 1 − y t − ˆ s km − 2 , 0 } # . This, together with ( EC.34 ), implies that v km ( t ) = v km ( t − 1 ) + ( x t − x t − 1 ) − ( x t − k − x t − k − 1 ) − ( C + ( k − m ) V )( y t − y t − 1 ) − V ( y t + m − y t ) + C ( y t − k − y t − k − 1 ) + ( C + kV − V )  max { y t − y t − 1 , 0 } − max { y t − ˆ s km − 1 − y t − ˆ s km − 2 , 0 }  − V  θ ( t − 1 ) − θ ( t − ˆ s km − 1 ) − ˆ s km max { y t − ˆ s km − 1 − y t − ˆ s km − 2 , 0 }  . Thus, for each k ∈ [ 1, ˆ k ] Z and m ∈ [ 0, k − 1 ] Z , the values of v km ( k + 1 ) , v km ( k + 2 ) , . . . , v km ( T − m ) can be determined r ecursively in O ( T ) time. Hence, the values of k , m , t and the set S corresponding to the largest possible violation of inequality ( 19 ) can be obtained in O ( T 3 ) time. Next, we consider inequality ( 20 ). Consider any given ( x , y ) ∈ R 2 T + . Let x ′ t = x T − t + 1 and y ′ t = y T − t + 1 for t ∈ [ 1, T ] Z . Inequality ( 20 ) becomes x ′ T − t + 1 − x ′ T − t − k + 1 ≤ ( C + ( k − m ) V ) y ′ T − t + 1 + V m ∑ i = 1 y ′ T − t + i + 1 − C y ′ T − t − k + 1 − ∑ s ∈ S ( C + ( k − s ) V − V ) ( y ′ T − t − s + 1 − y ′ T − t − s ) . Letting t ′ = T − t + 1, this inequality becomes x ′ t ′ − x ′ t ′ − k ≤ ( C + ( k − m ) V ) y ′ t ′ + V m ∑ i = 1 y ′ t ′ + i − C y ′ t ′ − k − ∑ s ∈ S ( C + ( k − s ) V − V ) ( y ′ t ′ − s − y ′ t ′ − s − 1 ) . (EC.35) Because the values of k , m , t and the set S corr esponding to the largest possible violation of inequality ( 19 ) can be obtained in O ( T 3 ) time, the values of k , m , t ′ and the set S corresponding to the lar gest possible violation of inequality ( EC.35 ) can be obtained in O ( T 3 ) time. Hence, the values of k , m , t and the set S corresponding to the lar gest possible violation of inequality ( 20 ) can be obtained in O ( T 3 ) time. □ C.11. Pro of of Prop osition 11 Proposition 11 . Consider any k ∈ [ 1, T − 1 ] Z such that C − C − k V > 0 , any m ∈ [ 0, k − 1 ] Z , and any S ⊆ [ 0, min { k − 1, L − m − 2 } ] Z . For any t ∈ [ k + 1, T − m − 1 ] Z , the inequality x t − x t − k ≤ ( C + ( k − m ) V − V ) y t + m + 1 + V m ∑ i = 1 y t + i + V y t − C y t − k − ∑ s ∈ S ( C + ( k − s ) V − V ) ( y t − s − y t − s − 1 ) ( 21 ) ec.38 is valid and facet-deﬁning for conv ( P ) . For any t ∈ [ m + 2, T − k ] Z , the inequality x t − x t + k ≤ ( C + ( k − m ) V − V ) y t − m − 1 + V m ∑ i = 1 y t − i + V y t − C y t + k − ∑ s ∈ S ( C + ( k − s ) V − V ) ( y t + s − y t + s + 1 ) ( 22 ) is valid and facet-deﬁning for conv ( P ) . Proof. W e ﬁrst prove that inequality ( 21 ) is valid and facet-deﬁning for conv ( P ) . For notational conve- nience, we deﬁne s max = max { s : s ∈ S } if S  = ∅ , and s max = − 1 if S = ∅ . Consider any t ∈ [ k + 1, T − m − 1 ] Z . T o prove that the linear inequality ( 21 ) is valid for conv ( P ) , it sufﬁces to show that it is valid for P . Consider any element ( x , y ) of P . W e show that ( x , y ) satisﬁes ( 21 ). W e divide the analysis into three cases: Case 1: y t = 0. In this case, by ( 2c ) and ( 2d ), − x t − k ≤ − C y t − k and x t = 0. Thus, the left-hand side of ( 21 ) is at most − C y t − k . Because y t = 0 and t ∈ [ 2, T ] Z , by Lemma 1 (i), y t − j − y t − j − 1 ≤ 0 for all j ∈ [ 0, min { t − 2, L − 1 } ] Z . Because S ⊆ [ 0, min { k − 1, L − m − 2 } ] Z , m ≥ 0, and t ≥ k + 1, we have S ⊆ [ 0, min { t − 2, L − 1 } ] Z . Thus, y t − s − y t − s − 1 ≤ 0 for all s ∈ S . Because m ≤ k − 1, S ⊆ [ 0, k − 1 ] Z , and C + V > V , the coefﬁcients “ C + ( k − m ) V − V ” and “ C + ( k − s ) V − V ” on the right-hand side of ( 21 ) are positive for any s ∈ S . Thus, the right-hand side of ( 21 ) is at least − C y t − k . Therefor e, in this case, ( x , y ) satisﬁes ( 21 ). Case 2: y t = 1 and y t − s ′ − y t − s ′ − 1 = 1 for some s ′ ∈ S . In this case, y t − s ′ = 1 and y t − s ′ − 1 = 0. Because y t = 1 and t ∈ [ 2, T ] Z , by Lemma 1 (ii), there exists at most one j ∈ [ 0, min { t − 2, L } ] Z such that y t − j − y t − j − 1 = 1. Because S ⊆ [ 0, min { k − 1, L − m − 2 } ] Z , m ≥ 0, and t ≥ k + 1, we have S ⊆ [ 0, min { t − 2, L } ] Z . Thus, y t − s − y t − s − 1 ≤ 0 for all s ∈ S \ { s ′ } . Because y t − s ′ − y t − s ′ − 1 = 1 and t − s ′ ∈ [ 2, T ] Z , by ( 2a ), we have y τ = 1 for all τ ∈ [ t − s ′ , min { T , t − s ′ + L − 1 } ] Z . Because S ⊆ [ 0, L − m − 2 ] Z , we have t − s ′ + L − 1 ≥ t + m + 1. Thus, y τ = 1 for all τ ∈ [ t − s ′ , t + m + 1 ] Z , which implies that ( C + ( k − m ) V − V ) y t + m + 1 + V ∑ m i = 1 y t + i + V y t = C + k V . Because S ⊆ [ 0, k − 1 ] Z and C + V > V , the coefﬁcient “ C + ( k − s ) V − V ” on the right-hand side of inequality ( 21 ) is positive for all s ∈ S . Hence, the right-hand side of ( 21 ) is at least C + k V − C y t − k − ( C + ( k − s ′ ) V − V ) = s ′ V + V − C y t − k . By ( 2e ), ∑ t τ = t − s ′ ( x τ − x τ − 1 ) ≤ ∑ t τ = t − s ′ Vy τ − 1 + ∑ t τ = t − s ′ V ( 1 − y τ − 1 ) , which implies that x t − x t − s ′ − 1 ≤ s ′ V + V . Because y t − s ′ − 1 = 0, by ( 2d ), x t − s ′ − 1 = 0. By ( 2c ), − x t − k ≤ − C y t − k . Thus, x t − x t − k ≤ s ′ V + V − C y t − k . Therefor e, in this case, ( x , y ) satisﬁes ( 21 ). Case 3: y t = 1 and y t − s − y t − s − 1 ≤ 0 for all s ∈ S . Because S ⊆ [ 0, k − 1 ] Z , m ∈ [ 0, k − 1 ] Z , and C + V > V , we have C + ( k − m ) V − V > 0 and C + ( k − s ) V − V > 0 for each s ∈ S . Hence, the terms “ ( C + ( k − m ) V − V ) y t + m + 1 ” and “ − ∑ s ∈ S ( C + ( k − s ) V − V ) ( y t − s − y t − s − 1 ) ” on the right-hand side of inequality ( 21 ) are nonnegative. W e divide our analysis into three subcases. Case 3.1: y τ = 0 for some τ ∈ [ t , t + m + 1 ] Z . Let t ′ = min { τ ∈ [ t , t + m + 1 ] Z : y τ = 0 } . Then, y τ = 1 for all τ ∈ [ t , t ′ − 1 ] Z . Thus, the right-hand side of ( 21 ) is at least ( t ′ − t − 1 ) V + V − C y t − k . By ( 2f ), ∑ t ′ τ = t + 1 ( x τ − 1 − x τ ) ≤ ∑ t ′ τ = t + 1 Vy τ + ∑ t ′ τ = t + 1 V ( 1 − y τ ) , which implies that x t − x t ′ ≤ ( t ′ − t − 1 ) V + V . Because y t ′ = 0, by ( 2d ), x t ′ = 0. By ( 2c ), − x t − k ≤ − C y t − k . Hence, x t − x t − k ≤ ( t ′ − t − 1 ) V + V − C y t − k . Thus, the left-hand side of ( 21 ) is less than or equal to the right-hand side. Case 3.2: y τ = 1 for all τ ∈ [ t , t + m + 1 ] Z and y τ = 0 for some τ ∈ [ t − k , t − 1 ] Z . In this case, the right- hand side of ( 21 ) is at least ( C + ( k − m ) V − V ) + mV + V − C y t − k = C + k V − C y t − k . Let t ′ = max { τ ∈ [ t − k , t − 1 ] Z : y τ = 0 } . Because t ′ ≥ t − k and C + V > V , the right-hand side of ( 21 ) is greater than V + ( t − t ′ − 1 ) V − C y t − k . By ( 2e ), ∑ t τ = t ′ + 1 ( x τ − x τ − 1 ) ≤ ∑ t τ = t ′ + 1 Vy τ − 1 + ∑ t τ = t ′ + 1 V ( 1 − y τ − 1 ) , which implies that x t − x t ′ ≤ V + ( t − t ′ − 1 ) V . Because y t ′ = 0, by ( 2d ), x t ′ = 0. By ( 2c ), − x t − k ≤ − C y t − k . Hence, x t − x t − k ≤ V + ( t − t ′ − 1 ) V − C y t − k . Thus, the left-hand side of ( 21 ) is less than the right-hand side. Case 3.3: y τ = 1 for all τ ∈ [ t − k , t + m + 1 ] Z . In this case, the right-hand side of ( 21 ) is at least k V . By ( 2e ), ∑ t τ = t − k + 1 ( x τ − x τ − 1 ) ≤ ∑ t τ = t − k + 1 Vy τ − 1 + ∑ t τ = t − k + 1 V ( 1 − y τ − 1 ) , which implies that x t − x t − k ≤ k V . Thus, the left-hand side of ( 21 ) is less than or equal to the right-hand side. In Cases 3.1–3.3, ( x , y ) satisﬁes ( 21 ). Summarizing Cases 1–3, we conclude that ( 21 ) is valid for conv ( P ) . Consider any t ∈ [ k + 1, T − m − 1 ] Z . T o prove that inequality ( 21 ) is facet-deﬁning for conv ( P ) , it sufﬁces to show that there exist 2 T afﬁnely independent points in conv ( P ) that satisfy ( 21 ) at equality . Because 0 ∈ conv ( P ) and 0 satisﬁes ( 21 ) at equality , it sufﬁces to create the remaining 2 T − 1 nonzero linearly independent points. W e denote these 2 T − 1 points as ( ¯ x r , ¯ y r ) for r ∈ [ 1, T ] Z \ { t − k } and ( ˆ x r , ˆ y r ) for r ∈ [ 1, T ] Z , and denote the q th component of ¯ x r , ¯ y r , ˆ x r , and ˆ y r as ¯ x r q , ¯ y r q , ˆ x r q , and ˆ y r q , respectively . Let ϵ = min { V − C , C − C − kV } > 0. W e divide these 2 T − 1 points into the following eight groups. ec.39 (A1) For each r ∈ [ 1, t − 1 ] Z \ { t − k } , we create a point ( ¯ x r , ¯ y r ) as follows: ¯ x r q =        C , for q ∈ [ 1, t − 1 ] Z \ { r } ; C + ϵ , for q = r ; V , for q = t ; 0, for q ∈ [ t + 1, T ] Z ; and ¯ y r q =  1, for q ∈ [ 1, t ] Z ; 0, for q ∈ [ t + 1, T ] Z . It is easy to verify that ( ¯ x r , ¯ y r ) satisﬁes ( 2a )–( 2f ). Thus, ( ¯ x r , ¯ y r ) ∈ conv ( P ) . Note that ¯ x r t = V , ¯ x r t − k = C , ¯ y r t = ¯ y r t − k = 1, ¯ y r t + m + 1 = 0, ∑ m i = 1 ¯ y r t + i = 0, and ¯ y r t − s − ¯ y r t − s − 1 = 0 for all s ∈ S . Hence, ( ¯ x r , ¯ y r ) satisﬁes ( 21 ) at equality . (A2) W e create the same point ( ¯ x t , ¯ y t ) as in group (A3) in the proof of Proposition 9 . Thus, ( ¯ x t , ¯ y t ) ∈ conv ( P ) . Note that ¯ x t t = C + k V + ϵ , ¯ x t t − k = C + ϵ , ¯ y t t = ¯ y t t − k = ¯ y t t + m + 1 = 1, ∑ m i = 1 ¯ y t t + i = m , and ¯ y t t − s − ¯ y t t − s − 1 = 0 for all s ∈ S . Hence, ( ¯ x t , ¯ y t ) satisﬁes ( 21 ) at equality . (A3) For each r ∈ [ t + 1, T ] Z , we create a point ( ¯ x r , ¯ y r ) as follows: ¯ x r q =      C , for q ∈ [ 1, t − k − 1 ] Z ; C + ( q − t + k ) V , for q ∈ [ t − k , t ] Z ; C + kV , for q ∈ [ t + 1, T ] Z \ { r } ; C + kV + ϵ , for q = r ; and ¯ y r q = 1 for all q ∈ [ 1, T ] Z . It is easy to verify that ( ¯ x r , ¯ y r ) satisﬁes ( 2a )–( 2d ). Note that ¯ x r q − ¯ x r q − 1 = 0 when q ∈ [ 2, t − k ] Z , ¯ x r q − ¯ x r q − 1 = V when q ∈ [ t − k + 1, t ] Z , and − ϵ ≤ ¯ x r q − ¯ x r q − 1 ≤ ϵ when q ∈ [ t + 1, T ] Z . Thus, − V ¯ y r q − V ( 1 − ¯ y r q ) ≤ ¯ x r q − ¯ x r q − 1 ≤ V ¯ y r q − 1 + V ( 1 − ¯ y r q − 1 ) for all q ∈ [ 2, T ] Z . Hence, ( ¯ x r , ¯ y r ) satisﬁes ( 2e ) and ( 2f ). Therefor e, ( ¯ x r , ¯ y r ) ∈ conv ( P ) . Note that ¯ x r t = C + k V , ¯ x r t − k = C , ¯ y r t = ¯ y r t − k = ¯ y r t + m + 1 = 1, ∑ m i = 1 ¯ y r t + i = m , and ¯ y r t − s − ¯ y r t − s − 1 = 0 for all s ∈ S . Hence, ( ¯ x r , ¯ y r ) satisﬁes ( 21 ) at equality . (A4) For each r ∈ [ 1, t − 1 ] Z , we create the same point ( ˆ x r , ˆ y r ) as in gr oup (A2) in the proof of Pr oposition 1 . Thus, ( ˆ x r , ˆ y r ) ∈ conv ( P ) . T o show that ( ˆ x r , ˆ y r ) satisﬁes ( 21 ) at equality , we ﬁrst consider the case where t − r − 1 / ∈ S . In this case, ˆ x r q = ˆ y r q = 0 for all q ∈ [ t , t + m + 1 ] Z . Because t − k ≤ t − s max − 1 ≤ r , we have ˆ x r t − k = C , and ˆ y r t − k = 1. Because t − s − 1  = r for all s ∈ S , we have ˆ y r t − s − ˆ y r t − s − 1 = 0 for all s ∈ S . Hence, ( ˆ x r , ˆ y r ) satisﬁes ( 21 ) at equality . Next, we consider the case where t − r − 1 ∈ S . In this case, ˆ x r t = V + ( t − r − 1 ) V and ˆ y r q = 1 for all q ∈ [ t , t + m + 1 ] Z . Because t − k ≤ t − s max − 1 ≤ r , we have ˆ x r t − k = ˆ y r t − k = 0. In addition, ˆ y r t − s − ˆ y r t − s − 1 = 1 when s = t − r − 1, and ˆ y r t − s − ˆ y r t − s − 1 = 0 when s  = t − r − 1. Hence, ( ˆ x r , ˆ y r ) satisﬁes ( 21 ) at equality . (A5) For each r ∈ [ t , t + m ] Z , we create a point ( ˆ x r , ˆ y r ) as follows: ˆ x r q =        C , for q ∈ [ 1, 2 t − r − 1 ] Z ; V + ( q + r − 2 t ) V , for q ∈ [ 2 t − r , t − 1 ] Z ; V + ( r − q ) V , for q ∈ [ t , r ] Z ; 0, for q ∈ [ r + 1, T ] Z ; and ˆ y r q =  1, for q ∈ [ 1, r ] Z ; 0, for q ∈ [ r + 1, T ] Z . It is easy to verify that ( ˆ x r , ˆ y r ) satisﬁes ( 2a )–( 2d ). Note that − V ≤ ˆ x r q − ˆ x r q − 1 ≤ V when q ∈ [ 2, r ] Z , ˆ x r q − ˆ x r q − 1 = − V when q = r + 1, and ˆ x r q − ˆ x r q − 1 = 0 when q ∈ [ r + 2, T ] Z . Thus, − V ˆ y r q − V ( 1 − ˆ y r q ) ≤ ˆ x r q − ˆ x r q − 1 ≤ V ˆ y r q − 1 + V ( 1 − ˆ y r q − 1 ) for all q ∈ [ 2, T ] Z . Hence, ( ˆ x r , ˆ y r ) satisﬁes ( 2e ) and ( 2f ). Ther efore, ( ˆ x r , ˆ y r ) ∈ conv ( P ) . Note that ˆ x r t = V + ( r − t ) V and ˆ y r t = 1. Note also that ˆ y r t + m + 1 = 0 and V ∑ m i = 1 y r t + i = ( r − t ) V . Because t − k ≤ t − m − 1 ≤ 2 t − r − 1, we have ˆ x r t − k = C and ˆ y r t − k = 1. For any s ∈ S , because t − s ≤ t ≤ r , we have ˆ y r t − s − ˆ y r t − s − 1 = 0. Thus, ( ˆ x r , ˆ y r ) satisﬁes ( 21 ) at equality . ec.40 (A6) W e create a point ( ˆ x t + m + 1 , ˆ y t + m + 1 ) as follows: ˆ x t + m + 1 q =    C , for q ∈ [ 1, t − k − 1 ] Z ; C + ( q − t + k ) V , for q ∈ [ t − k , t ] Z ; C + kV , for q ∈ [ t + 1, T ] Z ; and ˆ y t + m + 1 q = 1 for all q ∈ [ 1, T ] Z . It is easy to verify that ( ˆ x t + m + 1 , ˆ y t + m + 1 ) satisﬁes ( 2a )–( 2d ). Note that ˆ x t + m + 1 q − ˆ x t + m + 1 q − 1 = 0 when q ∈ [ 2, t − k ] Z , ˆ x t + m + 1 q − ˆ x t + m + 1 q − 1 = V when q ∈ [ t − k + 1, t ] Z , and ˆ x t + m + 1 q − ˆ x t + m + 1 q − 1 = 0 when q ∈ [ t + 1, T ] Z . Thus, − V ˆ y t + m + 1 q − V ( 1 − ˆ y t + m + 1 q ) ≤ ˆ x t + m + 1 q − ˆ x t + m + 1 q − 1 ≤ V ˆ y t + m + 1 q − 1 + V ( 1 − ˆ y t + m + 1 q − 1 ) for all q ∈ [ 2, T ] Z . Hence, ( ˆ x t + m + 1 , ˆ y t + m + 1 ) satisﬁes ( 2e ) and ( 2f ). Ther efore, ( ˆ x t + m + 1 , ˆ y t + m + 1 ) ∈ conv ( P ) . Note that ˆ x t + m + 1 t = C + k V , ˆ x t + m + 1 t − k = C , ˆ y t + m + 1 t + m + 1 = ˆ y t + m + 1 t = ˆ y t + m + 1 t − k = 1, ∑ m i = 1 y t + m + 1 t + i = m , and ˆ y t + m + 1 t − s − ˆ y t + m + 1 t − s − 1 = 0 for all s ∈ S , Thus, ( ˆ x t + m + 1 , ˆ y t + m + 1 ) satisﬁes ( 21 ) at equality . (A7) For each r ∈ [ t + m + 2, T ] Z , we create a point ( ˆ x r , ˆ y r ) as follows: ˆ x r q =  0, for q ∈ [ 1, r − 1 ] Z ; C , for q ∈ [ r , T ] Z ; and ˆ y r q =  0, for q ∈ [ 1, r − 1 ] Z ; 1, for q ∈ [ r , T ] Z . It is easy to verify that ( ˆ x r , ˆ y r ) satisﬁes ( 2a )–( 2f ). Thus, ( ˆ x r , ˆ y r ) ∈ conv ( P ) . Note that ˆ x r t = ˆ x r t − k = 0, ˆ y r t = ˆ y r t − k = ˆ y r t + m + 1 = 0, ∑ m i = 1 y r t + i = 0, and ˆ y r t − s − ˆ y r t − s − 1 = 0 for all s ∈ S , Hence, ( ˆ x r , ˆ y r ) satisﬁes ( 21 ) at equality . T able EC.9 shows a matrix with 2 T − 1 rows, wher e each row r epresents a point created by this pro- cess. This matrix can be transformed into the matrix in T able EC.10 via the following Gaussian elimination process: (i) For each r ∈ [ 1, t − 1 ] Z \ { t − k } , the point with index r in group (B1), denoted ( ¯ x r , ¯ y r ) , is obtained by setting ( ¯ x r , ¯ y r ) = ( ¯ x r , ¯ y r ) − ( ˆ x t , ˆ y t ) . Here, ( ¯ x r , ¯ y r ) is the point with index r in group (A1) , and ( ˆ x t , ˆ y t ) is the point with index t in group (A5) . (ii) The point in gr oup (B2), denoted ( ¯ x t , ¯ y t ) , is obtained by setting ( ¯ x t , ¯ y t ) = ( ¯ x t , ¯ y t ) − ( ˆ x t + m + 1 , ˆ y t + m + 1 ) . Here, ( ¯ x t , ¯ y t ) is the point in group (A2) , and ( ˆ x t + m + 1 , ˆ y t + m + 1 ) is the point in group (A6) . (iii) For each r ∈ [ t + 1, T ] Z , the point with index r in group (B3), denoted ( ¯ x r , ¯ y r ) , is obtained by set- ting ( ¯ x r , ¯ y r ) = ( ¯ x r , ¯ y r ) − ( ˆ x t + m + 1 , ˆ y t + m + 1 ) . Here, ( ¯ x r , ¯ y r ) is the point with index r in group (A3) , and ( ˆ x t + m + 1 , ˆ y t + m + 1 ) is the point in group (A6) . (iv) For each r ∈ [ 1, t − 1 ] Z , the point with index r in group (B4), denoted ( ˆ x r , ˆ y r ) , is obtained by setting ( ˆ x r , ˆ y r ) = ( ˆ x r , ˆ y r ) if t − r − 1 / ∈ S , and setting ( ˆ x r , ˆ y r ) = ( ˆ x t + m + 1 , ˆ y t + m + 1 ) − ( ˆ x r , ˆ y r ) if t − r − 1 ∈ S . Here, ( ˆ x r , ˆ y r ) is the point with index r in group (A4) , and ( ˆ x t + m + 1 , ˆ y t + m + 1 ) is the point in gr oup (A6) . (v) For each r ∈ [ t , t + m ] Z , the point with index r in group (B5), denoted ( ˆ x r , ˆ y r ) , is obtained by setting ( ˆ x r , ˆ y r ) = ( ˆ x r , ˆ y r ) . Here, ( ˆ x r , ˆ y r ) is the point with index r in group (A5) . (vi) The point in group (B6), denoted ( ˆ x t + m + 1 , ˆ y t + m + 1 ) , is obtained by setting ( ˆ x t + m + 1 , ˆ y t + m + 1 ) = ( ˆ x t + m + 1 , ˆ y t + m + 1 ) − ( ˆ x t + m + 2 , ˆ y t + m + 2 ) . Here, ( ˆ x t + m + 1 , ˆ y t + m + 1 ) is the point in gr oup (A6) , and ( ˆ x t + m + 2 , ˆ y t + m + 2 ) is the point with index t + m + 2 in group (A7) . (vii) For each r ∈ [ t + m + 2, T ] Z , the point with index r in group (B7), denoted ( ˆ x r , ˆ y r ) , is obtained by setting ( ˆ x r , ˆ y r ) = ( ˆ x r , ˆ y r ) − ( ˆ x r + 1 , ˆ y r + 1 ) if r  = T , and setting ( ˆ x r , ˆ y r ) = ( ˆ x r , ˆ y r ) if r = T . Here, ( ˆ x r , ˆ y r ) and ( ˆ x r + 1 , ˆ y r + 1 ) are the points with indices r and r + 1, respectively , in group (A7) . The matrix shown in T able EC.10 is lower triangular; that is, the position of the last nonzero component of a row of the matrix is greater than the position of the last nonzero component of the previous row . This implies that the 2 T − 1 points in gr oups (A1) – (A7) are linearly independent. Ther efore, inequality ( 21 ) is facet-deﬁning for conv ( P ) . ec.41 T able EC.9 A matrix with the rows r epresenting 2 T − 1 linearly independent points in conv ( P ) satisfying inequality ( 21 ) at equality Group Point Index r x y 1 · · · t − k − 1 t − k t − k + 1 · · · t − 1 t t + 1 · · · t + m t + m + 1 t + m + 2 · · · T 1 · · · t − 1 t t + 1 · · · t + m t + m + 1 t + m + 2 · · · T (A1) ( ¯ x r , ¯ y r ) 1 C + ϵ · · · C C C · · · C V 0 · · · 0 0 0 · · · 0 1 · · · 1 1 0 · · · 0 0 0 · · · 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . t − k − 1 C · · · C + ϵ C C · · · C V 0 · · · 0 0 0 · · · 0 1 · · · 1 1 0 · · · 0 0 0 · · · 0 t − k + 1 C · · · C C C + ϵ · · · C V 0 · · · 0 0 0 · · · 0 1 · · · 1 1 0 · · · 0 0 0 · · · 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . t − 1 C · · · C C C · · · C + ϵ V 0 · · · 0 0 0 · · · 0 1 · · · 1 1 0 · · · 0 0 0 · · · 0 (A2) t C · · · C C + ϵ C + V + ϵ · · · C + ( k − 1 ) V + ϵ C + kV + ϵ C + kV · · · C + k V C + kV C + k V · · · C + kV 1 · · · 1 1 1 · · · 1 1 1 · · · 1 (A3) t + 1 C · · · C C C + V · · · C + ( k − 1 ) V C + k V C + k V + ϵ · · · C + k V C + kV C + k V · · · C + k V 1 · · · 1 1 1 · · · 1 1 1 · · · 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . t + m C · · · C C C + V · · · C + ( k − 1 ) V C + k V C + kV · · · C + kV + ϵ C + k V C + kV · · · C + kV 1 · · · 1 1 1 · · · 1 1 1 · · · 1 t + m + 1 C · · · C C C + V · · · C + ( k − 1 ) V C + k V C + kV · · · C + k V C + kV + ϵ C + k V · · · C + k V 1 · · · 1 1 1 · · · 1 1 1 · · · 1 t + m + 2 C · · · C C C + V · · · C + ( k − 1 ) V C + k V C + kV · · · C + k V C + kV C + kV + ϵ · · · C + k V 1 · · · 1 1 1 · · · 1 1 1 · · · 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . T C · · · C C C + V · · · C + ( k − 1 ) V C + k V C + kV · · · C + k V C + kV C + k V · · · C + kV + ϵ 1 · · · 1 1 1 · · · 1 1 1 · · · 1 (A4) ( ˆ x r , ˆ y r ) 1 (See Note EC.9 -1) (See Note EC.9 -1) . . . t − 1 (A5) t C · · · C C C · · · C V 0 · · · 0 0 0 · · · 0 1 · · · 1 1 0 · · · 0 0 0 · · · 0 t + 1 C · · · C C C · · · V V + V V · · · 0 0 0 · · · 0 1 · · · 1 1 1 · · · 0 0 0 · · · 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . t + m C · · · C C (See Note EC.9 -2) · · · V + ( m − 1 ) V V + mV V + ( m − 1 ) V · · · V 0 0 · · · 0 1 · · · 1 1 1 · · · 1 0 0 · · · 0 (A6) t + m + 1 C · · · C C C + V · · · C + ( k − 1 ) V C + k V C + kV · · · C + k V C + kV C + kV · · · C + k V 1 · · · 1 1 1 · · · 1 1 1 · · · 1 (A7) t + m + 2 0 · · · 0 0 0 · · · 0 0 0 · · · 0 0 C · · · C 0 · · · 0 0 0 · · · 0 0 1 · · · 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . T 0 · · · 0 0 0 · · · 0 0 0 · · · 0 0 0 · · · C 0 · · · 0 0 0 · · · 0 0 0 · · · 1 Note EC.9 -1: For r ∈ [ 1, t − 1 ] Z , the x and y vectors in group (A4) are given as follows: ˆ x r = ( C , . . . , C | {z } r terms , 0, . . . , 0 | {z } T − r terms ) and ˆ y r = ( 1, . . . , 1 | {z } r terms , 0, . . . , 0 | {z } T − r terms ) if t − r − 1 / ∈ S ; ˆ x r = ( 0, . . . , 0 | {z } r terms , V , V + V , V + 2 V , . . . , V ( r − t − 1 ) V | {z } t − r terms , V + ( t − r − 1 ) V , . . . , V + ( t − r − 1 ) V | {z } T − t terms ) and ˆ y r = ( 0, . . . , 0 | {z } r terms , 1, . . . , 1 | {z } T − r terms ) if t − r − 1 ∈ S . Note EC.9 -2: In group (A5), ˆ x t + m t − k + 1 = C if m < k − 1, and ˆ x t + m t − k + 1 = V if m = k − 1. ec.42 T able EC.10 Lower triangular matrix obtained from T able EC.9 via Gaussian elimination Group Point Index r x y 1 · · · t − k − 1 t − k t − k + 1 · · · t − 1 t t + 1 · · · t + m t + m + 1 t + m + 2 · · · T 1 · · · t − 1 t t + 1 · · · t + m t + m + 1 t + m + 2 · · · T (B1) ( ¯ x r , ¯ y r ) 1 ϵ 0 · · · 0 0 · · · 0 0 0 · · · 0 0 0 · · · 0 0 · · · 0 0 0 · · · 0 0 0 · · · 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . t − k − 1 0 · · · ϵ 0 0 · · · 0 0 0 · · · 0 0 0 · · · 0 0 · · · 0 0 0 · · · 0 0 0 · · · 0 t − k + 1 0 · · · 0 0 ϵ · · · 0 0 0 · · · 0 0 0 · · · 0 0 · · · 0 0 0 · · · 0 0 0 · · · 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . t − 1 0 · · · 0 0 0 · · · ϵ 0 0 · · · 0 0 0 · · · 0 0 · · · 0 0 0 · · · 0 0 0 · · · 0 (B2) t 0 0 · · · ϵ ϵ · · · ϵ ϵ 0 · · · 0 0 0 · · · 0 0 · · · 0 0 0 · · · 0 0 0 · · · 0 (B3) t + 1 0 0 · · · 0 0 · · · 0 0 ϵ · · · 0 0 0 · · · 0 0 · · · 0 0 0 · · · 0 0 0 · · · 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . t + m 0 0 · · · 0 0 · · · 0 0 0 · · · ϵ 0 0 · · · 0 0 · · · 0 0 0 · · · 0 0 0 · · · 0 t + m + 1 0 0 · · · 0 0 · · · 0 0 0 · · · 0 ϵ 0 · · · 0 0 · · · 0 0 0 · · · 0 0 0 · · · 0 t + m + 2 0 0 · · · 0 0 · · · 0 0 0 · · · 0 0 ϵ · · · 0 0 · · · 0 0 0 · · · 0 0 0 · · · 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . T 0 0 · · · 0 0 · · · 0 0 0 · · · 0 0 0 · · · ϵ 0 · · · 0 0 0 · · · 0 0 0 · · · 0 (B4) ( ˆ x r , ˆ y r ) 1 1 · · · 0 0 0 · · · 0 0 0 · · · 0 . . . (Omitted) . . . . . . . . . . . . . . . . . . . . . . . . . . . t − 1 1 · · · 1 0 0 · · · 0 0 0 · · · 0 (B5) t 1 · · · 1 1 0 · · · 0 0 0 · · · 0 t + 1 1 · · · 1 1 1 · · · 0 0 0 · · · 0 . . . (Omitted) . . . . . . . . . . . . . . . . . . . . . . . . . . . t + m 1 · · · 1 1 1 · · · 1 0 0 · · · 0 (B6) t + m + 1 (Omitted) 1 · · · 1 1 1 · · · 1 1 0 · · · 0 (B7) t + m + 2 0 · · · 0 0 0 · · · 0 0 1 · · · 0 . . . (Omitted) . . . . . . . . . . . . . . . . . . . . . . . . . . . T 0 · · · 0 0 0 · · · 0 0 0 · · · 1 ec.43 Next, we show that inequality ( 22 ) is valid and facet-deﬁning for conv ( P ) . Denote x ′ t = x T − t + 1 and y ′ t = y T − t + 1 for t ∈ [ 1, T ] Z . Because inequality ( 21 ) is valid and facet-deﬁning for conv ( P ) for any t ∈ [ k + 1, T − m − 1 ] Z , the inequality x ′ T − t + 1 − x ′ T − t + k + 1 ≤ ( C + ( k − m ) V − V ) y ′ T − t − m + V m ∑ i = 1 y ′ T − t − i + 1 + V y ′ T − t + 1 − C y ′ T − t + k + 1 − ∑ s ∈ S ( C + ( k − s ) V − V ) ( y ′ T − t + s + 1 − y ′ T − t + s + 2 ) is valid and facet-deﬁning for conv ( P ′ ) for any t ∈ [ k + 1, T − m − 1 ] Z . Let t ′ = T − t + 1. Then, the inequality x ′ t ′ − x ′ t ′ + k ≤ ( C + ( k − m ) V − V ) y ′ t ′ − m − 1 + V m ∑ i = 1 y ′ t ′ − i + V y ′ t ′ − C y ′ t ′ + k − ∑ s ∈ S ( C + ( k − s ) V − V ) ( y ′ t ′ + s − y ′ t ′ + s + 1 ) is valid and facet-deﬁning for conv ( P ′ ) for any t ′ ∈ [ m + 2, T − k ] Z . Hence, by Lemma 2 , inequality ( 22 ) is valid and facet-deﬁning for conv ( P ) for any t ∈ [ m + 2, T − k ] Z . □ C.12. Pro of of Prop osition 12 Proposition 12 . For any given point ( x , y ) ∈ R 2 T + , the most violated inequalities ( 21 ) and ( 22 ) can be determined in O ( T 3 ) time if such violated inequalities exist. Proof. W e ﬁrst consider inequality ( 21 ). Consider any given ( x , y ) ∈ R 2 T + . For notational convenience, denote ˆ k = max { k ∈ [ 1, T − 1 ] Z : C − C − kV > 0 } , and denote ˆ s km = min { k − 1, L − m − 2 } for any k ∈ [ 1, ˆ k ] Z and m ∈ [ 0, k − 1 ] Z . For any t ∈ [ 1, T ] Z , let θ ( t ) = t ∑ τ = 2 max { y τ − y τ − 1 , 0 } . Then, for any k ∈ [ 1, ˆ k ] Z , m ∈ [ 0, k − 1 ] Z , and t ∈ [ k + 1, T − m − 1 ] Z , ˆ s km ∑ s = 1 max { y t − s − y t − s − 1 , 0 } = t − 1 ∑ τ = t − ˆ s km max { y τ − y τ − 1 , 0 } = θ ( t − 1 ) − θ ( t − ˆ s km − 1 ) . (EC.36) For any k ∈ [ 1, ˆ k ] Z , m ∈ [ 0, k − 1 ] Z , t ∈ [ k + 1, T − m − 1 ] Z , and S ⊆ [ 0, ˆ s km ] Z , let ˜ v km ( S , t ) = x t − x t − k − ( C + ( k − m ) V − V ) y t + m + 1 − V m ∑ i = 1 y t + i − V y t + C y t − k + ∑ s ∈ S ( C + ( k − s ) V − V ) ( y t − s − y t − s − 1 ) . If ˜ v km ( S , t ) > 0, then ˜ v km ( S , t ) is the amount of violation of inequality ( 21 ). If ˜ v km ( S , t ) ≤ 0, there is no violation of inequality ( 21 ). For any k ∈ [ 1, ˆ k ] Z , m ∈ [ 0, k − 1 ] Z , and t ∈ [ k + 1, T − m − 1 ] Z , let v km ( t ) = max S ⊆ [ 0, ˆ s km ] Z { ˜ v km ( S , t ) } . If v km ( t ) > 0, then v km ( t ) is the lar gest possible violation of inequality ( 21 ) for this combination of k , m , and t . If v km ( t ) ≤ 0, the largest possible violation of inequality ( 21 ) is zero for this combination of k , m , and t . Because C + V > V , we have C + ( k − s ) V − V > 0 for all k ∈ [ 1, ˆ k ] Z , s ∈ [ 0, ˆ s km ] Z , and m ∈ [ 0, k − 1 ] Z . Thus, for any k ∈ [ 1, ˆ k ] Z , m ∈ [ 0, k − 1 ] Z , and t ∈ [ k + 1, T − m − 1 ] Z , ˜ v km ( S , t ) is maximized when S contains all s ∈ [ 0, ˆ s km ] Z such that y t − s − y t − s − 1 > 0 (if any). If it does not exist any s ∈ [ 0, ˆ s ] Z such that y t − s − y t − s − 1 > 0, then ˜ v km ( S , t ) is maximized when S = ∅ , and v km ( t ) = x t − x t − k − ( C + ( k − m ) V − V ) y t + m + 1 − V ∑ m i = 1 y t + i − V y t + C y t − k . Hence, for any k ∈ [ 1, ˆ k ] Z , m ∈ [ 0, k − 1 ] Z , and t ∈ [ k + 1, T − m − 1 ] Z , v km ( t ) = x t − x t − k − ( C + ( k − m ) V − V ) y t + m + 1 − V m ∑ i = 1 y t + i − V y t + C y t − k + ˆ s km ∑ s = 0 ( C + ( k − s ) V − V ) max { y t − s − y t − s − 1 , 0 } . ec.44 Determining θ ( t ) for all t ∈ [ 1, T ] Z can be done recursively in O ( T ) time by setting θ ( 1 ) = 0 and setting θ ( t ) = θ ( t − 1 ) + max { y t − y t − 1 , 0 } for t = 2, . . . , T . Clearly , for each k ∈ [ 1, ˆ k ] Z and each m ∈ [ 0, k − 1 ] Z , the value of v km ( k + 1 ) can be determined in O ( T ) time. For any k ∈ [ 1, ˆ k ] Z , m ∈ [ 0, k − 1 ] Z , and t ∈ [ k + 2, T − m − 1 ] Z , v km ( t ) − v km ( t − 1 ) = ( x t − x t − 1 ) − ( x t − k − x t − k − 1 ) − ( C + ( k − m ) V − V ) ( y t + m + 1 − y t + m ) − V " m ∑ i = 1 y t + i − m ∑ i = 1 y t + i − 1 # − V ( y t − y t − 1 ) + C ( y t − k − y t − k − 1 ) + ( C + kV − V ) " ˆ s km ∑ s = 0 max { y t − s − y t − s − 1 , 0 } − ˆ s km ∑ s = 0 max { y t − s − 1 − y t − s − 2 , 0 } # − V " ˆ s km ∑ s = 0 s max { y t − s − y t − s − 1 , 0 } − ˆ s km ∑ s = 0 s max { y t − s − 1 − y t − s − 2 , 0 } # = ( x t − x t − 1 ) − ( x t − k − x t − k − 1 ) − ( C + ( k − m ) V − V ) ( y t + m + 1 − y t + m ) − V ( y t + m − y t ) − V ( y t − y t − 1 ) + C ( y t − k − y t − k − 1 ) + ( C + kV − V )  max { y t − y t − 1 , 0 } − max { y t − ˆ s km − 1 − y t − ˆ s km − 2 , 0 }  − V " ˆ s km ∑ s = 1 max { y t − s − y t − s − 1 , 0 } − ˆ s km max { y t − ˆ s km − 1 − y t − ˆ s km − 2 , 0 } # . This, together with ( EC.36 ), implies that v km ( t ) = v km ( t − 1 ) + ( x t − x t − 1 ) − ( x t − k − x t − k − 1 ) − ( C + ( k − m ) V − V ) ( y t + m + 1 − y t + m ) − V ( y t + m − y t ) − V ( y t − y t − 1 ) + C ( y t − k − y t − k − 1 ) + ( C + k V − V )  max { y t − y t − 1 , 0 } − max { y t − ˆ s km − 1 − y t − ˆ s km − 2 , 0 }  − V  θ ( t − 1 ) − θ ( t − ˆ s km − 1 ) − ˆ s km max { y t − ˆ s km − 1 − y t − ˆ s km − 2 , 0 }  . Thus, for each k ∈ [ 1, ˆ k ] Z and m ∈ [ 0, k − 1 ] Z , the values of v km ( k + 1 ) , v km ( k + 2 ) , . . . , v km ( T − m − 1 ) can be determined r ecursively in O ( T ) time. Hence, the values of k , m , t and the set S corresponding to the largest possible violation of inequality ( 21 ) can be obtained in O ( T 3 ) time. Next, we consider inequality ( 22 ). Consider any given ( x , y ) ∈ R 2 T + . Let x ′ t = x T − t + 1 and y ′ t = y T − t + 1 for t ∈ [ 1, T ] Z . Inequality ( 22 ) becomes x ′ T − t + 1 − x ′ T − t − k + 1 ≤ ( C + ( k − m ) V − V ) y ′ T − t + m + 2 + V m ∑ i = 1 y ′ T − t + i + 1 + V y ′ T − t + 1 − C y ′ T − t − k + 1 − ∑ s ∈ S ( C + ( k − s ) V − V ) ( y ′ T − t − s + 1 − y ′ T − t − s ) . Letting t ′ = T − t + 1, this inequality becomes x ′ t ′ − x ′ t ′ − k ≤ ( C + ( k − m ) V − V ) y ′ t ′ + m + 1 + V m ∑ i = 1 y ′ t ′ + i + V y ′ t ′ − C y ′ t ′ − k − ∑ s ∈ S ( C + ( k − s ) V − V ) ( y ′ t ′ − s − y ′ t ′ − s − 1 ) . (EC.37) Because the values of k , m , t and the set S corr esponding to the largest possible violation of inequality ( 21 ) can be obtained in O ( T 3 ) time, the values of k , m , t ′ and the set S corresponding to the lar gest possible violation of inequality ( EC.37 ) can be obtained in O ( T 3 ) time. Hence, the values of k , m , t and the set S corresponding to the lar gest possible violation of inequality ( 22 ) can be obtained in O ( T 3 ) time. □ App endix D: Supplement to Section 5 D.1. Results of the Third Exp eriment Under Diﬀerent Demand Settings For the third experiment, we consider two additional cases, namely a less congested demand setting and a more congested demand setting. For the mor e congested demand setting, we obtain the new demand ¯ d b t at ec.45 bus b ∈ B for each period t ∈ [ 1, T ] Z by incr easing the corr esponding demand d b t in the thir d experiment by 10%, while ensuring that ¯ d b t will not exceed the total generation capacity of these 54 thermal generators at the same time. Thus, we set ¯ d b t = min { 1.1 d b t , ∑ g ∈ G C g } . For the less congested demand setting, we obtain ¯ d b t by decreasing the demand d b t by 10%; that is, we set ¯ d b t = 0.9 d b t . T ables EC.11 and EC.12 present the computational results for the less congested demand setting. Under this setting, formulations F1 + and F1 + -X exhibit similar overall performance. Both formulations successfully solved 19 instances within the time limit. The integrality gaps for F1 + and F1 + -X are nearly identical acr oss all instances. This can be attributed to the decrease in demand for each time period, resulting in a smaller number of generators being started up, as well as a reduction in the number of ramp-ups and ramp-downs. Therefor e, few valid inequalities are applied in each solution pr ocess. Consequently , the r eduction in the integrality gap is small from F1 + to F1 + -X. For most instances, the solution time of F1 + is smaller than F1 + -X, because valid inequalities ( 5 )–( 10 ) are added as constraints in F1 + -X, making the subpr oblems at each node of the branching process mor e difﬁcult to solve. T ables EC.13 and EC.14 present the computational results for the more congested setting. Under this setting, formulation F1 + -X outperforms formulation F1 + in all instances. The integrality gaps of the F1 + -X formulation are much smaller than those of F1 + , and the “Pct. reduction” is approximately 90% for each instance. Eight instances are solved to optimality by formulation F1 + -X, and the TGaps for the remaining twelve instances ar e all within 0.05%. On the other hand, none of these instances ar e solved successfully by formulation F1 + , and the TGaps are all above 0.26%. For most instances, formulation F1 + -X explores less nodes than formulation F1 + . This is because as the demand for each time period increases, mor e generators are needed, and the number of ramp-ups and ramp-downs also increases. As a result, more user cuts are used during the solution process. D.2. Results of the Three Exp eriments b y Disabling the Smart F eatures of the Solver In this section, we run the three experiments in Section 5 by disabling the smart features of CPLEX, includ- ing the presolve, heuristics, and default cut generation, to examine the strength of our proposed strong valid T able EC.11 Performance of MIP Formulations in the Third Experiment Under a Less Congested Demand Setting Instance # var # bin var # cstr CPU time [TGap] # nodes # user cuts F1 + F1 + -X F1 + F1 + -X F1 + F1 + -X F1 + -X 1 6372 1296 36124 43576 307.8 212.0 19395 12293 88 2 174.7 233.8 13023 10430 95 3 166.4 522.7 11248 13803 96 4 588.9 342.1 36443 17313 98 5 62.0 106.6 6775 2699 70 6 **[0.11%] **[0.07%] 115728 103776 229 7 1116.6 848.6 43524 36756 126 8 1247.9 1985.2 63266 59530 217 9 821.9 1774.3 41451 56742 198 10 1881.1 2230.6 60109 69543 188 11 745.1 2793.1 40789 72982 182 12 87.8 226.9 9905 9968 93 13 35.5 66.7 2370 2702 45 14 1187.4 1043.2 49699 43317 143 15 272.5 818.3 21934 39631 173 16 306.6 206.8 21323 9596 99 17 200.1 614.2 18638 27580 128 18 82.3 543.2 9904 18321 88 19 70.6 119.5 6875 3618 73 20 81.8 148.5 10301 10550 66 ec.46 T able EC.12 The Strength of LP Relaxations of MIP Formulations in the Thir d Experiment Under a Less Congested Demand Setting IGap Instance 1 2 3 4 5 6 7 8 9 10 F1 + 0.57% 0.59% 0.50% 0.63% 0.53% 0.58% 0.58% 0.64% 0.53% 0.58% F1 + -X 0.57% 0.59% 0.50% 0.63% 0.53% 0.58% 0.58% 0.63% 0.52% 0.57% Instance 11 12 13 14 15 16 17 18 19 20 F1 + 0.60% 0.57% 0.47% 0.58% 0.58% 0.59% 0.60% 0.59% 0.47% 0.46% F1 + -X 0.60% 0.57% 0.47% 0.58% 0.58% 0.59% 0.60% 0.59% 0.47% 0.46% Pct. reduction Instance 1 2 3 4 5 6 7 8 9 10 F1 + -X 0.14% 0% 0% 0% 0% 0% 0% 1.36% 1.82% 1.17% Instance 11 12 13 14 15 16 17 18 19 20 F1 + -X 0% 0.17% 0% 0.04% 0.09% 0.13% 0% 0% 0.65% 0.11% T able EC.13 Computational Results of the Third Experiment Under a Mor e Congested Demand Setting Instance # var # bin var # cstr CPU time [TGap] # nodes # user cuts F1 + F1 + -X F1 + F1 + -X F1 + F1 + -X F1 + -X 1 6372 1296 36124 43576 **[0.27%] 616.8 409894 124616 809 2 **[0.30%] 503.2 893931 62034 847 3 **[0.42%] 471.4 49184 587395 581 4 **[0.28%] 928.7 579540 87807 652 5 **[0.33%] **[0.04%] 516315 860515 684 6 **[0.31%] **[0.03%] 588982 1021155 764 7 **[0.40%] 1195.5 602518 223840 669 8 **[0.53%] **[0.05%] 634564 523103 1087 9 **[0.51%] **[0.05%] 543776 289863 819 10 **[0.53%] **[0.02%] 589285 434974 925 11 **[0.39%] **[0.04%] 589812 432163 1088 12 **[0.45%] **[0.03%] 641832 609698 932 13 **[0.43%] 1714.3 610207 251633 897 14 **[0.45%] **[0.03%] 603593 286153 827 15 **[0.46%] **[0.04%] 834771 506532 797 16 **[0.35%] **[0.04%] 617076 432809 734 17 **[0.27%] **[0.02%] 838514 462097 778 18 **[0.27%] 3221.3 603741 562787 892 19 **[0.35%] 880.5 603631 310824 792 20 **[0.58%] **[0.02%] 639278 614575 962 inequalities to help solve UC formulations. As the numbers of variables and constraints remain unchanged for all tested formulations in these three experiments, they ar e not reported in the following tables. The results of the ﬁrst experiment are summarized in T ables EC.15 and EC.16 . Using formulation F1, CPLEX is unable to solve any of these 20 instances to optimality within the one-hour time limit. In T able EC.15 , the “—” rows denote the instances for which a feasible solution cannot be found by CPELX within the time limit using formulation F1. Using formulation F1-X, CPLEX can solve 2 instances to opti- mality within the time limit. The number of nodes explored by F1-X is much smaller than that of formu- lation F1. The number of user cuts added by F1-X in the solution pr ocess is smaller compar ed with the total number of constraints in formulations F1 and F1-X. For the instances that cannot be solved to opti- ec.47 T able EC.14 The Strength of LP Relaxations of MIP Formulations in the Thir d Experiment Under a More Congested Demand Setting IGap Instance 1 2 3 4 5 6 7 8 9 10 F1 + 2.33% 2.61% 2.53% 2.24% 2.43% 2.40% 2.48% 2.60% 2.56% 2.48% F1 + -X 0.21% 0.21% 0.26% 0.20% 0.21% 0.17% 0.23% 0.32% 0.29% 0.26% Instance 11 12 13 14 15 16 17 18 19 20 F1 + 2.59% 2.76% 2.61% 2.41% 2.56% 2.56% 2.35% 2.25% 2.41% 2.48% F1 + -X 0.24% 0.22% 0.23% 0.30% 0.26% 0.22% 0.24% 0.26% 0.25% 0.25% Pct. reduction Instance 1 2 3 4 5 6 7 8 9 10 F1 + -X 90.81% 91.93% 89.87% 91.24% 91.36% 93.04% 90.54% 87.50% 88.79% 89.52% Instance 11 12 13 14 15 16 17 18 19 20 F1 + -X 90.62% 91.88% 91.33% 87.41% 89.93% 91.36% 89.80% 88.40% 89.72% 89.76% T able EC.15 Performance of MIP Formulations in the First Experiment by Disabling Smart Features Instance CPU time [TGap] # nodes # user cuts F1 F1-X F1 F1-X F1-X 1 **[0.25%] **[0.04%] 3243082 1331163 428 2 **[0.33%] **[0.10%] 1687199 906171 768 3 **[0.35%] **[0.05%] 1398467 403775 937 4 **[0.30%] **[0.01%] 1217117 1592032 490 5 **[0.48%] **[0.03%] 1218680 494229 823 6 **[0.53%] **[0.02%] 2492793 837237 823 7 **[0.31%] **[0.06%] 1647817 576308 472 8 **[0.53%] **[0.05%] 1431881 547503 732 9 **[0.44%] **[0.04%] 1062217 481563 899 10 **[0.62%] **[0.03%] 1232270 552396 790 11 **[0.47%] 671.2 397815 17587 1050 12 **[0.66%] **[0.03%] 345767 132849 2423 13 **[0.71%] **[0.02%] 300637 50590 1907 14 — **[0.03%] — 70987 3195 15 — **[0.03%] — 17541 2050 16 **[0.70%] 935.7 262321 15358 1347 17 — **[0.02%] — 68966 2168 18 **[1.63%] **[0.01%] 155156 106438 2116 19 — **[0.02%] — 61927 2338 20 — **[0.02%] — 66462 1810 mality using formulation F1-X, the terminating gaps are all within 0.05% except for instances 2 and 7, and are much smaller than those using formulation F1. The integrality gaps generated by formulation F1-X are substantially smaller than those of F1, with reductions of at least 55%. These results demonstrate that the proposed strong valid inequalities signiﬁcantly tighten the single-binary formulation, thereby enhancing the efﬁciency of the solution pr ocess. T ables EC.17 and EC.18 present the r esults for the second experiment. Using formulation F2, no instance can be solved to optimality by CPLEX within the time limit. In contrast, using formulation F2-X, the two- binary formulation with our pr oposed valid inequalities, CPLEX can solve 5 instances to optimality within the time limit. Using formulation F2-Y , the two-binary formulation with valid inequalities from Pan and Guan ( 2016 ), CPLEX is able to solve 2 instances to optimality within the time limit. Using formulation F2-Z, ec.48 T able EC.16 The Strength of LP Relaxations of MIP Formulations in the First Experiment by Disabling Smart Features IGap Instance 1 2 3 4 5 6 7 8 9 10 F1 0.45% 0.39% 0.41% 0.36% 0.51% 0.62% 0.34% 0.55% 0.51% 0.64% F1-X 0.20% 0.14% 0.08% 0.06% 0.05% 0.05% 0.07% 0.06% 0.05% 0.05% Instance 11 12 13 14 15 16 17 18 19 20 F1 0.32% 0.32% 0.40% 0.39% 0.52% 0.36% 0.45% 0.42% 0.37% 0.44% F1-X 0.05% 0.02% 0.02% 0.03% 0.03% 0.08% 0.01% 0.01% 0.02% 0.02% Pct. reduction Instance 1 2 3 4 5 6 7 8 9 10 F1-X 55.7% 63.9% 81.6% 84.5% 90.2% 92.4% 78.7% 89.5% 89.5% 92.8% Instance 11 12 13 14 15 16 17 18 19 20 F1-X 83.1% 94.1% 95.1% 92.3% 94.0% 77.1% 96.0% 96.6% 94.2% 94.4% T able EC.17 Performance of MIP Formulations in the Second Experiment by Disabling Smart Features Instance CPU time [TGap] # nodes # user cuts F2 F2-X F2-Y F2-Z F2 F2-X F2-Y F2-Z F2-X F2-Y F2-Z 1 **[0.26%] 2429.2 184.4 **[0.03%] 3749277 4217040 266731 1948319 461 211 571 2 ** [0.29%] **[0.05%] **[0.09%] **[0.05%] 3496986 2166004 1841074 1683486 521 304 576 3 **[0.34%] **[0.06%] **[0.05%] **[0.05%] 1936293 1603280 1303901 520019 730 293 426 4 ** [0.26%] **[0.02%] **[0.02%] **[0.02%] 2231992 2161428 1690409 1537977 396 171 338 5 **[0.46%] **[0.04%] **[0.04%] **[0.03%] 1786842 1790680 1151283 715773 780 238 535 6 **[0.51%] 3241.8 **[0.01%] **[0.01%] 4239872 3331564 1556754 1175628 782 298 563 7 **[0.27%] **[0.05%] 332.8 **[0.06%] 1824764 1492391 161419 915783 504 177 450 8 ** [0.46%] **[0.05%] **[0.05%] **[0.05%] 1955325 1620740 703513 631110 542 220 444 9 **[0.43%] **[0.03%] **[0.05%] **[0.05%] 2294145 1603961 661026 403170 1007 351 608 10 **[0.60%] **[0.02%] **[0.04%] **[0.03%] 2024929 1983320 875849 950878 869 299 720 11 **[0.42%] **[0.05%] **[0.05%] **[0.05%] 647225 479026 448558 184850 1066 643 1392 12 **[0.38%] **[0.02%] **[0.06%] **[0.03%] 558586 410723 66363 42267 1983 639 1077 13 **[0.41%] **[0.02%] **[0.02%] **[0.02%] 562676 437129 148136 148136 1420 1650 1483 14 **[0.57%] **[0.04%] **[0.03%] **[0.03%] 478871 317574 412241 164262 2285 3048 1140 15 **[0.82%] 797.1 **[0.02%] **[0.03%] 617261 70838 181056 41661 1827 658 1739 16 **[0.48%] 1812.6 **[0.03%] **[0,02%] 521416 283080 236032 107088 1576 721 1372 17 **[0.80%] 788.0 **[0.02%] 2570.1 314340 85493 204695 34590 1446 563 1239 18 **[0.81%] **[0.01%] **[0.01%] **[0,02%] 482334 304661 195106 125472 1987 679 1501 19 **[0.69%] **[0.02%] **[0.02%] **[0.02%] 403289 370957 173920 94781 2124 600 1238 20 **[0.81%] **[0.01%] **[0.02%] **[0.04%] 452740 290555 149267 22281 1764 496 1315 the two-binary formulation with our and Pan and Guan ’s inequalities, CPLEX can solve only 1 instance to optimality within the time limit. This can be attributed to the increase in the size of the LP relaxation at each node during the branching pr ocess, resulting from the addition of a large number of user cuts, which adversely affects the solution time. Fewer nodes are explored by the three strong formulations F2-X, F2-Y , and F2-Z, compared to formulation F2. The number of user cuts added during the solution process by these three strong formulations is smaller compared to the total number of constraints in formulations F2, F2-X, F2-Y , and F2-Z. For nearly all of the unsolved instances, the terminating gaps of the three strong formula- tions are within 0.05%. The integrality gaps of these three strong formulations are also much smaller than those using formulation F2. Furthermore, formulations F1-X and F2-Y have similar performance, suggest- ing that the single-binary formulation with our proposed valid inequalities exhibits comparable results to ec.49 T able EC.18 The Strength of LP Relaxations of MIP Formulations in the Second Experiment by Disabling Smart Features IGap Instance 1 2 3 4 5 6 7 8 9 10 F2 0.37% 0.31% 0.25% 0.20% 0.26% 0.25% 0.25% 0.29% 0.27% 0.36% F2-X 0.20% 0.14% 0.08% 0.06% 0.05% 0.04% 0.07% 0.06% 0.06% 0.05% F2-Y 0.20% 0.14% 0.08% 0.05% 0.05% 0.05% 0.08% 0.06% 0.06% 0.05% F2-Z 0.20% 0.14% 0.08% 0.05% 0.05% 0.04% 0.07% 0.06% 0.06% 0.05% Instance 11 12 13 14 15 16 17 18 19 20 F2 0.23% 0.19% 0.20% 0.20% 0.26% 0.20% 0.22% 0.19% 0.17% 0.22% F2-X 0.05% 0.02% 0.02% 0.03% 0.03% 0.02% 0.03% 0.02% 0.02% 0.04% F2-Y 0.06% 0.03% 0.02% 0.03% 0.03% 0.02% 0.02% 0.02% 0.02% 0.03% F2-Z 0.05% 0.02% 0.02% 0.03% 0.03% 0.02% 0.02% 0.02% 0.02% 0.03% Pct. reduction Instance 1 2 3 4 5 6 7 8 9 10 F2-X 45.5% 54.8% 69.2% 69.6% 80.0% 83.4% 72.1% 78.0% 77.4% 86.9% F2-Y 45.5% 54.7% 67.9% 73.2% 81.8% 80.6% 69.6% 78.8% 78.6% 85.9% F2-Z 45.5% 54.8% 69.4% 73.2% 81.8% 84.1% 72.1% 80.3% 78.6% 86.9% Instance 11 12 13 14 15 16 17 18 19 20 F2-X 78.2% 87.5% 89.3% 86.9% 88.6% 91.4% 88.2% 89.7% 85.6% 83.2% F2-Y 76.2% 84.9% 89.0% 86.0% 89.3% 87.8% 89.3% 91.8% 88.1% 84.3% F2-Z 78.1% 87.7% 89.5% 87.0% 89.4% 91.4% 89.3% 91.8% 88.1% 84.4% T able EC.19 Performance of MIP Formulations in the Third Experiment by Disabling Smart Featur es Instance CPU time [TGap] # nodes # user cuts F1 + F1 + -X F1 + F1 + -X F1 + -X 1 **[1.07%] **[0.31%] 678548 267276 1636 2 **[1.14%] **[0.33%] 812095 204492 1470 3 **[0.85%] **[0.23%] 755989 276532 1500 4 **[1.04%] **[0.45%] 828645 157520 1421 5 **[1.32%] 3398.7 601218 298525 1943 6 **[1.23%] **[0.17%] 511022 324346 1594 7 **[1.42%] **[0.23%] 962694 491768 1712 8 **[1.20%] **[0.37%] 747792 294910 1424 9 **[1.10%] **[0.27%] 880796 215853 1512 10 **[0.91%] **[0.26%] 1046090 285761 1862 11 **[1.11%] **[0.31%] 870860 382853 1457 12 **[1.59%] **[0.26%] 765177 275495 1955 13 **[0.96%] **[0.22%] 902408 206672 1464 14 **[1.09%] **[0.47%] 902171 285107 1594 15 **[1.27%] **[0.35%] 704154 267693 1580 16 **[1.23%] **[0.30%] 850453 101352 1338 17 **[1.17%] **[0.38%] 569100 276125 1412 18 **[0.96%] **[0.27%] 710371 282015 1551 19 **[1.23%] **[0.27%] 639002 274153 1569 20 **[1.20%] **[0.25%] 608163 414109 1420 ec.50 T able EC.20 The Strength of LP Relaxations of MIP Formulations in the Thir d Experiment by Disabling Smart Features IGap Instance 1 2 3 4 5 6 7 8 9 10 F1 + 0.94% 1.10% 0.81% 1.04% 0.86% 0.89% 1.15% 1.08% 1.03% 0.88% F1 + -X 0.45% 0.46% 0.43% 0.63% 0.32% 0.37% 0.46% 0.58% 0.48% 0.46% Instance 11 12 13 14 15 16 17 18 19 20 F1 + 0.97% 1.23% 0.94% 1.07% 1.08% 0.93% 0.96% 0.89% 1.04% 0.90% F1 + -X 0.47% 0.53% 0.44% 0.59% 0.49% 0.42% 0.49% 0.48% 0.48% 0.40% Pct. reduction Instance 1 2 3 4 5 6 7 8 9 10 F1 + -X 52.2% 58.4% 46.6% 39.7% 62.3% 59.1% 59.7% 46.4% 52.8% 47.4% Instance 11 12 13 14 15 16 17 18 19 20 F1 + -X 51.6% 57.1% 53.5% 44.8% 54.5% 55.4% 48.9% 46.3% 54.0% 56.1% the two-binary formulation with strong valid inequalities. In addition, formulation F2-X outperforms both F2-Y and F2-Z, demonstrating that the effectiveness of our proposed valid inequalities in tightening the two-binary formulation is more signiﬁcant than that of valid inequalities from Pan and Guan ( 2016 ). The results for the thir d experiment are presented in T ables EC.19 and EC.20 . Using formulation F1 + , CPLEX cannot solve any of these 20 instances to optimality within the time limit while using formulation F1 + -X, CPLEX can solve 1 instance to optimality . For the instance that cannot be solved to optimality within the time limit, the terminating gaps of formulation F1 + -X are much smaller than those of formulation F1 + . The number of nodes explored by formulation F1 + -X is smaller than that of formulation F1 + . The number of user cuts added in the solution process is also smaller than the total number of constraints in formulations F1 + and F1 + -X. The integrality gaps of formulation F1 + -X are also much smaller than those using formu- lation F1 + , which indicates that our proposed valid inequalities can tighten the single-binary formulation signiﬁcantly .

A Polyhedral Study on Unit Commitment with a Single Type of Binary Variables

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