Mixed integer linear modeling

Transcription

1 Mied integer linear modeling Andrés Ramos Pedro Sánchez Sona Wogrin ESCUELA TÉCNICA SUPERIOR DE INGENIERÍA DEPARTAMENTO DE ORGANIZACIÓN INDUSTRIAL

2 CONTENTS PROBLEM CLASSIFICATION SEVERAL CHARACTERISTIC PROBLEMS FIXED COST PROBLEM LOGICAL PROPOSITIONS MINIMUM, MAXIMUM AND ABSOLUTE VALUE PIECEWISE LINEAR (master) CONVEX AND NONCONVEX REGION (master) SPECIAL ORDERED SETS (master) REFORMULATION (master) 2

3 PROBLEM CLASSIFICATION SEVERAL CHARACTERISTIC PROBLEMS FIXED COST PROBLEM LOGICAL PROPOSITIONS MINIMUM, MAXIMUM AND ABSOLUTE VALUE PIECEWISE LINEAR (master) CONVEX AND NONCONVEX REGION (master) SPECIAL ORDERED SETS (master) REFORMULATION (master) Problem Classification

4 IP Problem classification Linear problems where several or all the variables are integer. A particular case of integer variables are binary variables (0/).. PIP (pure integer programming) all integer 2. BIP (binary integer programming) all binary 3. MIP (mied integer programming) some integer o binary 4

5 Justification of optimization problem with integer variables Investments are discrete variables (generation or transmission epansion planning, singular equipment acquisition, people hiring) Decisions are binary variables (plant or store location) 5

6 Binary representation of discrete variables integer variable y i binary variable (0/) N = i 2 y 0 u i= 0 i 2 N u 2 N+ 6

7 2 PROBLEM CLASSIFICATION SEVERAL CHARACTERISTIC PROBLEMS FIXED COST PROBLEM LOGICAL PROPOSITIONS MINIMUM, MAXIMUM AND ABSOLUTE VALUE PIECEWISE LINEAR (master) CONVEX AND NONCONVEX REGION (master) SPECIAL ORDERED SETS (master) REFORMULATION (master) Several Characteristic Problems

8 Several LP and BIP characteristic problems They have been ehaustively studied. Limited importance in practice, but they may appear as part of other problems. Linear Programming LP Transportation Transshipment Assignment Binary Integer Programming BIP Knapsack Covering Packing Partitioning Traveling salesman 8

9 Transportation problem Minimize the total transportation cost of a certain product from origin to destination, satisfying the destination demand without eceeding the origin offer. a i product offer in origin i m origins b product demand in destination n destinations unitary transportation cost from i to c i a b a b 2 a m m n b n 9

10 Transportation problem formulation min i= = i m n c i i Offer available in each origin i Demand in each destination i 0 units of product transported from i to i, Hypothesis: offer equals demand of the product m n m i i= = If a > b add universal sink with zero cost i i= = m n n a = b If a < b add universal source with very high cost i i= = n = m i= = a i =,, m i i = b =,, n i 0

11 Transportation problem structure 2 n 2 22 m restricciones de oferta 2n m m2 mn n restricciones de demanda If a and b are integer i i are integer because the matri is totally unimodular (i.e., every square submatri has determinant 0, or )

12 Transshipment problem Determine in a network of n nodes the cheapest routes to carry product units from their origins to their destinations through intermediate transshipment locations. Each origin generates b i > 0 units. Each destination consumes b i < 0 units. Each transshipment neither generates nor consumes units b i = 0. c i transportation unit cost from i to in this direction. 2

13 Transshipment problem formulation min i= = i n n c i i Balance or flow conservation in each node i n n = b i =,, n i ki i = k= i, i 0 units of product transported from i to Hypothesis: offer equals demand n i= b i = 0 3

14 Transhipment Eample bi 0 0 Offer Nodes Transhipment Nodes Supply Nodes b i 4

15 Task assignment problem n tasks n persons (machines, etc.) to do them It is particular case of a transportation problem. Minimize the total cost of doing the tasks knowing that each person does task and each task is done by person. c i cost of doing task i by person i if task i is done by person = 0 otherwise Although it is not necessary to declare them as binary variables. 5

16 Task assignment problem formulation min i= = i n n c i Each task i is done by a person n = = i =,, n i Each person does one task n i= i = =,, n i i 0 i, 6

17 Task sequencing in one machine Given several tasks to do, their duration and an estimated problem due date, state the mied integer programming problem to find the sequence that minimizes the mean delay of the tasks, with the following data: Task A B C D Processing time Due date

18 Task sequencing in one machine d Let be the processing time of task and the due date of task. Define the problem variables as i if task is done in step i = 0 otherwise The obective function will minimize the mean delay min pi 4 Subect to these constraints: Each task is done once In each step only one task i i i i = = i r 8

19 Task sequencing in one machine For each step i a task is done and its due date is r On the other hand, task done in this step end at time n Variables iand pi, take into account if the task ends before due date (prompted) or after (delayed), therefore p i, is the delay, that appears in the obective function d + n p = r i k i i i k i n, p 0 0, i i i { } i d k i k 9

20 Knapsack problem n proects Maimize the total value of selecting a set of proects without eceeding the available budget. c cost of proect v value of proect b available budget if proect is done = 0 otherwise 20

21 Knapsack problem formulation ma n = v Limit on the available budget n = c b { 0,} 2

22 Set Covering Problem m characteristics (flights) n set of characteristics (sequences of flights). If a set is selected, then all characteristics of this set should be done. Minimize the total cost of the selected sets in such a way that all characteristics are covered at least once. c cost of selected set Membership matri a i if characteristic i belongs to set = 0 otherwise Decision variables if set is selected = 0 otherwise 22

23 Set Covering Formulation min n = c Each characteristic i should be selected at least once. n = a i =,, m i { } 0, =,, n 23

24 Set Covering Eample: Crew Assignment An airline company needs to assign crews to cover all its flights. Specifically, it requires to solve the set covering problem of three crews whose origin airport is San Francisco for all flights shown in the first column of the table. The rest of columns show 2 feasible flight sequences (sets) for any crew. It is necessary to choose three flight sequences (one for each crew) to cover all flights. It is possible to have more than one crew in the same flight (the etra crew is considered as passengers although the personnel is paid normally). The assignment cost of a crew to a flight sequence is given in thousands of Euros at the last row. The obective is to minimize the total assignment cost of the three crews to cover all flights. 24

25 Feasible Flight Sequences SF - LA SF - Denver SF - Seattle LA - Chicago LA - SF Chicago - Denver Chicago - Seattle Denver - SF Denver - Chicago Seattle - SF Seattle - LA Cost (M )

26 Crew Assignment Formulation min Flight Covering Three crews Assignment 2 = 3 { } = Solution 0, =, if the set is chosen = 0 otherwise 3 = 4 = = = 0 3, 4, cost = 8 M = 5 = 2 = = 0, 5, 2 cost = 8 M 26

27 Set Packing Problem m proects n proect sets. If a set is selected all proects of this set are done. Maimize the total benefit without doing one proect more than once. benefit of selecting the set c a i if the proect i is in the set = 0 otherwise if set is selected = 0 otherwise 27

28 Set Packing Formulation ma n = c Each proect i of all sets cannot be selected more than once n = a i =,, m i 0, =,, n { } 28

29 Set Partitioning Problem Eactly each characteristic (proect) of all sets should be chosen only once n = a = i =,, m i 29

30 Dot Graphs of Covering, Partitioning and Packing Problems COVERING PARTITIONING PACKING 30

31 Traveling Salesman Problem(TSP) Given a list of cities and the distances between each pair of cities, this problem consists of finding what is the shortest possible route that visits each city eactly once and returns to the origin city Formulation : i min i i i, U i i i i = = i { 0,} i i { } Card( U) U,..., n /2 Card( U) n 2 i c i if the path between and is included in the route = 0 otherwise 3

32 Traveling Salesman Problem(TSP) Formulation 2: ik if the path between i and is included in the route at stage k = 0 otherwise min ik, k i, k i, i,, k ik ik ik i = i = = k =, k { 0,} ik ik i r ik c rk + 32

33 3 PROBLEM CLASSIFICATION SEVERAL CHARACTERISTIC PROBLEMS FIXED COST PROBLEM LOGICAL PROPOSITIONS MINIMUM, MAXIMUM AND ABSOLUTE VALUE PIECEWISE LINEAR (master) CONVEX AND NONCONVEX REGION (master) SPECIAL ORDERED SETS (master) REFORMULATION (master) Fied Cost Problem

34 Fied Cost Problem Obective Cost Function: A binary variable y represents the binary decision on an activity realization Mathematical formulation: n { } 0 = 0 ( ) = k + c > 0 M should have the lowest possible value y n ( ) min f ( ) = k y + c = = 0 =,..., n y 0, =,..., n M y f > 0 = 0 = 0 f c k 34

35 Fied Cost Problem: Unit Commitment on Electric Systems Determine thermal generation units should be connected to the electric network each hour of the day (or week) in such a way that: Variable Generation Costs are minimized (including fuel costs and startup/shutdown costs). Demand is supplied each hour A specific level of spinning reserve is given Technical limits are fulfilled (minimum/maimum outputs, ramp up/down) 35

36 Unit Commitment Problem. Data and Variables DATA D h R a t b t ca t cp t P t P t rs t rb t demand during hour h [MW] spinning reserve ratio related with demand [p.u.] linear coefficient of fuel variable cost of unit t[ /MWh] fied coefficient of fuel variable cost of unit t[ /h] startup cost of unit t [ ] shutdown cost of unit t [ ] maimum output of unit t [MW] minimum output of unit t [MW] ramp up of unit t [MW/h] ramp down of unit t [MW/h] VARIABLES P ht A ht AR ht PR ht output of unit t during hour h [MW] commitment of unit t during hour h {0,} start up of unit t during hour h {0,} shut down of unit t during hour h {0,} 36

37 Unit Commitment Problem. Formulation min T H T h= t= t= T t= P ht ( a P + b A + ca AR + cp PR ) = t ht t ht t ht t ht D h ( PA P ) = RD t ht ht h P A P PA t ht ht t ht A A = AR PR ht h t ht ht P P rs ht h t t P P rb h t ht t P 0 A, AR, PR { 0,} ht H H ht ht ht 2HT ( H ) T ( H ) T ( H ) T 37

38 4 PROBLEM CLASSIFICATION SEVERAL CHARACTERISTIC PROBLEMS FIXED COST PROBLEM LOGICAL PROPOSITIONS MINIMUM, MAXIMUM AND ABSOLUTE VALUE PIECEWISE LINEAR (master) CONVEX AND NONCONVEX REGION (master) SPECIAL ORDERED SETS (master) REFORMULATION (master) Logical Propositions

39 Logical Propositions How to model the proposition: if the product A is produced then the product B should be produced too. The condition of the production of product is modelled as. Then this proposition is mathematically written as A B This proposition cannot be included directly (with arrows) into the linear problem. In this eample, one constraint B is included or not depending on the value of variable A (endogenous problem), modifying the structure of the problem. 39

40 Disunctive Propositions (i) A couple of constraints where only one (either of the two) should be met, while the other one is not neccesary. Then, it should meet one constraint but not neccesarily both. f ( ) 0 or g( ) 0 40

41 Eample of Disunctive Propositions (ii) One of these two constraints should be met or Adding M (high value constant) is equivalent to relaing the constraint (for positive variables with positive coefficients) Rela the constraint and fulfill the 2 Rela the constraint 2 and fulfill the M Using an auiliary binary variable one of both is fulfilled and the other one is relaed Mδ M( δ ) M 2 if constraint is relaed δ = 0 if constraint 2 is relaed 4

42 Fulfillment at least k of N constraints At leastk of N (k < N) constraints should be met f (,, ) d n f (,, ) d 2 n 2 f (,, ) d N n N Using k = and N = 2 is the disunctive case Formulation: f (,, ) d + Mδ n f (,, ) d + Mδ 2 n 2 2 f (,, ) d + Mδ N n N N N i= i { } δ 0, i =,..., N i δ = N k 42

43 Selecting one from N values The equation should fulfill one of the possible values Formulation: f (,, ) = d δ N i= δ = i { } n i i i= δ 0, i =,..., N i N f (,, ) n = d d 2 d N 43

44 Simple Propositions (i) Using the previous constraint of the fied cost Mδ M is an upper positive bound of and δ its associated binary variable. δ = M If the constraint is relaed and is met by default If δ = 0then 0 So this constraint allows to model the proposition δ = On the other hand, if > 0then δ =. If 0 the constraint does not imply anything > 0 δ = Both are equivalent propositions because P Q is equivalent to No Q No P δ = 0 0 Mδ > 0 δ =

45 Simple Propositions (ii) Analogously the constraint being m a negative lower bound of and δ the binary variable. δ = m If the constraint does not imply anything as is fulfilled by default. If δ = 0 then 0.. So, this constraint allows to model the proposition δ = 0 0 < 0 δ = 0 δ On the other hand, if then. If the constraint does not imply anything. < 0 = Both propositions are equivalent as P Q is equivalent to No Q No P mδ δ = 0 0 mδ < 0 δ = 45

46 Proposition of constraint (i) The proposition is modeled as a δ = b a b + M( δ) being M a upper bound of the constraint for any value of a b M If δ = the original constraint is formulated and if δ = 0the original constraint is relaed. Analogously to the previous case this constraint models the net proposition a > b δ = 0 46

47 Proposition of constraint (ii) The proposition can be replaced by or also by a b δ = δ = 0 a > b δ = 0 b + a ε Both are equivalent to a b + ε + ( m ε) δ Being m a lower bound of the constraint for any value of a b m 47

48 Proposition of constraint (i) Symmetrically propositions of upper or equal to can be modeled. δ = The proposition a b is equivalent to a b + m( δ ) being m a lower bound of the constraint for any value of. a b m If δ = the original constraint is formulated and if δ = 0 the original constraint is relaed. Analogously to the previous case this constraint models the net proposition a < b δ = 0 48

49 Proposition of constraint (ii) The proposition can be replaced by or also by Both are equivalent to a b δ = δ = 0 a < b δ = 0 a b ε a b ε + ( M + ε) δ Being M a upper bound of the constraint for any value of a b M 49

50 Proposition of = constraint (i) These equality constraints are replaced by constraints of upper than or equal to constraints and lower than and equal to constraints simultaneously. The proposition a b is equivalent to δ = = δ = a b δ = a b a b+ M( δ) The net two constraints model this a b m equality Effectively when δ =both constraints are fulfilled and when δ = 0 both constraints are relaed. + ( δ) 50

51 Proposition of = constraint (ii) The proposition is a combination of the previous cases and besides The resulting formulation: a = b δ = δ = y δ = δ = b δ = b δ = and additional constraint that models the fulfillment of previous constraints δ + δ δ a a a b + ε + ( m ε) δ a b ε + ( M + ε) δ 5

52 Double propositions δ = Double propositions are split into two simple propositions. δ = a b a b is equivalent to a b δ = and the same for other types of double propositions 52

53 Proposition Modeling Tables (i) δ = b a b+ M( δ) a a b δ = a b+ ε + ( m ε) δ δ = b a b+ m( δ) a a b δ = a b ε + ( M + ε) δ δ = = b a b+ M( δ) a a b+ m( δ) a = b δ = a b+ ε + ( m ε) δ a b ε + ( M + ε) δ δ + δ δ 53

54 Double Proposition Modeling Tables (ii) δ = b a b + M( δ ) a a a b + ε + ( m ε) δ δ = b a b + m( δ ) a a b ε + ( M + ε) δ δ = = b a b + M( δ ) a b + m( δ ) a b + ε + ( m ε) δ a b ε + ( M + ε) δ δ + δ δ 54

55 Equivalences between conditional and/or compounded propositions These equivalences are useful to transform implications before converting them into linear constraints P Q not P or Q P (Q and R) (P Q) and (P R) P (Q or R) (P Q) or (P R) (P and Q) R (P R) or (Q R) (P or Q) R (P R) and (Q R) not (P or Q) not (P and Q) not P and no Q not P or not Q 55

56 Proposition equivalence A proposition can be formulated using disunctive constraints f ( ) > 0 g( ) 0 This is equivalent to f ( ) 0 or g( ) 0 f() <= 0 g() <=0 F T T F F F T T T T F T f() > 0 g() <=0 T T T T F F F T T F F T 56

57 Simple Conditional and/or compounded Propositions X i constraint i, δ i binary variable that shows that constraint i is met The first row shows that constraint or 2 (or both) should be met. So, at least one of the two variables δ y δ 2 should be equal to. The linear equation is Besides there must be a constraint that says that if the constraint i is met then δ + δ2 i δ = > 0 δ = i i X o X 2 δ + δ 2 X y X 2 δ =, δ 2 = no X δ = 0 X X 2 δ - δ 2 0 X X 2 δ - δ 2 = 0 This proposition has been modeled for the fied cost problem and its modeling equation was δ = i 57

58 Comple Conditional and/or compounded Propositions Comple Propositions are split into a double proposition to obtain linear constraints directly For eample, ( X A o XB) ( XC o XD o XE) it is modeled as δ + δ δ + δ + δ A B C D E and is transformed into a double proposition δa + δb δ = δc + δd + δe which is equivalent to δ + δ δ = A Net, these equivalences are mathematically formulated B δ = δ + δ + δ C D E 58

59 Equivalence Formulation (Eample) If the product A is produced or the product B (or both) then at least one of the products C, D or E should be produced. X i constraint of production of product i δ i = binary variable to satisfy the constraint i ( X o X ) ( X o X o X ) A B C D E δ + δ δ = A B δ = δ + δ + δ C D E Formulation δ δ δ A B δ δ δ + δ 0 C D E δ + δ 2δ 0 A B δ δ δ + δ 0 C D E 59

60 Alternative Formulation (Eample) equivalent to Formulation: ( X o X ) ( X o X o X ) A B C D E δ δ + δ + δ A C D E δ δ + δ + δ B C D E [ ] [ X ( X o X o X ) y X ( X o X o X ) δ δ = δ + δ + δ A A C D E B C D E C D E δ δ = δ + δ + δ B X i A B C D E i δ δ δ + δ 0 i Mδ δ δ 0 δ δ 0 δ C D E { 0, }, δ { 0,} 60

61 Basket Team Problem A basket coach has 9 players that are ranked from to 3 based on their skills on ball handle, shot, rebound and defense Player Positions Handle Shot Rebound Defense Pivot Guard Pivot, Forward Forward, Guard Pivot, Forward Forward,Guard Pivot, Forward Pivot Forward

62 Basket Team (Logical Constraints) The team should be composed by 5 players that should have the maimum defense value satisfying the following conditions:. At least two players should be able to play as pivot, two as forward and one as guard. Each player only plays in one position. 2. Their average value of ball handle, shot and rebound should be upper or equal to If the player 3 is selected, then the player 6 cannot be selected. 4. If the player is selected, the player 4 or the player 5 should be selected but not both. If the player is not selected, players 4 and 5 may be selected. 5. The player 8 or 9, but not both, should be selected. Formulate a linear problem to optimize the basket team. 62

63 ma p 7a 7 Escuela Técnica Superior i, yi de { 0, Ingeniería } ICAI = y + y + y + 2 3p 5p 7 p 8 y + y + y + y + y + 2 3a 4a 5a 6a 7a 9 + y + y 2 4b 6b = = 0 is equivalent to 0 or y 3 y or alternatively + + = is equivalent to 0 ó ( + y + ) y + ( y ) or alternative ly + ( y ) + = 8 9 y + y = 0 3p 3a 3 y + y = 0 4a 4b 4 y + y = 0 5p 5a 5 y + y = 0 6a 6b 6 + = Two optimal solutions = 2 = 3 = 5 = 8 = = 4 = 6 = 7 = 9 = and the rest of players = 0 Notation equivalence: p:pivot a: Forward (alero) b: Point Guard (base) 63

64 Product of Binary Variables δ δ 2 = 0 δ 0, i δ δ 2 δ i δ δ { } { 0,} 0 { 0,} δ = 0 o δ 2 = 0 Reemplazar δ δ 2 por δ 3 δ 3 = δ = y δ 2 = Reemplazar δ por y δ = 0 y = 0 δ = y = δ + δ 2 δ + δ = δ + δ = 2 2 { } δ, δ 0, i i δ δ 3 δ δ 3 2 δ + δ + δ 2 3 δi y 0 { 0,} y Mδ + y 0 y + Mδ M M 64

65 5 PROBLEM CLASSIFICATION SEVERAL CHARACTERISTIC PROBLEMS FIXED COST PROBLEM LOGICAL PROPOSITIONS MINIMUM, MAXIMUM AND ABSOLUTE VALUE PIECEWISE LINEAR (master) CONVEX AND NONCONVEX REGION (master) SPECIAL ORDERED SETS (master) REFORMULATION (master) Minimum, Maimum and Absolute Value

66 Minimum or maimum of variables minz z= ma(, y) minz z z y maz z= min(, y) maz z z y 66

67 Modeling the absolute value z z + min min( + ) st.. st.. X = + X +, 0 67

68 6 PROBLEM CLASSIFICATION SEVERAL CHARACTERISTIC PROBLEMS FIXED COST PROBLEM LOGICAL PROPOSITIONS MINIMUM, MAXIMUM AND ABSOLUTE VALUE PIECEWISE LINEAR (master) CONVEX AND NONCONVEX REGION (master) SPECIAL ORDERED SETS (master) REFORMULATION (master) Piecewise Linear (Master)

69 Piecewise linear function This function can appear when linearizing a nonlinear function Needed when it is a concave/conve function in a minimization/maimization problem 69

70 Modeling a piecewise linear function Piecewise linear function defined as a set of segments s Point is over the piecewise function g( ) (equality constraint) It is assumed that the abscise of the first segment is the origin b 0 =0 g( ) f s c s b s- b s 70

71 Three possible approaches Approaches. Incremental 2. Multiple selection 3. Conve combination LP relaations of the three approaches are equivalent Any feasible solution of a relaation corresponds to a feasible solution of the other ones with the same cost 7

72 Incremental modeling s Define z as the load or usage of each segment s Total load or value will be =z s Segment s+ has load 0 unless the previous one is full. s If and only if z + s s s > 0 then z = b b Introduce binary variables y s s if z > 0 = 0 otherwise Problem formulation ( ) ( ) ˆ s s s s s s s being f = f + c b f + c b difference in cost in the intersection point of segments s- and s s s ( ˆ s s ) g( ) = c z + f y = s s s + ( ) ( ) y s z b b y z b b y s s s s s s s { } y = S+ 0,, 0 72

73 Multiple selection modeling s Define z as the total load if is in this segment s Total load or value will be =z s s If total load is in a segment, for this segment z = and s for the other ones z = 0 Introduce binary variables y s s if z > 0 = 0 otherwise Problem formulation s s s s ( ) g( ) = c z + f y = s s s s s s s s s s s z b y z b y y y { 0,} 73

74 Conve combination modeling Any point of the segment is a conve combination of s s their etremes with weights µ, λ Problem formulation s s s s ( µ λ ) ( ) ( ) g( ) = µ c b + f + λ c b + f s s s = b + b µ + λ = s s s y s y { } s s s µ, λ 0, y 0, s s s s s s s s 74

75 7 PROBLEM CLASSIFICATION SEVERAL CHARACTERISTIC PROBLEMS FIXED COST PROBLEM LOGICAL PROPOSITIONS MINIMUM, MAXIMUM AND ABSOLUTE VALUE PIECEWISE LINEAR (master) CONVEX AND NONCONVEX REGION (master) SPECIAL ORDERED SETS (master) REFORMULATION (master) Conve and Non-Conve Regions (Master)

76 Maimizing an obective function. Concave region Optimization problem with a concave feasible region z * z z b2 + a2 z b + a z b3 + a3 LP Formulation ma z z b + a z b + a 2 2 z b + a 3 3, z 0 * 76

77 Maimizing an obective function. Nonconcave region (i) Optimization problem with a nonconcave feasible region z * z z b2 + a2 z b3 + a3 z b4 + a4 z b + a * 77

78 Maimizing an obective function. Nonconcave region (ii) LP Formulation ma z z b + a z b + a 2 2 z b + a 3 3 z b + a 4 4, z 0 z z b2 + a2 z b3 + a3 z b + a * 4 4 z z b + a * 78

79 Maimizing an obective function. Nonconcave region (iii) Divide nonconcave feasible region in concave feasible sub-regions and use a binary variable (multiple selection) to choose the feasible sub-region y s if we are in sub-region s = 0 otherwise z z b2 + a2 z b3 + a3 z b + a z b4 + a4 s s + s b s s b b + y s = y s+ = 79

80 Maimizing an obective function. Nonconcave region (iv) If we are in s If we are in s+ ma z z b y + a + b y + a s s s+ s+ 3 3 z b y + a + b y + a = s s s+ s s s s s s s s b y b y s s y s { } s s, z, 0, y 0, s s =, = 0 + y y + ma z z b a s+ 2 2 s = 0 s s s s, z, 0 s z b + a = b b s Sub-region s s s y = 0, y + = 80 ma z z b + a s 3 3 z b + a = Sub-region s+ For every sub-region 4 4 s+ = 0 s+ b b s+ s s+ s+ s+,, 0 z

81 8 PROBLEM CLASSIFICATION SEVERAL CHARACTERISTIC PROBLEMS FIXED COST PROBLEM LOGICAL PROPOSITIONS MINIMUM, MAXIMUM AND ABSOLUTE VALUE PIECEWISE LINEAR (master) CONVEX AND NONCONVEX REGION (master) SPECIAL ORDERED SETS (master) REFORMULATION (master) Special Ordered Sets (Master)

82 SOS and SOS2 SOS: set of variables in which a single variable must be different from 0 SOS2: set of variables in which at most two variables must be different from 0 and must be consecutive Eample: maintenance scheduling of thermal units 82

83 Maintenance scheduling (i) Assumption: Scheduled maintenance of each unit lasts for an integer number of periods. p p p + p + 2 Maintenance scheduling involves inter-period variables and constraints: Decisions taken in period p affect adacent periods. 83

84 Maintenance scheduling (ii) Relevant information to decide the scheduled maintenance: M t Maintenance duration for each thermal unit t: (epressed in number of periods). Must be consecutive 84

85 Maintenance scheduling (iii) Inter-period variables: Unit unavailable due to maintenance: u pt unit t unav ailabl e due to maintenance in period = 0 otherwis e Startup (beginning) and shutdown (end) of the maintenance period: p su sd pt pt maintenance of unit t begins in p = 0 otherwise maintenance of unit t ends in p = 0 other wise 85

86 Maintenance scheduling (iv) Contiguity of maintenance periods: Formulation I: uqt Mtsupt p, t p q< p+ Mt upt up t supt p, t supt t p Eample: maintenance begins in p = 3 and lasts M t = 4 periods: su3 t = = = = = = uqt u3 t u4t u5 t u6t u3 t u4t u5 t u6 t 3 q 6 u su + u =0+0=0 u = 0 2t 2t t 2t u su + u =+0= u 3t 3t 2t 3t 86

87 Maintenance scheduling (v) Contiguity of maintenance periods: Formulation II: supt = sd p+ M t p, t upt up t = supt sd pt p, t ( su ) pt + sd pt 2 t p Eample: maintenance begins in p = 3 and lasts M t = 4 periods su3 t = su sd = 0 sd = 0 sd = 3t 7t 7t 7t u u2t u3 t + su3 t sd3 t = 0 u2t u3 t + 0 = 0 u 2t 3t = 0 = 87

88 Minimum uptime and downtime constraints unit t comitte d in period p upt = 0 otherwise stratup of unit t begins in p supt = 0 otherwise shutdown of unit t begins in p sd pt = 0 otherwise p UT + q p p DT + q p t t su u p, t qt pt su u DT p, t qt pt t 88

89 9 PROBLEM CLASSIFICATION SEVERAL CHARACTERISTIC PROBLEMS FIXED COST PROBLEM LOGICAL PROPOSITIONS MINIMUM, MAXIMUM AND ABSOLUTE VALUE PIECEWISE LINEAR (master) CONVEX AND NONCONVEX REGION (master) SPECIAL ORDERED SETS (master) REFORMULATION (master) Reformulation (Master)

90 Reformulation Most MIP problems can be formulated in different ways In MIP problems, a good formulation is crucial to solve the model Good MIP formulation measure Integrality gap difference between the obective function of the MIP and LP-relaation solutions Given two equivalent MIP formulations, one is stronger (better) than the other, if the feasible region of the linear relaation is strictly contained in the feasible region of the other one. Integrality gap is lower. 90

91 Warehouse location problem (no limits) (i) Choose where to locate warehouses among a set of locations and assign clients to the warehouses minimizing the total cost. No limits means that there are no limit in the number of clients assigned to a warehouse. Data Variables i c warehouse located in y = 0 other wise fraction of demand of client i met from i locations clients localization cost in h cost of satisfying the demand of client i from i 9

92 Warehouse location problem (no limits) (ii) Formulation I minc y + h i i = i y i y i i i { 0, }, [ 0,] Number of constraints: I+IJ i Formulation II minc y + h i y i i i i i = i My { 0, }, [ 0,] Number of constraints: I+J i Both formulations are MIP equivalent. However, formulation I is stronger Intuitively as many constraints the worse. That s true in LP. However, in many MIP problems the more constraints the better. 92

93 Production problem with fied and inventory costs (i) Data Variables t time period c fied cost, p variable cost, h inventory cost d t t t t t y t s t demand to produce = 0 not produce amount produced inventory at the end of the period Formulation I min ( c y + p + h s ) t t t t t 0 t t = s = 0 T t t t t t t s + = d + s t My t s { }, s 0, y 0, t t t Number of constraints: 2T Number of variables: 3T 93

94 Production problem with fied and inventory costs (ii) Variables to produce yt = 0 not produce q quantity produced in period i to met the demand in period t i it Formulation II min t i= T t T ( ) i i i+ t it t t t= i= t= it it t i it q = d t q d y it q t { } 0, y 0, t p + h + h + + h q + c y Number of constraints: T+T 2 /2 Number of variables: T+T 2 /2 Formulation II is better. However, it has greater number of constraints and variables. 94

95 Reformulation criteria It can be interesting increase the number of variables if they can be used in the branching strategy of B&B. For eample, artificial division of a zone in regions N, S, E and W to branch first in these zonal regions. Alternatively, introduce lazy constraints Avoid the use of big M parameters or put tight (lowest upper bound) values for the big M Alternative formulation for the Fied Cost problem using SOS variable (at most one of the variables can be 0) GAMS/CPLEX supports the use of an indicator variable + 0 0, + +=,, 0, 95

96 Some tips for MIP It can be interesting increase the number of variables if they can be used in the branching strategy of B&B. For eample, artificial division of a zone in regions N, S, E and W to branch first in these zonal regions. Alternatively, introduce lazy constraints (only in GAMS/GUROBI) Avoid the use of big M parameters or put tight (lowest upper bound) values for the big M GAMS/CPLEX/GUROBI supports the use of an indicator constraint + 0 0, , Write in the file cple.opt indic constraint$y 0 96

97 Tight and compact unit commitment G. Gentile, G. Morales-España and A. Ramos A Tight MIP Formulation of the Unit Commitment Problem with Start-up and Shut-down Constraints EURO Journal on Computational Optimization 5 (), March /s y G. Morales-España, C.M. Correa-Posada, A. Ramos Tight and Compact MIP Formulation of Configuration-Based Combined-Cycle Units IEEE Transactions on Power Systems 3 (2), , March /TPWRS G. Morales-España, J.M. Latorre, and A. Ramos Tight and Compact MILP Formulation for the Thermal Unit Commitment Problem IEEE Transactions on Power Systems 28 (4): , Nov /TPWRS G. Morales-España, J.M. Latorre, and A. Ramos Tight and Compact MILP Formulation of Start-Up and Shut-Down Ramping in Unit Commitment IEEE Transactions on Power Systems 28 (2): , May /TPWRS

98 Andrés Ramos Pedro Sánchez Sona Wogrin Alberto Aguilera, Madrid, Spain Tel Fa info-doi@doi.icai.upcomillas.es