Discrete Optimization

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1 Discrete Optimization [Chen, Batson, Dang: Applied integer Programming] Chapter 3 and by Johan Högdahl and Victoria Svedberg Seminar 2,

2 Todays presentation Chapter 3 Transforms using 0-1 variables 3.1 Transform logical expressions 3.2 Transform Non-binary to 0-1 variable 3.3 Transform Piecewise Linear Functions 3.4 Transform 0-1 Polynomial Functions 3.5 Transform Functions with Products of 0-1 and Cont. Variables 3.6 Transform Non-simultaneous Constraints Chapter 4 Better formulation by preprocessing 4.1 Better Formulation 4.2 Automatic Problem Preprocessing 4.3 Tightening bounds on Variables

3 3.1 Transform logical expressions It may be natural to formulate expressions as logical expressions Example: If we select project A then we will also select project B The logical relations is Conjuntion (A and B) Disjunction (A or B) Simple implication (if A then B) Double implication (A if and only if B) Negation (not A)

4 3.1 Transform logical expressions Consider the project planning problem. Decision variables: Let y j = 1 if we select project j, else 0. Formulated as logical statements A: project A is selected (y A = 1) or not selected (y A = 0) B: project B is selected (y B = 1) or not selected (y B = 0)

5 3.1 Transform logical expressions Conjunction (A and B) A and B => both A and B are selected Transforms into: y A = 1 and y B = 1 Alternative formulation: y A + y B = 2 Disjunction (A or B) A or B => either A or B are selected or both are selected Transforms into: y A + y B 1

6 3.1 Transform logical expressions Simple implication (if A then B) If A then B => if A are selected then B are selected, else B are either selected or not selected Transforms into: y A y B Double implication (A if and only if B) A if and only if B <=> (if A then B) and (if B then A) which is expressed as y A y B and y B y A Transforms into: y A = y B

7 3.1 Transform logical expressions Negation (not A) Not A reverses the statement A, that is not(y A = 1) => y A = 0. Relation between either/or and if/then statements If A then B <=> not A or B True since when A is true; not A is false so B must be true if the statement shall be true. In the other case when A is false the statement is true if independent of B. Multiple boolean operations on variables

8 3.2 Transform non-binary to 0-1 variable Non-binary variables General integer variables y {0,1,2,...} Discrete variables that takes on non-consecutive integer values. For example y {2,5,9,21} Other cases are easily transformed into one of the to categories above

9 3.2 Transform non-binary to 0-1 variable Transform integer variables Any finite upper bounded integer variable can be expressed by a set of 0-1 variables. Example: x 20 can be expressed as Where y j {0,1} for each j = 0,...,4 Generally we can express an integer x u as Where y j {0,1} for each j = 0,...,k

10 3.2 Transform non-binary to 0-1 variable How many binary variables do we need? The sequence of coefficients is given recursively by (1) With k coefficients u is bounded by Which implies that Taking log 2 of (2) gives (2) We need binary variables to represent a integer x u.

11 3.2 Transform non-binary to 0-1 variable Comments If -b z u and b > 0 then we just et z' = z + b so we have 0 z' u + b. The number of variables grows logarithmic. Transforming general integer variables is useful when There is a small number of variables and each having a low upper bound The proposed 0-1 algorithm is much more efficient than the existing general integer algorithm

12 3.2 Transform non-binary to 0-1 variable Transforming discrete variables If a variable is only allowed to take on one value in a list of integer numbers then we can replace that variable with a set of 0-1 variables. Example If z {1,5,7,9,23} then we may introduce y i = 1 if the i:th element of the list is chosen, else 0. That is z = y 1 + 5y 2 + 7y 3 + 9y y 5 y 1 + y 2 + y 3 + y 4 + y 5 = 1 (this is also called a multiple choice constraint) y 1 {0,1} for each i = 1,...,5

13 Transform piecewise linear functions f (x) = 10x if 0 x 100 (1) f (x) = x if 100 x 300 (2) f (x) = x if 300 x 500 (3)

14 Transform piecewise linear functions Every point x in the linesegment between the consecutive points a i and a i+1 can be described as: x = λ k a k + (1 λ k )a k+1 where 0 λ k 1 (4)

15 Transform piecewise linear functions f (x) is also a linesegment between f (a i ) and f (a i+1 ) it can therefore be described similarly: f (x) = λ k f (a k ) + (1 λ k )f (a k+1 ) where 0 λ k 1 (5)

16 General model: Transform piecewise linear functions x = λ 1a 1 + λ 2a λ r+1a r+1 (6) f (x) = λ 1f (a 1) + λ 2f (a 2) + + λ r+1f (a r+1) (7) λ 1 y 1 (8) λ 2 y 1 + y 2 (9) λ 3 y 2 + y 3 (10). (11) λ r y r 1 + y r (12) λ r+1 y r (13) r y k = 1 (14) k=1 r+1 λ k = 1 (15) k=1 y k 0 k (16) y k {0, 1}, 0 λ k 1 (17)

17 Transform concave piecewise linear functions Due to properties of a concave function, the model can be improved the following way: Each linesegment may be expressed as: The intercepts are formulated as: f (x) = b i + s i x (18) t 0 = 0, at a 0 = 0 (19) t i = t i 1 + s i 1 a i s i a i (20) The model can now be expressed as: (t i y i + s i x i ) (21) constraints i x = x i (22) i a i y i x a i+1 y i i i y i = 1 (23) y i {0, 1} i

18 Transform 0-1 polynomial functions Quadratic binary function f (y 1, y 2,..., y n) = j y 2 j + i k y i y k (24) f (y 1, y 2,..., y n) = j y j + i k y jk (25) 2y jk y j + y k y jk + 1 j k (26)

19 Transform 0-1 polynomial functions Binary function of general degree f (y 1, y 2,..., y n) = y j (27) j {1,...,n} f (y 1, y 2,..., y n) = y Q (28) Q y Q y j y Q + ( Q 1) j Q (29)

20 Transform functions with products of binary and continuous variables Bundle pricing problem Pricing of individual components and bundled components to maximize profit. If there are n components, there are also 2 n 1 bundling possibilities. General problem max s.t. ( ni j y ) ijx j i j y ij = 1 i (30) j (r ij x j )y ij r ij x j But this is nonlinear!

21 Transform functions with products of binary and continuous variables Bundle pricing problem Pricing of individual components and bundled components to maximize profit. If there are n components, there are also 2 n 1 bundling possibilities. General linear problem Replace y ij x j by z i j and add constraints: max s.t. ( i ni j z ) ij j y ij = 1 i j (r ijy ij z ij ) r ij x j z ij x j (31) z ij r ij y ij z ij x j (1 y ij )M j

22 3.6 Transform non-simultaneous constraints In order for a given problem formulation to be classified as a MIP each constraint must be satisfied simultaneous, according to the assumptions that we saw last week. But it may happen that we end up with non-simultaneous constraints when we model a problem.

23 3.6 Transform non-simultaneous constraints Either/or constraints A decision variable may be defined in disjunctive regions, for example out side the interval [a,b]. That is, x a or x b. Such constraints is transformed into x - a My -x + b M(1 - y) y {0,1}

24 3.6 Transform non-simultaneous constraints p out m constraints must hold (a generalization of the previous) If p out of m constraints must hold and we can choose any combination then we can transform that in the following way. Introduce y i = 1 if constraint i must hold, else 0, and write the constraints as f i (x) - b i My i y 1 + y y m = m - p y i {0,1}

25 3.6 Transform non-simultaneous constraints Disjunctive constraint sets (also a generalization) Suppose that either must one subset of constraints hold or else must another subset hold, but not both. That is Either subset 1: { a i^t*x - b i 0, i = 1...,m 1 } Or subset 2: { c i^t*x d i 0, i = 1...,m 2 } We transform it into simultaneous constraints with the 0-1 variable y by writing the constraints as a i^t*x - b i My for each i = 1...,m 1 c i^t*x d i M(1 - y) for each i = 1...,m 2 y {0,1}

26 3.6 Transform non-simultaneous constraints Negation of a constraint The negation of a constraint f(x) - b 0 is f(x) - b > 0 <=> -f(x) + b < 0 If/then constraints Since 3.2 we know that (if A then B) <=> (not A or B) Let A be f 1 (x) - b 1 0 and B be f 2 (x) - b 2 0 then Not A is -f 1 (x) + b 1 < 0 So, if f 1 (x) - b 1 0 then f 2 (x) - b 2 0 is equivalent to -f 1 (x) + b 1 < My f 2 (x) - b 2 M(1 - y) y {0,1}

27 4 Better formulation by preprocessing For every IP-problem there exists many, possibly infinite, alternative formulations and some of these are better then others. Better formulation = easier problem to solve. Some definitions are needed.

28 4.1 Better formulation First an example Consider the following three pure IP constraints IP1: 2y 1 + 2y 2 3 y 1, y 2 integer IP2: 3y 1 + 2y 2 3 y 1, y 2 integer IP3: y 1 + y 2 1 y 1, y 2 integer All three contain the same set of feasible points S = {(0,0),(0,1),(1,0)} The LP relaxations contains the same feasible solutions and are called alternative formulations of the set S

29 4.1 Better formulation Polyhedron Formulation S y is the set of feasible integer solutions. S xy is the set of feasible mixed integer solutions.

30 4.1 Better formulation Comments Any feasible region of a linear program is a polyhedron A polyhedron P is a formulation of S if it contains the exactly same set of feasible solutions as S Example Consider the constraint set

31 4.1 Better formulation Better formulation Ideal formulation

32 4.1 Better formulation Comments Which points do we mean with the extreme points? From definition 8.4: extreme points are those points on a convex set that cannot be represented as a strict (0 < t < 1) convex combination of two points. That means that the ideal formulation is the convex hull of S y. When P is a ideal formulation of S y then the optimal LP solution is the same as the optimal IP solution. Ideal formulation means that the integer program is easy.

33 Automatic problem preprocessing Tightening bounds on variables Fixing variables Eliminating redundant constraints Identifying feasibility Tightening constraints Decomposing the problem into independent subproblems Scaling the coefficient matrix

34 Tightening bounds on continuous variables min s.t. j c jx j j a ijx j b j i l j x j u j j (32) Isolate the variable one wants to investigate (x k ) and separate positive and negative a j. a ik x ik + a ij x j + a ij x j b i i (33) j k:a ij >0 j k:a ij <0 If a ik > 0, then an upper bound is defined to û k = 1 ( b i a ij l j a ik j k:a ij >0 j k:a ij <0 a ij u j ) (34) If a ik < 0, then a lower bound is defined to ˆlk = 1 ( b i a ij l j a ik j k:a ij >0 j k:a ij <0 a ij u j ) (35)

35 Tightening bounds on integer variables If x Z: if a ik > 0, then x k û k if a ik < 0, then x k ˆl k If x {0, 1}: Same method, but one knows that 0 x i 1 i.

36 Variable fixing, redundant constraints and infeasibility max s.t. j c jx j j a ijx j b j i l j x j u j j (36) If a ij > 0 i and c j < 0, fix x j at l j. If a ij < 0 i and c j > 0, fix x j at u j. If l k = u k when applying the bound tightening routine previously described x k can be fixed to l k.

37 Variable fixing, redundant constraints and infeasibility U i = a ij u j + a ij l j (37) j:a ij >0 j:a ij <0 L i = a ij l j + a ij u j (38) j:a ij >0 j:a ij <0 Check L i b i U i If b i U i then the ith constraint is redundant and can be removed. If b i L i then the ith constraint can not be satisfied and no feasible solution exists. If b i = L i then all x j with a ij > 0 can be fixed at x j = l j and all x j with a ij < 0 can be fixed at x j = u j.

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