Integer Programming. Wolfram Wiesemann. November 17, 2009

Transcription

1 Integer Programming Wolfram Wiesemann November 7, 2009

2 Contents of this Lecture Mixed and Pure Integer Programming Problems Definitions Obtaining a Pure Integer Progamming Problem Examples Capital Budgeting Depot Location Further Uses of Integer Variables Finite-Valued Variables Logical Operations Solving IP Problems: Cutting Plane Algorithm Outline: Cutting Plane Algorithm Gomory Cuts Example

4 Definitions Mixed Integer Programming Problem. subject to Ax = b x j 0 min x 0 = c T x for j N = {,...,n} x j Z for j Z N. Note: x j N \ Z are continuous, as before. Pure Integer Programming Problem. Z = N {x 0 }, i.e., all variables (including slack and objective value) are integral. Can be achieved by scaling.

6 Obtaining a Pure Integer Progamming Problem Consider the following problem: subject to min x 0 = x 2 x 2 2 x + x x 2 x 2 2 x,x 2 0 x,x 2 Z. This is not a pure integer programming problem: x 0 not integral. Slack variables not integral.

7 Obtaining a Pure Integer Progamming Problem Step. Scale the equations of the model. min x 0 = 2x x 2 ( 6) subject to 2x + x 2 4 x 9x 2 4 x,x 2 0 x,x 2 Z. ( ) ( 6)

8 Obtaining a Pure Integer Progamming Problem Step 2. Insert (integral) slack variables: min x 0 = 2x x 2 subject to 2x + x 2 + x = 4 x 9x 2 + x 4 = 4 x,x 2,x,x 4 0 x,x 2,x,x 4 Z.

10 Capital Budgeting Company has resources i {,...,m}. Resource i has limited availability b i. Company can undertake projects j {,...,n}. Project j requires a ij units of resource i and gives revenues c j. Which projects should be undertaken such that the resource availabilities are observed and the revenues maximised? subject to max x n c j x j j= n a ij x j b i j= x j {0,} i {,...,m} j {,...,n}

12 Depot Location () Company has m potential distribution sites i {,...,m}. Building a distribution centre at site i costs f i. There are n customers j {,...,n}, each of whose demands need to be satisfied from one or more distribution centres. Satisfying fraction x ij of customer j s demand from distribution centre i costs c ij, given that centre i is built. Which distribution centres should be built, and how should the customer demand s be satisfied, to minimise costs?

13 Depot Location (2) subject to min x,y m f i y i + i= m i= j= n c ij x ij m x ij = i= x ij y i x ij 0 y i {0,} j {,...,n} i {,...,m}, j {,...,n} i {,...,m}, j {,...,n} i {,...,m}

15 Finite-Valued Variables Assume a variable x j can only take a finite number of values: x j {p,...,p m }. We can introduce variables zj,...,zm j constraint {0,} and add the Now, we can substiute x j with z j z m j =. p z j p m z m j in the objective function and all constraints.

16 Finite-Valued Variables Example. x j {,,} can be modeled as zj + zj 2 + zj = zj,zj 2,zj {0,}. We then substitute x j everywhere by zj + zj 2 + zj. Exercise. Is it possible to save variables here?

18 Logical Operations We can model logical operations on the constraints via integer variables. For example, the expression can be expressed by: where M is a large number. a T x b d T x e a T x Mδ b d T x M( δ) e δ {0,},

19 Logical Operations Example. We want to model the following problem: min x subject to x [0,] x 2. Solution. This can be expressed as: min x subject to x + Mδ x 2 M( δ) x 0.

20 Logical Operations Example. We want to model the following problem: min x x 2 subject to x + x 2 4 x x 2 but not both x,x 2 > x,x 2 0.

21 Logical Operations Solution. This can be expressed as: min x + x 2 subject to x + x 2 4 x Mδ x 2 M( δ) x + M( δ) x 2 + Mδ x,x 2 0.

23 The Big Picture Want to solve a pure IP problem: Don t know how to solve IP problems. Know how to solve continuous problems (Simplex). Outline of a solution procedure: Solve a continuous relaxation. Contains all originally feasible solutions, plus others. If optimal solution is integral, we are done. Otherwise, tighten the relaxation and repeat. Continuous relaxation: x j Z x j R. Tightening: Add cutting planes (cut off current optimum).

24 The Big Picture A cutting plane algorithm to solve pure integer programming problems works as follows.. Solve the IP problem with continuous variables instead of discrete ones. 2. If the resulting optimal solution x is integral, stop optimal solution found.. Generate a cut, i.e., a constraint which is satisfied by all feasible integer solutions but not by x. 4. Add this new constraint, resolve problem, and go back to (2). Terminates after finite number of iterations in (2). Resulting x is integral and optimal.

25 Consider the following problem: max x 0 = x + 8x 2 subject to x + x 2 6 x + 9x 2 4 x,x 2 0 x,x 2 Z.

26 Step. Solve the IP problem with continuous variables instead of discrete ones. 6 o 4 o o o o o o 2 o o o o o o o o o o o 0 o o o o o o o

27 Step 2. If the resulting optimal solution x is integral, stop optimal solution found. 6 o 4 o o x o o o o 2 o o o o o o o o o o o 0 o o o o o o o Resulting solution is x = (2.2,.7) and hence not integral.

28 Step. Generate a cut, i.e., a constraint which is satisfied by all feasible integer solutions but not by x. 6 o 4 o o x o o o o 2 o o o o o o o o o o o 0 o o o o o o o We generate cut 2x + x 2.

29 Step 4. Add this new constraint, resolve problem, and go back to (2). 6 o 4 o o o o o o 2 o o o o o o o o o o o 0 o o o o o o o New optimal solution is x = (,). Note: previous x is not feasible anymore.

30 Step 2. If the resulting optimal solution x is integral, stop optimal solution found. 6 o 4 o o o o o o x 2 o o o o o o o o o o o 0 o o o o o o o x = (,) is integral optimal solution found.

31 Remark. The cut we introduced only removed non-integral solutions. Cuts never cut off feasible solutions of the original IP problem! 6 o 4 o o o o o o 2 o o o o o o o o o o o 0 o o o o o o o

33 Ralph E. Gomory (* 929) Outline of an Algorithm for Integer Solutions to Linear Programs Bulletin of the American Mathematical Society, Vol. 64, pp , 98.

34 Gomory Cut Assume x,...,x n 0 and integral. We show how to construct a Gomory Cut for a x a n x n = b, where a j,b R (not necessarily integral). Note that this can be written as ( a +[a a ] )x }{{} +...+( a n +[a n a n ] )x }{{} n = b +[b b ], }{{} f f where β = max {α Z : α β} (largest integer smaller than or equal to β). f n

35 Gomory Cut We separate fractional and integral terms: ( a + f )x ( a n + f n )x n = b + f f x f n x n f = b a x... a n x n. Observations.. As x j Z for all feasible x, right-hand side is integral. 2. Thus, for all feasible x, left-hand side must be integral, too.. As 0 f <, x 0 and left-hand side integral, left-hand side must be non-negative. Consequence. f x f n x n f 0 f x f n x n f for every feasible x.

36 Gomory Cut Suppose Step of our cutting plane algorithm gives non-integral x. Then there is row in Simplex tableau with x i + j / I y ij x j = y i0 with y i0 / Z. Gomory Cut. Setting f j := y ij y ij, f := y i0 y i0, we get: f j x j f. j / I ( ) is fulfilled for all feasible x but not for x : j / I f jx j = 0 < f. ( )

38 Consider the following problem: subject to max x + 4x 2 2 x + x 2 2 x 2 x 2 x,x 2 0 x,x 2 Z.

39 Step. Convert maximisation objective into minimisation. subject to min x 0 = x 4x 2 2 x + x 2 2 x 2 x 2 x,x 2 0 x,x 2 Z.

40 Step. Scale the equations of the problem. subject to min x 0 = x 4x 2 2 x + x 2 ( ) 2 x 2 x 2 ( ) x,x 2 0 x,x 2 Z.

41 Step. Scale the equations of the problem. min x 0 = x 4x 2 subject to 2x + x 2 2x 2x 2 x,x 2 0 x,x 2 Z.

42 Step. Insert integral slack variables. min x 0 = x 4x 2 subject to 2x + x 2 + x = 2x 2x 2 + x 4 = x,x 2,x,x 4 0 x,x 2,x,x 4 Z.

43 Step. Solve continuous relaxation of problem. BV x x 2 x x 4 RHS x x 2 x 4 2-2

46 Step. Solve continuous relaxation of problem. BV x x 2 x x 4 RHS x x 2 x x x 2 2 x 4 4 2

47 Step. Solve continuous relaxation of problem. BV x x 2 x x 4 RHS x x 2 2 x 4 4 2

50 Step. Solve continuous relaxation of problem. BV x x 2 x x 4 RHS x x 2 2 x x x x Solution optimal; Simplex stops.

51 Step. Generate cut based on x row. 7 x + 4 x ( 2x x 2 ) + 4 ( 2x 2x 2 ) 4 x Introduce new variable x with x = x + 4 x 4 = x. Add this cut to the problem and go back to Step ().

52 Step. Solve continuous relaxation of problem. BV x x 2 x x 4 x RHS x x x 7 ζ

53 Step. Solve continuous relaxation of problem. BV x x 2 x x 4 x RHS x x x ζ 7 4-4

54 Step. Solve continuous relaxation of problem. BV x x 2 x x 4 x RHS x x x 7 ζ x x 2 2 x 9 2 x 4 4 Solution optimal; Simplex stops.

55 Step. Generate cut based on x 2 row. x + x 4 ( 2x x 2 ) + ( x ) 4 x + x 2 4 Introduce new variable x 6 with x 6 = x + x 4 = 4 x x 2 Add this cut to the problem and go back to Step ().

56 Step. Solve continuous relaxation of problem. BV x x 2 x x 4 x x 6 RHS x x 2 2 x 2 x 4 4 ζ 9 4 -

57 Step. Solve continuous relaxation of problem. BV x x 2 x x 4 x x 6 RHS x x 2 2 x 9 2 x 4 4 ζ 4 -

58 Step. Solve continuous relaxation of problem. BV x x 2 x x 4 x x 6 RHS x x 2 2 x 2 x 4 4 ζ x x 2 2 x 4 x 4 4 x Solution optimal; Simplex stops

59 Step. Generate cut based on x 2 row. x + x 6 ( 2x x 2 ) + (4 x x 2 ) x + 2x 2 6 Introduce new variable x 7 with x 7 = x + x 6 = 6 x 2x 2 Add this cut to the problem and go back to Step ().

60 Step. Solve continuous relaxation of problem. BV x x 2 x x 4 x x 6 x 7 RHS x x 2 2 x 4 x 4 4 x ζ

61 Step. Solve continuous relaxation of problem. BV x x 2 x x 4 x x 6 x 7 RHS x x 2 2 x 4 x 4 4 x ζ -

62 Step. Solve continuous relaxation of problem. BV x x 2 x x 4 x x 6 x 7 RHS x x 2 x 2 x x ζ x x 2-2 x 2-2 x x -2 x -