linear integer programming and its application - isical.ac.inacmu/ss2017/school17.pdf · 2017/7/9 8...

Linear integer programming and its application

Presented by

Dr. Sasthi C. Ghosh Associate Professor

Advanced Computing & Microelectronics Unit

Indian Statistical Institute

Kolkata, India

Outline

Introduction to Integer programming

Linear relaxation

Rounding in linear relaxation

Converting finite-valued integer variables to binary variable

Implementing logical conditions using binary variable

Converting product of binary variable to linear form

Converting non-integer coefficients to integer coefficients

Dealing with probabilistic constraints

Solving integer programs: Branch and bound

Implementation tool: CPLEX

Conclusion

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Introduction

Pure integer programming

All variables are integer.

Mixed integer programming

Some variables are restricted to be integer but some are not.

Binary integer programming

When variables are restricted to 0 or 1. It could be pure or mixed.

Maximize 8x1 + 11x2 + 6x3 + 4x4

Subject to

5x1 + 7x2 + 4x3 + 3x4 14

xj {0, 1} j = 1, 2, …, 4

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Linear relaxation

Linear relaxation

Can be solved in polynomial time. Polynomial in number of variables and constraints.

Formed by dropping the integrality restrictions.

Maximize 8x1 + 11x2 + 6x3 + 4x4

Subject to

5x1 + 7x2 + 4x3 + 3x4 14

0 xj 1, j = 1, 2, …, 4

It uses 0 xj 1 instead of xj {0, 1}

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Rounding does not give optimal solution

Maximize 8x1 + 11x2 + 6x3 + 4x4

Subject to

5x1 + 7x2 + 4x3 + 3x4 14

xj {0, 1} j = 1, 2, …, 4

Solution of linear relaxation

x1=x2=1, x3=0.5, x4=0 and optimal objective: 22

Rounding x3 to 0 gives, optimal objective: 19

Rounding x3 to 1 does not satisfy the constraint

Optimal integer solution

x1 = 0, x1 = x3 = x4 = 1 and optimal objective: 21

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Rounding may not be possible at all

Maximize 8x1 + 11x2 + 6x3 + 4x4

Subject to

5x1 + 7x2 + 4x3 + 3x4 = 14

xj {0, 1} j = 1, 2, …, 4

Solution of linear relaxation

x1=x2=1, x3=0.5, x4=0 and optimal objective: 22

Rounding x3 to 0 or 1 does not satisfy the constraint

Optimal integer solution

x1 = 0, x2=x3=x4=1 and optimal objective: 21


Finite-valued integer variables

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Converting finite-valued integer variables to binary

Assume a variable xj can only take a finite number of values: xj ∈ {p1, . . . , pm}. Introduce variables z1

j , . . . , zm

j ∈ {0, 1} and add the constraint z1

j + . . . + zmj = 1.

Substitute xj with p1 z

1j + . . . + pm z

mj

in the objective function and all constraints.

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Converting finite-valued integer variables to binary

Maximize 8x1 + 11x2 + 6x3 + 4x4 5x1 + 7x2 + 4x3 + 3x4 <= 14 x1 ∈ {1, 2, 3} xj ∈ {0, 1}, j=2, 3, 4.

Solution: x1 = 2, x2 = 0, x3 = 1, x4 = 0, Objective=22

Example: x1 ∈ {1, 2, 3} can be modelled as z1

1 + z21 + z3

1 = 1 where z11 , z

21 , z

31 ∈ {0, 1}

Substituting x1 everywhere by 1 z11 + 2 z2

1 + 3 z31 gives:

Maximize 8 z1

1 + 16 z21 + 24 z3

1 + 11x2 + 6x3 + 4x4 5 z1

1 + 10 z21 + 15 z3

1 + 7x2 + 4x3 + 3x4 <= 14 z1

1 + z21 + z3

1 = 1

Solution: z1

1 = 0, z21 = 1, z3

1 =0, x2 = 0, x3 = 1, x4 = 0, Objective=22 z1

1 = 0, z21 = 1, z3

1 =0 x1=2

Finite-valued integer variables

Implementing logical conditions

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Implementing logical conditions using binary variable:

At most one of A and B: a + b ≤ 1 At least one of A and B: a + b ≥ 1 If A then B: b ≥ a If A then not B: 1 - b ≥ a a + b ≤ 1 If not A then B: b ≥ 1 – a a + b ≥ 1 If A then B, and if B then A: a = b If A then B and C: b ≥ a and c ≥ a If A then B or C: b + c ≥ a If B or C then A: a ≥ b and a ≥ c If B and C then A: a ≥ b + c – 1

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Converting product of binary variables into linear form without any approximation

c = ab a ≥ c, b ≥ c, and c ≥ a + b -1

If C then A and B: a ≥ c and b ≥ c (ab ≥ c)

If A and B then C: c ≥ a + b – 1 (c ≥ ab)

d = abc a ≥ d, b ≥ d, c ≥ d and d ≥ a + b + c -2


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Implementing “either/or” logical conditions using binary variable:

Maximize x1 + 3x2 subject to either x1 + x2 ≤ 6 (c1) or x1 + 2x2 ≤ 7 (c2) x1 ≥ 0 and x2 ≥ 0

Can be expressed as x1 + x2 - M ≤ 6 and x1 + 2x2 - M(1 - ) ≤ 7 {0, 1} and M is a large number

=0: x1 + x2 ≤ 6, x1 + 2x2 ≤ +∞ c1 is satisfied =1: x1 + x2 ≤ +∞, x1 + 2x2 ≤ 7 c2 is satisfied


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Implementing “either/or” logical conditions using binary variable:

Maximize 2x1 + 3x2 subject to either x1 + 2x2 6 (c1) or 2x1 + x2 9 (c2) x1 ≥ 0 and x2 ≥ 0

Can be expressed as x1 + 2x2 + M 6 and 2x1 + x2 + M(1 - ) 9 {0, 1} and M is a large number

=0: x1 + 2x2 6, 2x1 + x2 -∞ c1 is satisfied =1: x1 + 2x2 -∞, 2x1 + x2 9 c2 is satisfied


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Implementing “either/or but not both” logical conditions using binary variable:

Minimize x1 - x2

subject to x1 + x2 ≤ 4 x1 ≥ 1 ∨ x2 ≥ 1 but not both x1, x2 > 1 x1, x2 ≥ 0. This can be expressed as:

Minimize x1 - x2 subject to x1 + x2 ≤ 4 x1 ≥ 1 −M x2 ≥ 1 −M(1 − ) x1 ≤ 1 +M(1 − ) x2 ≤ 1 +M x1, x2 ≥ 0, ∈ {0, 1} =0: x1 ≥ 1 and x2 ≤ 1 =1: x1 ≤ 1 and x2 ≥ 1


So, either x1 ≥ 1 or x2 ≥ 1 but not both x1, x2 > 1

Converting to integer coefficient

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Converting non-integer coefficients to integer coefficients:

Minimize y = −(1/3)x1 − (1/2)x2 subject to (2/3)x1 + (1/3)x2 ≤ (4/3) (1/2)x1 − (3/2)x2 ≤ (2/3) x1, x2 ≥ 0, x1, x2 ∈ Z. Can be expressed as Minimize y = −2x1 − 3x2 (*6) subject to 2x1 + x2 ≤ 4 (*3) 3x1 − 9x2 ≤ 4 (*6) x1, x2 ≥ 0, x1, x2 ∈ Z.

Solving integer programs

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Solving integer programs (IP)

Branch and bound: dividing the problem into a number of smaller problems.

Cutting plane: adding constraints to force integrality

Relationship to linear relaxation (LR)

Since LR is less constrained than IP, the following is immediate:

For maximization, the optimal objective of LR is greater than or equal to the optimal objective of IP.

For minimization, the optimal objective of LR is less than or equal to the optimal objective of IP.

If LR is infeasible, then so is IP

If LR is optimized by integer values, then that solution is optimal for IP.

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If objective function coefficients are all integers, then

For maximization, the optimal objective for IP is less than or equal to the round down value of optimal objective of LR.

For minimization, the optimal objective for IP is greater than or equal to the round up value of optimal objective of LR.

Solving LR gives the following information:

A bound on the optimal value and if lucky, may give the optimal solution to IP.

Rounding the solution of LR will not in general give the optimal solution of IP. In fact, rounding may not be feasible for many cases.


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Branch and bound:

Solve the LR of the IP. If solution is integer, we are done. Otherwise, create two subproblems by branching on a fractional variable.

Choose an active subproblem and branch on a fractional variable. Repeat untill there are no active subproblems.

A subproblem is not active when any of the following holds:

You used the subproblem to branch on,

All variables in the solution are integer,

The subproblem is infeasible,

You can mark an subproblem not active by a bounding argument.

Branching rule for binary variable:

x=0 and x=1 for fractional value of x

Branching rule for non-binary variable

x 4 and x 4 for say x=4.2

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Branch and bound

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Conclusion

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Integer programming is an effective means to solve many real world problems.

Linear relaxation only gives bounds.

Rounding produces non-optimal results and sometimes may not be possible at all.

Finite-valued integer variable can be converted to binary.

Non-integer coefficients can be converted to integer coefficients.

Many logical conditions can be formulated using integer programming.

Some form of non-linear constraints can be converted to equivalent linear constraints without any approximation while some others can be done so with approximation.

linear integer programming and its application - isical.ac.inacmu/ss2017/school17.pdf · 2017/7/9 8...

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