ip lecture 1

7/29/2019 Ip Lecture 1

1/62

Integer Programming

Wolfram Wiesemann

November 17, 2009
http://goforward/http://find/http://goback/


2/62

Contents of this Lecture

Mixed and Pure Integer Programming Problems

DefinitionsObtaining a Pure Integer Progamming Problem

ExamplesCapital Budgeting

Depot Location

Further Uses of Integer VariablesFinite-Valued VariablesLogical Operations

Solving IP Problems: Cutting Plane AlgorithmOutline: Cutting Plane AlgorithmGomory CutsExample
http://find/


3/62





Depot Location


http://find/


4/62

Definitions

Mixed Integer Programming Problem.

min x0 = cTx

subject to

Ax = b

xj 0 for j N = {1, . . . , n}

xj Z for j Z N.

Note: xj N \ Z are continuous, as before. Pure Integer Programming Problem. Z = N {x0}, i.e.,

all variables (including slack and objective value) are integral.Can be achieved by scaling.
http://find/


5/62





Depot Location


http://find/http://goback/


6/62

Obtaining a Pure Integer Progamming Problem

Consider the following problem:

min x0 = 13

x1 12

x2

subject to

2

3 x1 +

1

3 x2

4

31

2x1

3

2x2

2

3x1, x2 0

x1, x2 Z.

This is not a pure integer programming problem:

x0 not integral.

Slack variables not integral.
http://find/


7/62


Step 1. Scale the equations of the model.

min x0 = 2x1 3x2 (6)

subject to

2x1 + x2 4 (3)

3x1 9x2 4 (6)

x1, x2 0

x1, x2 Z.
http://find/


8/62


Step 2. Insert (integral) slack variables:

min x0 = 2x1 3x2

subject to

2x1 + x2 + x3 = 4

3x1 9x2 + x4 = 4

x1, x2, x3, x4 0

x1, x2, x3, x4 Z.
http://find/


9/62





Depot Location


http://find/


10/62

Capital Budgeting

Company has resources i {1, . . . ,m}. Resource i has limitedavailability bi.

Company can undertake projects j {1, . . . ,n}. Project jrequires aij units of resource i and gives revenues cj.

Which projects should be undertaken such that the resourceavailabilities are observed and the revenues maximised?

maxx

nj=1

cjxj

subject to

nj=1

aijxj bi i {1, . . . ,m}

xj {0, 1} j {1, . . . ,n}
http://find/


11/62





Depot Location


http://find/


12/62

Depot Location (1)

Company has m potential distribution sites i {1, . . . ,m}.Building a distribution centre at site i costs fi.

There are n customers j {1, . . . , n}, each of whose demands

need to be satisfied from one or more distribution centres.Satisfying fraction xij of customer js demand fromdistribution centre i costs cij, given that centre i is built.

Which distribution centres should be built, and how should

the customer demands be satisfied, to minimise costs?
http://find/


13/62

Depot Location (2)

minx,y

mi=1

fiyi +m

i=1

nj=1

cijxij

subject to

mi=1

xij = 1 j {1, . . . , n}

xij yi i {1, . . . , m} , j {1, . . . , n}

xij 0 i {1, . . . , m} , j {1, . . . , n}yi {0, 1} i {1, . . . , m}
http://find/


14/62





Depot LocationFurther Uses of Integer Variables

Finite-Valued VariablesLogical Operations

http://find/


15/62

Finite-Valued Variables

Assume a variable xj can only take a finite number of values:

xj {p1, . . . , pm} .

We can introduce variables z1j , . . . , zmj {0, 1} and add the

constraint

z1j + . . . + zmj = 1.

Now, we can substiute xj with

p1z1j + . . . + pmzmj

in the objective function and all constraints.
http://find/


16/62

Finite-Valued Variables

Example. xj {1, 3, 11} can be modeled as

z1j + z2

j + z3

j = 1

z1j , z2

j , z3

j {0, 1} .

We then substitute xj everywhere by

1z1j + 3z2

j + 11z3

j .

Exercise. Is it possible to save variables here?


17/62





Depot LocationFurther Uses of Integer Variables

Finite-Valued VariablesLogical Operations


O


18/62

Logical Operations

We can model logical operations on the constraints via integervariables. For example, the expression

aTx b dTx e

can be expressed by:

aTx M b

dTx M(1 ) e

{0, 1} ,

where M is a large number.

L i l O i


19/62

Logical Operations

Example. We want to model the following problem:

min x

subject tox [0, 1] x 2.

Solution. This can be expressed as:

min x

subject to

x 1 + M

x 2 M(1 )

x 0.

L i l O i


20/62

Logical Operations

Example. We want to model the following problem:

min x1 x2

subject to

x1 + x2 4

x1 1 x2 1 but not both x1, x2 > 1

x1, x2 0.

L i l O ti
http://find/


21/62

Logical Operations

Solution. This can be expressed as:

min x1 + x2

subject to

x1 + x2 4x1 1 M

x2 1 M(1 )

x1 1 + M(1 )

x2 1 + M

x1, x2 0.

C t t f thi L t
http://find/


22/62


Mixed and Pure Integer Programming ProblemsDefinitionsObtaining a Pure Integer Progamming Problem

ExamplesCapital BudgetingDepot Location



The Big Pict e
http://find/


23/62

The Big Picture

Want to solve a pure IP problem:

Dont 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: xj Z xj R.Tightening: Add cutting planes (cut off current optimum).

The Big Picture


24/62

The Big Picture

A cutting plane algorithm to solve pure integer programmingproblems works as follows.

1. Solve the IP problem with continuous variables instead ofdiscrete ones.

2. If the resulting optimal solution x

is integral, stop optimalsolution found.

3. Generate a cut, i.e., a constraint which is satisfied by allfeasible 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 isintegral and optimal.

Example


25/62

Example


max x0 = 5x1 + 8x2

subject to

x1 + x2 6

5x1 + 9x2 45

x1, x2 0

x1, x2 Z.

Example
http://find/


26/62

Example

Step 1. Solve the IP problem with continuous variables instead ofdiscrete ones.

0

1

2

3

4

5

6

0 1 2 3 4 5 6 7 8

o o o o o o o

o o o o o o

o o o o o

o o o o

o o

o

Example
http://find/


27/62

Example

Step 2. If the resulting optimal solution x is integral, stop

optimal solution found.

0

1

2

3

4

5

6

0 1 2 3 4 5 6 7 8

o o o o o o o

o o o o o o

o o o o o

o o o o

o o

o

x

Resulting solution is x = (2.25, 3.75) and hence not integral.

Example
http://find/


28/62

Example

Step 3. Generate a cut, i.e., a constraint which is satisfied by all

feasible integer solutions but not by x

.

0

1

2

3

4

5

6

0 1 2 3 4 5 6 7 8

o o o o o o o

o o o o o o

o o o o o

o o o o

o o

o

x

We generate cut 2x1 + 3x2 15.

Example
http://find/


29/62

Example

Step 4. Add this new constraint, resolve problem, and go back to(2).

0

1

2

3

4

5

6

0 1 2 3 4 5 6 7 8

o o o o o o o

o o o o o o

o o o o o

o o o o

o o

o

New optimal solution is x = (3, 3).Note: previous x is not feasible anymore.

Example


30/62

Example

Step 2. If the resulting optimal solution x is integral, stop

optimal solution found.

0

1

2

3

4

5

6

0 1 2 3 4 5 6 7 8

o o o o o o o

o o o o o o

o o o o o

o o o o

o o

o

x

x = (3, 3) is integral optimal solution found.

Example
http://find/


31/62

Example

Remark. The cut we introduced only removed non-integral

solutions. Cuts never cut off feasible solutions of the original IPproblem!

0

1

2

3

4

5

6

0 1 2 3 4 5 6 7 8

o o o o o o o

o o o o o o

o o o o o

o o o o

o o

o

http://find/


32/62


Mixed and Pure Integer Programming ProblemsDefinitionsObtaining a Pure Integer Progamming Problem




Ralph E. Gomory (* 1929)
http://find/


33/62

p y ( 9 9)

Outline of an Algorithm for Integer Solutions to Linear ProgramsBulletin of the American Mathematical Society, Vol. 64,

pp. 275-278, 1958.

Gomory Cut
http://find/


34/62

y

Assume x1, . . . , xn 0 and integral. We show how to construct aGomory Cut for

a1x1 + . . . + anxn = b,

where aj, b R (not necessarily integral). Note that this can be

written as

(a1+[a1 a1] f1

)x1+. . .+(an+[an an] fn

)xn = b+[b b] f

,

where = max { Z : } (largest integer smaller than orequal to ).

Gomory Cut
http://find/


35/62

y

We separate fractional and integral terms:

(a1 + f1)x1 + . . . + (an + fn)xn = b + f

f1x1 + . . . + fnxn f = b a1 x1 . . . an xn.

Observations.

1. As xj Z for all feasible x, right-hand side is integral.

2. Thus, for all feasible x, left-hand side must be integral, too.

3. As 0 f< 1, x 0 and left-hand side integral, left-hand sidemust be non-negative.

Consequence. f1x1 + . . . + fnxn f 0 f1x1 + . . . + fnxn ffor every feasible x.

Gomory Cut
http://find/


36/62

y

Suppose Step 1 of our cutting plane algorithm gives non-integralx. Then there is row in Simplex tableau with

xi +

j/I

yijx

j = yi0

with yi0 / Z.Gomory Cut. Setting fj := yij yij, f := yi0 yi0, we get:

j/I

fjxj f. ()

() is fulfilled for all feasible x but not for x:

j/I fjx

j = 0 < f.



37/62

Mixed and Pure Integer Programming ProblemsDefinitions





Example
http://find/


38/62


max 3x1 + 4x2

subject to

25

x1 + x2 3

2

5x1

2

5x2 1

x1, x2 0

x1, x2 Z.

Example


39/62

Step 1. Convert maximisation objective into minimisation.

min x0 = 3x1 4x2

subject to

25

x1 + x2 3

2

5x1

2

5x2 1

x1, x2 0

x1, x2 Z.

Example
http://find/


40/62

Step 1. Scale the equations of the problem.

min x0 = 3x1 4x2

subject to

25

x1 + x2 3 (5)

2

5x1

2

5x2 1 (5)

x1, x2 0

x1, x2 Z.

Example
http://find/


41/62

Step 1. Scale the equations of the problem.

min x0 = 3x1 4x2

subject to

2x1 + 5x2 15

2x1 2x2 5

x1, x2 0

x1, x2 Z.

Example
http://find/


42/62

Step 1. Insert integral slack variables.

min x0 = 3x1 4x2

subject to

2x1 + 5x2 + x3 = 15

2x1 2x2 + x4 = 5

x1, x2, x3, x4 0

x1, x2, x3, x4 Z.

Example
http://find/


43/62

Step 1. Solve continuous relaxation of problem.

BV x1 x2 x3 x4 RHS

x0 3 4 0x3 2 5 1 15x4 2 -2 1 5

Example
http://find/


44/62


BV x1 x2 x3 x4 RHS

x0 3 4 0x3 2 5 1 15x4 2 -2 1 5

Example
http://find/


45/62


BV x1 x2 x3 x4 RHS

x0 3 4 0x3 2 5 1 15x4 2 -2 1 5

Example
http://find/


46/62


BV x1 x2 x3 x4 RHS

x0 3 4 0x3 2 5 1 15

x4 2 -2 1 5

x07

54

5-12

x22

51 1

53

x4

14

5

2

5 1 11

Example
http://find/


47/62


BV x1 x2 x3 x4 RHS

x07

5

4

5 -12x2

2

51 1

53

x414

5

2

51 11

Example
http://find/


48/62


BV x1 x2 x3 x4 RHS

x07

5 4

5 -12x2

2

51 1

53

x414

5

2

51 11

Example


49/62


BV x1 x2 x3 x4 RHS

x07

5

4

5

-12

x22

51 1

53

x414

5

2

51 11

Example


50/62


BV x1 x2 x3 x4 RHS

x07

54

5-12

x22

51 1

53

x4 1452

51 11

x0 -1 1

2-352

x2 11

71

7

10

7

x1 11

7

5

14

55

14

Solution optimal; Simplex stops.

Example


51/62

Step 3. Generate cut based on x1 row.

1

7x3 +

5

14x4

13

14

1

7

(15 2x1 5x2) +5

14

(5 2x1 2x2) 13

14 x1 3

Introduce new variable x5 with

x5 = 1314

+ 17

x3 + 514

x4 = 3 x1.

Add this cut to the problem and go back to Step (1).

Example
http://find/


52/62


BV x1 x2 x3 x4 x5 RHS

x0 -1 1

2-352

x2 1 17 17

107

x1 11

7

5

14

55

14

17

5

14-1 13

14

Example
http://find/


53/62



x0 -1 1

2-352

x2 1 17 17 107

x1 11

7

5

14

55

14

17

5

14-1 13

14

Example


54/62



x0 -1 1

2-352

x2 11

71

7

10

7

x1 11

7

5

14

55

14

17

5

14-1 13

14

x0 4

57

581

5

x2 11

52

5

9

5

x1 1 1 3

x42

51 14

5

13

5

Solution optimal; Simplex stops.

Example
http://find/


55/62


1

5x3 +

3

5x5

4

5

1

5

(15 2x1 5x2) +3

5

(3 x1) 4

5 x1 + x2 4


x6 = 15

x3 + 35

x5 45

= 4 x1 x2


Example
http://find/


56/62


BV x1 x2 x3 x4 x5 x6 RHS

x0 4

57

581

5

x2 11

5 2

5

9

5

x1 1 1 3

x42

51 14

5

13

5

15

3

5-1 4

5

Example
http://find/


57/62


BV x1 x2 x3 x4 x5 x6 RHS

x0 4

57

581

5

x2 11

5

25

9

5

x1 1 1 3

x42

51 14

5

13

5

15

3

5-1 4

5


58/62

Example


59/62


1

3x3 +

1

3x6

1

3

1

3

(15 2x1 5x2) +1

3

(4 x1 x2) 1

3 x1 + 2x2 6


x7 = 13

x3 + 13

x6 13

= 6 x1 2x2


Example
http://goforward/http://find/http://goback/


60/62


BV x1 x2 x3 x4 x5 x6 x7 RHS

x0 1

37

343

3

x2 11

32

3

7

3

x1 1 1

3

5

3

5

3

x44

31 14

3

19

3

x51

31 5

3

4

3

13 13 -1 13

Example
http://find/


61/62



x0 1

37

343

3

x2 11

32

3

7

3

x1 1 1353

53

x44

31 14

3

19

3

x51

31 5

3

4

3

13 13 -1 13

ExampleStep 1. Solve continuous relaxation of problem.
http://find/


62/62


x0 1

3

7

3

43

3

x2 11

3

2

3

7

3

x1 1 1

3

5

3

5

3

x44

31

14

3

19

3

x5

1

31

5

3

4

3

13

1

3-1

1

3

x0 -2 -1 -14

x2 1 -1 1 2

x1 1 2 -1 2

x4 1 -6 4 5

x5 1 -2 1 1

x3 1 1 -3 1
http://find/

ip lecture 1

Documents