Integer Linear Programming. Introduction Integer L P problem (P) F(P) = feasible domain of P...

Integer Linear Programming

Introduction

• Integer L P problem

(P)

• F(P) = feasible domain of P

• Exemple

1

1

Min

s. t. 1, ,

0, integer 1, ,

n

j jj

n

ij j ij

j

c x

a x b i m

x j n

1 2

1 2

1

1 2

Min 5

s. t. 10 20

2

, 0, integer

z x x

x x

x

x x

Introduction


(P)


• Exemple

2010 21 xx

21 x

( ) 0,0 , 1,0 , 2,0 , 0,1 , 1,1 , 2,1 , 0,2F P

1

1

Min

s. t. 1, ,

0, integer 1, ,

n

j jj

n

ij j ij

j

c x

a x b i m

x j n

1 2

1 2

1

1 2

Min 5

s. t. 10 20

2

, 0, integer

z x x

x x

x

x x

Introduction


(P)


• denote the relaxtion of (P) where the integrality constraints are relaxed

• Exemple

1 2

1 2

1

1 2

Min 5

s.t. 10 20

2

, 0, integer

z x x

x x

x

x x

P

2010 21 xx

21 x

P1

1

Min

s. t. 1, ,

0, integer 1, ,

n

j jj

n

ij j ij

j

c x

a x b i m

x j n

Introduction


(P)


• denote the relaxtion of (P) where the integrality constraints are relaxed

• Exemple

1 2

1 2

1

1 2

Min 5

s.t. 10 20

2

, 0,entiers

z x x

x x

x

x x

P

2010 21 xx

21 x

PF

P 1

1

Min

s. t. 1, ,

0, integer 1, ,

n

j jj

n

ij j ij

j

c x

a x b i m

x j n

Introduction


(P)

• How to solve the problem? Why dont we solve the relaxed

problem, and take the rounded solution?

• Exemple

Solution of the relaxed problem:

(2, 9/5) et z = –11

Rounded solution: (2, 1) et z = –7

1 2

1 2

1

1 2

Min 5

s.t. 10 20

2

, 0,entiers

z x x

x x

x

x x

2010 21 xx

21 x

PF

But (0, 2) is feasible with z = –10

1

1

Min

s. t. 1, ,

0, integer 1, ,

n

j jj

n

ij j ij

j

c x

a x b i m

x j n

Solution methods

• Foundamental principle

Generate a set of linear constraints

to be added to (P)

• Exemple

entiers,0,

2

2010 àSujet

5Min

21

1

21

21

xx

x

xx

xxz

2010 21 xx

21 x

PF

42 21 xx

Méthodes de résolution

• Foundamental principle

Generate a set of linear constraints

to be added to (P) in order to

generate a new problem (PR) such that

Furthermore, when solving the problem , the variables take integer values, and hence it is optimal for (P).

• Exemple

2010 21 xx

21 x

PF

42 21 xx

1 2

1 2

1

1 2

1 2

Min 5

s.t. 10 20

2

2 4

, 0, integer

z x x

x x

x

x x

x x

F PR F P

F PR F P

PR

Cutting plane algorithm

• General statement of the cutting plane algorithm

Introduce new linear constraints to the problem in order to reduce the feasible domain of the relaxed problem but keep all the feasible points of the problem.

• In this procedure, we solve a sequence of relaxed problems until an integer optimal solution is reached

• Any problem of the sequence is obtained by adding an additional linear constraint (a cut) to the current problem.

n

j 1

1

Consider the following integer linear programming problem:

( ) Min

s. t. 1, ,

0, integer, 1, ,

j j

n

ij j ij

j

P c x

a x b i m

x j n

How to generate a Gomory cut.

Let be an optimal basis of ( ), and let be the basic

variable in the row of the optimal tableau that is not integer.k

th

B P x

i

1

1

1 2

11 12 1 1 1

1 2

1 2

basic r.h.s

var.

0 1 0 0

1 0 0 0

0

m

m

k j j j n

j j n

k i i ij in i

j m m

x x x x x x x z

x t t t t b

x t t t t b

x t t

1 2

0 1 0

0 0 0 1

mj mn m

j n

t t b

z c c c c z

The optimal tableau is as follows:

The cooresponding row of the tableau is as follows:

1

where : is an index of a non basic variable

and is not integer.

ik ij jj J

i

x t x b

J j j

b

and consequently. (2)ik ij j k ij j

j J j J

x t x x t x b


1


and is not integer.

ik ij jj J

i

x t x b

J j j

b

Let the largest integer (floor value) .d d

Since 0 , thenj

ij j ij jj J j J

x jt x t x

If we account for the fact that the variables must take

integer values, it follows from (2) that

. (3)

Hence all solutions of ( ) satisfy (3).

j

ik ij jj J

x

x t x b

P

Now consider the relation obtained by substracting

(1) from (3):

4

Note that

0 et 0

i

i

iij ij jj J

iij ij

t t x b b

t t b b

Since any solution of ( ) satisfies (1) and (3), then it also satifies (4),

and we can introduce this constraint in ( ) without eliminating any solution of ( ).

P

P P


1


and is not integer.

ik ij jj J

i

x t x b

J j j

b

Let the largest integer (floor value) .d d

Since 0 , thenj

ij j ij jj J j J

x jt x t x

and consequently. (2)ik ij j k ij j

j J j J

x t x x t x b

But the current optimal solution of the relaxed problem ( ) where

0 does not satisfies constraint (4), and thus introducing (4)

reduces the size of the feasible domain of the relaxed problem ( ).

j

P

x j J

P

Now consider the relation obtained by substracting

(1) from (3):

4

Note that

0 et 0

i

i

iij ij jj J

iij ij

t t x b b

t t b b

Since any solution of ( ) satisfies (1) and (3), then it also satifies (4),

and we can introduce this constraint in ( ) without eliminating any solution of ( ).

P

P P

To continue the solution process, we introduce the constraint

where is a slack variable having a cost equal to 0, in the last simplex

tableau

i ii iij ij j ij ij jj J j J

t t x b b t t x x b b

x

to generate a basic solution for the new problem by looking at

as the basic variable in the new row of the tableau.x

This basic solution is not feasible since 0.

Then we continue the solution process using the dual simplex method.

i ix b b

1

1

1 2

11 12 1 1 1

1 2

1

basic r.h.s.

var.

0 1 0 0 0

1 0 0 0 0

m

m

k j j j n

j j n

k i i ij in i

j m

x x x x x x x x z

x t t t t b

x t t t t b

x t t

2

1 1 2 2

1 2

0 0 1 0 0

0 0 0 1 0

0 0 0 0 1

i

m mj mn m

ii i i i ij ij in in

j n

t t b

x t t t t t t t t b b

z c c c c z

To continue the solution process, we introduce the constraint

where is a slack variable having a cost equal to 0, in the last simplex

tableau

i ii iij ij j ij ij jj J j J


x

to generate a basic solution for the new problem by looking at

as the basic variable in the new row of the tableau.x

This basic solution is not feasible since 0.

Then we continue the solution process using the dual simplex method.

i ix b b

Remarks:

1) If (i.e., is integer) , et si is not integer, then

1

indicates that ( ) is not feasible since the left hand side take an integer

value for all f

iij ij ij

ik ij jj J

t t t j J b

x t x b

P

easible solution of ( ) while the right hand element

is not integer.

2) The same process is applied at each iteration.

iP b

1 2

1 2 3

1 2 3

Consider the following exemple

Min 21 11

s.t. 7 4 13

, , 0, integer

x x

x x x

x x x

1 2 3

2 3 4

Iteration 1:Basic solution of ( )

4 1 13

7 7 7optimal value 39

New constraint:4 4 1 1 13 13

7 7 7 7 7 7

P

x x x

x x x

2 3 2 3

3 1 2

2 1 2 1

Graphic interpretation:

4 1 64 6.

7 7 7Mais 13 7 4 . Then

4 13 7 4 6 1

x x x x

x x x

x x x x

i iij ij j i ij ij j ij J j J


1 2

1 2 3

1 2 3

Min 21 11s.t. 7 4 13

, , 0, integer

x xx x x

x x x

1 2

1 2 3

2 3 4

1 2 3 4

Iteration 2:

Solve the relaxed problem

Min 21 11

s.t. 7 4 13

4 1 6

7 7 7 , , , 0, integer

x x

x x x

x x x

x x x x

1 4

2 3 4

The result:

1

1 7 3

4 4 21

optimal value 372

x x

x x x

nd2 3 4

3 4 5

3 4 5

1 7 3New constraint specified from the 2 row :

4 4 21 1 7 7 3 3

4 4 4 4 2 21 1 1

4 4 2

x x x

x x x

x x x

3 4

4

2 3 4

3 2 3

3

3 1 2

1 2 2

Graphic interpretation2

Replace by its value obtainedfrom the last constraint added

4 1 6,

7 7 7to obtain

4 1 62.

7 7 7replace

13 7 4to obtain

8 32 4 613 8 2

7 7 7 7

8

x xx

x x x

x x x

xx x x

x x x

1 2 1 28 6

4 2 13 2 37 7

x x x x

iij ij j ij J

iij ij j ij J

t t x b b

t t x x b b

1 2

1 2 3

2 3 4

1 2 3 4

Min 21 11Sujet à 7 4 13

4 1 6

7 7 7, , , 0, entiers

x xx x x

x x x

x x x x

1 3 5

1 2 5

1 4

We obtain

1

2 3

1

x x x

x x x

x x

1 2

1 4

2 3 4

3 4 5

1 2 3 4 5

Iteration 3:

Solve the problem

Min 21 11

s.t. 1

1 7 3

4 4 21 1 1

4 4 2, , , , 0, integer

x x

x x

x x x

x x x

x x x x x

1 2 3

Then the optimal solution is integer:

0, 3, 1

optimal value 33

x x x

Reference:

A. Schrijver, Theory of Linear and Integer Programming,

Wiley & Sons, 1986, 354 - 357

Convergence of Gomory cutting plane algorithm

Under some assumptions for selecting the row of the

tableau to specify the next cut, this author shows that:

"... the cutting plane method terminates"

Branch & Bound method

• Here also we solve a sequence of relaxed problems.

• First we solve . If the optimal solution is integer, then this solution is optimal for the original problem (P).

• Otherwise we use a variable which is not integer.

• Consider two new constraints

P x

having value i ix x

(floor of )

and

(ceeling of )

ii i

ii i

x x x

x x x

1

1

Min

( ) s.t. 1, ,

0, integer 1, ,

n

j jj

n

ij j ij

j

c x

P a x b i m

x j n

1

1

Min

( ) s.t. 1, ,

0, 1, ,

n

j jj

n

ij j ij

j

c x

P a x b i m

x j n

66611 21 xx

4510 21 xx1 2

1 2

1 2

1 2

Min 5

s.t. 11 6 66

10 45

, 0, integer

z x x

x x

x x

x x

1 2

opt. sol. of the relaxed problem:

3.75, 4.125

24.375

x x

z

1

1

New constraints to consider:

3.75 3

and

3.75 4

x

x

With these two new constraints - feasible point of are maintained - a slice of the feasible domain of the relaxed problem is eliminated.

P

66611 21 xx

4510 21 xx1 2

1 2

1 2

1 2

Min 5

s.t. 11 6 66

10 45

, 0, integer

z x x

x x

x x

x x

1 2

Opt. sol. of the relaxed problem:

3.75, 4.125

24.375

x x

z

1

1


3.75 3

and

3.75 4

x

x

Slice of

eliminated

F P

66611 21 xx

4510 21 xx1 2

1 2

1 2

1 2

Min 5

s.t. 11 6 66

10 45

, 0, integer

z x x

x x

x x

x x

2PF 3PF

2

3

But what is left of nis not connected since two subsets

and are left, and their intersection is empty.

F P F P

F P

1 2

Opt. sol. of the relaxed problem:

3.75, 4.125

24.375

x x

z

1

1


3.75 3

and

3.75 4

x

x

Slice of

eliminated

F P

66611 21 xx

4510 21 xx

2PF 3PF

2 1 2

1 2

1 2

1

1 2

Min 5

s.t. 11 6 66

10 45

4

, 0, integer

P z x x

x x

x x

x

x x

3 1 2

1 2

1 2

1

1 2

Min 5

s.t. 11 6 66

10 45

3

, 0, integer

P z x x

x x

x x

x

x x

22

33

Continue the process by defining the problem associated with

and associated with .

P F P

P F P

2 3The best optimal solution of and is the optimal solution of .

P PP

1 2

1 2

1 2

1 2

Min 5

s.t. 11 6 66

10 45

, 0, integer

z x x

x x

x x

x x

1

1


3.75 3

and

3.75 4

x

x

• Next iteration

Select one of the two problems (P2) or (P3 )

Solve this selected problem as we did for (P).

• In our exemple, we select the problem (P3)

2

2

New constraints to considered:

4.2 4

and

4.2 5

x

x

66611 21 xx

4510 21 xx

2PF 3PF

3 1 2

1 2

1 2

1

1 2

Min 5

s.t. 11 6 66

10 45

3

, 0, integer

P z x x

x x

x x

x

x x

3

1 2

Opt. sol. of the relaxed prob. :

3, 4.2

24

P

x x

z

3Slice of

eliminated

F P

3 1 2

1 2

1 2

1

1 2

Min 5

s.t. 11 6 66

10 45

3

, 0, integer

P z x x

x x

x x

x

x x

66611 21 xx

4510 21 xx

2PF 5F P

5 1 2

1 2

1 2

1

2

1 2

Min 5

s.t. 11 6 66

10 45

3

4

, 0, integer

P z x x

x x

x x

x

x

x x

4 1 2

1 2

1 2

1

2

1 2

Min 5 s.t. 11 6 66

10 4535

, 0, integer

P z x xx x

x xxx

x x

2

2

New constraints to considered:

4.2 4

and

4.2 5

x

x

• Next iteration

Select a problem in the set which has not been solved yet.



5 4 2, ,P P P

66611 21 xx

4510 21 xx

2PF 5PF

5 1 2

1 2

1 2

1

2

1 2

Min 5

s.t. 11 6 66

10 45

3

4

, 0, integer

P z x x

x x

x x

x

x

x x

5

1 2

Opt. sol. of the relaxed problem :

3, 4

23

P

x x

z

Since the optimal solution of the relaxed sub problem is integer, then itis a feasible solution of (P).Do not generate new sub problems since we have identified the best feasible solution in this part of the feasible domain of (P). During this solution process, we keep the best feasible integer solution found so far,And its value becomes an upper bound BS for the optimal value in (P).

• Next iteration




4 2,P P

66611 21 xx

4510 21 xx

2PF

4 1 2

1 2

1 2

1

2

1 2

Min 5

s.t. 11 6 66

10 45

3

5

, 0, integer

P z x x

x x

x x

x

x

x x

4

4

Problem is not feasible

P

F P

Stop looking for feasible solutions in the part since it is empty.

• Next iteration




2P

2 1 2

1 2

1 2

1

1 2

Min 5

s.t. 11 6 66

10 45

4

, 0, integer

P z x x

x x

x x

x

x x

66611 21 xx

4510 21 xx

2PF 2

1 2

Opt. sol. of the relaxed problem :

4, 3.667

22.333

P

x x

z

The optimal solution of the relaxed problem a is not integer but its value z = –22.333 > BS = – 23Stop looking for an integer solution in the part of the feasible domain of (P) since it is not possible to find one having a value smaller than the BS = – 23.

• The procedure stops when all the relaxed problems have been solved.

• The integer solution having its value equal to BS in an optimal solution of (P).

Summary of the Branch & Bound approach

• Iterative approach.• At each iteration, - a list of candidate problems is available to be solved. When the procedure starts, the list includes aoly the problem (P). - a candidate problem is selected, and the correponding relaxed problem is solved - the optimal solution of the relaxed allows to update the list of candidate problems or the upper bound and the best integer solution found so far.

e

The Branch & Bound method

• Initialisation The list of candidate problem includes only the problem (P) BS = ∞ Go to Step 2.• Étape 1 Si la liste est vide, terminer. La solution optimale est celle associée à BS, a moins que BS = ∞ dans lequel cas le problème P n’as pas de solution.• Step 2 Select the first candidate problem (PC) on the top of the list.

If the list is empty, stop. The optimal solution is the one

associated with , unless = ,and in this case the problem

has

n

Ste

o s

p 1

olution.

BS BS

Select the first candidate problem (PC) on the top of t

he

Step

l

2

ist.

To analyse , solve the relaxe problem .

If , go to Step1 .

If , go to Step 1.

If the optimal solution

Ste

of is integer,

p 3

then

PC PC

F PC

v PC BS

PC

if , then : ,v PC BS BS v PC

go to Step1.


If , go to Step1 .

If , go to Step 1.


Ste

of is integer,

p 3

then

PC PC

F PC

v PC BS

PC

if , then : ,v PC BS BS v PC go to Step1.

Select a variable which is no

St

t in

ep

te

er.

4

gjx

Generate a first new problem by including the constraint

to the problem and place it on the top of the listj jx x PC

Generate a second new problem by including the constraint

to the problem , and place it on the top of the listj jx x PC Go to Step 2.

If the list is empty, stop. The optimal solution is the one

associated with , unless = ,and in this case the problem

has

n

Ste

o s

p 1

olution.

BS BS

Select the first candidate problem (PC) on the top of t

he

Step

l

2

ist.


If , go to Step1 .

If , go to Step 1.


Ste

of is integer,

p 3

then

PC PC

F PC

v PC BS

PC

if , then : ,v PC BS BS v PC

go to Step1.

Enumeration three of the Branch & Bound

P

2P 3P

4P 5P

41 x 31 x

42 x52 x

Int. sol.BS = – 23

Infeasible solution

Non int. sol.z = – 22.333 > BS

Optimal solution

2.42 x

75.31 x1 2

1 2

1 2

1 2

Min 5 s.t. 11 6 66

10 45, 0,

z x xx x

x xx x

3 1 2

1 2

1 2

1

1 2

Min 5 s.t. 11 6 66

10 453

, 0,

P z x xx x

x xx

x x

5 1 2

1 2

1 2

1

2

1 2

Min 5 s.t. 11 6 66

10 4534

, 0,

P z x xx x

x xxx

x x

4 1 2

1 2

1 2

1

2

1 2

Min 5 s.t. 11 6 66

10 4535

, 0,

P z x xx x

x xxx

x x

2 1 2

1 2

1 2

1

1 2

Min 5 s.t. 11 6 66

10 454

, 0,

P z x xx x

x xx

x x

Candidate problem

P

2P

3P 4P

5P

How to search in the three (how to select the next candidate problem)

:

decend rapidly in the three in order t

a) Depth first search (select the last genera

o reach as soon as

possible a feasible i

ted candidate problem)

nteger solution

require to complete some iterations to solve the candidate

b) Search using the best node (

problem at each node in order t

us

o

ing t

ident

he best ca

ify the no

ndidate pr

de having

oblem):

the best potentiel to improve the upper bound

Selection of the separating variable ,j j j j jx x x x x The variable such that

a) is the largest

j

j j

x

x x b) is the smallestj jx x

c) is closer to 0.5j jx x

The optimal solution of the current relaxation of the candidate pro

Solving the relaxed candidate problem using the dual simplex method

blem

(at the current node) is not feasible for the relaxed problem obtained

by adding a constraint of the type ou .

Then we can use the dual simplex method to solve the

relaxed candidate problem using the optimal solution

of the current problem to g

j j j jx x x x

enerate the initial basic solution.

Reference.

A. Atamturk, M.W.P. Savelsbergh, "Integer-Programming Software

Systems", Annals of Operations Research 140, 67-124, 2005.

Integer Linear Programming. Introduction Integer L P problem (P) F(P) = feasible domain of P...

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