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Integer Optimization (University of Jordan) CUTTING PLANE METHOD 29-04-2018 1 / 38

The University of JordanDepartment of Mathematics

Cutting Plane Method

Supervised and edited by : Dr. Baha ALZALG

Prepared and presented by :

Nadjiba MANSOURIIlhem KADRISamia MAZOUZAhmed BOUCHENAK

Soraya ABDELAZIZMohamed BENRABIASaid ATLAOUIIbtissem ATTAFI

Integer and Combinatorial Optimization (0301973)

Spring 2018

Sommaire

1 Introduction

2 The Cutting Plane MethodGeneral SchemeAn Example (with the graphical method)

3 The Gomory Cutting Plane MethodThe Algorithm StepsExamples (with the simplex method)How to avoid cycling ?

4 Conclusion


Introduction

Classification

An integer programming problem is a mathematical optimization orfeasibility program in which some or all of the variables are restrictedto be integers which is very difficult to solve. However, there arethree different types of algorithms :

1 Exact algorithms.

2 Approximation algorithms.

3 Heuristic algorithms.


Introduction

Exact Algorithms

Examples of exact algorithms designed and used to solvecombinatorial optimization problems :

1 Cutting Plane Method.

2 Branch and Bound Method.

3 Branch and Cut Method.

4 Dynamic Programming.

In this work, we present a method called theCutting Plane Method.


Introduction

Relaxation

Let P be a mixed integer linear program :min cTx + dTy + eT zs.t. Ax + By + Cz ≤ b

x , y , z > 0 (P)y ∈ Z.

The linear programming relaxation is :min cTx + dTy + eT zs.t. Ax + By + Cz ≤ b (P ′)

x , y > 0.

Problem (P ′) is called linear relaxation of (P).


The Cutting Plane Method General Scheme

General Scheme of the Cutting Plane Method

The integer programming problem

min cTxs.t. Ax ≤ b

x ≥ 0 integer.

The linear programming relaxation

min cTxs.t. Ax ≤ b

x ≥ 0.


The Cutting Plane Method General Scheme

General Scheme of the Cutting Plane Method

The main idea of the Cutting Plane Method is to solve the integerprogramming problem by solving a sequence of linear programmingproblems, as follows :

1 Solve the linear programming relaxation.

2 If the resulting optimal solution x? is integral, stop (the optimalsolution is found).

3 If not, generate a cut by using the Gomory cutting planealgorithm, i.e, a constraint which is satisfied by all feasibleinteger solution but not by x?.

4 Add this new constraint, resolve problem, and go back to (2).This terminates after a finite number of iterations in (2) and theresulting x? is integral and optimal.


The Cutting Plane Method An Example (with the graphical method)

Example 1



(We will discuss later how to generate such cuts.)


The Gomory Cutting Plane Method

The Gomory Cutting Plane Algorithm

The first finitely terminating algorithm for integer programming was acutting plane algorithm proposed by Ralph Gomory in 1958 at IBM.


The Gomory Cutting Plane Method The Algorithm Steps

The Gomory Cutting Plane AlgorithmStep 1

We solve the standard form linear programming problem with thesimplex method.Let x? be the optimal basic feasible solution.

We partition x into two subvectors : x =

(xBxN

), where

xB is the subvector corresponding the basic variables, andxN is the subvector corresponding the nonbasic variables.Assume for simplicity that

xB(i) = xi ,∀i = 1, . . . ,m (i.e., the first m variables are basic).



From the optimal tableau

B−1Ax = B−1b, · · · · · · · · · (∗)

where• A is the coefficient matrix of the problem in standard form.• B is the submatrix of A whose columns are ai for i in the set ofindices of basic variables.From (∗) we obtain

xB + B−1ANxN = B−1b

where N is the set of indices of nonbasic variables and AN is thesubmatrix of A whose columns are ai for i ∈ N .(Therefore, A = [B AN ]).



Step 2

We consider one equality from the optimal tableau

xi +∑j∈N

aijxj = ai0 ,

where

aij = (B−1Aj)i and ai0 = (B−1b)i and ai0 is fractional,

where [t] is floor of t. Since xj ≥ 0 ∀j , we have

xi +∑j∈N

[aij ]xj ≤ ai0, as [aij ] ≤ aij .

Since xj must be integer, we obtain

xi +∑j∈N

[aij ]xj ≤ [ai0].

This inequality is valid for all integer sols, but is not satisfied by x?.Integer Optimization (University of Jordan) CUTTING PLANE METHOD 29-04-2018 17 / 38

The Gomory Cutting Plane Method Examples (with the simplex method)

Example 2 (a minimization integer program)

Consider the integer programming problem

min x1 − 2x2s.t. −4x1 + 6x2 ≤ 9

x1 + x2 ≤ 4x1, x2 ≥ 0 and both integers.

The problem in the standard form becomes

min x1 − 2x2s.t. −4x1 + 6x2 + x3 = 9

x1 + x2 + x4 = 4x1, x2, x3, x4 ≥ 0.



Solve the linear programming relaxation by using the simplexmethod : Note that

A =

(−4 6 1 01 1 0 1

).

We partition x into two subvectors x = (xB , xN), where

xB = (x1, x2) (basic variables),

xN = (x3, x4) (nonbasic variables).

It follows that

B =

(−4 61 1

)and AN =

(1 00 1

).



The system B−1Ax = B−1b can be written as

x1 − 110x3 + 3

5x4 = 15

10, · · · · · · · · · (1)

x2 + 110x3 + 2

5x4 = 25

10. · · · · · · · · · (2)

So the optimal solution is x1 =(1510, 2510

).

While the two equations have the same fractional part in theright-hand side values (which is 5/10), equation (2) has the biggestright-hand side value in this system.As a result, choosing equation (2) and applying the Gomory cuttingplane, we get

x2 +

[1

10

]x3 +

[2

5

]x4 ≤

[25

10

]. This leads to x2 ≤ 2.



We augment the linear programming relaxation by adding theconstraint x2 + x5 = 2 to obtain

min x1 − 2x2s.t. −4x1 + 6x2 + x3 = 9

x1 + x2 + x4 = 4x2 + x5 = 2x1, x2, x3, x4, x5 ≥ 0.

Applying the simplex method, we get



Iteration 1

x1 x2 x3 x4 x5 c Ratiox3 -4 6 1 0 0 9 3

2

x4 1 1 0 1 0 4 4x5 0 1 0 0 1 2 2z 1 -2 0 0 0 0 0

Iteration 2

x1 x2 x3 x3 x5 c Ratiox2

−23

1 16

0 0 32

-1x4

53

0 −16

1 0 52

32

x523

0 −16

0 1 12

34

z −13

0 13

0 0 3

Iteration 3

x1 x2 x3 x4 x5 c Ratiox2 0 1 0 0 1 2x4 0 0 3

121 −5

2−52

x1 0 0 −14

0 32

34

z 0 0 12

0 1 132



So, the new optimal solution is x2 = (34, 2), which is not integer.

The equations in the optimal tableau are

x1 − 14x3 + 3

2x5 = 3

4, · · · · · · · · · (3)

x2 + x5 = 2. · · · · · · · · · (4)

Choosing equation (3) (because it has the biggest fractional partamong all the right-hand side values in the system (3/4 > 0)) andapplying the Gomory cutting plane method, we get

x1 +

[−1

4

]x3 +

[3

2

]x5 ≤

[3

4

]. This leads to x1 − x3 + x5 ≤ 0.

Since x2 + x5 = 2, we have x5 = 2− x2. In term of the originalvariables x1, x2, we get

−3x1 + 5x2 ≤ 7.



We add this constraint, together with the previously added constraint.

Applying the simplex method, we get the new optimal solutionx3 = (1, 2), which is integer ! So, it is the optimal solution of theoriginal problem.



Example 3 (a maximization integer program)

Consider the following integer programming problem

max 3x1 + 4x2s.t. 2x1 + 5x2 ≤ 15

2x1 − 2x2 ≤ 5x1, x2 ≥ 0 and both integers.

The standard form is

max 3x1 + 4x2s.t. 2x1 + 5x2 + x3 = 15

2x1 − 2x2 + x4 = 5x1, x2, x3, x4 ≥ 0.



We solve the LP relaxation by using the simplex method. We have

A =

(2 5 1 02 −2 0 1

),B =

(2 52 −2

), and b =

(155

).

B−1Ax = B−1b ⇐⇒ x1 + 17x3 + 5

14x4 = 55

14,

x2 + 17x3 − 1

7x4 = 10

7.

The optimal solution is

x1 =(55

14,

10

7

).



Choosing the first equation and applying the Gomory cutting planemethod, we obtain

x1 +[1

7

]x3 +

[ 5

14

]x4 ≤

[55

14

], which leads to x1 ≤ 3.

Introducing a new slack variable to get x1 + x5 = 3, x5 ≥ 0.Adding this cut to the linear problem and solving by the simplexmethod, we find the new optimal solution x2 = (3, 9

5), which is not

integer.



We choose an equation from the optimal tableau, which is

x2 +1

5x3 −

2

5x5 =

9

5,

and apply the Gomory cutting plane method and find the cut

x1 + x2 ≤ 4 or x2 + x1 + x6 = 4, x6 ≥ 0.

Add this cut to the problem and go back to step 1. The new optimalsolution is x3 = (5

3, 73), which is still not integer.



From the optimal tableau, we choose the equation

x2 +1

3x3 −

2

3x6 =

7

3,

and after applying the Gomory cutting plane method, we find the cut

x1 + 2x2 ≤ 6 or x1 + 2x2 + x7 = 6, x7 ≥ 0.

We add this cut to the problem and go back step 1. Finally, find thenew optimal solution x4 = (2, 2), which is integer ! It is an optimalsolution to the original problem.



A graph illustrates the three cuts generated in Example 3.


The Gomory Cutting Plane Method How to avoid cycling ?

Another cut to avoid the cycling

We apply the following cut to avoid the cycling :∑j∈N

fjxj ≥ f , · · · · · · · · · (∗∗)

wherefj = aij − [aij ] and f = ai0 − [ai0].

This inequality is satisfied by all feasible x but it is not satisfied by x?.



Example 4 (on avoiding cycling)

max z = 2x1 + 3x2s.t. 2x1 + 2x2 ≤ 7

x1 ≤ 2x2 ≤ 2x1, x2 ≥ 0 and both integers.

The standard form is

max z = 2x1 + 3x2s.t. 2x1 + 2x2 + x3 = 7

x1 + x4 = 2x2 + x5 = 2x1, x2, x3, x4, x5 ≥ 0.



After solving the relaxation by using the simplex method, the optimaltableau is

cj 2 3 0 0 0CBi basic variable x1 x2 x3 x4 x5 Sol

2 x1 1 0 12

0 -1 32

0 x4 0 0 −12

1 1 12

3 x2 0 1 0 0 1 2zj 2 3 1 0 1 9

cj − zj 0 0 -1 0 -1

The optimal solution is

x1 =

(3

2, 2

),

but it is not integer.Integer Optimization (University of Jordan) CUTTING PLANE METHOD 29-04-2018 33 / 38


We choose the following equation from the optimal tableau

x1 +1

2x3 − x5 =

3

2,

and apply the Gomory cutting plane method to obtain the cut

x1 + x2 ≤ 3.

If we add this constraint to the LPP and use the simplex method, weget the same cut in each sequence, which will lead to cycling !

cj 2 3 0 0 0 0CBi basic variable x1 x2 x3 x4 x5 x6 Sol

2 x1 1 0 12

0 -1 0 32

0 x4 0 0 −12

1 1 0 12

3 x2 0 1 0 0 1 0 23 x6 1 1 0 0 0 1 3

zj 2 3 1 0 1 0 9cj − zj 0 0 -1 0 -1 0



So, we use the cut (∗∗) in our example for solve the ILP. The newcut is

1

2x3 ≥

1

2or−1

2x3 ≤

−1

2,

which is equivalent to

−1

2x3 + x6 =

−1

2, x6 ≥ 0.

Now, we use the dual simplex method to solve the resulting problem.



Iteration 1


2 x1 1 0 12

0 -1 0 32

0 x4 0 0 −12

1 1 0 12

3 x2 0 1 0 0 1 0 23 x6 0 0 −1

20 0 1 −1

2

zj 2 3 1 0 1 0 9cj − zj 0 0 -1 0 -1 0



Iteration 2


2 x1 1 0 0 0 -1 1 10 x4 0 0 0 1 1 -1 13 x2 0 1 0 0 1 0 23 x6 0 0 1 0 0 -2 2

zj 2 3 0 0 1 2 8cj − zj 0 0 0 0 -1 -2

The tableau is now optimal and the optimal solution is x = (1, 2),which is integer !


Conclusion

Conclusion

The Cutting Plane Method is very useful for solving integerprogramming problems, but there is a difficulty lies in the choice ofinequalities which represent the cut of only a very small piece of thefeasible set of the linear programming relaxation.


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