integer programming, gomory

Integer Programming Exclusive Including Gomory Method

Edited by Avinash Juriani

Management Science

Operations Research

by

Assoc. Prof. Sami Fethi

© 2007 Pearson Education

Ch 5: Integer programmimg

Operations Research © 2010/11, Sami Fethi, EMU, All Right Reserved. 2

Chapter Topics

Integer Programming (IP) Models

Examples for Integer Programming (IP) Models

Integer Programming Graphical Solution

Traditional approaches to solving integer

programming problems

Branch and Bound Method

Gomory cutting plane Method

Examples for Gomory cutting plane Method



Why used?

The implicit assumption was that solutions could be

fractional or real numbers (i.e., non-integer) in the linear

programming models. However, non-integer solutions

are always practical.

When only integer solutions are practical or logical, non-

integer solution values can be rounded off to the nearest

feasible integer values.

For example, if the case is nail, it will cause little

concern considering 8000.4 nails as 8000 nails. Round

off makes the case cost only a few cents apeice.

For example, if the case is a production of jet aircraft,

round off could affect profit or cost by million dollar

considering 7.4 jet airliners as 7 or 8.



Integer Programming Models: Types of Models

Total Integer Model: All decision variables required to have

integer solution values.

0-1 Integer Model: All decision variables required to have

integer values of zero or one.

Mixed Integer Model: Some of the decision variables (but

not all) required to have integer values.



Example 1: A Total Integer Model (1 of 2)

Machine shop obtaining new presses and lathes.

Marginal profitability: each press $100/day; each lathe

$150/day.

Resource constraints: $40,000, 200 sq. ft. floor space.

Machine purchase prices and space requirements:

Machine

Required Floor Space (sq. ft.)

Purchase Price

Press Lathe

15

30

$8,000

4,000



A Total Integer Model (2 of 2)

Integer Programming Model:

Maximize Z = $100x1 + $150x2

subject to:

8,000x1 + 4,000x2 $40,000

15x1 + 30x2 200 ft2

x1, x2 0 and integer

x1 = number of presses

x2 = number of lathes



Example 2: A Total Integer Model (1 of 1)


Maximize Z = 6x1 + 5x2 + 2x3

subject to:

10 x1 + 4 x2 + 2 x3 600

2 x1 + 5 x2 + 2 x3 800

x1, x2 ,,x3 0 and integer

x1 = 22

x2 = 0

x3 = 378

Z=888



Recreation facilities selection to maximize daily usage by

residents.

Resource constraints: $120,000 budget; 12 acres of land.

Selection constraint: either swimming pool or tennis center

(not both).

Data:

Recreation Facility

Expected Usage (people/day)

Cost ($) Land

Requirement (acres)

Swimming pool Tennis Center Athletic field Gymnasium

300 90 400 150

35,000 10,000 25,000 90,000

4 2 7 3

Example 3 : A 0 - 1 Integer Model (1 of 2)




Maximize Z = 300x1 + 90x2 + 400x3 + 150x4

subject to:

$35,000x1 + 10,000x2 + 25,000x3 + 90,000x4 $120,000

4x1 + 2x2 + 7x3 + 3x4 12 acres

x1 + x2 1 facility

x1, x2, x3, x4 = 0 or 1

x1 = construction of a swimming pool

x2 = construction of a tennis center

x3 = construction of an athletic field

x4 = construction of a gymnasium







subject to:

10 x1 + 4 x2 + 2 x3 600

2 x1 + 5 x2 + 2 x3 800

x1, x2 ,,x3 0 or 1

x1 = 1

x2 = 1

x3 = 1

Z=13



Example 5: A Mixed Integer Model (1 of

2)

$250,000 available for investments providing greatest

return after one year.

Data:

Condominium cost $50,000/unit, $9,000 profit if sold

after one year.

Land cost $12,000/ acre, $1,500 profit if sold after one

year.

Municipal bond cost $8,000/bond, $1,000 profit if sold

after one year.

Only 4 condominiums, 15 acres of land, and 20

municipal bonds available.




Maximize Z = $9,000x1 + 1,500x2 + 1,000x3

subject to:

50,000x1 + 12,000x2 + 8,000x3 $250,000

x1 4 condominiums

x2 15 acres

x3 20 bonds

x2 0


x1 = condominiums purchased

x2 = acres of land purchased

x3 = bonds purchased

Example 5: A Mixed Integer Model (2 of 2)



Example 6 : A Mixed Integer Model (1 of 1)



subject to:

10 x1 + 4 x2 + 2 x3 600

2 x1 + 5 x2 + 2 x3 800

x1, x2 ,,x3 0

x1 = 22.2

x2 = 0

x3 = 377.8

Z=888.9



Integer Programming Example

Graphical Solution of Maximization Model

Maximize Z = $100x1 + $150x2

subject to:

8,000x1 + 4,000x2 $40,000

15x1 + 30x2 200 ft2


Optimal Solution:

Z = $1,055.56

x1 = 2.22 presses

x2 = 5.55 lathes

Figure 1 Feasible Solution Space with Integer Solution Points



Rounding non-integer solution values up to the

nearest integer value can result in an infeasible

solution.

A feasible solution is ensured by rounding down non-

integer solution values but may result in a less than

optimal (sub-optimal) solution.

Integer Programming Graphical Solution



Traditional approaches to solving integer programming problems

Branch and Bound Method

Based on principle that total set of feasible solutions

can be partitioned into smaller subsets of solutions.

Smaller subsets evaluated until best solution is found.

Method is a tedious and complex mathematical

process.

This method can be used more easier If Excel or QM is

conducted.




Gomory cutting plane Method

In 1958 Ralph Gomory was the first individual to

develop a systematic (algorithmic) approach for solving

linear integer programming problems.

His method is an algebraic approach based on the

systematic addition of new constraints (or cuts), which

are satisfied by an integer solution but not by a

continous variable solution.

The idea of a cut is a new contstraint permits the new

feasible region to include all the feasible integer solution

for the original constraints, but it does not include the

optimal noninteger solution originally found.



Example 1- Gromory cutting plane Method

Integer Programming Model: Gromory’s fractional cut

Maximize Z = $ 3x1 + $ 5x2

subject to:

1 x1 + 4 x2 9

2 x1 + 3 x2 4

Non-negativity

x1, x2 0

Given the optimal noninteger solution to the problem:




Find the optimal integer solution using Gomory’s cutting

plane method

3 5 0 0

C Pmix Quan X1 X2 S1 S2

5 X2 7/5 0 1 2/5 -1/5

3 X1 17/5 1 0 -3/5 4/5

Z 86/5 3 5 1/5 7/5

C-Z 0 0 -1/5 -7/5




Steps in Gomory cutting plane Method

To find the cut, we arbitrarily choose one of

noninteger variables in the optimal solution.

We choose X2 and need to rewrite this row with any

noninteger value in it express as the sum of an

integer and a non negative fraction less than 1.

The numbers in such process should be rewritten:

4/3=1+1/3

5/4=1+1/4

2/3=0+2/3

-2/3=-1+1/3





Initially, we write these values in the C row (1, 2/5, -1/5, 7/5)

as the sum of an integer and a nonnegative fraction less

than 1.

(1+0) X2+ (0+2/5) S1 + (-1+4/5) S2= (1+2/5)

Then we take all the integer coefficient to the right-hand

side:

2/5 S1 +4/5 S2= (2/5+1-1X2+1S2)

2/5 S1 +4/5 S2= (2/5+some integer )

Pmix Quan X1 X2 S1 S2

X2 7/5 0 1 2/5 -1/5





Omitting some integer part and write the cut eqn as follow:

The eqn must be greater than or equal 2/5.

2/5 S1 +4/5 S2 ≥ 2/5

For the sake of simplicity, we multiply the eqn by -1 and we

avoid having to deal with subtracted slack variables which

require an artificial variable.

- 2/5 S1 - 4/5 S2 ≤ - 2/5

- 2/5 S1 - 4/5 S2 + S3 = - 2/5 -This is the required cut.


X2 7/5 0 1 2/5 -1/5




We add the required cut in the new simplex tableau. Since C-Z

values are zero or negative, it is difficult to determine which

variable to bring into the solution. However, if there is any

negative number within the quantity column, the number should

leave from the system so S3 should be taken into account.

3 5 0 0 0

C Pmix Quan X1 X2 S1 S2 S3

5 X2 7/5 0 1 2/5 -1/5 0

3 X1 17/5 1 0 -3/5 4/5 0

S3 -2/5 0 0 -2/5 -4/5 1

Z 86/5 3 5 1/5 7/5 0

C-Z 0 0 -1/5 -7/5 0




Now divide the negative values in the S3 row into the

corresponding values in C-Z row and bring in that variable which

has the smallest quotient:

S1=(- 1/5)/ (-2/5)= ½, S2=(- 7/5)/ (-1/4)= 7/4, S1 is the smallest

one so S1 comes in whilst S3 goes out.

3 5 0 0 0


5 X2 7/5 0 1 2/5 -1/5 0

3 X1 17/5 1 0 -3/5 4/5 0

S3 -2/5* 0 0 -2/5 -4/5 1 Min

Z 86/5 3 5 1/5 7/5 0

C-Z 0 0 -1/5* -7/5* 0




S3 is leaving from the system and S1 is entering

into the system.

New S1 =( 1, 0, 0, 1, 2, -5/2)

New X2 = old X2- key * New S1

1= 7/5-2/5*1

0= 0-2/5*0

1= 1-2/5*0

0= 2/5-2/5*1

−1= -1/5-2/5*2

1= 0-2/5*-5/2




S3 is leaving from the system and S1 is entering into

the system

New S1 =( 1, 0, 0, 1, 2, -5/2)


4= 17/5-(-3/5)*1

1= 1-(-3/5)*0

0= 0-(-3/5)*0

0= -3/5-(-3/5)*1

2= 4/5-(-3/5)*2

-3/2= 0-(-3/5)*-5/2




Now, we know that the optimal solution has been reached. It

manufactures 4 tables and 1 chair and this uses all the hours

available except for 1 hour in department 1 since S1 has the

value 1 in the final solution. The profit earned by this optimal

integer solution is $ 17 which is only slightly lower than $86/5 ($

17.20) profit earned by the optimal noninteger solution in the first

Table:

3 5 0 0 0


5 X2 1 0 1 0 -1 1

3 X1 4 1 0 0 2 -3/2

S1 1 0 0 1 2 -5/2

Z 17 3 5 0 1 1/2

C-Z 0 0 0 -1 -1/2




Integer Programming Model: Gromory’s fractional cut

Maximize Z = $ 2x1 + $ 1x2

subject to:

2 x1 + 5 x2 17

3 x1 + 2 x2 10

Non-negativity


(a) Find the optimal integer solution using simplex method

(b) Find the optimal integer solution using Gomory’s cutting

plane method




Find the optimal integer solution using both simplex and

Gomory’s cutting plane methods

2 1 0 0

C Pmix Quan X1 X2 S1 S2 Q/XS

0 S1 17 2 5 1 0 17/2

0 S2 10 1 0 0 1 10/1

Z 0 0 0 0 0 0

C-Z 2 1 0 0 0

Max

Min




S2 is leaving from the system and X1 is entering into the

system.

New X1 =( 10/3, 3/3, 2/3, 0, 1/3)

New S1 = old S1- key * New X1

31/3= 17-2*10/3

0= 2-2*3/3

11/3= 5-2*2/3

1= 1-2*0

−2/3= 0-2*1/3




This is the optimal solution, however the values in Quantity

column are noninteger. It seems that the simplex technique

applied in part (a) is not good enough so we need to use

Gromory cutting plane method in part (b).

2 1 0 0

C Pmix Quan X1 X2 S1 S2 Q/XS

0 S1 31/3 0 11/3 1 -2/3 -

2 X1 10/3 1 2/3 0 1/3 -

Z 20/3 2 4/3 0 2/3 -

C-Z 0 -1/3 0 -2/3 -





(1+0) X1+ (0+2/3) X2 + (0+1/3) S2= (3+1/3)

X1+2/3X2+1/3S2= 3+1/3

2/3 X2 +1/3 S2= 1/3+(3-1X1)

2/3 X2 +1/3 S2 = (1/3+some integer ) (multiply by -1

and add S3 variable).

-2/3 X2 -1/3 S2 + S3= -1/3


X1 10/3 1 2/3 0 1/3




Now divide the negative values in the S3 row into the

corresponding values in C-Z row and bring in that variable

which has the smallest quotient:

X2=(- 1/3)/ (-2/3)= ½, S2=(- 2/3)/ (-1/3)= 2, X2 is the smallest

one so X2 comes in whilst S3 goes out.

2 1 0 0 0


2 X1 10/3 1 2/3 0 1/3 0

0 S1 31/3 0 11/3 1 -2/3 0

0 S3 -1/3 0 -2/3 0 -1/3 1

Z 20/3 2 4/3 0 2/3 0

C-Z 0 -1/3 0 -2/3 0

Min

Min Min




S3 is leaving from the system and X2 is entering into the

system.

New X2 =( 1/2, 0, 1, 0, ½, -3/2)

New S1 = old S1- key * New X2

17/2= 31/3-11/3*1/2

0= 0-11/3*0

0= 11/3-11/3*1

1= 1-11/3*0

−5/2= -2/3-11/3*1/2

11/2= 0-11/3*-3/2




This is the optimal solution, however the values in Quantity

column are noninteger so we need to use Gromory cutting

plane method in choosing the smallest value in Quantity

column. We therefore selected X2

2 1 0 0 0


1 X2 1/2 0 1 0 1/2 -3/2

2 X1 3 1 0 0 0 1

0 S1 17/2 0 0 1 -5/2 11/2

Z 13/2 2 1 0 1/2 13/2

C-Z 0 0 0 -1/2 -1/2





(0) X1+ (1+0) X2 + (0) S1+ (0+1/2) S2+ (-1+1/2) S3 =(0+1/2)

X2+1/2 S2-S3-1/2 S3= 1/2

1/2 S2 -1/2 S3= (or ≥) 1/2+(1S3-1X2)

1/2 S2 -1/2 S3 ≥ (1/2+some integer ) (multiply by -1 and

add S4 variable).

-1/2 S2 +1/2 S3 ≤ -1/2

-1/2 S2 + 1/2 S3 + S4 = -1/2

Pmix Quan X1 X2 S1 S2 S3

X2 1/2 0 1 0 ½ -3/2




Now divide the negative values in the S4 row into the corresponding values in C-Z row

and bring in that variable which has the smallest quotient:

S2=(- 1/2)/ (-1/2)= 1, S3=(- 1/2)/ (1/2)= -1, S2 is the smallest one so S2 comes in whilst

S4 goes out.

2 1 0 0 0 0

C Pmix Quan X1 X2 S1 S2 S3 S4

1 X2 1/2 0 1 0 1/2 -3/2 0

2 X1 3 1 0 0 0 1 0

0 S1 17/2 0 0 1 -5/2 11/2 0

0 S4 -1/2 0 0 0 -1/2 1/2 1 Min

Z 13/2 2 1 0 1/2 1/2 0

C-Z 0 0 0 -1/2 -1/2 0

Min




S2 is leaving from the system and S4 is entering into the

system.

New S2 =( 1, 0, 0, 0, 1,-1 -2)


0= 1/2-1/2*1

0= 0-1/2*0

1= 1-1/2*0

0= 0-1/2*0

0= 1/2-1/2*(1)

-½ = -3/2-1/2*(-1)

1= 0-1/2*(-2)





system.

New S2 =( 1, 0, 0, 0, 1,-1 -2)


3= 3-0*1

1= 1-0*0

0= 0-0*0

0= 0-0*0

0= 0-0*(1)

1 = 1-0*(-1)

0= 0-0*(-2)





system.

New S2 =( 1, 0, 0, 0, 1,-1 -2)

New S1 = old S1- key * New S2

11= 17/2-(-5/2)*1

0= 0-(-5/2) *0

0= 0-(-5/2) *0

1= 1- (-5/2) *0

0= -5/2- (-5/2) *(1)

3 = 11/2-(-5/2) *(-1)

-5= 0-(-5/2) *(-2)




2 1 0 0 0 0

C Pmix Quan X1 X2 S1 S2 S3 S4

1 X2 0 0 1 0 0 -1/2 1

2 X1 3 1 0 0 0 1 0

0 S1 11 0 0 1 0 3 -5

0 S2 1 0 0 0 1 -1 -2

Z 6 2 1 0 0 3/2 1

C-Z 0 -2 0 0 -3/2 -1




Now, we know that the optimal solution has been reached. It

manufactures 3 mugs and 0 bowl and this uses all the hours and

materials available except for 11 hours of labor as well as 1 lb of

clay since S1 and S2 have the value 11 and 1 in the final solution

row. The profit earned by this optimal integer solution is $ 6 which

is only slightly lower than $20/3 ($ 6.66) profit earned by the

optimal noninteger solution in the first Table:



End of chapter

integer programming, gomory

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