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MEPM 531 Operation Research Assignment - Integer Linear Programming, American University of Ras Al Khaimah, UAE

Q1. Solve the following integer linear programming problem by using:

a) Branch and Bound Method

Problem is: Max Z = 3x1 + 4x2

Subject to 2x1+ x2 ≤ 6

2x1 + 3x2 ≤ 9

X1, X2 ≥ 0, and integers

Solution in 4 parts:

Part 1: To draw constraint 2x1+ x2 ≤ 6 → (1)

2x1+ x2 ≤ 6

Treat it as 2x1+ x2 = 6

When x1 = 0 then x2 =?

2(0) + x2 = 6

x2 = 6

When x2 = 0 then x1 =?

2x1 + 0 = 6

x1 = 6/2 = 3

x1

0

3

x2

6

0

Part 2: To draw constraint 2x1+ 3x2 ≤ 9 → (2)

2x1 + 3x2 ≤ 9

Treat it as 2x1 + 3x2 = 9

When x1 = 0 then x2 = ?

2(0) + 3 x2 = 9

3 x2 = 9

x2 = 9/3 = 3

When x2 = 0 then x1 =?

2x1 + 3 (0) = 9

2x1 = 9

x2 = 9/2 = 4.5

x1

0

4.5

x2

3

0

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Part 3: To draw constraint >= 0 → (3)

Treat it as = 0

Equation is not possible.

Part 4: To draw constraint >= 0 → (4)

Treat it as = 0

Equation is not possible.

Operation Research Assignment figure.jpg

The value of the objective function at each of these extreme points is as follow:

Extreme Point Coordinates (x1, x2)

Line through Extreme Point

Objective function value Z = 3x1+ 4x2

O(0,0)

5→ x1 >= 0

6→ x2 >= 0

3(0) + 4(0) = 0

A(3, 0)

1→ 2 x1 + x2 <= 6

6→ x2 >= 0

3(3) + 4(0) = 9

B(9/4, 3/2)

1 → 2 x1 + x2 <= 6

2 → 2 x1 + 3x2 <= 9

3(9/4) + 4(3/2) = 51/4

C(0,3)

2 → 2 x1 + 3x2 <= 9

5 → x1 >= 0

3(0) + 4(3) = 12

The maximum value of the extreme function Z = 51/4 occurs at the extreme point (9/4, 3/2). Hence, the optional solution to the given LP problem is: x1 = 94, x2 =32 and max Z = 51/4.

b) Fractional Cut Method

Problem is:

Max Z = 3x1+ 4x2

Subject to 2x1+ x2 ≤ 6

2x1+ 3x2 ≤ 9

X1, X2 ≥ 0, and integers

Standard form

Z = 3x1+ 4x2 + 0S1 + 0S2

S. t.

2x1+ x2 + S1 = 6

2x1+ 3x2 + S2 = 9

X1, X2, S1, S2 >= 0

IBFS

S1 = 6 and S2 = 9

Cj (3 4 0 0)

CB

YB

XB

Y1*

Y2

Y3

Y4

0

Y3

6

2

2

1

0

0

Y4

9

2

3

0

1

CB X Y1 CB X Y2 + CB X Y3

Zj 0X2 = 0, 0 X 2 + 0X3 = 0; 0 0

0 0 0 0

Net Evaluation Zj - Cj -3 -4 0 0

Net Evaluation not >= 0

Most negative -4, arbitrary we select Y1 enters basis.

Min {XB/Y1j} = min {6/2, 9/2} 6/2 Y3 Leave basis

Cj (3 4 0 0)

CB

YB

XB

Y1*

Y2

Y3

Y4

1

Y1

6/6 = 1

2

2

1

0

0

Y4

9

2

3

0

1

CB X Y1 CB X Y2 + CB X Y3

Zj 0X2 = 0, 0 X 2 + 0X3 = 0; 0 0

0 0 0 0

Net Evaluation Zj - Cj -3 -4 0 0

Net Evaluation not >= 0

Most negative -4, arbitrary we select Y1 enters basis.

Min {XB / Y1j} = min {6/2, 9/2} 6/2 Y3 Leave basis

CB

YB

XB

Y1*

Y2

Y3

Y4

1

Y1

6/6 = 1

2/2 = 1

2/2 = 1

½

0

0

Y4

9/3 = 3

2/3

3/3 = 3

0/3 = 0

1/3

CB X Y1 CB X Y2 + CB X Y3

Zj 0X2 = 0, 0 X 2 + 0X3 = 0; 0 0

0 0 0 0

Net Evaluation Zj - Cj 1 3/2 0 0

Net Evaluation not >= 0

Most negative -4, arbitrary we select Y1 enters basis.

Min {XB/Y1j} = min {1, 3/2} 6/2 Y4 Leave basis.

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Q2. Solve the following integer linear problem by:

a) Branch and Bound Method

Max Z = 4x1 + 6x2+ 2x3

s.t. 4x1 - 4x2 ≤ 5

-x1 + 6x2 ≤ 5

-x1 + x2 + x3 ≤ 5

x1, x2, x3 ≥ 0 and integers

Part 1: To draw constraint 4x1 - 4x2 ≤ 5 → (1)

4x1 - 4x2 ≤ 5

Treat it as 4x1 - 4x2 = 5

When x1 = 0 then x2 =?

4(0) - 4x2 = 5

- 4x2 = 5

x2 = -5/4 = -1.25

When x2 = 0 then x1 =?

4x1 - 4(0) = 5

4x1 = 5

x1 = 5/4 = 1.25

x1

0

5/4

x2

-5/4

0

Part 2: To draw constraint -x1+ 6x2 ≤ 5 → (2)

-x1 + 6x2 ≤ 5

Treat it as -x1 + 6x2 = 5

When x1 = 0 then x2 =?

-1(0) + 6x2 = 5

6x2 = 5

x2 = 5/6

When x2 = 0 then x1 = ?

-x1 + 6(0) = 5

-x1 = 5

x1 = -5

x1

0

5/6

x2

-5/6

0

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Part 3: To draw constraint -x1+ x2+ x3 ≤ 5 → (3)

-x1+ x2+ x3 ≤ 5

Treat it as -x1+ x2+ x3 = 5

When x1 = 0 then x2 = -4/5 from equation (1)

Then -x1+ x2+ x3 = 5

-1(0) + -4/5 + x3 = 5

x3 = 5 + 4/5 = 29/5

When x2 = 0 then x1 = 5/4 from equation (1)

-x1+ x2+ x3 = 5

-(5/4) + 1(0) + x3 = 5

-(5/4) + x3 = 5

x3 = 5 + 5/4 = 25/4

When x1 = 0 then x2 = 5/6 from equation (2)

Then -1 (0)+ 5/6 + x3 = 5

5/6 + x3 = 5

x3 = 5 - 5/6 = 5

When x2 = 0 then x1 = 5/6 from equation (2)

-5/6 + 1(0) + + x3 = 5

-(5/6) + x3 = 5

x3 = 5 -(5/6) = 25/6

x1 = 0, x2 = -4/5 then

x3

29/5

x2 = 0 , x1 =5/4 then

x3

25/4

x1 = 0, x2 = 5/6 then

x3

5

x2 = 0 x1 = 5/6

x3

25/6

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b) Fractional Cut Method

Max Z = 4x1+ 6x2 + 2x3

s.t. 4x1 - 4x2 ≤ 5

-x1 + 6x2 ≤ 5

-x1 + x2+ x3 ≤ 5

x1, x2, x3 ≥ 0 and integers

Standard form

Z = 4x1 + 6x2 + 2x3

s.t. 4x1- 4x2 ≤ 5

-x1 + 6x2 ≤ 5

-x1 + x2+ x3 ≤ 5

x1, x2, x3 ≥ 0 and integers

IBFS

S1 = 5 and S2 = 5

Cj (4 6 0 0)

CB

YB

XB

Y1*

Y2

Y3

Y4

0

Y3

5

4

-2

2

0

0

Y4

5

3

3

0

1

CB X Y1 CB X Y2 + CB X Y3

Zj 0X2 = 0, 0 X 2 + 0X3 = 0; 0 0

0 0 0 0

Net Evaluation Zj - Cj -8 -7 0 0

Net Evaluation not >= 0

Most negative -4, arbitrary we select Y1 enters basis.

Min {XB / Y1j} = min {6/2, 9/2} 6/2 Y3 Leave basis

Cj (3 4 0 0)

CB

YB

XB

Y1*

Y2

Y3

Y4

1

Y1

6/7

2

2

1

0

0

Y4

8/7

2

3

0

1

CB X Y1 CB X Y2 + CB X Y3

Zj 0X2 = 0, 0 X 2 + 0X3 = 0; 0 0

0 0 0 0

Net Evaluation Zj - Cj -3 -4 0 0

Net Evaluation not >= 0

Most negative -4, arbitrary we select Y1 enters basis.

Min {XB / Y1j} = min {5/2, 3/2} 5/2 Y3 Leave basis

CB

YB

XB

Y1*

Y2

Y3

Y4

1

Y1

6/3 = 3

2/2 = 1

2/2 = 1

1/2

0

0

Y4

9/3 = 3

2/3

3/3 = 3

0/3 = 0

1/3

CB X Y1 CB X Y2 + CB X Y3

Zj 0X2 = 0, 0 X 2 + 0X3 = 0; 0 0

0 0 0 0

Net Evaluation Zj - Cj 1 3/2 0 0

Net Evaluation not >= 0

Most negative -4, arbitrary we select Y1 enters basis.

Min {XB / Y1j} = min {7, 5/2} 5/2 Y4 Leave basis.

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3) Solve the following integer linear problem by:

a) Branch and Bound Method

Max Z = 3x1 + x2 + 3x3

s. t. - x1 + 2x2 + x3 ≤ 4

4x2 - 3x3 ≤ 2

x1 - 3x2 + 2x3 ≤ 3

x1, x2, x3 ≥ 0 and integers

Solution: To apply branch and bound method, the three constraints to be added to LP model.

x1 <= 1

x2 <= 1

x3 <= 1

Solution Steps by Big M method:

Max Z = 3x1 + x2 + 3x3

s. t. - x1 + 2x2 + x3 ≤ 4

4x2 - 3x3 ≤ 2

x1 - 3x2 + 2x3 ≤ 3

x1, x2, x3 ≥ 0

Solution is:

Max Z A = 7 (x1= 1, x2 = 1, x3 = 1)

and ZL = 7 (x1= 1, x2 = 1, x3 = 1) obtained by the rounded off solutions value.

This problem has integer solution, so no further branching is required.

The branch and bound diagram.

A

x1= 1, x2 = 1, x3 = 1

ZA = 7

ZL = 7

Solution steps by Big M method.

The 0-1 Integer Programming problem algorithm thus terminated and the optimal integer solution is: ZA = 7 and x1= 1, x2 = 1, x3 = 1

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b) Fractional Cut Method

Standard form

Max Z = 3x1 + x2 + 3x3

s. t. - x1 + 2x2 + x3 ≤ 4

4x2 - 3x3 ≤ 2

x1 - 3x2 + 2x3 ≤ 3

x1, x2, x3 ≥ 0 and integers

IBFS

S1 = 0 and S2 = 3

Cj (4 6 0 0)

CB

YB

XB

Y1*

Y2

Y3

Y4

0

Y3

0

2

4

2

0

0

Y4

3

1

3

3

1

CB X Y1 CB X Y2 + CB X Y3

Zj 0X2 = 0, 0 X 2 + 0X3 = 0; 0 0

0 0 0 0

Net Evaluation Zj - Cj 10 13 0 0

Net Evaluation not >= 0

Most negative -4, arbitrary we select Y1 enters basis.

Min {XB / Y1j} = min {1/2, 4/2} 1/2 Y3 Leave basis

Cj (3 4 0 0)

CB

YB

XB

Y1*

Y2

Y3

Y4

1

Y1

3/2

3

2

1

0

0

Y4

6/3

5

3

0

1

CB X Y1 CB X Y2 + CB X Y3

Zj 0X2 = 0, 0 X 2 + 0X3 = 0; 0 0

0 0 0 0

Net Evaluation Zj - Cj -3 -4 0 0

Net Evaluation not >= 0

Most negative -4, arbitrary we select Y1 enters basis.

Min {XB / Y1j} = min {5/2, 3/2} 5/2 Y3 Leave basis

CB

YB

XB

Y1*

Y2

Y3

Y4

1

Y1

6/3 = 3

2/2 = 1

2/2 = 1

1/2

0

0

Y4

9/3 = 3

2/3

3/3 = 3

0/3 = 0

1/3

CB X Y1 CB X Y2 + CB X Y3

Zj 0X2 = 0, 0 X 2 + 0X3 = 0; 0 0

0 0 0 0

Net Evaluation Zj - Cj 1 3/2 0 0

Net Evaluation not >= 0

Most negative -4, arbitrary we select Y1 enters basis.

Min {XB / Y1j} = min {8, 3/5} 3/5 Y4 Leave basis.

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