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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
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
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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.
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 + S_{1} = 6
2x1+ 3x2 + S_{2} = 9
X1, X2, S1, S2 >= 0
IBFS
S1 = 6 and S2 = 9
C_{j} (3 4 0 0)
C_{B}

Y_{B}

X_{B}

Y_{1}*

Y_{2}

Y_{3}

Y_{4}

0

Y_{3}

6

2

2

1

0

0

Y_{4}

9

2

3

0

1

C_{B} X Y_{1} C_{B} X Y_{2} + C_{B} X Y_{3}
Z_{j} 0X2 = 0, 0 X 2 + 0X3 = 0; 0 0
0 0 0 0
Net Evaluation Z_{j}  C_{j} 3 4 0 0
Net Evaluation not >= 0
Most negative 4, arbitrary we select Y_{1} enters basis.
Min {X_{B}/Y_{1j}} = min {6/2, 9/2} 6/2 Y_{3} Leave basis
C_{j} (3 4 0 0)
C_{B}

Y_{B}

X_{B}

Y_{1}*

Y_{2}

Y_{3}

Y_{4}

1

Y_{1}

6/6 = 1

2

2

1

0

0

Y_{4}

9

2

3

0

1

C_{B} X Y_{1} C_{B} X Y_{2} + C_{B} X Y_{3}
Z_{j} 0X2 = 0, 0 X 2 + 0X3 = 0; 0 0
0 0 0 0
Net Evaluation Z_{j}  C_{j} 3 4 0 0
Net Evaluation not >= 0
Most negative 4, arbitrary we select Y_{1} enters basis.
Min {X_{B} / Y_{1j}} = min {6/2, 9/2} 6/2 Y_{3} Leave basis
C_{B}

Y_{B}

X_{B}

Y_{1}*

Y_{2}

Y_{3}

Y_{4}

1

Y_{1}

6/6 = 1

2/2 = 1

2/2 = 1

½

0

0

Y_{4}

9/3 = 3

2/3

3/3 = 3

0/3 = 0

1/3

C_{B} X Y_{1} C_{B} X Y_{2} + CB X Y_{3}
Z_{j} 0X2 = 0, 0 X 2 + 0X3 = 0; 0 0
0 0 0 0
Net Evaluation Z_{j}  C_{j} 1 3/2 0 0
Net Evaluation not >= 0
Most negative 4, arbitrary we select Y_{1} enters basis.
Min {X_{B}/Y_{1j}} = min {1, 3/2} 6/2 Y_{4} 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
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
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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
C_{j} (4 6 0 0)
C_{B}

Y_{B}

X_{B}

Y_{1}*

Y_{2}

Y_{3}

Y_{4}

0

Y_{3}

5

4

2

2

0

0

Y_{4}

5

3

3

0

1

C_{B} X Y_{1} C_{B} X Y_{2} + C_{B} X Y_{3}
Z_{j} 0X2 = 0, 0 X 2 + 0X3 = 0; 0 0
0 0 0 0
Net Evaluation Z_{j}  C_{j} 8 7 0 0
Net Evaluation not >= 0
Most negative 4, arbitrary we select Y_{1} enters basis.
Min {X_{B} / Y_{1j}} = min {6/2, 9/2} 6/2 Y_{3} Leave basis
C_{j} (3 4 0 0)
C_{B}

Y_{B}

X_{B}

Y_{1}*

Y_{2}

Y_{3}

Y_{4}

1

Y_{1}

6/7

2

2

1

0

0

Y_{4}

8/7

2

3

0

1

C_{B} X Y_{1} C_{B} X Y_{2} + C_{B} X Y_{3}
Z_{j} 0X2 = 0, 0 X 2 + 0X3 = 0; 0 0
0 0 0 0
Net Evaluation Z_{j}  C_{j} 3 4 0 0
Net Evaluation not >= 0
Most negative 4, arbitrary we select Y_{1} enters basis.
Min {X_{B} / Y_{1j}} = min {5/2, 3/2} 5/2 Y_{3} Leave basis
C_{B}

Y_{B}

X_{B}

Y_{1}*

Y_{2}

Y_{3}

Y_{4}

1

Y_{1}

6/3 = 3

2/2 = 1

2/2 = 1

1/2

0

0

Y_{4}

9/3 = 3

2/3

3/3 = 3

0/3 = 0

1/3

C_{B} X Y_{1} C_{B} X Y_{2} + C_{B} X Y_{3}
Z_{j} 0X2 = 0, 0 X 2 + 0X3 = 0; 0 0
0 0 0 0
Net Evaluation Z_{j}  C_{j} 1 3/2 0 0
Net Evaluation not >= 0
Most negative 4, arbitrary we select Y_{1} enters basis.
Min {X_{B} / Y_{1j}} = min {7, 5/2} 5/2 Y_{4} 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 01 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
C_{j} (4 6 0 0)
C_{B}

Y_{B}

X_{B}

Y_{1}*

Y_{2}

Y_{3}

Y_{4}

0

Y_{3}

0

2

4

2

0

0

Y_{4}

3

1

3

3

1

C_{B} X Y_{1} C_{B} X Y_{2} + C_{B} X Y_{3}
Z_{j} 0X2 = 0, 0 X 2 + 0X3 = 0; 0 0
0 0 0 0
Net Evaluation Z_{j}  C_{j} 10 13 0 0
Net Evaluation not >= 0
Most negative 4, arbitrary we select Y_{1} enters basis.
Min {X_{B} / Y_{1j}} = min {1/2, 4/2} 1/2 Y_{3} Leave basis
C_{j} (3 4 0 0)
C_{B}

Y_{B}

X_{B}

Y_{1}*

Y_{2}

Y_{3}

Y_{4}

1

Y_{1}

3/2

3

2

1

0

0

Y_{4}

6/3

5

3

0

1

C_{B} X Y_{1} C_{B} X Y_{2} + C_{B} X Y_{3}
Z_{j} 0X2 = 0, 0 X 2 + 0X3 = 0; 0 0
0 0 0 0
Net Evaluation Z_{j}  C_{j} 3 4 0 0
Net Evaluation not >= 0
Most negative 4, arbitrary we select Y_{1} enters basis.
Min {XB / Y_{1j}} = min {5/2, 3/2} 5/2 Y_{3} Leave basis
C_{B}

Y_{B}

X_{B}

Y_{1}*

Y_{2}

Y_{3}

Y_{4}

1

Y_{1}

6/3 = 3

2/2 = 1

2/2 = 1

1/2

0

0

Y_{4}

9/3 = 3

2/3

3/3 = 3

0/3 = 0

1/3

C_{B} X Y_{1} C_{B} X Y_{2} + C_{B} X Y_{3}
Z_{j} 0X2 = 0, 0 X 2 + 0X3 = 0; 0 0
0 0 0 0
Net Evaluation Z_{j}  C_{j} 1 3/2 0 0
Net Evaluation not >= 0
Most negative 4, arbitrary we select Y_{1} enters basis.
Min {X_{B} / Y_{1j}} = min {8, 3/5} 3/5 Y_{4} Leave basis.
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