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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:
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Extreme Point Coordinates (x1, x2)
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Line through Extreme Point
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Objective function value Z = 3x1+ 4x2
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O(0,0)
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5→ x1 >= 0
6→ x2 >= 0
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3(0) + 4(0) = 0
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A(3, 0)
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1→ 2 x1 + x2 <= 6
6→ x2 >= 0
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3(3) + 4(0) = 9
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B(9/4, 3/2)
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1 → 2 x1 + x2 <= 6
2 → 2 x1 + 3x2 <= 9
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3(9/4) + 4(3/2) = 51/4
|
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C(0,3)
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2 → 2 x1 + 3x2 <= 9
5 → x1 >= 0
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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
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XB
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Y1*
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Y2
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Y3
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Y4
|
|
0
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Y3
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6
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2
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2
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1
|
0
|
|
0
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Y4
|
9
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2
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3
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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
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YB
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XB
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Y1*
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Y2
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Y3
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Y4
|
|
1
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Y1
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6/6 = 1
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2
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2
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1
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0
|
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0
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Y4
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9
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2
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3
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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
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XB
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Y1*
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Y2
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Y3
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Y4
|
|
1
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Y1
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6/6 = 1
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2/2 = 1
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2/2 = 1
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½
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0
|
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0
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Y4
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9/3 = 3
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2/3
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3/3 = 3
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0/3 = 0
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1/3
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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
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
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x1 = 0, x2 = -4/5 then
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x3
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29/5
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x2 = 0 , x1 =5/4 then
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x3
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25/4
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x1 = 0, x2 = 5/6 then
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x3
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5
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x2 = 0 x1 = 5/6
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x3
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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
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-2
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2
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0
|
|
0
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Y4
|
5
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3
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3
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0
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1
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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
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Y4
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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
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0
|
|
0
|
Y4
|
9/3 = 3
|
2/3
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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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