Operations Research: Linear Programming and Decision Analysis

This Operations Research assignment solution covers a range of topics, starting with linear programming problems solved using the Simplex method, including artificial variables and slack variables. It then delves into sensitivity analysis, examining how changes in profit coefficients affect the optimal solution. The solution further addresses transportation problems, utilizing methods like the Hungarian algorithm to determine optimal assignments and minimize costs. Finally, the assignment includes decision analysis, using expected monetary value (EMV) to evaluate different investment strategies and identify the most profitable decision. The solution is comprehensive, showing each step and calculation in detail, making it a valuable resource for students studying operations research.

Running head: Operations Research
Operations Research
Name of the student
Name of the university
Author’s note

1Operations Research
Table of Contents
Answer 1..........................................................................................................................................2
Part a............................................................................................................................................2
Part b............................................................................................................................................4
Answer 2........................................................................................................................................11
Part a..........................................................................................................................................11
Part b..........................................................................................................................................11
Part c..........................................................................................................................................12
Part d..........................................................................................................................................12
Part e..........................................................................................................................................13
Part f...........................................................................................................................................13
Part g..........................................................................................................................................14
Part h..........................................................................................................................................14
Answer 3........................................................................................................................................15
Answer 4........................................................................................................................................18
Answer 5........................................................................................................................................19
Part a..........................................................................................................................................19
Par b...........................................................................................................................................20

2Operations Research
Answer 1
Part a
For the equation, 3 x+ y=5we introduce the Artificial variable A1
Thus 3 x+ y+ A1=5
For the equation, 4 x+3 y ≥ 6we subtract the Slack variable S1 and Artificial variable A2 is used.
Thus 4 x+3 y −S1+ A1=6
For the equation, x +2 y ≤ 4we add the Slack variable S3
Thus x +2 y + S2=4
Thus, we have
3x + y + A1 = 3
4x + 3y −S1 + A2 = 6
x + 2y + S2 = 4
Where x , y , S1 , S2 , A1 , A2 ≥ 0
Iteration 1
C j 4 3 0 0 M M
B CB X B x y S1 S2 A1 A2 MinRatio
XB
x
A1 M 3 (3) 1 0 0 1 0 3
3 =1

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A2 M 6 4 3 -1 0 0 1 6
4 =3
2
S1 0 4 1 2 0 1 0 0 4
1 =4
Z = 0 Z j 7M 4M -M 0 M M
C j−Z j -7M+4 -4M+3 M 0 0 0
Negative minimum C j−Z jis -7M+4 and its column index is 1. Thus the entering variable is x1.
The minimum ration is 1 and its row index is 1. So, the leaving basis variable is A1.
Thus, the pivot element is 3.
Hence,
R1 ( new ) = R1 ( old )
5
R2 ( new ) =R2 ( old ) −4 R1 ( new )
R3 ( new ) =R2 ( old )−4 R1 ( new )
Iteration 2
C j 4 3 0 0 M
B CB X B x y S1 S2 A2 MinRatio
XB
x
x1 4 1 1 1
3
0 0 0 3
3 =1
A2 M 2 0 ( 5
3 ) -1 0 1 6
4 =3
2
S1 0 3 0 5
3
0 1 0 4
1 =4
Z = 4 Z j 4 5 M
3 + 4
3
-M 0 M
C j−Z j 0 −5 M
3 +5
3
M 0 0

4Operations Research
Negative minimum C j−Z jis −5 M
3 +5
3 and its column index is 2. Thus the entering variable is x2
.
The minimum ration is 6
5 and its row index is 2. So, the leaving basis variable is A2.
Thus, the pivot element is 5
3.
Hence,
R2 ( new ) =R2 ( old ) x ( 3
5 )
R1 ( new ) =R1 ( old )− 1
3 R2 ( new )
R3 ( new ) =R3 ( old ) − 5
3 R2 ( new )
Iteration 3
C j 4 3 0 0
B CB X B x y S1 S2 MinRatio
XB
x
x1 4 3
5
1 0 1
5
0
x2 3 6
5
0 1 −3
5
0
S1 0 1 0 0 1 1
Z = 6 Z j 4 3 -1 0
C j−Z j 0 0 1 0
Since, all C j−Z j ≥ 0

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Hence the optimal solution is given by
x= 3
5 , y= 6
5
Min Z = 6
Part b
For the equation, 3 x1−2 x2 +x3 ≤ 5 we introduce the Slack variable S1
Thus 3 x1−2 x2 + x3+ S1=5
For the equation, x1+ 3 x2−4 x3 ≤ 9 the Slack variable S2 is used.
Thus x1+ 3 x2−4 x3+S2=9
For the equation, x2+ 5 x3 ≥1 we subtract the Slack variable S3and add the artificial variable A1
Thus x2+ 5 x3−S3+ A1=1
For the equation, x1+ x2 +x3=6 the artificial variable A2is added.
Thus, x1+ x2 + x3 +A2=6
Thus, we have
3 x1 −2 x2 + x3 +S1 = 5
x1 +3 x2 −4 x3 + S2 = 9
x2 +5 x3 + S3 + A1 = 1
x1 + x2 + x3 + A2 = 6
Where x1 , x2 , x3 , S1 , S2 , S3 , A1 , A2 ≥ 0

6Operations Research
Iteration 1
Negative minimum C j−Z jis -6M+1 and its column index is 3. Thus the entering variable is x3.
The minimum ration is 1
5 and its row index is 3. So, the leaving basis variable is A1.
Thus, the pivot element is 5.
Hence,
R3 ( new ) = R3 ( old )
5
R1 ( new ) =R1 ( old ) −R3 ( new )
R2 ( new ) =R2 ( old ) +4 R3 ( new )
R4 ( new ) =R4 ( old ) −R3 ( new )
Iteration 2

7Operations Research
Negative minimum C j−Z jis -M+2 and its column index is 3. Thus the entering variable is x1.
The minimum ration is 8
5 and its row index is 1. So, the leaving basis variable is S1.
Thus, the pivot element is 3.
Hence,
R1 ( new ) = R1 ( old )
3
R2 ( new ) =R2 ( old ) −R1 ( new )
R3 ( new ) =R3 ( old )
R4 ( new ) =R4 ( old ) −R1 ( new )

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Iteration 3
Negative minimum C j−Z jis −23 M
15 − 26
15 and its column index is 2. Thus the entering variable is
x2.
The minimum ration is 1 and its row index is 3. So, the leaving basis variable is x3.
Thus, the pivot element is 1
5.
Hence,
R3 ( new ) =R3 ( old ) x 5
R1 ( new ) =R1 ( old ) + 11
15 R3 ( new )
R2 ( new ) =R2 ( old ) − 68
15 R3 ( new )
R4 ( new ) =R4 ( old ) − 23
15 R3 ( new )

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Iteration 4
Negative minimum C j−Z jis −5 M
3 − 5
3 and its column index is 6. Thus the entering variable is
S3
The minimum ration is 1 and its row index is 2. So, the leaving basis variable is S2.
Thus, the pivot element is 11
3 .
Hence,
R2 ( new ) =R2 ( old ) x ( 3
11 )

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R1 ( new ) =R1 ( old ) + 2
3 R2 ( new )
R3 ( new ) =R3 ( old ) + R2 ( new )
R4 ( n ew ) =R4 ( old ) − 5
3 R2
( new )
Iteration 5
Negative minimum C j−Z jis −29 M
11 − 18
11 and its column index is 3. Thus the entering variable is
x3.
The minimum ration is 11
29 and its row index is 4. So, the leaving basis variable is A2.

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Thus, the pivot element is 29
11.
Hence,
R4 ( new )=R4 ( old ) x( 11
29 )
R1 ( new ) =R1 ( old ) + 5
11 R4 ( new )
R2 ( new ) =R2 ( old ) + 68
11 R4 ( new )
R3 ( new ) =R3 ( old ) + 13
11 R4 ( new )
Iteration 6