SMU MBA II Sem Operation Research May 2014
Q. A paper mill produces two grades of paper viz., X and Y. Because of raw material restrictions, it cannot produce more than 400 tons of grade X paper and 300 tons of grade Y paper in a week. There are 160 production hours in a week. It requires 0.20 and 0.40 hours to produce a ton of grade X and Y papers. The mill earns a profit of Rs. 200 and Rs. 500 per ton of grade X and Y paper respectively. Formulate this as a Linear Programming
Let x1 for paper X and x2 for paper Y
Objective Function (Z): Profit of both X and Y Paper is given.
Max (Z) = 200×1+500×2
Constrains: two, one for production hours and second for raw material;
Subject to:
X1<=400
X2<=300
0.20×1+0.40×2<=160
x1, x2 >=0
Q. A Company produces 150 cars. But the production rate varies with the distribution.
Production Rate  147  148  149  150  151  152  153 
Probability  0.05  0.10  0.15  0.20  0.30  0.15  0.05 
At present the track will hold 150 cars. Using the following random numbers determine the average number of cars waiting for shipment in the company and average number of empty space in the truck. Random Numbers 82, 54, 50, 96, 85, 34, 30, 02, 64, 47.
Solution:
Production Rate  Probability  Cumulative probability  RN Range 
147  0.05  0.05  00 –04 
148  0.10  0.15  05 – 14 
149  0.15  0.30  1529 
150  0.20  0.50  30 – 49 
151  0.30  0.80  50 – 79 
152  0.15  0.95  80 – 94 
153  0.05  1.00  95 – 99 
Simulation for 10 days using the given random numbers
Days  RN  Production Rate  Car Waiting  Space in the truck 
1  82  152  2  
2  54  150  –  
3  50  150  –  
4  96  153  3  
5  85  152  2  
6  34  150  –  
7  30  150  –  
8  02  147  –  3 
9  64  151  1  
10  47  150  2 
Total 8 3
Therefore, Avg number of cars waiting =8/10= 0.8 /day
Avg number of empty space =3/10= 0.3/day
Q. Find an optimal solution to an assignment problem with the following cost matrix:
J1 J2 J3 J4
M1 10 9 7 8
M2 5 8 7 7
M3 5 4 6 5
M4 2 3 4 5
To illustrate this consider the following assignment problem
J1 J2 J3 J4
M1 10 9 7 8
M2 5 8 7 7
M3 5 4 6 5
M4 2 3 4 5
The solution is as follows.
First, the minimum element in each row is subtracted from all the elements in that row.
This gives the following reducedcost matrix.
J1 J2 J3 J4
M1 3 2 0 1
M2 0 3 2 2
M3 1 0 2 1
M4 0 1 2 3
Since both the machines M2 and M4 have a zero cost corresponding to job J1 only, a
feasible assignment using only cells with zero costs is not possible. To get additional
zeros subtract the minimum element in the fourth column from all the elements in that
column. The result is shown in the table below.
J1 J2 J3 J4
M1 3 2 0 0
M2 0 3 2 1
M3 1 0 2 0
M4 0 1 2 2
Only three jobs can be assigned using the zero cells, so a feasible assignment is still not
possible. In such cases, the procedure draws a minimum number of lines through some
selected rows and columns in such a way that all the cells with zero costs are covered by
these lines. The minimum number of lines needed is equal to the maximum number of
jobs that can be assigned using the zero cells. 65
In the current example, this can be done with three lines as follows.
J1 J2 J3 J4
M1 3 2 0 0
M2 0 3 2 1
M3 1 0 2 0
M4 0 1 2 2
Now select the smallest element, which is not covered by the lines. In the current
example, it is 1. Subtract this number from all the elements, which are not covered. Then
add this number to all those covered elements that are at the intersection of two lines.
This gives the following reduced cost matrix.
J1 J2 J3 J4
M1 4 2 0 0
M2 0 2 1 0
M3 2 0 2 0
M4 0 0 1 1
A feasible assignment is now possible and an optimal solution is assigned
M1 J3
M2 J1
M3 J4
M4 J2
The total cost is given by: 7 +5+ 5+ 3 = 20
An alternate optimal solution is:
M1 J3
M2 J4
M3 J2
M4 J1
In case a feasible set could not be obtained at this step, one has to repeat the step of
drawing lines to cover the zeros and continue until a feasible assignment is obtained.
SMU Short Notes: Useful for MBA Assignments and SMU Exams
MBA IInd (2nd) Operation Research Solved Assignment 2014
For complete solution you can go to:
http://www.researchingtheworld.com/category/smuassignment2/smuassignment/
Solution 2b)
Let us variables X1 & X2 respectively for black & white
The profit per unit on these products in Rs. 50 & Rs. 40 respectively
So this is the case for profit maximisation
Max(Z) = 50*X1 + 40*X2
Total capacity = 50,000 man hours
Taken time to produced one unit of black = 3 hours
Taken time to produced one unit of white = 2 hours
3*X1 + 2*X2 ≤ 50,000
Market max demand for black = 8,000 units
Market max demand for white = 10,000 units
X1 ≤ 8,000
X2 ≤ 10,000
Solution 3b)
1) Identify the smallest and the next to smallest costs. Determine the difference between them for each row & column
2) Identify the row or column with the largest difference among all the rows and columns.
Factories 
C1 
C2 
C3 
C4 
Supply 

F1 
3 
2(40) 
7 
6 
5040 
1 
F2 
7 
5 
2 
3 
60 
1 
F3 
2 
5 
4 
5 
25 
2 
Requirements 
60 
40 
20 
15 

1 
3 
2 
2 
Factories 
C1 
C2 
C3 
C4 
Supply 

F1 
3(10) 
2(40) 
7 
6 
10 
1 
3 
F2 
7 
5 
2 
3 
60 
1 
1 
F3 
2 
5 
4 
5 
25 
2 
2 
Requirements 
6010 
40 
20 
15 

1 
3 
2 
2 

1 
X 
2 
2 
Factories 
C1 
C2 
C3 
C4 
Supply 

F1 
3(10) 
2(40) 
7 
6 
10 
1 
3 
x 
F2 
7 
5 
2 
3 
60 
1 
1 
1 
F3 
2(25) 
5 
4 
5 
25 
2 
2 
2 
Requirements 
5025 
40 
20 
15 

1 
3 
2 
2 

1 
X 
2 
2 

5 
X 
2 
2 
Factories 
C1 
C2 
C3 
C4 
Supply 

F1 
3(10) 
2(40) 
7 
6 
10 
1 
3 
x 
F2 
7(25) 
5 
2 
3 
6025 
1 
1 
1 
F3 
2(25) 
5 
4 
5 
25 
2 
2 
2 
Requirements 
25 
40 
20 
15 

1 
3 
2 
2 

1 
x 
2 
2 

5 
x 
2 
2 
Factories 
C1 
C2 
C3 
C4 
Supply 

F1 
3(10) 
2(40) 
7 
6 
10 
1 
3 
x 
F2 
7(25) 
5 
2(20) 
3 
3520 
1 
1 
1 
F3 
2(25) 
5 
4 
5 
25 
2 
2 
2 
Requirements 
25 
40 
20 
15 

1 
3 
2 
2 

1 
x 
2 
2 

5 
x 
2 
2 
Factories 
C1 
C2 
C3 
C4 
Supply 

F1 
3(10) 
2(40) 
7 
6 
10 
1 
3 
x 
F2 
7(25) 
5 
2(20) 
3(15) 
15 
1 
1 
1 
F3 
2(25) 
5 
4 
5 
25 
2 
2 
2 
Requirements 
25 
40 
20 
15 

1 
3 
2 
2 

1 
x 
2 
2 

5 
x 
2 
2 
Total cost = 3*10 + 2*40 + 7*25 + 2*20 + 3*15 + 2*25
= 30 + 80 + 175 + 40 + 45 + 50
= 420
Solution 4b)
1) 1) Check the number rows equal to the number columns
Rows = 4
Columns = 4
Both rows & columns are equal, so move to next steps
2) 2) Row reduced matrix –
J1  J2  J3  J4  
M1  107  97  87  77 
M2  33  43  53  63 
M3  21  11  11  11 
M4  43  33  53  63 
J1  J2  J3  J4  
M1  5  2  1  0 
M2  0  1  2  3 
M3  1  0  0  0 
M4  1  0  2  3 
3)3) Column reduced matrix –
J1  J2  J3  J4  
M1  5  2  1  0 
M2  0  1  2  3 
M3  1  0  0  0 
M4  1  0  2  3 
444) Assign zero rowwise if there is only one zero in the row and cross (X) or cancel other zeros in that column.
J1  J2  J3  J4  
M1  5  2  1  [0] 
M2  [0]  1  2  3 
M3  1  (x)  [0]  (x) 
M4  1  [0]  2  3 
Since the number of assignments is 4 which is equal to the number of rows
M1 to J4 => 7
M2 to J1 => 3
M3 to J3 => 1
M4 to J2 => 3
Total cost => 14
Therefore, the optimum assignment schedule is M1 to J4, M2 to J1, M3 to J3, M4 to J2;
SMU Solved Assignment For MBA (20132014): Statistics (Numerical Part only)
For complete solved assignment in clear image please click:
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Q.2 (b) In a sample of 120 workers in a factory, the mean and standard deviation of wages
were Rs. 11.35 and Rs.3.03 respectively. Find the percentage of workers getting wages
between Rs.9 and Rs.17 in the whole factory assuming that the wages are normally
distributed.
Q.4 (b) Calculate Karl Pearson’s coefficient of correlation between X series and Y series.
X 110 120 130 120 140 135 155 160 165 155
Y 12 18 20 15 25 30 35 20 25 10
Q.6 Construct Fisher’s Ideal Index for the given information and check whether Fisher’s
formula satisfies Time Reversal and Factor Reversal Tests.
Items P0 Q0 P1 Q1
A 16 5 20 6
B 12 10 18 12
C 14 8 16 10
D 20 6 22 10
E 80 3 90 5
F 40 2 50 5
Formula of Fishers Ideal Index
Computation of Fisher’s Ideal Index
Fisher’s formula satisfies Time Reversal Test
Fisher’s formula satisfies Factor Reversal Test