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 15-29
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 reduced-cost 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.

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