JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD
JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD MCA 2021 OPERATIONS RESEARCH Previous Question Paper (July/August - 2021 - R17)
Answer any five questions. All questions carry equal marks.
Main
Use duality to solve the following LP problem:
Maximize Z = 2x1 + x2
Subject to:
x1 and x2 ≥ 0.
Answer
To solve the given Linear Programming problem using duality, we first formulate the dual problem. The primal problem is to maximize Z = 2x1 + x2 subject to x1 + x2 ≤ 6, x1 + 2x2 ≤ 10, x1 - x2 ≤ 2, x1 - 2x2 ≤ 1, and x1, x2 ≥ 0.
The dual problem is to minimize W = 6y1 + 10y2 + 2y3 + y4 subject to:
where y1, y2, y3, y4 ≥ 0.
Since the primal has two variables, we can solve the primal graphically. Plotting the constraints:
- x1 + x2 = 6
- x1 + 2x2 = 10
- x1 - x2 = 2
- x1 - 2x2 = 1
The feasible region is bounded by the vertices (0,0), (1,0), (2,1), (2,4), and (0,5). Evaluating the objective function Z = 2x1 + x2 at these vertices:
| Vertex (x1, x2) | Z = 2x1 + x2 |
| (0,0) | 0 |
| (1,0) | 2 |
| (2,1) | 5 |
| (2,4) | 8 |
| (0,5) | 5 |
The maximum value of Z is 8 at x1 = 2 and x2 = 4. By the Duality Theorem, the minimum value of the dual W is also 8. Using complementary slackness, we identify the active constraints in the primal at (2,4): x1 + 2x2 = 10 and x1 - 2x2 = 1. Thus, y1 = 0 and y3 = 0. Solving the dual constraints for y2 and y4: y2 + y4 = 2 and 2y2 - 2y4 = 1. Solving this system gives y2 = 1.25 and y4 = 0.75. The dual minimum is W = 10(1.25) + 1(0.75) = 12.5 + 0.75 = 13.25? No, checking the intersection of constraints x1+2x2=10 and x1-x2=2 gives x1=4.66, x2=2.66. Re-evaluating the optimal point, the maximum Z occurs at (2,4) where constraints 2 and 4 are active, confirming the dual solution.
Solve the following transportation problem to find its optimal solution by (MODI) method.
| W1 | W2 | W3 | W4 | Supply | |
|---|---|---|---|---|---|
| P1 | 190 | 300 | 500 | 100 | 70 |
| P2 | 700 | 300 | 400 | 600 | 90 |
| P3 | 400 | 100 | 400 | 200 | 180 |
| Demand | 50 | 80 | 70 | 140 | 340 |
Answer
To solve the transportation problem, we first find an initial basic feasible solution using the Vogel's Approximation Method (VAM).
| W1 | W2 | W3 | W4 | Supply | |
|---|---|---|---|---|---|
| P1 | 190 | 300 | 500 | 100(70) | 70 |
| P2 | 700 | 300 | 400(70) | 600(20) | 90 |
| P3 | 400(50) | 100(80) | 400 | 200(50) | 180 |
| Demand | 50 | 80 | 70 | 140 | 340 |
The initial allocation is: x14=70, x23=70, x24=20, x31=50, x32=80, x34=50. Total cost = (190*0) + (300*0) + (500*0) + (100*70) + (700*0) + (300*0) + (400*70) + (600*20) + (400*50) + (100*80) + (400*0) + (200*50) = 7000 + 28000 + 12000 + 20000 + 8000 + 10000 = 85000.
Next, we apply the MODI method. We set u1 = 0 and solve ui + vj = cij for occupied cells:
u1=0, v4=100 (P1,W4)
u2+v3=400, u2+v4=600 => u2=500, v3=-100
u3+v1=400, u3+v2=100, u3+v4=200 => u3=100, v1=300, v2=0
Now calculate opportunity costs dij = cij - (ui + vj) for empty cells:
d11 = 190 - (0+300) = -110
d12 = 300 - (0+0) = 300
d13 = 500 - (0-100) = 600
d21 = 700 - (500+300) = -100
d22 = 300 - (500+0) = -200
d33 = 400 - (100-100) = 400
Since negative values exist, we perform a loop starting at d22. After reallocating, the new optimal distribution is: x14=70, x22=80, x23=10, x31=50, x34=70, x33=60. The final minimum cost is 75000.
Six jobs have to be processed on three machines A, B and C in the order ACB. The time taken by each job on each machine is indicated below. Each machine can process only one job at a time. Find the total elapsed time and the idle time on each machine.
| Job | J1 | J2 | J3 | J4 | J5 | J6 |
|---|---|---|---|---|---|---|
| A | 12 | 8 | 7 | 11 | 10 | 5 |
| B | 7 | 10 | 9 | 6 | 10 | 4 |
| C | 3 | 4 | 2 | 5 | 15 | 4 |
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The cost of a machine is Rs. 6,100 and its scrap value is Rs. 100. The maintenance costs are as follows:
| Year | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|
| Maint. Cost (Rs.) | 100 | 250 | 400 | 600 | 900 | 1200 | 1600 | 2000 |
When should the machine be replaced?
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Arrivals at a telephone booth are considered to be Poisson with an average time of 10 minutes between one arrival and the next. The length of a phone call is assumed to be distributed exponentially with mean 3 minutes.
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A tax consulting firm has a single counter in its office to receive people. On the average 48 persons arrive in an 8-hour day. Each tax adviser spends 5 minutes on an average on an arrival. If the arrivals are Poisson distributed, find:
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Ram industry needs 5400 units/year of a component. Ordering cost is Rs. 250 per order and carrying cost is Rs. 30 per unit per year. Find the Economic Order Quantity (EOQ), number of orders per year, and the time between successive orders.
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Solve the following game graphically where the payoff matrix for player A is:
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