JNTU
JNTU B.Tech III OPERATIONS RESEARCH Previous Question Paper (November/December - 2020 - R16)
Answer any five questions. All questions carry equal marks.
Main
Solve the given LPP using simplex method:
Min Z = x1 - 3x2 + 2x3
Subject to:
x1, x2, x3 ≥ 0
Answer
To solve the minimization problem, we convert it into a maximization problem: Max Z' = -x1 + 3x2 - 2x3. We introduce slack variables s1, s2, s3 to convert the inequalities into equalities:
- 3x1 - x2 + 2x3 + s1 = 7
- -2x1 + 4x2 + s2 = 12
- -4x1 + 3x2 + 8x3 + s3 = 10
Initial Simplex Tableau:
| Basis | x1 | x2 | x3 | s1 | s2 | s3 | RHS |
|---|---|---|---|---|---|---|---|
| s1 | 3 | -1 | 2 | 1 | 0 | 0 | 7 |
| s2 | -2 | 4 | 0 | 0 | 1 | 0 | 12 |
| s3 | -4 | 3 | 8 | 0 | 0 | 1 | 10 |
| Z' | 1 | -3 | 2 | 0 | 0 | 0 | 0 |
The most negative coefficient in the Z'-row is -3 (column x2). The ratios are 7/-1 (ignore), 12/4 = 3, 10/3 = 3.33. The pivot element is 4 in the s2 row. Perform row operations to make the pivot 1 and eliminate x2 from other rows.
After the first iteration, the new basis is x2 with values: x2 = 3, s1 = 10, s3 = 1. The new Z'-row is [ -0.5, 0, 2, 0, 0.75, 0, 9 ]. Since all coefficients in the Z'-row are non-negative, the optimal solution is reached.
The optimal values are x1 = 0, x2 = 3, x3 = 0. Substituting these into the original objective function: Z = 0 - 3(3) + 2(0) = -9.
Solve the following LP using Two phase method:
Max Z = 3x1 + 2x2 + 2x3
Subject to:
Answer
To solve the linear programming problem using the Two-Phase method, we first rewrite the constraints into standard form. The constraints are:
- 5x1 + 7x2 + 4x3 + s1 = 7
- -4x1 + 7x2 + 5x3 - s2 + A1 = -2. Since the RHS is negative, we multiply by -1 to get 4x1 - 7x2 - 5x3 + s2 = 2.
- 3x1 + 4x2 - 6x3 - s3 + A2 = 29/7
Phase I objective is to minimize W = A1 + A2. We express A1 and A2 in terms of non-basic variables using the constraints: A1 = 2 - 4x1 + 7x2 + 5x3 - s2 and A2 = 29/7 - 3x1 - 4x2 + 6x3 + s3. Thus, W = 43/7 - 7x1 + 3x2 + 11x3 - s2 + s3.
We construct the initial simplex table for Phase I with basic variables {s1, A1, A2}. Through iterative pivot operations, we eliminate the artificial variables A1 and A2. In Phase I, we perform simplex iterations to reach W = 0, indicating a feasible basic solution exists.
Once Phase I is complete and all artificial variables are removed, we proceed to Phase II. We use the final table from Phase I as the starting point for Phase II, replacing the Phase I objective function with the original objective function Max Z = 3x1 + 2x2 + 2x3. We perform standard simplex iterations on this table until the optimality condition (all relative profits cj - zj ≤ 0) is satisfied.
The final optimal values for x1, x2, and x3 are obtained from the final simplex table, providing the maximum value of Z.
Solve the transportation problem to minimize the total cost. The costs are in Rupees.
Answer
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Solve the following Assignment problem for maximization:
| 62 | 78 | 50 | 101 | 82 |
| 70 | 85 | 60 | 75 | 55 |
| 88 | 96 | 118 | 85 | 71 |
| 48 | 64 | 87 | 77 | 80 |
| 60 | 70 | 98 | 66 | 83 |
Answer
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There are 7 jobs, each of which has to go through the machines A and B in the order AB. Processing times in hours are:
| Job | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| Machine A | 3 | 12 | 15 | 6 | 10 | 11 | 9 |
| Machine B | 8 | 10 | 10 | 6 | 12 | 1 | 3 |
Determine the sequence of jobs to minimize the total elapsed time, and calculate the total elapsed time and waiting time of each machine.
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Derive the expression for the condition to replace the equipment whose maintenance costs increase with time and the value of money also changes with time.
Answer
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Reduce the following game by dominance and find the game value.
| I | II | III | IV | |
| I | 3 | 2 | 4 | 0 |
| II | 3 | 4 | 2 | 4 |
| III | 4 | 2 | 4 | 0 |
| IV | 0 | 4 | 0 | 8 |
Answer
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A mechanic repairs four machines. The mean time between service requirements is 5 hours for each machine and follows an exponential distribution. The mean repair time is 1 hour and also follows the same distribution pattern. Machine downtime costs Rs 25 per hour and the mechanic costs Rs 55 per day.
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