JNTU
JNTU MCA 2020 OPERATIONS RESEARCH Previous Question Paper (October / November - 2020 - R15)
Answer any five questions. All questions carry equal marks.
Main
A company manufactures two kinds of machines, each requiring a different manufacturing technique. The deluxe machine requires 18 hours of labour, 8 hours of testing and yields a profit of Rs.400. The standard machine requires 3 hours of labour, 4 hours of testing and yields a profit of Rs.200. There are 800 hours of labour and 600 hours of testing available each month. A marketing forecast has shown that the monthly demand for the standard machine is to be more than 150. The management wants to know the number of each model to be produced monthly that will maximize total profit. Formulate and solve this as a linear programming problem.
Answer
Let x be the number of deluxe machines and y be the number of standard machines produced monthly. The objective is to maximize the total profit Z.
Subject to the constraints based on resource availability and market demand:
- Labour: 18x + 3y ≤ 800
- Testing: 8x + 4y ≤ 600
- Demand: y > 150
- Non-negativity: x ≥ 0
To solve, we analyze the constraints. From the demand constraint y > 150, we substitute y = 150 into the other constraints to find the feasible region boundaries:
- For labour: 18x + 3(150) = 800 ⇒ 18x = 350 ⇒ x = 19.44
- For testing: 8x + 4(150) = 600 ⇒ 8x = 0 ⇒ x = 0
The feasible region for x is 0 ≤ x ≤ 19.44. Since we want to maximize Z = 400x + 200y, and the coefficient of x (400) is greater than the coefficient of y (200), we should maximize x for any given y. However, y is constrained by the testing limit: 4y ≤ 600 - 8x. As x increases, the maximum possible y decreases. At x = 0, y can be up to 150 (testing limit 600/4). At x = 19.44, y = 150 (labour limit). Testing the vertices of the feasible region:
- Point (0, 150): Z = 400(0) + 200(150) = 30,000
- Point (19.44, 150): Z = 400(19.44) + 200(150) = 7,776 + 30,000 = 37,776
Rounding to the nearest whole machine, producing 19 deluxe machines and 150 standard machines yields a maximum profit of Rs. 37,600.
A nutrition scheme for babies is proposed by a committee of doctors. Babies can be given two types of food (I and II) which are available in standard sized packets weighing 50 grams. The cost per packet of these foods are Rs.2 and Rs.3. The Vitamin availability in each type of food per packet and the minimum vitamin requirement for each type of vitamin are summarized in table below:
| Vitamin | Vitamin availability in type I food | Vitamin availability in type II food | Minimum daily required vitamin |
|---|---|---|---|
| 1 | 1 | 1 | 6 |
| 2 | 7 | 1 | 14 |
| Cost/packet | Rs. 2 | Rs. 3 | - |
Develop a linear programming model to determine the optimal combination of food types with the minimum cost such that the minimum requirement of vitamin in each type is satisfied solve it by simplex method.
Answer
To minimize the cost of the nutrition scheme, we define the decision variables as follows: Let x₁ be the number of packets of food type I and x₂ be the number of packets of food type II. The objective is to minimize the total cost function:
Subject to the constraints based on vitamin requirements:
To solve using the Simplex Method (Big-M method), we introduce surplus variables s₁, s₂ and artificial variables A₁, A₂:
The modified objective function is Minimize Z = 2x₁ + 3x₂ + MA₁ + MA₂. Converting to maximization: Maximize -Z = -2x₁ - 3x₂ - MA₁ - MA₂. Substituting A₁ = 6 - x₁ - x₂ + s₁ and A₂ = 14 - 7x₁ - x₂ + s₂, the objective function becomes:
Performing the simplex iterations:
- Iteration 1: Entering variable is x₁ (most positive coefficient in row 0). Ratio test: 6/1=6, 14/7=2. Pivot element is 7.
- After pivoting, x₁ becomes a basic variable.
- Iteration 2: The objective row coefficients are checked. The entering variable is x₂.
- The final optimal solution is obtained at x₁ = 4/3 and x₂ = 14/3.
The minimum cost is calculated as:
A company as three factories at Amethi, Baghpat and Gwalior having production capacity of 5,000, 6,000 and 2,500 tonnes respectively. Four distribution centres at Allahabad, Bombay, Kolkata and Delhi requiring 6,000 tonnes, 4,000 tonnes, 2,000 tonnes and 1,500 tonnes respectively of the product. The transportation costs in thousands of rupees per tonne from different factories to different centres are given as below:
| Factories | Allahabad | Bombay | Kolkata | Delhi |
|---|---|---|---|---|
| Amethi | 3 | 2 | 7 | 6 |
| Bagphat | 7 | 5 | 2 | 3 |
| Gwalior | 2 | 3 | 4 | 5 |
Suggest an optimum schedule and find the minimum cost of transportation.
Answer
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Consider the problem of assigning four sales persons to four different sales regions as shown below such that the total sales are maximized.
| Sales man | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| A | 5 | 11 | 8 | 9 |
| B | 5 | 7 | 9 | 7 |
| C | 7 | 8 | 9 | 9 |
| D | 6 | 8 | 11 | 12 |
The cell entries represent annual sales figures in crore of rupees. Find the optimal allocation of the sales persons to different regions.
Answer
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Find the sequence that minimizes the total time required in performing the following jobs on three machines in the order ABC. Processing time in hours are given in the following table:
| Job | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| M/c A | 8 | 10 | 6 | 7 | 11 |
| M/c B | 5 | 6 | 2 | 3 | 4 |
| M/c C | 4 | 9 | 8 | 6 | 5 |
What is the total processing time for all the 5 jobs? Also find the idle time on each machine.
Answer
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A company is planning to replace an equipment by a new equipment whose first cost is Rs.1,00,000. The operating and maintenance cost of the equipment during its first year of operation is Rs.10,000 and it increases by Rs.2,000 every year thereafter. The resale value of the equipment at the end of the first year of its operation is Rs.65,000 and it decreases by Rs.10,000 every year thereafter. Find the economic life of the equipment by assuming the interest rate as 12%.
Answer
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Consider 4 × 4 game played by Players A and B and solve it optimally:
| Player A / B | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| A1 | 6 | 2 | 4 | 8 |
| A2 | 2 | -1 | 1 | 12 |
| A3 | 2 | 3 | 3 | 9 |
| A4 | 5 | 2 | 6 | 10 |
Answer
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In a railway marshalling yard, goods trains arrive at a rate of 30 trains per day. Assuming that the inter-arrival time follows an exponential distribution and the service time (the time taken to hump the train) distribution is also exponential with an average of 36 minutes. Calculate:
Expected queue size (line length)
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Probability that the queue size exceeds 10.
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If the input of trains increases to an average of 33 per day, what will the change in a and b?
Answer
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