1. The manager of an oil refinery must decide on the optimal mix of two possible

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1. The manager of an oil refinery must decide on the optimal mix of two possible

1. The manager of an oil refinery must decide on the optimal mix of two possible blending processes of which the input and output per production run are given as follows:
Process Units
Input
Crude A
Crude B
Output
Gasoline X
Gasoline Y
1
5
3
5
8
2
4
5
4
4
The maximum amount available of crude A and B are 200 units and 150 units respectively. Market requirements show that at least 100 units of gasoline X and 80 units of gasoline Y must be produced. The profit per production run from process 1 and process 2 are Rs. 300 and Rs. 400 respectively. Formulate this problem as a linear programming model.
2. A city police department has the following minimal daily requirement for policeman. Note, you are to consider period 1 as following immediately after period 6. Each policeman works eight consecutive hours. Let X denote the number of men starting work in period t everyday. The police department seeks a daily manpower schedule that employs the least number of policemen, provided that each of the above requirements is met. Formulate linear programming model to find an optimal schedule.
Time of Day
Period
Minimal number of police required during a period
2 – 6
1
20
6 – 10
2
50
10 – 14
3
80
14 – 18
4
100
18 – 22
5
40
22 – 2
6
30
3. A car dealer selects his cars for sale very carefully so as to ensure the optimization of his profits. He deals in 4 types of cars A, B, F and G. The purchase value of the cars range at Rs. 60,000, 150,000, 55,000 and 220,000 and the sales value is fixed at Rs. 80,000, 175,000, 75,000 and 250,000 respectively. The probability of sale are 0.8, 0.9, 0.6 and 0.50 respectively during a period of six months. In order to invest Rs. 20,00,000 in his deals, he wishes to maintain the rates of purchase of cars as 3 : 1 : 2 : 4. Work out how and how much he should buy. Formulate this problem as LP model.
4. Use graphical method to solve the following LP problems
Max. Z = 3×1 + 4×2
Subject to
2×1 + x2 ≤ 10
x1 + 3×2 ≤ 12
x1 + x2 ≤ 6
x1, x2 ≥ 0

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