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Linear Programming - Model Formulation & Graphical Method |
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Diet Problem
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| Nutrient | Food Item | |
|---|---|---|
| A | B | |
| Thiamin | 0.12 mg/oz | 0.10 mg/oz |
| Phosphorus | 0.75 mg/oz | 1.70 mg/oz |
| Iron | 1.20 mg/oz | 1.10 mg/oz |
The cost of food items A and B is Rs. 2 per oz and Rs. 1.70 per oz respectively. The minimum daily requirements of these nutrients are atleast 1.00 mg of thiamin, 7.50 mg of phosphorus, and 10.00 mg of iron. Formulate this problem in the linear programming form.
Let x1 and x2 be the number of units (ounces) of A and B respectively. The objective here is to minimize the total cost of the food items, which is given by the linear function
Minimize z = 2x1 + 1.7x2
0.12x1 + 0.10x2 ³
1.0
0.75x1 + 1.70x2 ³
7.5
1.20x1 + 1.10x2 ³
10.0
x1 ³ 0, x2 ³ 0
The manger of Deep Sea Oil Refinery must decide on the optimal mix of two possible blending processes of which the inputs and outputs per production run are given in the following table:
| Process | Input (units) | Output (units) | Profit | ||
|---|---|---|---|---|---|
| Crude P | Crude Q | Gasoline S | Gasoline T | ||
| A | 6 | 2 | 6 | 8 | 400 |
| B | 3 | 6 | 4 | 5 | 500 |
| Total availabilities | 300 | 250 | 120 | 100 | |
Formulate the blending problem as a linear programming problem.
Let x1 and x2 be the number of production runs of process A & B respectively. The objective here is to maximize the profit. The decision problem can be formulated as
Maximize z = 400x1 + 500x2
subject to
6x1 + 3x2 £ 300
2x1 + 6x2 £ 250
6x1 + 4x2 ³ 120
8x1 + 5x2 ³ 100
x1 ³ 0, x2 ³ 0
Although all the above illustrations were highly simplified, the decision problems shown have close counterparts in actual companies. The effort required to formulate a real situation depends on the complexity of the problem.
By studying the above examples and answering the formulation exercises given at the end of this chapter, you will acquire considerable experience of formulating a model.
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"Experience is a dim lamp, which only lights the one who bears it." -Louis Ferdinand |