Question 5 OR (b)
Minimum-cost blend of two raw materials by the simplex (Big-M) method
12 marks
- Time to write
- About 24 minutes
Question
Question 5 OR (b) · 12 marksA firm's final product must weigh exactly 6 kg. The two raw materials used to make it are A, at a cost of ₹6 per kg, and B, at a cost of ₹10 per kg. At least 3 kgs of B and not more than 5 kg of A must be used. How much of each raw material should be used per unit of the final product if the cost has to be minimized? Use the simplex method.
Reworded from the university paper: names and figures differ, the task is the same. The original paper belongs to the University of Delhi. See DU's question bank.
Before you write
Read off the question and its marks
Read from the question
- At least 3The question sets a minimum: give 3 or more.
- How muchFormula first, then the working, then the answer with its unit (₹, %, units).
What the paper leaves open
Second part of the alternative to Q5 (with the part before it, 6 + 12 = 18 marks).
Write it in this order
- Read the data; note exactly what is asked 2 min
- 1 Formulation and standard form slack, surplus, artificial 3 marks 5 min
- 2 Initial tableau and first iteration 3 marks 5 min
- 3 Second iteration 3 marks 5 min
- 4 Optimal solution and interpretation 3 marks 5 min
- Check: arithmetic, units (₹, %), final answer underlined 2 min
Try it first. About 24 minutes for 12 marks. Write it out, then open the solution.
Solution
Hover or tap a marked figure, like 72,000, to see how it is made and where else it appears.
Formulation. Let x₁ kg of A and x₂ kg of B go into one unit of the final product.
Minimise Z = 6x₁ + 10x₂ subject to x₁ + x₂ = 6 (weight exactly 6 kg) x₁ ≤ 5 (not more than 5 kg of A) x₂ ≥ 3 (at least 3 kg of B) x₁, x₂ ≥ 0.
This has mixed constraints (=, ≤, ≥), a case the DU guideline includes for the simplex method.
Standard form. Add a slack S₁ to the ≤ constraint, subtract a surplus S₂ from the ≥ constraint, and add artificial variables A₁ (equation) and A₂ (≥ constraint), each with cost M (a very large number):
Minimise Z = 6x₁ + 10x₂ + 0S₁ + 0S₂ + MA₁ + MA₂ x₁ + x₂ + A₁ = 6; x₁ + S₁ = 5; x₂ − S₂ + A₂ = 3.
The simplex method moves from one basic feasible solution to a better one until no improvement is possible, in a finite number of steps. For minimisation the entering variable has the most negative Cⱼ − Zⱼ; the solution is optimal when every Cⱼ − Zⱼ ≥ 0.
| Cᵦ | Basis | x₁ (6) | x₂ (10) | S₁ (0) | S₂ (0) | A₁ (M) | A₂ (M) | Quantity | Ratio |
|---|---|---|---|---|---|---|---|---|---|
| M | A₁ | 1 | 1 | 0 | 0 | 1 | 0 | 6 | 6/1 = 6 |
| 0 | S₁ | 1 | 0 | 1 | 0 | 0 | 0 | 5 | — |
| M | A₂ | 0 | 1 | 0 | −1 | 0 | 1 | 3 | 3/1 = 3 ← |
| Cⱼ − Zⱼ | 6 − M | 10 − 2M | 0 | M | 0 | 0 | Z = 9M |
x₂ enters (most negative, 10 − 2M); A₂ leaves (smallest ratio 3); key element 1.
| Cᵦ | Basis | x₁ | x₂ | S₁ | S₂ | A₁ | A₂ | Quantity | Ratio |
|---|---|---|---|---|---|---|---|---|---|
| M | A₁ | 1 | 0 | 0 | 1 | 1 | −1 | 3 | 3/1 = 3 ← |
| 0 | S₁ | 1 | 0 | 1 | 0 | 0 | 0 | 5 | 5/1 = 5 |
| 10 | x₂ | 0 | 1 | 0 | −1 | 0 | 1 | 3 | — |
| Cⱼ − Zⱼ | 6 − M | 0 | 0 | 10 − M | 0 | 2M − 10 | Z = 3M + 30 |
x₁ enters (6 − M is the most negative); A₁ leaves (ratio 3 against 5).
| Cᵦ | Basis | x₁ | x₂ | S₁ | S₂ | A₁ | A₂ | Quantity |
|---|---|---|---|---|---|---|---|---|
| 6 | x₁ | 1 | 0 | 0 | 1 | 1 | −1 | 3 |
| 0 | S₁ | 0 | 0 | 1 | −1 | −1 | 1 | 2 |
| 10 | x₂ | 0 | 1 | 0 | −1 | 0 | 1 | 3 |
| Cⱼ − Zⱼ | 0 | 0 | 0 | 4 | M − 6 | M − 4 | Z = 48 |
Every Cⱼ − Zⱼ ≥ 0 (4, M − 6, M − 4 are non-negative for large M), and no artificial variable is in the basis, so this tableau is optimal.
Optimal solution: x₁ = 3 kg of A, x₂ = 3 kg of B, minimum cost Z = 6(3) + 10(3) = ₹48 per unit of product. S₁ = 2: the 5 kg ceiling on A is not reached (only 3 kg is used). The non-basic S₂ has Cⱼ − Zⱼ = 4: each extra kg of B required above 3 kg would raise the cost by ₹4 (replacing a kg of A at ₹6 by a kg of B at ₹10).
Check by substitution: 3 + 3 = 6 kg; A = 3 ≤ 5; B = 3 ≥ 3; cost 18 + 30 = ₹48. The other corner of the feasible segment, x₁ = 0, x₂ = 6, costs ₹60, and x₁ cannot exceed 3 because x₂ ≥ 3.
Where the 12 marks sit
| # | Component | Marks |
|---|---|---|
| 1 | Formulation and standard form (slack, surplus, artificial) | 3 |
| 2 | Initial tableau and first iteration | 3 |
| 3 | Second iteration | 3 |
| 4 | Optimal solution and interpretation | 3 |
| Total | 12 |
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