Discussion
A dairy firm has three plants \(P_{1}, P_{2}, P_{3}\) located in a state. They daily milk production at each plant (in thousand liters) is \(P_{1} \to 6\), \(P_{2} \to 1\), \(P_{3} \to 10\). Each day, the firm must fulfil the needs of its four distribution centers \(D_{1}, D_{2}, D_{3}, D_{4}\). The minimum requirement at each center (in thousand liters) is \(D_{1} \to 7\), \(D_{2} \to 5\), \(D_{3} \to 3\), \(D_{4} \to 2\). Cost in hundred of rupees of shipping one thousand liters from each plant to each distribution center is given below :
| Plant | D1 | D2 | D3 | D4 |
|---|---|---|---|---|
| P1 | 2 | 3 | 11 | 7 |
| P2 | 1 | 0 | 6 | 1 |
| P3 | 5 | 8 | 15 | 9 |
Then the initial basic feasible by North-West Corner method is :
Answer: D
The North-West Corner solution is degenerate, with cost ₹ 11,600.
Supply is 6 + 1 + 10 = 17 and demand is 7 + 5 + 3 + 2 = 17, so the problem is balanced. Starting at the top-left cell:
- P1 to D1: 6 (P1 exhausted, D1 needs 1 more)
- P2 to D1: 1 (P2 exhausted and D1 satisfied at the same time)
- P3 to D2: 5, P3 to D3: 3, P3 to D4: 2
\[ \begin{aligned} &6(2) + 1(1) + 5(8) + 3(15) + 2(9) \\ &= 12 + 1 + 40 + 45 + 18 = 116 \end{aligned} \]
The cost is 116 hundred rupees, ₹ 11,600. A basic solution needs \(m + n - 1 = 3 + 4 - 1 = 6\) occupied cells, but only 5 are used, because a row and a column were satisfied together at the second step. So the solution is degenerate, and the options calling it non-degenerate or costing ₹ 12,000 are wrong.
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