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प्रश्न
Determine basic feasible solution to the following transportation problem using North west Corner rule.
| Sinks | |||||||
| A | B | C | D | E | Supply | ||
| P | 2 | 11 | 10 | 3 | 7 | 4 | |
| Origins | Q | 1 | 4 | 7 | 2 | 1 | 8 |
| R | 3 | 9 | 4 | 8 | 12 | 9 | |
| Demand | 3 | 3 | 4 | 5 | 6 | ||
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उत्तर
For the given problem, total supply is 4 + 8 + 9 = 21 and total demand is 3 + 3 + 4 + 5 + 6 = 21.
Since the total supply equals total demand, it is a balanced problem and we can find a feasible solution by the North West Comer rule.
First allocation:
| A | B | C | D | E | (ai) | |
| P | (3)2 | 11 | 10 | 3 | 7 | 4/1 |
| Q | 1 | 4 | 7 | 2 | 1 | 8 |
| R | 3 | 9 | 4 | 8 | 12 | 9 |
| (bj) | 3/0 | 3 | 4 | 5 | 6 |
Second allocation:
| B | C | D | E | (ai) | |
| P | (1)11 | 10 | 3 | 7 | 1/0 |
| Q | 4 | 7 | 2 | 1 | 8 |
| R | 9 | 4 | 8 | 12 | 9 |
| (bj) | 3/2 | 4 | 5 | 6 |
Third allocation:
| B | C | D | E | (ai) | |
| Q | (2)4 | 7 | 2 | 1 | 8/6 |
| R | 9 | 4 | 8 | 12 | 9 |
| (bj) | 2/0 | 4 | 5 | 6 |
Fourth allocation:
| C | D | E | (ai) | |
| Q | (4)7 | 2 | 1 | 6/2 |
| R | 4 | 8 | 12 | 9 |
| (bj) | 4/0 | 5 | 6 |
Fifth allocation:
| D | E | (ai) | |
| Q | (2)2 | 1 | 2/0 |
| R | 8 | 12 | 9 |
| (bj) | 5/3 | 6 |
Final allocation:
| D | E | (ai) | |
| R | (3)8 | (6)12 | 9/6/0 |
| (bj) | 3/0 | 6/0 |
First, we allow 3 units to (R, D) cell.
Then balance 6 to (R, E) cell.
Thus we have the following allocations:
| A | B | C | D | E | Supply | |
| P | (3)2 | (1)11 | 10 | 3 | 7 | 4/1 |
| Q | 1 | (2)4 | (4)7 | (2)2 | 1 | 8 |
| R | 3 | 9 | 4 | (3)8 | (6)12 | 9 |
| Demand | 3 | 3 | 4 | 5 | 6 |
Transportation schedule:
P → A
P → B
Q → B
Q → C
Q → D
R → D
R → E
i.e x11 = 3
x12 = 1
x22 = 2
x23 = 4
x24 = 2
x34 = 3
x35 = 6
Total cost = (3 × 2) + (1 × 11) + (2 × 4) + (4 × 7) + (2 × 2) + (3 × 8) + (6 × 12)
= 6 + 11 + 8 + 28 + 4 + 24 + 72
= 153
Thus the minimum cost of the transportation problem by Northwest Comer rule is ₹ 153.
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