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प्रश्न
Find an initial basic feasible solution of the following problem using the northwest corner rule.
D1 | D2 | D3 | D4 | Supply | |
O1 | 5 | 3 | 6 | 2 | 19 |
O2 | 4 | 7 | 9 | 1 | 37 |
O3 | 3 | 4 | 7 | 5 | 34 |
Demand | 16 | 18 | 31 | 25 |
उत्तर
Given the transportation table is
D1 | D2 | D3 | D4 | Supply (ai) |
|
O1 | 5 | 3 | 6 | 2 | 19 |
O2 | 4 | 7 | 9 | 1 | 37 |
O3 | 3 | 4 | 7 | 5 | 34 |
Demand (bj) |
16 | 18 | 31 | 25 | 90 |
Total supply = Total Demand = 90.
The given problem is a balanced transportation problem.
Hence there exists a feasible solution to the given problem.
First allocation:
D1 | D2 | D3 | D4 | (ai) | |
O1 | (16)5 | 3 | 6 | 2 | 19/3 |
O2 | 4 | 7 | 9 | 1 | 37 |
O3 | 3 | 4 | 7 | 5 | 34 |
(bj) | 16/0 | 18 | 31 | 25 | 90 |
Second allocation:
D1 | D2 | D3 | D4 | (ai) | |
O1 | (16)5 | (3)3 | 6 | 2 | 19/3/0 |
O2 | 4 | 7 | 9 | 1 | 37 |
O3 | 3 | 4 | 7 | 5 | 34 |
(bj) | 16/0 | 18/15 | 31 | 25 | 90 |
Third allocation:
D1 | D2 | D3 | D4 | (ai) | |
O1 | (16)5 | (3)3 | 6 | 2 | 19/3/0 |
O2 | 4 | (15)7 | 9 | 1 | 37/22 |
O3 | 3 | 4 | 7 | 5 | 34 |
(bj) | 16/0 | 18/15/0 | 31 | 25 | 90 |
Fourth allocation:
D1 | D2 | D3 | D4 | (ai) | |
O1 | (16)5 | (3)3 | 6 | 2 | 19/3/0 |
O2 | 4 | (15)7 | (22)9 | 1 | 37/22/0 |
O3 | 3 | 4 | 7 | 5 | 34 |
(bj) | 16/0 | 18/15/0 | 31/9 | 25 | 90 |
Fifth allocation:
D1 | D2 | D3 | D4 | (ai) | |
O1 | (16)5 | (3)3 | 6 | 2 | 19/3/0 |
O2 | 4 | (15)7 | (22)9 | 1 | 37/22/0 |
O3 | 3 | 4 | (9)7 | 5 | 34/25 |
(bj) | 16/0 | 18/15/0 | 31/9/0 | 25 | 35 |
Final allocation:
D1 | D2 | D3 | D4 | (ai) | |
O1 | (16)5 | (3)3 | 6 | 2 | 19/3/0 |
O2 | 4 | (15)7 | (22)9 | 1 | 37/22/0 |
O3 | 3 | 4 | (9)7 | (25)5 | 34/25/0 |
(bj) | 16/0 | 18/15/0 | 31/9/0 | 25/0 | 35 |
Transportation schedule:
O1 → D1
O1 → D2
O2 → D2
O2 → D3
O3 → D3
O3 → D4.
i.e x11 = 16
x12 = 3
x22 = 15
x23 = 22
x33 = 9
x34 = 25.
Total transportation cost = (16 × 5) + (3 × 3) + (15 × 7) + (22 × 9) + (9 × 7) + (25 × 5)
= 80 + 9 + 105 + 198 + 63 + 125
= 580
Thus the minimum cost is ₹ 580 using the northwest comer rule.
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