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प्रश्न
Determine an initial basic feasible solution to the following transportation problem by using north west corner rule
| Destination | Supply | ||||
| D1 | D2 | D3 | |||
| S1 | 9 | 8 | 5 | 25 | |
| Source | S2 | 6 | 8 | 4 | 35 |
| S3 | 7 | 6 | 9 | 40 | |
| Requirement | 30 | 25 | 45 | ||
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उत्तर
Total supply = 25 + 35 + 40 = 100 = `sum"a"_"i"`
Total requirement = 30 + 25 + 45 = 100 = `sum"b"_"j"`
Since `sum"a"_"i" = sum"b"_"j"`
The given transportation problem is balanced and we can find an initial basic feasible solution.
North West Corner Rule
First allocation:
| D1 | D2 | D3 | (ai) | |
| S1 | (25)9 | 8 | 5 | 25/0 |
| S2 | 6 | 8 | 4 | 35 |
| S3 | 7 | 6 | 9 | 40 |
| (bj) | 30/5 | 25 | 45 |
Second allocation:
| D1 | D2 | D3 | (ai) | |
| S2 | (5)6 | 8 | 4 | 35/30 |
| S3 | 7 | 6 | 9 | 40 |
| (bj) | 5/0 | 25 | 45 |
Third allocation:
| D2 | D3 | (ai) | |
| S2 | (25)8 | 4 | 30/5 |
| S3 | 6 | 9 | 40 |
| (bj) | 25/0 | 45 |
Fourth allocation:
| D3 | (ai) | |
| S2 | (5)4 | 5/0 |
| S3 | (40)9 | 40/0 |
| (bj) | 45/40/0 |
We first allow 5 units to cell (S2, D3).
Then balance 40 units we allot to cell (S3, D3).
The final allotment is given as follows
| D1 | D2 | D3 | Supply | |
| S1 | (25)9 | 8 | 5 | 25 |
| S2 | (5)6 | (25)8 | (5)4 | 35 |
| S3 | 7 | 6 | (40)9 | 40 |
| Requirement | 30 | 25 | 45 |
Transportation schedule:
S1 → D1
S2 → D1
S2 → D2
S2 → D3
S3 → D3
i.e x11 = 25
x21 = 5,
x22 = 25
x23 = 5
x33 = 40
Total cost is = (25 × 9) + (5 × 6) + (25 × 8) + (5 × 4) + (40 × 9)
= 225 + 30 + 200 + 20 + 360
= 835
Hence the minimum cost is ₹ 835 by NWC method.
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