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Question
Determine an initial basic feasible solution to the following transportation problem by using least cost method
| 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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Solution
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.
Least cost method (LCM)
First allocation:
| D1 | D2 | D3 | (ai) | |
| S1 | 9 | 8 | 5 | 25 |
| S2 | 6 | 8 | (35)4 | 35/0 |
| S3 | 7 | 6 | 9 | 40 |
| (bj) | 30 | 25 | 45/10 |
Second allocation:
| D1 | D2 | D3 | (ai) | |
| S1 | 9 | 8 | (10)5 | 25/15 |
| S3 | 7 | 6 | 9 | 40 |
| (bj) | 30 | 25 | 10/0 |
Third allocation:
| D1 | D2 | (ai) | |
| S1 | 9 | 8 | 15 |
| S3 | 7 | (25)6 | 40/15 |
| (bj) | 30 | 25/0 |
Fourth allocation:
| D1 | (ai) | |
| S1 | (15)9 | 15/0 |
| S3 | (15)7 | 15/0 |
| (bj) | 30/15/0 |
We first allow 15 units to cell (S3, D1) since it has the least cost.
Then we allow the balance 15 units to cell (S1, D1).
The final allotment is given as follows.
| D1 | D2 | D3 | Supply | |
| S1 | (15)9 | 8 | (10)5 | 25 |
| S2 | 6 | 8 | (35)4 | 35 |
| S3 | (15)7 | (25)6 | 9 | 40 |
| Requirement | 30 | 25 | 45 |
Transportation schedule:
S1 → D1
S1 → D3
S2 → D3
S3 → D1
S3 → D2
i.e x11 = 15
x13 = 10
x23 = 35
x31 = 15
x32 = 25
Total cost is = (15 × 9) + (10 × 5) + (35 × 4) + (15 × 7) + (25 × 6)
= 136 + 50 + 140 + 105 + 150
= 580
The optimal cost by LCM is ₹ 580.
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