Tamil Nadu Board of Secondary EducationHSC Commerce Class 12th

# 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 - Business Mathematics and Statistics

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Sum

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

#### 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.

Concept: Transportation Problem
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