Determine an initial basic feasible solution to the following transportation problem by using least cost method
Destination | Supply | ||||
D_{1} | D_{2} | D_{3} | |||
S_{1} | 9 | 8 | 5 | 25 | |
Source | S_{2} | 6 | 8 | 4 | 35 |
S_{3} | 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:
D_{1} | D_{2} | D_{3} | (a_{i}) | |
S_{1} | 9 | 8 | 5 | 25 |
S_{2} | 6 | 8 | ^{(35)}4 | 35/0 |
S_{3} | 7 | 6 | 9 | 40 |
(b_{j}) | 30 | 25 | 45/10 |
Second allocation:
D_{1} | D_{2} | D_{3} | (a_{i}) | |
S_{1} | 9 | 8 | ^{(10)}5 | 25/15 |
S_{3} | 7 | 6 | 9 | 40 |
(b_{j}) | 30 | 25 | 10/0 |
Third allocation:
D_{1} | D_{2} | (a_{i}) | |
S_{1} | 9 | 8 | 15 |
S_{3} | 7 | ^{(25)}6 | 40/15 |
(b_{j}) | 30 | 25/0 |
Fourth allocation:
D_{1} | (a_{i}) | |
S_{1} | ^{(15)}9 | 15/0 |
S_{3} | ^{(15)}7 | 15/0 |
(b_{j}) | 30/15/0 |
We first allow 15 units to cell (S_{3}, D_{1}) since it has the least cost.
Then we allow the balance 15 units to cell (S_{1}, D_{1}).
The final allotment is given as follows.
D_{1} | D_{2} | D_{3} | Supply | |
S_{1} | ^{(15)}9 | 8 | ^{(10)}5 | 25 |
S_{2} | 6 | 8 | ^{(35)}4 | 35 |
S_{3} | ^{(15)}7 | ^{(25)}6 | 9 | 40 |
Requirement | 30 | 25 | 45 |
Transportation schedule:
S_{1} → D_{1}
S_{1} → D_{3}
S_{2} → D_{3}
S_{3} → D_{1}
S_{3} → D_{2}
i.e x_{11} = 15
x_{13} = 10
x_{23} = 35
x_{31} = 15
x_{32} = 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.