Advertisement Remove all ads
Advertisement Remove all ads
Advertisement Remove all ads
Chart
Sum
Find the initial basic feasible solution of the following transportation problem:
I | II | III | Demand | |
A | 1 | 2 | 6 | 7 |
B | 0 | 4 | 2 | 12 |
C | 3 | 1 | 5 | 11 |
Supply | 10 | 10 | 10 |
Using North West Corner rule
Advertisement Remove all ads
Solution
Total demand (a_{i}) = 7 + 12 + 11 = 30 and total supply (b_{j}) = 10 + 10 + 10 = 30.
`sum"a"_"i" = sum"b"_"j"` = Σb_{j} ⇒ the problem is a balanced transportation problem and we can find a basic feasible solution.
North West Comer rule (NWC)
First allocation:
I | II | III | (a_{i}) | |
A | ^{(7)}1 | 2 | 6 | 7/0 |
B | 0 | 4 | 2 | 12 |
C | 3 | 1 | 5 | 11 |
(b_{j}) | 10/3 | 10 | 10 |
Second allocation:
I | II | III | (a_{i}) | |
B | ^{(3)}0 | 4 | 2 | 12/9 |
C | 3 | 1 | 5 | 11 |
(b_{j}) | 3/0 | 10 | 10 |
Third allocation:
II | III | (a_{i}) | |
B | ^{(9)}4 | 2 | 9/0 |
C | 1 | 5 | 11 |
(b_{j}) | 10/1 | 10 |
Fourth allocation:
II | III | (a_{i}) | |
C | ^{(1)}1 | ^{(10)}5 | 11/10/0 |
(b_{j}) | 10/1 | 10/0 |
We first allot 1 unit to (C, II) cell and then the balance 10 units to (C, III) cell.
Thus we have the following allocations:
I | II | III | Demand | |
A | ^{(7)}1 | 2 | 6 | 7 |
B | ^{(3)}0 | ^{(9)}4 | 2 | 12 |
C | 3 | ^{(1)}1 | ^{(10)}5 | 11 |
Supply | 10/3 | 10 | 10 |
Transportation schedule:
A → I
B → I
B → II
C → II
C → III
i.e x_{11} = 7
x_{21} = 3
x_{22} = 9
x_{32} = 1
x_{33} = 10
Total cost = (7 × 1) + (3 × 0) + (9 × 4) + (1 × 1) + (10 × 5)
= 7 + 0 + 36 + 1 + 50
= ₹ 94
Concept: Transportation Problem
Is there an error in this question or solution?