Advertisements
Advertisements
Question
Find an initial basic feasible solution of the following problem using the northwest corner rule.
| D1 | D2 | D3 | D4 | Supply | |
| O1 | 5 | 3 | 6 | 2 | 19 |
| O2 | 4 | 7 | 9 | 1 | 37 |
| O3 | 3 | 4 | 7 | 5 | 34 |
| Demand | 16 | 18 | 31 | 25 |
Solution
Given the transportation table is
| D1 | D2 | D3 | D4 | Supply (ai) |
|
| O1 | 5 | 3 | 6 | 2 | 19 |
| O2 | 4 | 7 | 9 | 1 | 37 |
| O3 | 3 | 4 | 7 | 5 | 34 |
| Demand (bj) |
16 | 18 | 31 | 25 | 90 |
Total supply = Total Demand = 90.
The given problem is a balanced transportation problem.
Hence there exists a feasible solution to the given problem.
First allocation:
| D1 | D2 | D3 | D4 | (ai) | |
| O1 | (16)5 | 3 | 6 | 2 | 19/3 |
| O2 | 4 | 7 | 9 | 1 | 37 |
| O3 | 3 | 4 | 7 | 5 | 34 |
| (bj) | 16/0 | 18 | 31 | 25 | 90 |
Second allocation:
| D1 | D2 | D3 | D4 | (ai) | |
| O1 | (16)5 | (3)3 | 6 | 2 | 19/3/0 |
| O2 | 4 | 7 | 9 | 1 | 37 |
| O3 | 3 | 4 | 7 | 5 | 34 |
| (bj) | 16/0 | 18/15 | 31 | 25 | 90 |
Third allocation:
| D1 | D2 | D3 | D4 | (ai) | |
| O1 | (16)5 | (3)3 | 6 | 2 | 19/3/0 |
| O2 | 4 | (15)7 | 9 | 1 | 37/22 |
| O3 | 3 | 4 | 7 | 5 | 34 |
| (bj) | 16/0 | 18/15/0 | 31 | 25 | 90 |
Fourth allocation:
| D1 | D2 | D3 | D4 | (ai) | |
| O1 | (16)5 | (3)3 | 6 | 2 | 19/3/0 |
| O2 | 4 | (15)7 | (22)9 | 1 | 37/22/0 |
| O3 | 3 | 4 | 7 | 5 | 34 |
| (bj) | 16/0 | 18/15/0 | 31/9 | 25 | 90 |
Fifth allocation:
| D1 | D2 | D3 | D4 | (ai) | |
| O1 | (16)5 | (3)3 | 6 | 2 | 19/3/0 |
| O2 | 4 | (15)7 | (22)9 | 1 | 37/22/0 |
| O3 | 3 | 4 | (9)7 | 5 | 34/25 |
| (bj) | 16/0 | 18/15/0 | 31/9/0 | 25 | 35 |
Final allocation:
| D1 | D2 | D3 | D4 | (ai) | |
| O1 | (16)5 | (3)3 | 6 | 2 | 19/3/0 |
| O2 | 4 | (15)7 | (22)9 | 1 | 37/22/0 |
| O3 | 3 | 4 | (9)7 | (25)5 | 34/25/0 |
| (bj) | 16/0 | 18/15/0 | 31/9/0 | 25/0 | 35 |
Transportation schedule:
O1 → D1
O1 → D2
O2 → D2
O2 → D3
O3 → D3
O3 → D4.
i.e x11 = 16
x12 = 3
x22 = 15
x23 = 22
x33 = 9
x34 = 25.
Total transportation cost = (16 × 5) + (3 × 3) + (15 × 7) + (22 × 9) + (9 × 7) + (25 × 5)
= 80 + 9 + 105 + 198 + 63 + 125
= 580
Thus the minimum cost is ₹ 580 using the northwest comer rule.
APPEARS IN
RELATED QUESTIONS
Write mathematical form of transportation problem
What do you mean by balanced transportation problem?
Obtain an initial basic feasible solution to the following transportation problem by using least-cost method.
| D1 | D2 | D3 | Supply | |
| O1 | 9 | 8 | 5 | 25 |
| O2 | 6 | 8 | 4 | 35 |
| O3 | 7 | 6 | 9 | 40 |
| Demand | 30 | 25 | 45 |
Explain Vogel’s approximation method by obtaining initial feasible solution of the following transportation problem.
| D1 | D2 | D3 | D4 | Supply | |
| O1 | 2 | 3 | 11 | 7 | 6 |
| O2 | 1 | 0 | 6 | 1 | 1 |
| O3 | 5 | 8 | 15 | 9 | 10 |
| Demand | 7 | 5 | 3 | 2 |
Consider the following transportation problem.
| D1 | D2 | D3 | D4 | Availability | |
| O1 | 5 | 8 | 3 | 6 | 30 |
| O2 | 4 | 5 | 7 | 4 | 50 |
| O3 | 6 | 2 | 4 | 6 | 20 |
| Requirement | 30 | 40 | 20 | 10 |
Determine initial basic feasible solution by VAM.
Obtain an initial basic feasible solution to the following transportation problem by north west corner method.
| D | E | F | C | Available | |
| A | 11 | 13 | 17 | 14 | 250 |
| B | 16 | 18 | 14 | 10 | 300 |
| C | 21 | 24 | 13 | 10 | 400 |
| Required | 200 | 225 | 275 | 250 |
Choose the correct alternative:
In a non – degenerate solution number of allocation is
Choose the correct alternative:
The Penalty in VAM represents difference between the first ______
Choose the correct alternative:
In an assignment problem the value of decision variable xij is ______
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 | ||
