Advertisements
Advertisements
Question
Determine an initial basic feasible solution of the following transportation problem by north west corner method.
| Bangalore | Nasik | Bhopal | Delhi | Capacity | |
| Chennai | 6 | 8 | 8 | 5 | 30 |
| Madurai | 5 | 11 | 9 | 7 | 40 |
| Trickly | 8 | 9 | 7 | 13 | 50 |
| Demand (Units/day) |
35 | 28 | 32 | 25 |
Advertisements
Solution
Let B, N, Bh, D represent the destinations Bangalore, Nasik, Bhopal, and Delhi respectively.
Let C, M, T represent the starting places Chennai, Madurai, and Trichy respectively.
The given transportation table is
| Bangalore | Nasik | Bhopal | Delhi | Capacity (ai) |
|
| Chennai | 6 | 8 | 8 | 5 | 30 |
| Madurai | 5 | 11 | 9 | 7 | 40 |
| Trickly | 8 | 9 | 7 | 13 | 50 |
| Demand (bj) |
35 | 28 | 32 | 25 | 120 |
Total capacity = Total Demand = 120.
So the given problem is a balanced transportation problem.
Hence there exists a feasible solution to the given problem.
First allocation:
| Bangalore | Nasik | Bhopal | Delhi | Capacity (ai) |
|
| Chennai | (30)6 | 8 | 8 | 5 | 30/0 |
| Madurai | 5 | 11 | 9 | 7 | 40 |
| Trickly | 8 | 9 | 7 | 13 | 50 |
| Demand (bj) |
35/5 | 28 | 32 | 25 | 120 |
Second allocation:
| Bangalore | Nasik | Bhopal | Delhi | (ai) | |
| Chennai | (30)6 | 8 | 8 | 5 | 30/0 |
| Madurai | (5)5 | 11 | 9 | 7 | 40/35 |
| Trickly | 8 | 9 | 7 | 13 | 50 |
| (bj) | 35/5/0 | 28 | 32 | 25 | 120 |
Third allocation:
| Bangalore | Nasik | Bhopal | Delhi | (ai) | |
| Chennai | (30)6 | 8 | 8 | 5 | 30/0 |
| Madurai | (5)5 | (28)11 | 9 | 7 | 40/35/7 |
| Trickly | 8 | 9 | 7 | 13 | 50 |
| (bj) | 35/5/0 | 28/0 | 32 | 25 | 120 |
Fourth allocation:
| Bangalore | Nasik | Bhopal | Delhi | (ai) | |
| Chennai | (30)6 | 8 | 8 | 5 | 30/0 |
| Madurai | (5)5 | (28)11 | (7)9 | 7 | 40/35/7/0 |
| Trickly | 8 | 9 | 7 | 13 | 50 |
| (bj) | 35/5/0 | 28/0 | 32/25 | 25 | 120 |
Fifth allocation:
| Bangalore | Nasik | Bhopal | Delhi | (ai) | |
| Chennai | (30)6 | 8 | 8 | 5 | 30/0 |
| Madurai | (5)5 | (28)11 | (7)9 | 7 | 40/35/7/0 |
| Trickly | 8 | 9 | (25)7 | 13 | 50/25 |
| (bj) | 35/5/0 | 28/0 | 32/25/0 | 25 | 120 |
Final allocation:
| Bangalore | Nasik | Bhopal | Delhi | (ai) | |
| Chennai | (30)6 | 8 | 8 | 5 | 30/0 |
| Madurai | (5)5 | (28)11 | (7)9 | 7 | 40/35/7/0 |
| Trickly | 8 | 9 | (25)7 | (25)13 | 50/25/0 |
| (bj) | 35/5/0 | 28/0 | 32/25/0 | 25/0 | 120 |
Transportation schedule: Chennai to Bangalore, Madurai to Bangalore, Madurai to Nasik, Madurai to Bhopal, Trichy to Bhopal, Trichy to Delhi.
i.e x11 = 30
x21 = 5
x22 = 28
x23 = 7
x33 = 25
x34 = 25
The total transportation cost = (30 × 6) + (5 × 5) + (28 × 11) + (7 × 9) + (25 × 7) + (25 × 13)
= 180 + 25 + 308 + 63 + 175 + 325
= 1076
Thus the minimum cost is ₹ 1076 by the northwest comer method.
APPEARS IN
RELATED QUESTIONS
Write mathematical form of transportation problem
What do you mean by balanced transportation problem?
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 |
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.
Determine basic feasible solution to the following transportation problem using North west Corner rule.
| Sinks | |||||||
| A | B | C | D | E | Supply | ||
| P | 2 | 11 | 10 | 3 | 7 | 4 | |
| Origins | Q | 1 | 4 | 7 | 2 | 1 | 8 |
| R | 3 | 9 | 4 | 8 | 12 | 9 | |
| Demand | 3 | 3 | 4 | 5 | 6 | ||
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 Least Cost method
Choose the correct alternative:
The transportation problem is said to be unbalanced if ______
Choose the correct alternative:
In an assignment problem the value of decision variable xij is ______
Consider the following transportation problem
| Detination | Availabiity | ||||
| D1 | D2 | D3 | D4 | ||
| O1 | 5 | 8 | 3 | 6 | 30 |
| O2 | 4 | 5 | 7 | 4 | 50 |
| O3 | 6 | 2 | 4 | 6 | 20 |
| Requirement | 30 | 40 | 20 | 10 | |
Determine an initial basic feasible solution using Least cost method
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 | ||
