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Shvoong Home>Science>Mathematics>Transportation Problem With Least Cost Entry Method. Summary

# Transportation Problem With Least Cost Entry Method.

Academic Paper Summary   by:Ashraful     Original Author: Md. Ashraful Babu

Question:- Transportation Problem with least cost entry method.

Factory Warehouse Availability

A B C D
F1 3 2 5 2 15

F2 2 1 4 4 24

F3 2 3 4 3 21

Requirement 13 12 16 19 60

Least cost method finds a better starting solution by the concentrating on the cheapest routes. This method assigns as much as possible to the variable with the smallest unit transportation cost in the entire tableau. Cross out the satisfied row or column and amounts of supply and demand are adjusted accordingly. Again we search for the next lowest unit transportation cost and the process continues till total supply and demand is exhausted.

Following steps are given below
Transportation model

Factory Warehouse Availability
(s)
A B C D
F1 3
2
5 2
15

F2 2 1
12 4
4 24-12=12

F3 2
3 4
3
21

Requirement (d) 13 12-12=0 16 19 48

1. The lowest cost is 1 is in x22, where the demand 12 ≤ supply 24. So we allocate 12 units in x22. The demand for B is exhausted.

Factory Warehouse Availability
(s)
A B C D
F1 3

2

5 2
15 15-15=0

F2 2 1
12 4
4 12

F3 2
13 3 4
3
21-13=8

Requirement (d) 13-13=0 0 16 19-15=4 20

2. The next lowest cost is 2 in x12, x14, x21 and x31. But we cannot allocate to x12 as the demand for B is exhausted. Now, we see the demand of x21 i.e. 13 is more than available supply of F2 is (24 – 12) = 12 units. For x14 demand 19 ≥ supply 15 and for x31 demand 13 ≤ supply 21. 15 and 13 units can be allocated to x14 and x31 respectively. So we allocate 15 units in x14 and 13 units in x31. Supply of the factory F1 is exhausted. So no allocation can be made in remaining of first row.

Factory Warehouse Availability
(s)
A B C D
F1 3

2

5 2
15 0

F2 2 1
12 4
4 12

F3 2
13
3 4
3
4 8-4=4

Requirement (d) 0 0 16 4-4=0 16

3. Next lowest cost is 3 in x34, where we can allocate 4 units.

Final transportation model

Factory Warehouse Availability
(s)
A B C D
F1 3

2

5 2
15 0

F2 2 1
12 4
12
4 12

F3 2
13
3 4
4 3
4 8

Requirement (d) 0 0 16 4 20

4. The demand of warehouse C can be met from F2 and F3. So 12 and 4 units are allocated in x23 and x33 respectively and to complete the allocation plan.

So, it can be observed that the number of occupied cells in the solution is (m + n - 1) = (3 + 4 – 1) = 6. Hence the initial solution is feasible.

The total transportation cost is
z = x22 × c22 + x14 × c14 + x31 × c31 + x34 × c34 + x23 × c23 + x33 × c33
=12 × 1 + 15 × 2 + 13 ×2 + 4 × 3 + 12 × 4+ 4 × 4 = 144
Published: May 29, 2010
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1. Answer   Question  :    real life situations of least cost method
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