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Shvoong Home>Science>Mathematics>The Problem Ofnorthwest Corner Method and Solutrion Summary

# The Problem Ofnorthwest Corner Method and Solutrion

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

Example-1: The following table indicates the three factories and four warehouses, unit transportation cost in taka, quantities available at each of the factories and requirements at each of the warehouses. Determine the initial feasible solution by northwest corner 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

Solution: Here s1 =15, s2 = 24, s3 =21
d1 = 13, d2 =12, d3 =16, d4 =19
and = = 60
Hence the problem is balanced.

Following steps are given below

Transportation model

Factory Warehouse Availability
(s)
A B C D
F1 3
13
2
2 5 2 15-15=0

F2 2 1
4
4 24

F3 2 3 4
2 3
19 21

Requirement (d) 13-13=0 12-2=10 16 19 60

1. Start allocation from northwest corner i.e. x11. Adjust demand and supply. So allocate 13 units in x11. Move to x12, Now remaining units from factory F1= (15 – 13) = 2 units can be allocated to x12, which complete the supply from F1.

Factory Warehouse Availability
(s)
A B C D
F1 3
13
2
2
5 2 0

F2 2 1
10
4
14 4 24-24=0

F3 2 3 4
3
21

Requirement (d) 0 12-12=0 16 19 60

2. Now, we move to x22, Here the availability supply is 24 units and the requirement at warehouse B remains only 12 – 2 =10 units (2 units already supplied to B from factory F1). Now 10 ≤ 24. So, 10 units are allocated to x22. This exhausted the demand of B. Now, the demand at C is 16 which is more than remaining supply of 14 units from F2. So 24 – 10 =14 units are allocated to x23 to exhaust the supply from F2.

Thus the final transportation model

Factory Warehouse Availability
(s)
A B C D
F1 3
13
2
2
5 2 15

F2 2 1
10
4
14
4 24

F3 2 3 4
2
3
19 21

Requirement (d) 13 12 16 19 60

3. Now we move to x33. The remaining demand of warehouse C is (16 – 14) = 2 units which is less than the supply of factory F3. So 2 units are allocated to x33 to exhaust the demand at C. Now we proceed to x34. To allocate the remaining supply from F3 is (21 – 2) = 19 in x34, which is also equal to the demand at D.

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 = x11 × c11 + x12 × c12 + x22 × c22 + x23 × c23 + x33 × c33 + x34 × c34
=13 × 3 + 2 × 2 + 10 × 1+ 14 × 4 + 2 × 4 + 19 × 3
= 174
Published: May 29, 2010
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