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1. north-west corner method example ( Enter your problem )
Algorithm and examples
  1. Introduction
  2. Algorithm & Example-1
  3. Example-2
  4. Unbalanced supply and demand example
Other related methods
  1. north-west corner method
  2. least cost method
  3. vogel's approximation method
  4. Row minima method
  5. Column minima method
  6. Russell's approximation method
  7. Heuristic method-1
  8. Heuristic method-2
  9. modi method (optimal solution)
  10. stepping stone method (optimal solution)

1. Introduction
(Previous example)
3. Example-2
(Next example)

2. Algorithm & Example-1





Algorithm
North-West Corner Method (NWCM) Steps (Rule)
Step-1: Select the upper left corner cell of the transportation matrix and allocate min(s1, d1).
Step-2: a. Subtract this value from supply and demand of respective row and column.

b. If the supply is 0, then cross (strike) that row and move down to the next cell.

c. If the demand is 0, then cross (strike) that column and move right to the next cell.

d. If supply and demand both are 0, then cross (strike) both row & column and move diagonally to the next cell.
Step-3: Repeat this steps until all supply and demand values are 0.

Example-1
1. Find Solution using North-West Corner method
D1D2D3D4Supply
S1193050107
S2703040609
S3408702018
Demand58714


Solution:
TOTAL number of supply constraints : 3
TOTAL number of demand constraints : 4
Problem Table is
`D_1``D_2``D_3``D_4`Supply
`S_1`193050107
`S_2`703040609
`S_3`408702018
Demand58714



The rim values for `S_1`=7 and `D_1`=5 are compared.

The smaller of the two i.e. min(7,5) = 5 is assigned to `S_1` `D_1`

This meets the complete demand of `D_1` and leaves 7 - 5=2 units with `S_1`

Table-1
`D_1``D_2``D_3``D_4`Supply
`S_1`19(5)3050102
`S_2`703040609
`S_3`408702018
Demand08714


Move horizontally,

The rim values for `S_1`=2 and `D_2`=8 are compared.

The smaller of the two i.e. min(2,8) = 2 is assigned to `S_1` `D_2`

This exhausts the capacity of `S_1` and leaves 8 - 2=6 units with `D_2`

Table-2
`D_1``D_2``D_3``D_4`Supply
`S_1`19(5)30(2)50100
`S_2`703040609
`S_3`408702018
Demand06714


Move vertically,

The rim values for `S_2`=9 and `D_2`=6 are compared.

The smaller of the two i.e. min(9,6) = 6 is assigned to `S_2` `D_2`

This meets the complete demand of `D_2` and leaves 9 - 6=3 units with `S_2`

Table-3
`D_1``D_2``D_3``D_4`Supply
`S_1`19(5)30(2)50100
`S_2`7030(6)40603
`S_3`408702018
Demand00714


Move horizontally,

The rim values for `S_2`=3 and `D_3`=7 are compared.

The smaller of the two i.e. min(3,7) = 3 is assigned to `S_2` `D_3`

This exhausts the capacity of `S_2` and leaves 7 - 3=4 units with `D_3`

Table-4
`D_1``D_2``D_3``D_4`Supply
`S_1`19(5)30(2)50100
`S_2`7030(6)40(3)600
`S_3`408702018
Demand00414


Move vertically,

The rim values for `S_3`=18 and `D_3`=4 are compared.

The smaller of the two i.e. min(18,4) = 4 is assigned to `S_3` `D_3`

This meets the complete demand of `D_3` and leaves 18 - 4=14 units with `S_3`

Table-5
`D_1``D_2``D_3``D_4`Supply
`S_1`19(5)30(2)50100
`S_2`7030(6)40(3)600
`S_3`40870(4)2014
Demand00014


Move horizontally,

The rim values for `S_3`=14 and `D_4`=14 are compared.

The smaller of the two i.e. min(14,14) = 14 is assigned to `S_3` `D_4`

Table-6
`D_1``D_2``D_3``D_4`Supply
`S_1`19(5)30(2)50100
`S_2`7030(6)40(3)600
`S_3`40870(4)20(14)0
Demand0000


Initial feasible solution is
`D_1``D_2``D_3``D_4`Supply
`S_1`19 (5)30 (2)50 10 7
`S_2`70 30 (6)40 (3)60 9
`S_3`40 8 70 (4)20 (14)18
Demand58714


The minimum total transportation cost `=19 xx 5+30 xx 2+30 xx 6+40 xx 3+70 xx 4+20 xx 14=1015`

Here, the number of allocated cells = 6 is equal to m + n - 1 = 3 + 4 - 1 = 6
`:.` This solution is non-degenerate


This material is intended as a summary. Use your textbook for detail explanation.
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1. Introduction
(Previous example)
3. Example-2
(Next example)





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