The solution to Transportation problem involves two phases. The first phase is to obtain the initial basic feasible solution; while obtaining the optimal solution is the second phase.
In the first phase, there are various methods of obtaining the initial basic feasible solution. However, this chapter is concerned with:
(i) North-West corner Rule, (ii) Least cost Method, and
(iii) Vogel‟s Approximation (or penalty) Method.
Note: that any of the three methods must satisfy the following conditions in order to obtain the initial solution:
(i) The problem must be balanced (i.e. ∑ ai = ∑ bj)
(ii) The number of cell‟s allocation must be equal to m + n-1, where m and n are numbers of rows and columns respectively.
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Any solution satisfying the two conditions is termed “Non-degenerate Basic Feasible solution” otherwise, it is called “Degenerate solution”.
(i) North-West Corner Rule (NWCR)
The method is the simplest but most inefficient as it has the highest total transportation cost in comparison to all other methods. The main reason that can be attributed to this is that the method does not take into account the cost of transportation for all the possible alternative routes.
The steps needed to solve a transportation problem by NWCR are:
Step 1: Begin by allocating to the North-West cell (i.e. variable X11) of transportation matrix the allowable minimum of the supply and demand capacities of that cell. i.e.
Min(a1, b1)
Step 2: Check if allocation made in the first step is equal to the supply (demand) available at the first row (column), then cross-out the exhausted row (column) so that no further assignment can be made to the said row (column). Move vertically (horizontally) to the next cell and apply Step1.
In case a1= b1, move to cell (2,2). (i.e. variable X22) and apply Step1.
Step 3: Step 2 should be continued until exactly one row or column is left uncrossed in the transportation matrix.
Then make allowable allocation to that row or column and stop. Otherwise, return to Step1.
ILLUSTRATION 21-1
SAO company has 3 plants or locations (A, B, C) where its goods can be produced with production capacity of 50, 60, 50 per month respectively for a particular product. These units are to be distributed to 4 points (X, Y, W, Z) of consumption with the demand of 50, 70, 30 and 10 per month respectively.
The following table gives the transportation cost (in Naira) from various plants to the various points of consumption:
Destination X Y W Z Source/Plant A
B C
21 18 27 22 19 18 24 20 24 25 28 25
Obtain the Initial Basic Feasible solution by NWCR.
SUGGESTED SOLUTION 21-1
The NWCR algorithm applied to this problem resulted to the following table:
Destination
Plant X Y W Z Supply
A
50 0 B
60 0 C
50 40 10 0 Demand 50
0 70
10 30
0 10
0 160
Total cost = 50(21) + 60(18) + 10(25) + 30(28) + 10(25) = N 3470
Note that in all transportation tables, figures in small boxes are the transportation costs from destinations to destinations while the circled figures represent the allocation.
Explanation to the above allocation
Beginning from North-West (i.e Cell X11) corner
Allocate 50 to cell X11 in order to satisfy the minimum of demand and supply capacities. Zero balance is left for both demand and supply. Therefore, row 1 and column 1 are crossed out
Move to cell X22 and allocate 60. The supply balance is zero while the demand balance is 10. Therefore, row 2 is crossed out.
50 21 0 18 27 22
0
19 18 24 20
25 28
24 25
60
10 30 1
0
475
Next, move to cell X32 and allocate 10 giving the balance of zero for demand and 40 for the supply. Column 2 is therefore crossed out.
Then move to cell X33 and allocate 30 to exhaust the demand and having 10 balance for supply. Therefore, column 3 is crossed out.
Finally, 10 is allocated to cell X34
(ii) Least Cost Method (LCM)
In this method the cheapest route is always the focus for allocation. It is a better method compared to NWCR because costs are considered for allocation. The algorithm is stated thus:
Step 1: Assign as much as possible to the smallest unit cost (ties are broken arbitrarily). Also bear in mind the idea of allowable minimum of supply and demand capacities as done in NWCR.
Step 2: Cross-out the exhausted row or column and adjust the supply and demand accordingly. If both row and column are exhausted simultaneously, only one is crossed-out (in order to avoid degenerating case).
Step 3: Look for the smallest cost in the uncrossed row or column and assign the allowable quantity. Repeat this process until left with exactly one uncrossed row or column.
ILLUSTRATION 21-2
Use the data in illustration 21-1 to determine the Initial Basic Feasible solution by Least Cost Method.
SUGGESTED SOLUTION 21-2
Using the above algorithm, gives the following table:
Destination
Plant X Y W Z Supply
A 50 50 0
B 40 20 60 40 0
C 10
30 10 50 40 30 0
Demand 50 10 0
70 20 0
30 0 10
0 160
Total Cost = 50(18) + 20(18) + 40(19) + 10(24) + 10(25) + 30(28)
= N3350
Explanation to the above allocation
The lowest cost cell is X12 with 18. Then 50 is allocated to that cell in order to satisfy minimum of the demand and supply. Zero balance is left for supply while that of demand is 20. Therefore, row one is crossed out.
For the remaining un-crossed cells, cell X22 has lowest cost of 18.
Then assign 20 to this cell in order to give zero balance for demand and 40 balance for supply. Therefore, column two is crossed out.
The next cell with lowest cost, for the uncrossed cells, is X21 with 19. Allocate 40 to this cell in order to give zero balance to supply and 10 to demand. Therefore, cross out row two.
Move to cell X31 because it has lowest cost. (24) among the uncrossed cells and then allocate 10. The balance of zero is obtained for demand while that of supply is 40. Column one is exhausted and it is crossed out.
Lastly, allocate 30 and 10 respectively to the remaining two cells X33 and X34.
21 18 27 22
19 18 24 20
24 25 28 25
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(iii) Vogel‟s Approximation Method (VAM)
VAM, which is also called penalty method, is an improvement on the LCM method that generates a better initial solution.
It makes use of opportunity cost (penalty) principles in order to make allocation to various cells by minimizing the penalty cost.
The steps in this method are:
Step 1: Compute for each row(column) the penalty by subtracting the smallest unit (Cij) from the next smallest unit cost in the same row (column).
Step 2: Select the row or column with the highest penalty and then allocate as much as possible to the variable with least cost in the selected row or column. Any ties should be handled arbitrarily.
Step 3: Adjust the supply and demand and cross-out the exhausted row or column. If a row and a column are simultaneously exhausted, only one of the two is crossed-out and the other assigned zero supply (demand).
Step 4: Compute the next penalties by considering uncrossed rows (columns) and go to step 2.
Step 5: Exactly when one row (column) remains uncrossed, then allocate the leftover and stop.
ILLUSTRATION 21-3
Use the data in illustration 21-1 to obtain the initial solution by VAM.
SUGGESTED SOLUTION 21-3
Applying the above steps, gives the following tables:
Destination
Plant X Y W Z Supply
A 50 50 0
B 40 20 60 40 0
C
10 30 10 50 40 30 0
Demand 50 10 0
70 20 0
30
0 10
0 160
Penalty Table Iteration Rows
A B C Columns
X Y W Z Allocation 1 2
3 4
3 1 1
* 1 1
* 1 1
* * 1
2 0 3 2 5 7 4 5 5 * 4 5 - * - -
X12 = 50 X22 = 20 X21 = 40 X31 = 10 X33 = 30 X34 = 10
In the Penalty Table, the circled figures represent highest penalty in each iteration; * represents the crossed-out row or column, and (-) stands for penalty not possible.
Note that in the penalty table, 3 is obtained under row A(in iteration 1) by the difference of costs 21 and 18. This principle is continually applied to other rows (B and C) to obtain 1 and 1 respectively.
For the column also, 2 is obtained under column x (in iteration 1) by the difference of costs 21 and 19.
Total cost = 50(18) + 40(19) + 20(18) + 10(24) + 30(28) + 10(25) = N 3350
21 18 27 22
19 18 24 20
24 25 28 25
The principle of obtaining the difference between the two least costs along row (column) is applied throughout the penalty table in order to make allocation as stated in step 2
479 ILLUSTRATION 21-4
A company with three factories (X,Y,Z) and five warehouses (A,B,C,D,E) in different locations has the transportation costs (in Naira) from factories to warehouses. Factory capacities and warehouse requirements are stated below:
Factories Warehouses Factory
Capacities
A B C D E
X Y Z
5 4 8
8 7 4
6 8 7
4 6 5
3 5 6
800 1100 600 Warehouse
Requirements 350 425 500 650 575 2500 Determine the initial feasible solution by
(b) North West Corner Method (b) Least Cost Method
(b) Vogel‟s Approximation Method SUGGESTED SOLUTION 21-4
(a) North West Corner Method Total Cost
=(350x5)+(425x8)+(25x6)+(475x8)+(125x6)+(525x5)+(575x6)
= N15,925
5 8 6 4 3 800 450
350 425 25 25 0
4 7 8 6 5 600 800
475 125 0
8 4 7 5 6 1100
525 575 575 0
Factories
x
y
Warehouses
A B C D E
475 525 0
z
350 425
0 0
Warehouse Requirements
Factory Capacities
2500
500 650 575
0 0
Explanation to the above allocation
Beginning from North-West (i.e cell X11) corner,
Allocate 350 to cell X11 in order to satisfy the minimum of factory capacity and warehouse requirement. Zero balance is left for warehouse requirement while that of factory capacity along X11 is 450. Therefore, column one is crossed out.
Move to cell X12 and allocate 425. The warehouse balance along cell X12 is zero while we have the balance of 25 for the factory capacity. Therefore, column two is crossed out.
Next, move to cell X13 and allocate 25 giving the balance of zero for factory capacity and 475 for the warehouse along that cell (i.e. X13). Row one is therefore crossed out.
Move to cell X23 to allocate 475. It then gives a balance of zero for the warehouse while that of factory capacity is 125. Column three is crossed out.
We also move to cell X24 and allocate 125 to that cell (X24) giving the balance of zero for the factory capacity and 525 for the warehouse. Then cross-out row two.
Finally, allocate 525 and 575 respectively to the remaining two cells X34
and X35.
LCM
Total Cost =
(225x4)+(575x3)+(350x4)+(250x8)+(425x4)+(250x7)+(425x5)
= N11,600
Explanation to the above allocation
The cell that has lowest cost is X15. Allocate 575 to that cell giving the balance of zero for warehouse requirement and 225 for factory capacity. Column five is crossed out.
5 8 6 4 3 800 225
225 575 0
4 7 8 6 5 600 250
475 125 0
8 4 7 5 6 1100 675
425 250 425 250 0
Factories Warehouses Factory
Capacities
A B C D E
425 500 650
x
y
z Warehouse Requirements
575
0 0 250 425 0 2500
0 0
350
481
For the remaining uncrossed cells, cells X14, X21 and X32 have the lowest cost of 4. Arbitrarily, cell X32 is allocated with 425 giving the balance of zero for warehouse requirement and 675 for the factory capacity. Column two is crossed-out.
Next, cell X21 is allocated 350 giving the balance of zero for warehouse requirement and 250 for the factory capacity. Column 1 is also crossed-out.
We allocate 225 to cell X14 to give the balance of zero for factory capacity and 425 for the warehouse requirement. Row one is crossed out.
The left-over cells are X23, X24, X33 and X34. We allocate 425 to cell X34 to give the balance of zero for warehouse requirement and 250 for the factory capacity. Therefore, column four is crossed out.
Finally, the remaining two cells X23 and X33 are respectively allocated 250 each (C) VAM
Penalty Table
Iteration Rows Columns
Allocation
X Y Z A B C D E
1 2 3 4 5
1 1 1 2 *
1 1 2 2 2
1 1 2 2 2
1 1 1
* - 3
* - - -
1 1 1 1 1
1 1 1 1 1
2 2 * - -
X32 = 425 X15 = 575 X21 = 350 X14 = 225 X24 = 250 X33 = 500 X34 = 175 The principle of obtaining the difference between the two least costs along row (column) is applied throughout the penalty table in order to make allocation as stated in step 2.
5 8 6 4 3 800 225
256 575 0
4 7 8 6 5 600 250
350 125 250 0
8 4 7 5 6 1100 675
425 500 175 500 0
Factories Warehouses Factory
Capacities
A B C D E
x
y
z
Warehouse Requirements
350 425 500 650 575
0 0 250 425 0 2500
0 0
175
In the Penalty Table, the circled figures represent highest penalty in each iteration.
(*) represents the crossed out row or column, and (-) stands for penalty not possible. Also, note in the penalty table that 1 is obtained under row X (in iteration 1) by the difference between 4 and 3. Similarly for row Y, 1 is obtained from the difference between 5 and 4. The difference computations continued in this manner for both rows and columns in each iteration.
Total Cost =
(225x4)+(575x3)+(350x4)+(250x6)+(425x4)+(500x7)+(175x5)
= N11,600.