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Operations Research (MTH601)
168
Table 4
Distribution center
Plant
1
2
3
4
5
Supply
1
20
25
27
20
15
40
2
18
21
22
24
20
70
3
19
17
20
18
19
90
4
0
0
0
0
0
30
Demand
30
40
60
40
60
230
FINDING AN INITIAL BASIC FEASIBLE SOLUTION
An initial basic feasible solution to a transportation problem can be found by any one of the three
following methods:
(i) North west corner rule (NWC)
(ii) Least cost method (LCM)
(iii) Vogel's approximation method (VAM)
North West Corner Rule
STEP 1 Start with the cell in the upper left hand corner (North West Corner).
STEP 2 Allocate the maximum feasible amount.
STEP 3 Move one cell to the right if there is any remaining supply. Otherwise, move one cell down. If both are
impossible, stop or go to step (2).
An example is considered at this juncture to illustrate the application of NWC rule.
Example 2
Table 5
Destinations
Origin
P
Q
R
Supply
A
5
7
8
70
B
4
4
6
30
C
6
7
7
50
Demand
65
42
43
150
In table 5 supply level, demands at various destinations, and the unit cost of transportation are given. Use NWC
rule to find the initial solution.
Solution:
The solution obtained by the North West Corner Rule is a basic feasible solution. In this method we do
not consider the unit cost of transportation. Hence the solution obtained may not be an optimal solution. But this
will serve as an initial solution, which can be improved.
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169
We give below the procedure for the solution of the above problem by NWC rule.
The North West Corner cell (AP) is chosen for allocation. The origin A has 70 items and the
destination P requires only 65 items. Hence it is enough to allot 65 items from A to P. The origin A which is
alive with 5 more items can supply to the destination to the right is alive with 5 more items can supply to the
destination to the right of P namely Q whose requirement is 42. So, we supply 5 items to Q thereby the origin A
is exhausted. Q requires 37 items more. Now consider the origin B that has 30 items to spare. We allot 30 items
to the cell (BQ) so that the origin B is exhausted. Then move to origin C and supply 7 more items to the
destination Q. Now the requirement of the destination Q is complete and C is left with 43 items and the same
can be alloted to the destination R. Now the origin C is emptied and the requirement at the destination R is also
complete. This completes the initial solution to the problem.
The above calculations are performed conveniently in a table 6 as shown below:
Table 6
Destinations
Origin
P
Q
R
A
65
5
70
65
0
B
30
30
0
C
7
43
50
7
43
0
65
42
43
0
37
0
0
The total cost of transportation by this method will be
65 5 + 5 7 + 30 4 + 7 7 + 43 7 = Rs. 830.
As the solution obtained by the North West Corner Rule may not be expected to be particularly close to
the optimal solution, we have to explore a promising initial basic feasible solution, so that we can teach the
optimal solution to the problem with minimum number of iterations.
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Least Cost Method
STEP 1: Determine the least cost among all the rows of the transportation table.
STEP 2:Identify the row and allocate the maximum feasible quantity in the cell corresponding to the least cost
in the row. Then eliminate that row (column) when an allocation is made.
STEP 3: Repeat steps 1 and 2 for the reduced transportation table until all the available quantities are
distributed to the required places. If the minimum cost is not unique, the tie can be broken arbitrarily.
To illustrate, consider the example repeated in table 7
Table 7
Destination
Origin
P
Q
R
Supply
A
5
7
8
70
B
4
4
6
30
0
C
6
7
7
50
Demand
65
42
43
55
We examine the rows A, B and C, 4 is the least cost element in the cell (B,P) and (B, Q) and the tie can be
broken arbitrarily. Select (B, P). The origin B can supply 30 items to P and thus origin B is exhausted. P still
requires 35 more units. Hence, deleting the row B, we have the reduced matrix as in the table 8
Table 8
Destination
Origin
P
Q
R
Supply
35
A
5
7
8
70
35
C
6
7
7
50
Demand
35
42
43
0
In the reduced matrix (table 8) we observe that 5 is the least element in the cell (A, P) and examine the supply at
A and demand at P. The destination P requires 35 items and this requirement is satisfied from A so that the
column P is deleted next. So we have the reduce matrix as in table 9
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Table 9
Destination
Origin
Q
R
Supply
35
A
7
8
35
0
C
7
7
50
Demand
42
43
7
In the reduced matrix (table 9) we choose 7 as least element corresponding to the cell (A. Q). We
supply 35 units from A to Q so have the reduced matrix in table10.
Table 10
Destination
Origin
Q
R
Supply
7
43
A
7
8
50
0
Demand
7
43
0
0
Now, only one row is left behind. Hence, we allow 7 items from C to Q and 43 items C to R.
We now have the allotment as per the least cost method as shown in the table 11
Table 11
Destination
Origin
P
Q
R
Supply
A
35
35
70
B
30
30
C
7
43
50
Demand
65
42
43
The cost of the allocation by the least cost method is 35 x 5 + 35 x 7 + 30 x 4 + 7 x 7 + 43 x 7 = Rs. 890
Vogel's Approximation Method (VAM)
This method is based on the 'difference' associated with each row and column in the matrix giving unit
cost of transportation cij. This 'difference' is defined as the arithmetic difference between the smallest and next
to the smallest element in that row or column. This difference in a row or column indicates the minimum unit
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172
penalty incurred in failing to make an allocation to the smallest cost cell in that row or column. This
difference also provides a measure of proper priorities for making allocations to the respective rows and column.
In other words, if we take a row, we have to allocate to the cell having the least cost and if we fail to do so, extra
cost will be incurred for a wrong choice, which is called penalty. The minimum penalty is given by this
difference. So, the procedure repeatedly makes the maximum feasible allocation in the smallest cost cell of the
remaining row or column, with the largest penalty. Once an allocation is fully made in a row or column, the
particular row or column is eliminated. Hence and allocation already made cannot be changed. Then we have a
reduced matrix. Repeat the same procedure of finding penalty of all rows and columns in the reduced matrix,
choosing the highest penalty in a row or column and allotting as much as possible in the least cost cell in that
row or column. Thus we eliminate another fully allocated row or column, resulting in further reducing the size
of the matrix. We repeat till all supply and demand are exhausted.
A summary of the steps involved in Vogel's Approximation Method is given below:
STEP 1: Represent the transportation problem in the standard tabular form.
STEP 2: Select the smallest element in each row and the next to the smallest element in that row. Find the
difference. This is the penalty written on the right hand side of each row. Repeat the same for each column. The
penalty is written below each column.
STEP 3: Select the row or column with largest penalty. If there is a tie, the same can be broken arbitrarily.
STEP 4: Allocate the maximum feasible amount to the smallest cost cell in that row or column.
STEP 5: Allocate zero else where in the row or column where the supply or demand is exhausted.
STEP 6: Remove all fully allocated rows or columns from further consideration. Then proceed with the
remaining reduced matrix till no rows or columns remain.
Let us apply Vogel's Approximation Method to the above example as given below in table12
Table 12
Origin
Destination
Supply
Row
difference
P
Q
R
A
5
7
8
70
(2)
30
6
30
0
(0)
B
4
4
C
6
7
7
50
(1)
Demand
65
42
43
12
Column difference (1)
(3)
(1)
Largest Penalty
The difference between the smallest and next to the smallest element in each row and in each column is
calculated. This is indicated in the parenthesis. We choose the maximum from among the differences. The first
individual allocation will be to the smallest cost of a row or column with the largest difference. So we select the
column Q (penalty = 3) for the first individual allocation, and allocate to (B, Q) as much as we can, since this
cell has the least cost location. Thus 30 units from B are allocated to Q. This exhausts the supply from B.
However, there is still a demand of 12 units from Q. The allocations to other cells in that column are 0, as
indicated. The next step is to write down the reduced matrix (as in table 13) eliminating row B (as it is
exhausted).
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Table 13
Origin
Destination
Supply
Row
difference
P
Q
R
65
7
8
70
5
(2)
A
5
C
6
7
7
50
(1)
Demand
65
12
43
0
Column difference (1)
(0)
(1)
From the table 13, (2) is the largest unit difference corresponding to the row A. This leads to an allocation in the
corresponding minimum cost location in row A, namely cell (A, P). The maximum possible allocation is only 65
as required by P from A and allocation of 0 to others in the row A. Column P is thus deleted and the reduced
matrix is given in table 14.
Maximum difference is 1 in row A and in column C. Select arbitrarily A and allot the least cost cell (A, Q) 5
units. Delete row A.
Now, we have only one row C and two columns Q and R (Table 15) indicating that all the available amount
from C has to be moved to Q and R as per their requirements. Hence we have the table 15
Table 14
Origin
Destination
Supply
Row
difference
P
R
5
8
5
0
(1)
A
7
C
7
7
50
Demand
12
43
7
Column difference
(0)
(1)
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Table 15
Destination
Supply
Origin
Q
R
7
43
50
0
C
7
7
7
43
0
0
Table 16
Destination
Origin
P
Q
R
Supply
A
65
5
70
B
30
30
C
7
43
50
Demand
65
42
43
We obtain as our basic feasible solution by re-tracking various positive allocations in successive stages.
We have the solution by Vogel's Approximation Method as shown in the table 16
The cost of allocation by Vogel's Approximation Method will be
65 5 + 5 7 + 30 4 + 7 7 + 43 7
= 325 + 35 + 120 + 49 + 301 = Rs. 830.
Note:
Cost of allocation for the same problem with three methods:
NWC method - Rs. 830/-
Least cost method - Rs.890/-
Vogel's Approximation Method - Rs. 830/-
Generally VAM gives a better initial solution.
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REVIEW QUESTIONS
1.
What do you mean by transportation problem?
2.
What do you mean by feasible solution and basic feasible solution of transportation problem?
3.
Describe Transportation problem. Give a method of finding an initial feasible solution.
4.
Explain in brief with examples.
(i) North West Corner rule.
(ii) Vogel's Approximation Method.
5.
Explain the classical transportation problem and write down its mathematical formulation and show
that it is a particular case of a linear programming problem.
6.
A petroleum company has three major oil fields and five oil refineries. The shipping costs from the
fields to the refineries, fields are as shown in the table.
Refinery Capacity
Production
Refinery
Barrels per day
Field
Barrels per day
A
10,000
1
20,000
B
13,000
2
25,000
C
13,000
3
30,000
D
16,000
E
18,000
Cost per Barrel (Rs.)
Field
A
B
C
D
E
1
400
300
330
380
360
2
350
350
380
320
350
3
370
300
400
350
340
Determine the optimum scheme using the North West Corner Rule.
Determine the optimum scheme using the Least Cost Method.
Determine the optimum scheme using Vogel's Approximation Method.
Compare the computations
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Table of Contents:
  1. Introduction:OR APPROACH TO PROBLEM SOLVING, Observation
  2. Introduction:Model Solution, Implementation of Results
  3. Introduction:USES OF OPERATIONS RESEARCH, Marketing, Personnel
  4. PERT / CPM:CONCEPT OF NETWORK, RULES FOR CONSTRUCTION OF NETWORK
  5. PERT / CPM:DUMMY ACTIVITIES, TO FIND THE CRITICAL PATH
  6. PERT / CPM:ALGORITHM FOR CRITICAL PATH, Free Slack
  7. PERT / CPM:Expected length of a critical path, Expected time and Critical path
  8. PERT / CPM:Expected time and Critical path
  9. PERT / CPM:RESOURCE SCHEDULING IN NETWORK
  10. PERT / CPM:Exercises
  11. Inventory Control:INVENTORY COSTS, INVENTORY MODELS (E.O.Q. MODELS)
  12. Inventory Control:Purchasing model with shortages
  13. Inventory Control:Manufacturing model with no shortages
  14. Inventory Control:Manufacturing model with shortages
  15. Inventory Control:ORDER QUANTITY WITH PRICE-BREAK
  16. Inventory Control:SOME DEFINITIONS, Computation of Safety Stock
  17. Linear Programming:Formulation of the Linear Programming Problem
  18. Linear Programming:Formulation of the Linear Programming Problem, Decision Variables
  19. Linear Programming:Model Constraints, Ingredients Mixing
  20. Linear Programming:VITAMIN CONTRIBUTION, Decision Variables
  21. Linear Programming:LINEAR PROGRAMMING PROBLEM
  22. Linear Programming:LIMITATIONS OF LINEAR PROGRAMMING
  23. Linear Programming:SOLUTION TO LINEAR PROGRAMMING PROBLEMS
  24. Linear Programming:SIMPLEX METHOD, Simplex Procedure
  25. Linear Programming:PRESENTATION IN TABULAR FORM - (SIMPLEX TABLE)
  26. Linear Programming:ARTIFICIAL VARIABLE TECHNIQUE
  27. Linear Programming:The Two Phase Method, First Iteration
  28. Linear Programming:VARIANTS OF THE SIMPLEX METHOD
  29. Linear Programming:Tie for the Leaving Basic Variable (Degeneracy)
  30. Linear Programming:Multiple or Alternative optimal Solutions
  31. Transportation Problems:TRANSPORTATION MODEL, Distribution centers
  32. Transportation Problems:FINDING AN INITIAL BASIC FEASIBLE SOLUTION
  33. Transportation Problems:MOVING TOWARDS OPTIMALITY
  34. Transportation Problems:DEGENERACY, Destination
  35. Transportation Problems:REVIEW QUESTIONS
  36. Assignment Problems:MATHEMATICAL FORMULATION OF THE PROBLEM
  37. Assignment Problems:SOLUTION OF AN ASSIGNMENT PROBLEM
  38. Queuing Theory:DEFINITION OF TERMS IN QUEUEING MODEL
  39. Queuing Theory:SINGLE-CHANNEL INFINITE-POPULATION MODEL
  40. Replacement Models:REPLACEMENT OF ITEMS WITH GRADUAL DETERIORATION
  41. Replacement Models:ITEMS DETERIORATING WITH TIME VALUE OF MONEY
  42. Dynamic Programming:FEATURES CHARECTERIZING DYNAMIC PROGRAMMING PROBLEMS
  43. Dynamic Programming:Analysis of the Result, One Stage Problem
  44. Miscellaneous:SEQUENCING, PROCESSING n JOBS THROUGH TWO MACHINES
  45. Miscellaneous:METHODS OF INTEGER PROGRAMMING SOLUTION