Download Solving Transportation Problems - Prof. Kumar and more Lecture notes Production and Operations Management in PDF only on Docsity!
UNIT 4 TRANSPORTATION PROBLEM
Structure
4.1 Introduction
Objectives
4.2 Mathematical Formulation of the Transportation Problem
4.3 Methods of Finding Initial Basic Feasible Solution
North-West Corner Rule Least Cost Method Vogel’s Approximation Method
4.4 Methods of Finding Optimal Solution
Stepping Stone Method Modified Distribution (MODI) Method
4.5 Some Special Cases
Unbalanced Transportation Problem Degeneracy Alternative Optimal Solutions Maximisation Transportation Problem Prohibited Routes
4.6 Summary
4.7 Solutions / Answers
4.1 INTRODUCTION
In Units 2 and 3 of this block, we have discussed linear programming problems and methods of solving them. You have studied the graphical and simplex methods of solving LPPs in Unit 2 and Unit 3, respectively. Transportation problem is also one of the sub-classes of Linear Programming problems in which the objective is to transport various quantities (goods) initially stored at various origins/plants/factories to different destinations/distribution centres/warehouses in such a way that the total transportation cost is minimised. To achieve this objective, we must know the quantity of available supplies and the quantities demanded. Also, we must know the costs of shipping a unit from various origins to various destinations. Solving LPPs by either of the methods discussed in Units 2 and 3 in this block involves a lot of computational work and is quite cumbersome and difficult for solving the transportation problems. So in this unit, we discuss the methods, which are specifically applied for solving transportation problems.
Objectives
After studying this unit, you should be able to:
define a transportation problem;
determine the basic feasible solution of a transportation problem using North-West Corner Rule, Least Cost and Vogel’s Approximation methods;
state the conditions for performing optimality test;
apply the Stepping Stone and Modified Distribution (MODI) methods of obtaining the optimal solution of a transportation problem; and
Optimisation Techniques-I solve the transportation problems for special cases such as unbalanced transportation problem, case of degeneracy, case of alternative solutions, maximisation transportation problem, problems with prohibited routes.
4.2 MATHEMATICAL FORMULATION OF THE
TRANSPORTATION PROBLEM
Suppose a manufacturer of an item has m origins (plants/factories) situated at places O1, O2,.. ., Oi,.. ., Om and suppose there are n destinations (warehouses/distribution centres) at D1, D2,... Dj... Dn. The aggregate of the capacities of all m origins is assumed to be equal to the aggregate of the requirements of all n destinations. Let Cij be the cost of transporting one unit from origin i to destination j. Let ai be the capacity/availability of items at origin i and bj , the requirement/demand of the destination j. Then this transportation problem can be expressed in a tabular form as follows:
Origin Destinations D 1 D 2... Dj... Dn
Availability / Capacity O 1
O 2 . . . Oi . . . Om
C 11 C 12 … C 1 j … C1n
C 21 C 22 … C 2 j … C2n
.... .... .... Ci1 Ci2 … Cij … Cin .... .... .... Cm1 Cm2 … Cmj… Cmn
a 1
a 2 . . . ai . . . am
Requirement/ Demand
b 1 b 2 … bj… bn Total
The condition for the existence of a feasible solution to a transportation problem is given as m n i i j 1 j 1
a b
^ …(1)
Equation (1) tells us that the total requirement equals the total capacity. If it is not so, a dummy origin or destination is created to balance the total capacity and requirement.
Now let xij be the number of units to be transported from origin i to destination j and Cij the corresponding cost of transportation. Then the total
transportation cost is
m n ij ij i 1 j 1
C x
. It means that the product of the number of
units allocated to each cell with the transportation cost per unit for that cell is calculated and added for all cells. Our objective is to allocate the units in such a way that the total transportation cost is least. The problem can also be stated as a linear programming problem as follows: We are to minimise the total transportation cost
Z =
m n ij ij i 1 j 1
C x
subject to the constraints:
Optimisation Techniques-I equal to the requirement at First Distribution Centre/Destination (or the quantity available at First Plant/Origin). At this stage, both Column 1 as well as Row 1 is exhausted. We cross them out and proceed to the north- west corner of the resulting matrix, i.e., cell (2, 2).
We continue in this manner, until we reach the south-east corner of the original matrix.
Let us take up an example to further explain NWC rule. Example 1: Apply the North-West Corner rule for finding the basic feasible solution of the following transportation problem:
Solution: We start from the North West corner, i.e., the Factory A and Warehouse D. The quantity needed at the First Warehouse (Warehouse D) is 80, which is less than the quantity available (160) at the First Factory A. Therefore, a quantity equal to the requirement at Warehouse D is to be allocated to the cell (A, D). Thus, the requirement of Warehouse D is met by Factory A. So we cross out column 1 and reduce the capacity of Factory A by 80. Then we go to cell (A, E), which is the North-West corner of the resulting matrix.
Warehouse Factory
D E F G Capacity
A
B
C
Requirement 80 90 110 220 500
The resulting matrix is Warehouse Factory
E F G Capacity
A
B
C
Requirement 90 110 220
Now, the quantity needed at the Second Warehouse (Warehouse E) is 90, which is greater than the quantity available (80) at the First Factory A. Therefore, we allocate a quantity equal to the capacity at Factory A, i.e., 80 to the cell (A, E). The requirement of warehouse E is reduced to 10. The capacity of Factory A is exhausted and has to be removed from the matrix. Therefore, we cross out row 1 and proceed to cell (B, E). This is the North- West corner of the resulting matrix.
Warehouse Factory
D E F G Capacity
A
B
C
Requirement 80 90 110 220 500
80
0
80
Transportation Problem Warehouse Factory
E F G Capacity
A
B
C
Requirement 90 110 220
The resulting matrix is
Warehouse Factory
E F G Capacity
B C
Requirement 10 110 220
Now, the quantity needed at the Second Warehouse (Warehouse E) is 10, which is less than the quantity available at the Second Factory B, which is
- Therefore, the quantity 10 equal to the requirement at Warehouse E is allocated to the cell (B, E). Hence, the requirement of Warehouse E is met and we cross out column 1. We reduce the capacity of Factory B by 10 and proceed to cell (B, F), which is the new North-West corner of the resulting matrix.
Warehouse Factory
E F G Capacity
B C
Requirement 10 110 220
The resulting matrix is
Warehouse Factory
F G Capacity
B C
Requirement 110 220
Again, the quantity needed at the Third Warehouse (Warehouse F) is 110. It is less than the quantity available at the Second Factory (Factory B), which is 140. Therefore, a quantity equal to the requirement at Warehouse F is allocated to the cell (B, F). Since, the requirement of warehouse F is met, we cross out Column 1 and reduce the capacity of Factory B by 110. Then we proceed to cell (B, G), which is the new North-West corner of the resulting matrix given below:
Warehouse Factory
G Capacity
B C
Requirement 220
Now, the quantity needed at the Fourth Warehouse (Warehouse G) is 220, which is greater than the quantity available at the Second Factory (Factory B). Therefore, we allocate the quantity equal to the capacity at Factory B to
80 0
10
(^110 )
0
10 140
0
Transportation Problem or column for which the capacity is exhausted or requirement is satisfied is removed from the transportation table. The process is repeated with the reduced matrix till all the requirements are satisfied. If there is a tie for the lowest cost cell while making any allocation, the choice may be made for a row or a column by which maximum requirement is exhausted. If there is a tie in making this allocation as well, then we can arbitrarily choose a cell for allocation.
Let us explain this method with the help of an example.
Example 2: Apply the Least Cost method for finding the Basic Feasible solution of the transportation problem of Example 1.
Solution: Here, the least cost is 37 in the cell (A, G). The requirement of the Warehouse G is 220 and the capacity of Factory A is 160. Hence, the maximum number of units that can be allocated to this cell is 160. Thus, Factory A is exhausted and the row has to be removed from the next matrix. The requirement of Warehouse G is reduced by 160.
Warehouse Factory
D E F G Capacity
A
B
C
Requirement 80 90 110 220 500
The reduced matrix, therefore, is
Warehouse Factory
D E F G Capacity
B
C
Requirement 80 90 110 60
Now, the least cost is 38, which is in the cell (C,E). The requirement of the Warehouse E is 90 and the capacity of Factory C is 190. Hence, the maximum number of units that can be allocated to this cell is 90. Thus, Warehouse E is exhausted and has to be removed for the next matrix. So we cross out this column. Moreover, we reduce the capacity of factory C by 90.
Warehouse Factory
D E F G Capacity
B
C
Requirement 80 90 110 60
The reduced matrix, therefore, is
Warehouse Factory
D F G Capacity
B C
Requirement 80 110 60
(^00)
060
160
0
(^00)
(^090 )
Optimisation Techniques-I The least cost in this matrix is 39, which is in the cell (C, D). The requirement of Warehouse D is 80 and the capacity of Factory C is 100. Hence, the maximum number of units that can be allocated to this cell is 80. Thus, the requirement of Warehouse D is exhausted and this column is removed for the next matrix. The capacity of Factory C is also reduced by 80. Warehouse Factory
D F G Capacity
B C
Requirement 80 110 60
Thus, the reduced matrix is Warehouse Factory
F G Capacity
B
C
Requirement 110 60
The least cost in this matrix is 40 which is in the cell (C, F). The requirement of Warehouse F is 110 and the capacity of Factory C is 20. Hence, the maximum number of units that can be allocated to this cell is 20. Thus, Factory C is exhausted and we remove the row for the next matrix. The requirement of Warehouse F is reduced by 20. It is now 90 in the reduced matrix. Warehouse Factory
F G Capacity
B
C
Requirement 110 60 The reduced matrix is Warehouse Factory
F G Capacity
B 52 51 150
Requirement 90 60
The least cost is 51 in the cell (B, G) and the requirement of warehouse G is 60 units. So we allocate 60 units to cell (B, G) and the remaining 90 units to the cell (B, F). Thus, the allocations given using Least Cost method are as shown in the following matrix along with the cost per unit of transportation:
Warehouse Factory
D E F G Capacity
A 42 48 38 37 160
B 40 49 52 51 150
C 39 38 40 43 190
Requirement 80 90 110 220 500
80
90
90
160
20
(^00)
(^080 )
(^00) (^090)
020
60
0
60 90
Optimisation Techniques-I Example 3: Apply the Vogel’s Approximation Method for finding the Basic Feasible Solution for the transportation problem of Example 1. Solution: In the first row, the least and the second least costs are 37 and 38 and their difference is 1. We write 1 in a new column created on the right. It is labelled Penalty. Similarly, the differences between the least and the second least costs in the second and third row, respectively, are 49−40= and 39−38=1. So we write the values (differences), i.e., 9 and 1 in the penalty column. Next we find the differences of the least and second least elements of each of the columns D, E, F and G. These are 40−39=1, 48−38=10, 40−38=2 and 43 −37=6 , respectively. We write them in a newly created penalty row at the bottom of the table. We now select the largest of these differences in the penalty row and column, which is 10 in this case. This value (10) corresponds to the second column (Column E) and the least cost in this column is 38. Hence the allocation of 90 units (the maximum requirement of Warehouse E) is to be made in the cell (C, E) from Factory C. Since the column corresponding to E is exhausted, it is removed for the next reduced matrix and the capacity of C is reduced by 90.
Warehouse Factory
D E F G Capacity Penalty
A
B
C
Requirement 80 90 110 220 500 Penalty 1 10 2 6
The reduced matrix, therefore, is Warehouse Factory
D F G Capacity Penalty
A
B
C
Requirement 80 110 220 410 Penalty 1 2 6
We now take the differences between the least and second least cost for each row and column of the reduced matrix. In the first row, the least and the second least costs are 37 and 38 and their difference is 1. We write it in the newly created penalty column. Similarly, we write the second difference element 51−40 = 11 and third difference element 40−39 = 1 in the second and third row of this column. Likewise, the differences of the smallest and second smallest elements of each of the columns D, F and G are 40−39 = 1, 40 −38 = 2 and 43−37 = 6, respectively. We write these in a newly created penalty row at the bottom of the table. Now, we select the largest of these differences in the penalty row and column, which is 11 in this case. This value (11) corresponds to Row B. Since the least cost in row B is 40, we allocate 80 units (the maximum requirement of Warehouse D) to the cell (B, D). Thus, the requirement of
0
90 100
Transportation Problem
Warehouse D is exhausted and we can remove it. We also reduce the capacity of Factory B by 80 in the next reduced matrix.
Warehouse Factory
D F G Capacity
A
B
C
Requirement 80 110 220 330
The reduced matrix is
Warehouse Factory
F G Capacity Penalty
A
B
C
Requirement 110 220 330 Penalty 2 6
Again, in the first row, the least and the second least costs are 37 and 38 and their difference is 1. We write it to the right of this row in the newly created penalty column. Similarly, the second and third elements in the second and third rows of this column are 52−51 = 1 and 43−40 = 3, respectively. Likewise, the differences of the least and second least elements of each of the columns F and G are 40−38 = 2 and 43−37 = 6, respectively. We write these in a newly created penalty row at the bottom of the table. Now, we select the largest of these differences, which is 6 in this case. It corresponds to Column G and the least cost in this column is 37. Hence, we allocate 160 units (the maximum capacity of Factory A) to the cell (A, G). Since Row A is exhausted, it is removed for the next reduced matrix. We also reduce the requirement of Warehouse G by 160 units.
Warehouse Factory
F G Capacity
A B C
Requirement 110 220 170
The reduced matrix is
Warehouse Factory
F G Capacity Penalty
B C
Requirement 110 60 170 Penalty 12 8
Once again, the difference of the least and the second least costs in the first row is 52−51=1. We write it in the newly created penalty column. Similarly, for the second row, the difference is 43−40=3. Likewise, the differences of the least and second least elements of each of the columns F and G are 52 −40=12 and 51−43=8, respectively. We write them in the newly created
0
8 0 70
160 0
60
Transportation Problem Origin / Distribution Centre
1 2 3 4 5 6 Availability
Requirement 8 8 16 3 8 21
Before studying the next section, match your answer with the answer given in Sec. 4.7.
4.4 METHODS OF FINDING OPTIMAL
SOLUTION
Once the initial basic feasible solution is determined, the optimality test is performed to find whether the obtained feasible solution is optimal or not. The optimality test is performed by applying one of the following methods:
i) Stepping Stone Method ii) Modified Distribution (MODI) Method
These methods reveal whether the initial basic feasible solution is optimal or not. They also improve the solution until the optimal solution is obtained. The Stepping Stone method is applied to a problem of small dimension as its application to a problem of large dimension is quite tedious and cumbersome. The MODI method is usually preferred over the Stepping Stone method.
Before discussing the methods of performing optimality test and finding optimal solutions, we explain the concept of independent allocations. We also state the conditions for performing optimality test.
Independent and Non-Independent Allocations
If it is not possible to form any closed loop through the allocations under consideration then the allocations are said to be independent. Here, formation of closed loop means that it is possible to travel from any allocation, back to itself by a series of horizontal and vertical jumps from one occupied cell (i.e., the cell containing allocation) to another, without a direct reversal of route (see Fig. 4.1a). If such a loop can be formed using some or all of the allocations under consideration, the allocations are known as non-independent.
● ● ● ● ● ● ●
(a) (b) Fig. 4.1: Example of a) Independent Allocations; b) Non-Independent Allocations.
Note: Every loop has an even number of cells.
Optimisation Techniques-I Conditions for Performing Optimality Test An optimality test can be applied to that feasible solution which satisfies the following conditions:
- It contains exactly m + n−1 allocations where m and n represent the number of rows and columns, respectively, of the transportation table.
- These allocations are independent. Let us now discuss the methods of performing optimality test and hence finding optimal solutions for each of these methods.
4.4.1 Stepping Stone Method
In the Stepping Stone method for obtaining the optimal solution of a transportation problem, you should follow the steps given below:
- Determine an initial basic feasible solution.
- Evaluate all unoccupied cells for the effect of transferring one unit from an occupied cell to the unoccupied cell as follows:
a) Select an unoccupied cell to be evaluated.
b) Starting from this cell, form a closed path (or loop) through at least three occupied cells. The direction of movement is immaterial because the result will be the same in both directions. Note that except for the evaluated cell, all cells at the corners of the loop have to be occupied.
c) At each corner of the closed path, assign plus (+) and minus (−) sign alternatively, beginning with the plus sign for the unoccupied cell to be evaluated.
d) Compute the net change in cost with respect to the costs associated with each cell traced in the closed path.
e) Repeat steps 2(a) to 2(d) until the net change in cost has been calculated for all occupied cells.
- If all net changes are positive or zero, an optimal solution has been arrived at. Otherwise go to step 4.
- If some net changes are negative, select the unoccupied cell having the most negative net change. If two negative values are equal, select the one that results in moving more units into the selected unoccupied cell with the minimum cost.
- Assign as many units as possible to this unoccupied cell.
- Go to Step 2 and repeat the procedure until all unoccupied cells are evaluated and the value of net change, i.e., net evaluation is positive or zero.
Let us now take up an example of transportation problem and apply the Stepping Stone method for finding the optimal solution.
Example 4: A company is spending `1000 on transportation of its units from three plants to four distribution centres. The availability of unit per plant and requirement of units per distribution centre, with unit cost of transportation are given as follows:
Optimisation Techniques-I Unoccupied Cell
Closed Path Net Change in Cost ( ` ) (P 1 , D 2 ) (^) (P 1 , D 2 ) (P 1 , D 4 ) (P 3 , D 4 ) (P 3 , D 2 ) (^30) – 12+20–10 = 28 (P 1 , D 3 ) (P 1 , D 3 ) (P 1 , D 4 ) (P 2 , D 4 ) (P 2 , D 3 ) (^50) – 12+60–40 = 58 (P 2 , D 1 ) (^) (P 2 , D 1 ) (P 1 , D 1 ) (P 1 , D 4 ) (P 2 , D 4 ) (^70) – 19+12–60 = 3 (P 2 , D 2 ) (^) (P 2 , D 2 ) (P 3 , D 2 ) (P 3 , D 4 ) (P 2 , D 4 ) (^30) – 10+20–60 = – 20 (P 3 , D 1 ) (^) (P 3 , D 1 ) (P 1 , D 1 ) (P 1 , D 4 ) (P 3 , D 4 ) (^40) – 19+12–20 = 13 (P 3 , D 3 ) (P 3 , D 3 ) (P 2 , D 3 ) (P 2 , D 4 ) (P 3 , D 4 ) (^60) – 40+60–20 = 60
The cell (P 2 , D 2 ) has the most negative opportunity cost (net change in cost). Therefore, transportation cost can be reduced by making allocation to this unoccupied cell. This means that if one unit is shifted to this unoccupied cell through the loop shown in the fourth row of the table, then `20 can be saved (see Fig. 4.3a). Hence, we shall shift as many units as possible to the cell (P 2 , D 2 ) through this loop. The maximum number of units that can be allocated to (P 2 , D 2 ) through this loop is 3.
Fig. 4.3: Closed Loop. This is because the shifting can be done only from the corners of the loop and more than 3 units cannot be shifted to (P 2 , D 2 ) as explained below: The allocations at other corners of the loops are 3, 10 and 8. If we try to shift more than 3 units, say, 4 units from the corner (P 2 , D 4 ), then 4 units will have to be subtracted from the corner (P 3 , D 2 ) so that the total of the column D 4 remains unchanged. But this will give 3 – 4= – 1 allocations to the cell (P 2 , D 4 ), which is impossible as negative allocations cannot be made. We obtain the maximum number of units that can be allocated to the cell (P 2 , D 2 ) through the loop as follows:
- First we assign (+) sign to the unoccupied cell (P 2 , D 2 ) to be evaluated and then (–) and (+) signs alternatively to other corners of the closed loop (moving in one direction) as shown in Fig. 4.3b.
- Then we take the minimum of the values at the corners that have been assigned the negative sign. In this case, the maximum number of units that can be allocated to the cell (P 2 , D 2 ) through the mentioned loop is the minimum of 3 and 8. It is 3. So we write it as:
2 4 3 2
the no. of units in P , D 3 min. the no. of units in P , D 8
The new table with these changes becomes:
D 1 D 2 D 3 D 4 P 1 P 2 P 3
● ● (P2,D 2 ) ● ●(P2,D 4 )
●(P 3 ,D 2 ) ●(P3,D 4 ) (^810)
3
2
13
5 3 7 5
(a) (b)
Transportation Problem In the above table, note that we have also allocated 3 units from (P 3 , D 2 ) to (P 3 , D 4 ) so that column D 4 remains unchanged. This leaves 5 units in the cell (P 3 , D 2 ) and there are 13 units in (P 3 , D 4 ). The total transportation cost associated with this solution is
Total cost = 19 5 + 12 2 + 30 3 + 40 7 + 10 5 + 20 13
= 95 + 24 + 90 + 280 + 50 + 260 = ` 799
Now, we repeat the optimality test to see if further allocation can be made to reduce the total transportation cost. The computation for the unoccupied cells is as follows:
Unoccupied Cell Closed Path Net Change in Cost ( ` ) (P 1 , D 2 ) (^) (P 1 , D 2 ) (P 1 , D 4 ) (P 3 , D 4 ) (P 3 , D 2 ) 30 – 12+20–10 = 28 (P 1 , D 3 ) (^) (P 1 , D 3 ) (P 1 , D 4 ) (P 3 , D 4 ) (P 3 , D 2 ) (P 2 , D 2 ) (P 2 , D 3 )
(P 2 , D 1 ) (^) (P 2 , D 1 ) (P 1 , D 1 ) (P 1 , D 4 ) (P 3 , D 4 ) (P 3 , D 2 ) (P 2 , D 2 )
(P 2 , D 4 ) (P 2 , D 4 ) (P 3 , D 4 ) (P 3 , D 2 ) (P 2 , D 2 ) 60 – 20+10–30 = 20
(P 3 , D 1 ) (P 3 , D 1 ) (P 1 , D 1 ) (P 1 , D 4 ) (P 3 , D 4 ) 40 – 19+12–20 = 13
(P 3 , D 3 ) (P 3 , D 3 ) (P 2 , D 3 ) (P 2 , D 2 ) (P 3 , D 2 ) 60 – 10+30–40 = 40
Since all opportunity costs in the unoccupied cells are non-negative, the current solution is an optimal solution with total transportation cost 799. Hence the maximum saving by optimum distribution is(1000 799) = `201.
Now, you should try to solve the following exercise:
E4) Perform the optimality test using the Stepping Stone method on the solution of the problem given in Example 1.
Before studying the next method, match your answer with the answer given in Sec. 4.7.
Note that the Stepping Stone method should be applied only to problems of small dimensions. For large dimensions, this method becomes quite tedious. For instance, if a problem involving five factories and seven warehouses has to be solved by the Stepping Stone method, an initial solution may involve 5+71=11 occupied cells and hence 5×711=24 unoccupied cells. To check whether this solution is an optimal solution or not, we shall have to find the opportunity cost for each of the 24 unoccupied cells by making 24 separate loops: one for each case. Then we will have to proceed as in the above problem. This will be very tedious and cumbersome. So, for problems of large dimensions, we use another method known as the Modified Distribution method (MODI). This method may be conveniently applied to problems of small dimension as well. We now describe the MODI method.
4.4.2 Modified Distribution Method (MODI)
The modified distribution method (MODI) is an improved form of the Stepping Stone method for obtaining an optimal solution of a transportation problem. The difference between the two methods is that in the Stepping Stone method, closed loops are drawn for all unoccupied cells for determining their opportunity costs. However, in the MODI method, the opportunity costs of all the unoccupied cells are calculated and the cell with
Transportation Problem
Let us denote the row numbers by u 1 , u 2 , u 3 and column numbers by v1, v2, v 3
and v 4 such that
u 1 + v 1 = 19, u 1 + v 4 = 12, u 2 + v 3 = 40,
u 2 + v 4 = 60, u 3 + v 2 = 10, u 3 + v 4 = 20.
Taking u 1 =0, we have v 1 =19 and v 4 =12 from the first two equations.
Putting the value of v 4 in the fourth equation, we have u 2 =6012=48.
Similarly, u 3 =2012=8, v 2 =108= 2 and v 3 =4048= 8.
Using these values, we fill the vacant cells of the above table using Cij = ui+vj and put dots in the already filled cells so that these cells are not considered again.
v 1 = 19 v 2 = 2 (^) v 3 = 8 v 4 = 12
u 1 = 0 ●^2 8 ●
u 2 = 48 67 50 ●^ ●
u 3 = 8 27 ●^0 ●
Now, subtracting these values from the corresponding values of the original cost matrix, we have the net evaluations, i.e., ij = Cij ( ui + vj )
● 30 2= 28 50 (8) = 58 ●
70 67= 3 30 50= 20 ●^ ● 40 27= 13 ●^60 0= 60 ●
Note that the cell (P 2 , D 2 ) has the most negative opportunity cost (net change in cost). Therefore, the transportation cost can be reduced by making allocation to this unoccupied cell. This means that if one unit is shifted to this unoccupied cell through the closed loop formed, beginning from this cell and using allocated cells, `20 can be saved. We form the loop beginning from this cell, i.e., the cell (P 2 , D 2 ) (see Fig. 4.4). We shift units to this unoccupied cell through the loop and add and subtract from the cells at the other corners of the loop which are assigned ‘+’ and ‘’ signs. So we get Fig. 4.4:
3
Fig. 4.4: Closed loop for cell (P 2 , D 2 ).
The maximum number of units that can be allocated to the cell (P 2 , D 2 ) through this loop is given by the minimum of the solution of the equations 3 =0 and 8 =0, i.e.,
2 4 3 2
the no. of units in P , D 3 min. the no. of units in P , D 8
So with the improved allocations, the table now becomes:
Optimisation Techniques-I 19 30 50 12
70 30 40 60
40 10 60 20
Thus, the total cost of transportation for this set = 19 5 + 12 2 + 30 3 + 40 7 + 10 5 + 20 13 = 95 + 24 + 90 + 280 + 50 + 260 = 799 Now, let us apply the optimality test to the improved solution. Proceeding in the same way as in the first iteration, first of all, we write the cost matrix for only allocated cells:
19 12 u 1
30 40 u 2
10 20 u 3
v 1 v 2 v 3 v 4
Let us denote the row numbers by u 1 , u 2 , u 3 and column numbers by v1, v2, v 3
and v 4 such that u 1 + v 1 = 19, u 1 + v 4 = 12, u 2 + v 3 = 40,
u 2 + v 2 = 30, u 3 + v 2 = 10, u 3 + v 4 = 20.
Taking u 1 =0, we have v 1 =19, v 4 =12 from the first two equations.
Putting the value of v 4 in the sixth equation, we have u 3 =2012= 8.
Similarly, u 2 =3002=28, v 2 =108= 2, v 3 =4028= 12.
Using these values, we fill all the vacant (unoccupied) cells of the table using cij= ui + vj for each unoccupied cell and put dot in already filled cells so that these cells are not considered again.
v 1 = 19 v 2 = 2 v 3 = 12 v 4 = 12 u 1 = 0 ● 2 12 ● u 2 = 28 47 ● ● 40 u 3 = 8 27 ● 20 ●
Now, subtracting these values from the corresponding values of the original cost matrix, we have the net evaluations ij = Cij ( ui + vj ) as:
● 28 38 ● 23 ● ● 20 13 ● 40 ●
Since none of the net evaluations is negative, this solution is optimal. Thus, the total minimum transportation cost is `799 and the maximum saving
= (1000 799) =201. Now, you should try to solve the following exercise.
5 3
5
7
2
13
