OPERATIONS RESEARCH Chapter 2 Transportation and Assignment Problems Prof. Bibhas C. Giri Department of Mathematics Jadavpur University Kolkata, India Email: bcgiri.jumath@gmail.com
MODULE - 2: Optimality Test by Stepping Stone Method and MODI Method, and Some Special Cases of Transportation Problem 2.1 Optimality Test Once an initial BFS is obtained, the next step is to check its optimality. An optimal solution is one where there is no other set of transportation route (allocations) that will further reduce the total transportation cost. To test the optimality, the following methods can be used: (i) Stepping stone method (ii) Modified distribution (MODI) method 2.1.1 Stepping Stone Method This is a procedure for determining the potential of improving upon each of the nonbasic variables in terms of the objective function. To determine this potential, each of the non-basic variables is considered one by one. For each such cell, we find what effect on the total cost would be if one unit is assigned to this cell. We then come to know whether the current solution is optimal or not. If not, we improve that solution further. We summarize the Stepping stone method in the following steps: 1. For all unoccupied cells, evaluate the effect of transferring one unit from an occupied cell to the unoccupied cell. This transfer is made by forming a closed path that retains the rim conditions of the problem. 2
2. Check the signs of net changes in the unit transportation costs. If the net changes are non-negative, then an optimal solution is obtained; otherwise, go to step 3. 3. Select the unoccupied cell with most negative net change. 4. Assign as many units as possible to the unoccupied cell satisfying rim conditions. The maximum number of units to be assigned is equal to the smaller oval boxed number among the occupied cells with the negative value in a closed path. 5. Go to step 1, and repeat the process until all unoccupied cells are evaluated and the net changes result in non-negative values. Example 2.1: Consider the transportation problem presented in Table 2.1. Determine the optimal solution of the problem using Stepping stone method. Origin D 1 D 2 D 3 Supply O 1 2 1 5 10 7 3 4 25 6 5 3 20 Demand 15 22 18 Table 2.1: Transportation table for Example 2.1 Solution: We compute an initial BFS of the problem by Matrix minimum method as shown in Table 2.2. The transportation cost associated with this solution is Rs.(1 10+ 7 13 + 3 12 + 6 2 + 3 18) = Rs.203. The Stepping stone method starts with evaluation of each of the unoccupied cells viz., (O 1,D 1 ), (O 1,D 3 ), (,D 3 ), (,D 2 ) to examine whether it would improve the initial solution by introducing any of these cells into the current solution basis. To demonstrate this method, let us evaluate the unoccupied cell (O 1,D 1 ) in Table 2.2 and assign 1 unit to this cell. For this allocation, Rs. 2 to is added to the cost. In order to make this adjustment and to satisfy the supply restriction at origin O 1 and demand restriction of D 1, we subtract 1 unit from (O 1,D 2 ), add 1 unit to (,D 2 ) and again subtract 1 unit from (,D 1 ). These adjustments are shown in Table 2.3. Following the closed path, the increase in the transportation cost per unit quantity of reallocation is observed as 2 1 + 3 7 = 3. This indicates that shift of 1 unit into unoccupied cell (O 1,D 1 ) decreases the transportation cost by Rs. 3. Obviously, cell (O 1,D 1 ) is to be included in the assignment. Now assign as much as possible in this
O 1 10 2 1 5 10 13 12 7 3 4 25 2 18 6 5 3 20 b j 15 22 18 Table 2.2: Initial BFS by Matrix minima method O 1 10 (+1) ( 1) 2 1 5 10 13 12 ( 1) (+1) 7 3 4 25 2 18 6 5 3 20 b j 15 22 18 Table 2.3 2 1 5 O 1 10 10 7 3 4 3 22 25 6 5 3 2 18 20 b j 15 22 18 Table 2.4 O 1 10 ( 1) (+1) 2 1 5 10 13 12 ( 1) (+1) 7 3 4 25 2 18 (+1) ( 1) 6 5 3 20 b j 15 22 18 Table 2.5 cell. The maximum amount that can be allocated to (O 1,D 1 ) is 10 and this will make the current basic variable corresponding to cell (O 1,D 2 ) non basic. Table 2.4 shows the optimal results after reallocation. Similarly, we can draw a closed path for the other unoccupied cells (O 1,D 3 ), (,D 3 ), (,D 2 ) which are shown in Tables 2.5, 2.6 and 2.7, respectively. The corresponding changes in cost for allocations in unoccupied cells are shown in Table 2.8. Since costs increase or remain unaltered for other unoccupied cells, after reallocation of a unit allocation, no optimal solution can not be obtained for these reallocations. Thus the improved solution is obtained as x 11 = 10, x 21 = 3, x 22 = 22, x 31 = 2 and x 33 = 18, see Table 2.4, and the corresponding minimum transportation cost is Rs 173.
O 1 10 2 1 5 10 13 12 ( 1) (+1) 7 3 4 25 2 18 (+1) ( 1) 6 5 3 20 b j 15 22 18 Table 2.6 O 1 10 2 1 5 10 13 12 (+1) ( 1) 7 3 4 25 2 18 ( 1) (+1) 6 5 3 20 b j 15 22 18 Table 2.7 Unoccupied cell Change in cost Remarks (O 1,D 3 ) +5 1 + 3 7 + 6 3 = 3 Cost increases (,D 3 ) +4 7 + 6 3 = 0 Neither increase nor decrease (,D 2 ) +5 3 + 7 6 = 3 Cost increases Table 2.8 2.1.2 Modified Distribution (MODI) Method The modified distribution method, also known as MODI method or (u, v) method provides a minimum cost solution to the transportation problem. In the stepping stone method, we have to draw as many closed paths as equal to the unoccupied cells for their evaluation. To the contrary, in MODI method, only closed path for the unoccupied cell with highest opportunity cost is drawn. The method is outlined in the following : 1. After getting an initial BFS, determine the values of u i and v j so that u i + v j = c ij for the occupied cells. 2. For the unoccupied cells, compute the opportunity cost d ij = c ij (u i + v j ). 3. If all d ij 0, the solution is optimal. On the other hand, if at least one d ij < 0, the solution is not optimal and further saving in transportation cost is possible. 4. Select the unoccupied cell with the smallest negative opportunity cost as this cell is to be included in the next solution.
5. Draw a loop for the unoccupied cell selected in the previous step. The right angle turn in this path is permitted only at occupied cells and at the original unoccupied cell. 6. Assign alternate + and - signs on the corner points of the closed path, starting from the unoccupied selected cell. 7. Determine the maximum number of units that should be shipped to this unoccupied cell. The smallest value with a negative position on the loop indicates the number of units that can be shipped to the entering cell. Now, add this quantity to all the cells on the corner points of the loop marked with + sign, and subtract it from those cells marked with - sign. In this way, an unoccupied cell becomes an occupied cell. 8. Repeat the whole procedure until an optimal solution is obtained. Example 2.2: Solve the following transportation problem: Origin D 1 D 2 D 3 D 4 Availability O 1 19 30 50 12 7 70 30 40 60 10 40 10 60 20 18 Demand 5 8 7 15 Table 2.9: Transportation table for Example 2.2 Solution: An initial BFS obtained by Matrix minima method is shown in Table 2.10. O 1 19 30 50 12 7 3 70 30 40 60 10 2 8 8 40 10 60 20 18 b j 5 8 7 15 Table 2.10: Initial BFS by Matrix minima method
We now test the optimality by MODI method. For this, we calculate the values of u i s (i = 1,2,3) and v j s (j = 1,2,3,4) for each occupied cell such that u i +v j = c ij. We assign an arbitrary value u 1 = 0 and get u 1 + v 4 = c 14 v 4 = 12 u 3 + v 4 = c 34 u 3 = 8 u 3 + v 2 = c 32 v 2 = 2 u 3 + v 1 = c 31 v 1 = 32 u 2 + v 1 = c 21 u 2 = 38 u 2 + v 3 = c 23 v 3 = 2 Now, we calculate the opportunity cost d ij = c ij (u i + v j ) for the unoccupied cells. Unoccupied cell Opportunity cost d ij (O 1,D 1 ) c 11 (u 1 + v 1 ) = 13 (O 1,D 2 ) c 12 (u 1 + v 2 ) = 28 (O 1,D 3 ) c 13 (u 1 + v 3 ) = 48 (,D 2 ) c 22 (u 2 + v 2 ) = 10 (,D 4 ) c 24 (u 2 + v 4 ) = 10 (,D 3 ) c 33 (u 3 + v 3 ) = 50 u i -13 28 48 O 1 19 30 50 12 7 0-10 10 3 70 30 40 60 10 38 50 2 8 8 40 10 60 20 18 8 b j 5 8 7 15 v j 32 2 2 12 Table 2.11: Optimality test by MODI method The opportunity costs are plotted in the transportation table within shadow boxes as shown in Table 2.11. Since all d ij 0, the current solution is not optimal. Now, we choose the most negative value (-13) from opportunity cost and draw a closed loop starting from (O 1,D 1 ) to this cell back through the cells (,D 1 ), (,D 4 ) and (O 1,D 4 ) as shown in Table 2.12.
O 1-13 28 48 (+) ( ) 19 30 50 12 7-10 10 3 70 30 40 60 10 50 2 8 8 ( ) (+) 40 10 60 20 18 b j 5 8 7 15 Table 2.12: Loop for improving the solution u i 28 61 2 5 O 1 19 30 50 12 7 0-23 -3 3 70 30 40 60 10 51 13 63 8 10 40 10 60 20 18 8 b j 5 8 7 15 v j 19 2-11 12 Table 2.13: Revised solution and test for optimality The revised solution is shown in Table 2.13. The optimality test by MODI method shows that all d ij 0. So, again we form a closed loop for the most negative opportunity cost (-23) as shown in Table 2.14 and trace out the smallest possible allocation amongst the occupied cell in the closed loop, which is 3. We then obtain the further improved solution as shown in Table 2.15. The optimality test by MODI method shows that all the current opportunity costs are nonnegative (see the shadow boxes in Table 2.15). Hence the optimality allocation is x 11 = 5, x 14 = 2, x 22 = 3, x 23 = 7, x 32 = 5, x 34 = 13 and the corresponding minimum transportation cost is Rs.(19 5 + 12 2 + 30 3 + 40 7 + 10 5 + 20 13) = Rs.799.
O 1 2 28 61 5 (+) ( ) 19 30 50 12 7-23 -3 3 ( ) (+) 70 30 40 60 10 13 63 8 10 ( ) (+) 40 10 60 20 18 b j 5 8 7 15 Table 2.14: Loop for further improving the solution u i O 1 28 38 5 2 19 30 50 12 7 0 23 20 3 70 30 40 60 10 28 13 40 5 13 40 10 60 20 18 8 b j 5 8 7 15 v j 19 2 12 12 Table 2.15: Improved solution and test for optimality by MODI method 2.2 Some Special Cases of Transportation Problem 2.2.1 Degeneracy in Transportation Problem If the basic feasible solution of a TP with m origins and n destinations has fewer than (m + n 1) positive x ij (occupied cells) then the problem is said to be a degenerate TP. To resolve degeneracy, we assign a small arbitrary quantity (ϵ) to that unoccupied cell which has the minimum transportation cost. Consider the TP given in Table 2.16. An initial BFS (see Table 2.17) obtained by Matrix minimum method shows that the number of basic variables = 5 < m + n 1 = 6. Hence, it is a degenerate TP. To resolve
Origin D 1 D 2 D 3 D 4 Supply O 1 2 2 2 4 1000 4 6 4 3 700 3 2 1 0 900 Demand 900 800 500 400 Table 2.16: Transportation table 2 2 2 4 O 1 900 100 1000 4 6 4 3 700 700 3 2 1 0 500 400 900 b j 900 800 500 400 Table 2.17: Initial BFS by Matrix minima method degeneracy, we assign a very small quantity ϵ(> 0) to that unoccupied cell which has the minimum transportation cost. In Table 2.17, there is a tie in selecting the smallest unoccupied cell. So, we choose the cell (,D 2 ) arbitrarily, see Table 2.18. 2 2 2 4 O 1 900 100 1000 4 6 4 3 700 700 3 2 1 0 ϵ 500 400 900 + ϵ b j 900 800 + ϵ 500 400 Table 2.18: Resolving degeneracy Now, we use the stepping stone method to find an optimal solution. We calculate the opportunity cost for all unoccupied cells as given below:
Unoccupied cell Increase in cost per unit Remarks of reallocation (O 1,D 3 ) +2 2 + 2 1 = 1 Cost increases (O 1,D 4 ) +4 2 + 2 0 = 4 Cost increases (,D 1 ) +4 6 + 2 2 = 2 Cost decreases (,D 3 ) +4 6 + 2 1 = 1 Cost decreases (,D 4 ) +3 6 + 2 0 = 1 Cost decreases (,D 1 ) +3 2 + 2 2 = 1 Cost increases The cell (,D 1 ) is having the maximum improvement potential, which is equal to 2. The maximum amount that can be allocated to (,D 1 ) is 700 and this will make the current basic variable corresponding to cell (,D 2 ) non-basic. The improved solution is shown in Table 2.19. 2 2 2 4 O 1 200 800 1000 4 6 4 3 700 700 3 2 1 0 ϵ 500 400 900 + ϵ b j 900 800 + ϵ 500 400 Table 2.19: Improved solution As no further improvement is possible, the optimal solution is x 11 = 200, x 12 = 800, x 21 = 700, x 32 = ϵ, x 33 = 500, x 34 = 400 and the corresponding transportation cost is Rs.(2 200+2 800+4 700+2 ϵ +1 500+0 400) = Rs.(5300+2ϵ). Since ϵ is a very small quantity, it can be neglected. Thus, the net transportation cost is Rs.5300. 2.2.2 Unbalanced Transportation Problem A transportation problem is called unbalanced if m i=1 a i n j=1 b j. The problem can be made balanced by introducing dummy source with zero costs of transporting from this source to all destinations, or dummy destination with zero costs of transporting to this destination. Example 2.3: Consider the transportation problem presented in the following table.
Origin D 1 D 2 D 3 Supply O 1 28 17 26 500 19 12 16 300 Demand 250 250 500 Table 2.20: Transportation table for Example 2.3 Solution: The given transportation problem is unbalanced as 2 i=1 a i = 800 < 3 j=1 b j = 1000. To solve this problem, we introduce a dummy row with transportation cost zero and availability 200, see Table 2.21. Using Matrix minima method, an initial BFS is obtained as shown in Table 2.22. The corresponding transportation cost is Rs.(50 28 + 450 26 + 250 12 + 50 16 + 200 0) = Rs.16900. For optimality test, one can use MODI method as explained before. Further, improvement in the solution can be done by forming loop, if needed. D 1 D 2 D 3 Supply O 1 28 17 26 500 Origin 19 12 16 300 Dummy 0 0 0 200 Demand 250 250 500 Table 2.21: Unbalanced transportation table 28 17 26 O 1 50 450 500 19 12 16 250 50 300 Dummy 0 0 0 200 200 b j 250 250 500 Table 2.22: Initial BFS 2.2.3 Prohibited Transportation Routes Sometimes there may be situations (e.g., road construction, bad road condition, strike, unexpected flood, local traffic rule, etc.) where it is not possible to use certain routes in a transportation problem. To handle such type of problems, a very large cost M or is assigned to each of such routes which are not available. Consider the following transportation problem in which the allocations in the cells (O 1,D 2 ) and (,D 2 ) are prohibited. We assign a cost to each of these cells and determine the initial BFS by Matrix minima method as shown in Table 2.24.
Origin D 1 D 2 D 3 Supply O 1 16 12 200 14 8 18 160 26 16 90 Demand 180 120 150 Table 2.23: Example of prohibited transportation route 16 12 O 1 50 150 200 14 8 18 40 120 160 26 16 90 90 b j 180 120 150 Table 2.24: Initial basic feasible solution 2.2.4 Maximization Transportation Problem There are certain types of transportation problem where the objective function is to be maximized instead of being minimized. These problems can be solved by converting the maximization problem into a minimization problem. For this, replace each cost element by the difference from the maximum cost element of the transportation table. Example 2.4: Suppose that three factories X, Y, and Z supply goods to four dealers spread all over the country. The production capacities of these factories are 200, 500 and 300 units per month, respectively. Transportation costs from the factories to the dealers are given in the following Table 2.25. Determine a suitable allocation to maximize the total net return. Solution: Maximization transportation problem can be converted into minimization transportation problem by subtracting each transportation cost from the maximum transportation cost. Here, the maximum transportation cost is 25. So, we subtract each cost from 25. The revised transportation problem is shown in Table 2.26. The initial BFS obtained by Matrix minima method is given in Table 2.27.
Origin D 1 D 2 D 3 D 4 Availability X 12 18 6 25 200 Y 8 7 10 18 500 Z 14 3 11 20 300 Demand 180 320 100 400 Table 2.25: Transportation table for Example 2.4 Origin X 13 7 19 0 200 Y 17 18 15 7 500 Z 11 22 14 5 300 b j 180 320 100 400 Table 2.26: Revised transportation table X 13 7 19 0 200 200 Y 17 18 15 7 80 320 100 500 Z 11 22 14 5 100 200 300 b j 180 320 100 400 Table 2.27: Initial BFS by Matrix minima method 2.2.5 Alternative Optimal Solution If d ij = 0 for an unoccupied cell in an optimal solution then an alternative optimal solution exists. The alternative optimal solution can be obtained by bringing such an unoccupied cell in the basic solution without increasing the total transportation cost.