unbalanced transportation problem

unbalanced transportation problem

1. UN BALANCED TRANSPORTATION PROBLEM MANEESH P DEPT. OF APPLIED ECONOMICS

2. The basic transportation problem was developed in 1941 by F.I. Hitchaxic. However it could be solved for optimally as an answer to complex business problem only in 1951,when Geroge B. Dantzig applied the concept of Linear Programming in solving the Transportation models. Transportation problems are primarily concerned with the optimal (best possible) way in which a product produced at different factories or plants (called supply origins) can be transported to a number of warehouses (called demand destinations). INTRODUCTION

3. Transportation problem is a special kind of LP problem in which goods are transported from a set of sources to a set of destinations subject to the supply and demand of the source and the destination respectively, such that the total cost of transportation is minimized. The objective in a transportation problem is:-To fully satisfy the destination requirements within the operating production capacity constraints at the minimum possible cost.

4. Whenever there is a physical movement of goods from the point of manufacture to the final consumers through a variety of channels of distribution (wholesalers, retailers, distributors etc.), there is a need to minimize the cost of transportation so as to increase the profit on sales. Transportation problems arise in all such cases. .

5. It aims at providing assistance to the top management in ascertaining how many units of a particular product should be transported from each supply origin to each demand destinations to that the total prevailing demand for the company’s product is satisfied, while at the same time the total transportation costs are minimized

6. i.e.; The total supply available at the plants exactly matches the total demand at the destinations. Hence, there is neither excess supply nor excess demand. Such type of problems where supply and demand are exactly equal are known as Balanced Transportation Problem. Supply (from various sources) are written in the rows, while a column is an expression for the demand of different warehouses. In general, if a transportation problem has m rows an n columns, then the problem is solvable if there are exactly (m + n –1) basic variables

7. . UNBALANCED TRANSPORTATION PROBLEM : A transportation problem is said to be unbalanced if the supply and demand are not equal. Two situations are possible:- 1. If Supply < demand, a dummy supply variable is introduced in the equation to make it equal to demand. 2. If demand < supply, a dummy demand variable is introduced in the equation to make it equal to supply.

8. Then before solving we must balance the demand & supply. When supply exceeds demand, the excess supply is assumed to go to inventory. A column of slack variables is added to the transportation table which represents dummy destination with a requirement equal to the amount of excess supply and the transportation cost equal to zero. When demand exceeds supply, balance is restored by adding a dummy origin. The row representing it is added with an assumed total availability equal to the difference between total demand & supply and with each cell having a zero unit cost.

9. Demand Less Than Supply Suppose that a plywood factory increases its rate of production from 100 to 250 desks The firm is now able to supply a total of 850 desks each period Warehouse requirements remain the same (700) so the row and column totals do not balance We add a dummy column that will represent a fake warehouse requiring 150 desks This is somewhat analogous to adding a slack variable We use the northwest corner rule and either vogel ’s approximation method or MODI to find the optimal solution

10. FROM TO A B C TOTAL AVAILABLE I 5 4 3 250 II 8 4 3 300 III 9 7 5 300 WAREHOUSE REQUIREMENTS 300 200 200 850 700

11. FROM TO A B C D TOTAL AVAILABLE I 5 4 3 250 II 8 4 3 300 III 9 7 5 300 WAREHOUSE REQUIREMENTS 300 200 200 150 850= 850

12. FROM TO A B C D TOTAL AVAILABLE I 5 4 3 250 II 8 4 3 300 III 9 7 5 300 WAREHOUSE REQUIREMENTS 300 200 200 150 (1) (1) (2) (3) (0) (0) (0) 250 300-250=50

13. FROM A B C D TOTAL AVAILABLE II 8 4 3 300 III 9 7 5 300 WAREHOUSE REQUIREMENTS 50 200 200 150 TO (1) (2) (1) (3) (2) (0) 200 300-200=100

14. FROM A C D TOTAL AVAILABLE II 8 3 100 III 9 5 300 WAREHOUSE REQUIREMENTS 50 200 150 TO 100 (5) (4) (1) (2) (0) 200-100=100

15. FROM A C D TOTAL AVAILABLE III 9 5 300 WAREHOUSE REQUIREMENTS 50 100 150 TO (4) (9) (5) (0) 50 300-50=250

16. FROM C D TOTAL AVAILABLE III 5 250 WAREHOUSE REQUIREMENTS 100 150 TO (5) (5) (0) 150 250-150=100

17. FROM C TOTAL AVAILABLE III 5 100 WAREHOUSE REQUIREMENTS 100 TO 100 100-100=0

18. FROM A B C D I 5 4 3 II 8 4 3 III 9 7 5 TO WAREHOUSE REQUIREMENTS TOTAL AVAILABLE 300 200 200 150 250 300 300 250 200 100 50 100 150 m+n-1 = 3+4-1= 6

19. FROM A B C D I 5 4 3 II 8 4 3 III 9 7 5 TO WAREHOUSE REQUIREMENTS TOTAL AVAILABLE 300 200 200 150 250 300 300 250 200 100 50 100 150 u1 u2 U3 v1 v2 v3 v4

20. U1+V1=5 U2+V2=4 U2+V3=3 U3+V1=9 U3+V3=5 U3+V4=0 ASSUMING U3=0 0+V1=9 V1=9 0+V3=5 V3=5 0+V4=0 V4=0 U1+V1=5 U1+9=5 U1= 5-9 = -4 U2+V3=3 U2+5=3 U2= 3-5= -2 (V3=5) U3+V4=0 U3+0=0 U3=0 U2+V2=4 -2+V2=4 V2= 4-(-2) = 6 (V4=0) (U2= -2)

21. CIJ TABLE 4 3 8 7 V1 V2 V3 V4 U1 U2 U3 -4 -2 5 6 9

22. u I-VJ TABLE 2 1 -4 7 -2 7 V1 V2 V3 V4 U1 U2 U3

23. DIJ TABLE 2 2 4 1 2 V1 V2 V3 V4 U1 U2 U3

24. Total cost = 250*5+200*4+100*3+50*9+100*5+150*0 1250+800+300+450+500+0 = 3,300 Initial basic feasible solution Here all the dij values are positive, therefore the solution is optimal.

25. Demand Greater than Supply The second type of unbalanced condition occurs when total demand is greater than total supply In this case we need to add a dummy row representing a fake factory The new factory will have a supply exactly equal to the difference between total demand and total real supply

26. FROM A B C PLANT SUPPLY W 6 4 9 200 X 10 5 8 175 Y 12 7 6 75 WAREHOUSE DEMAND 250 100 150 450 TO 500 Totals do not balance

27. FROM A B C PLANT SUPPLY W 6 4 9 200 X 10 5 8 175 Y 12 7 6 75 Z 50 WAREHOUSE DEMAND 250 100 150 500 TO 500

28. FROM A B C PLANT SUPPLY W 6 4 9 200 X 10 5 8 175 Y 12 7 6 75 Z 50 WAREHOUSE DEMAND 250 100 150 TO 200 100 75 50 75

29. CIJ TABLE 4 9 12 7 V1 V2 V3 U1 U2 U3 U4 -4 -2 -10 10 5 8

30. U I+VJ TABLE 1 4 8 3 -5 -2 V1 V2 V3 U1 U2 U3 U4

31. DIJ TABLE 3 5 4 4 5 2 V1 V2 V3 U1 U2 U3 U4 DIJ= CIJ-(UI+VJ)

32. Here, all the dij values are positive, therefore the solution is optimal. TC= 200*6+0*10+100*5+75*8+75*6+50*0 1200+500+600+450= 2750

33. CONCLUSION In real-life problems, total demand is frequently not equal to total supply These unbalanced problems can be handled easily by introducing dummy sources or dummy destinations If total supply is greater than total demand, a dummy destination (warehouse), with demand exactly equal to the surplus, is created If total demand is greater than total supply, we introduce a dummy source (factory) with a supply equal to the excess of demand over supply

34. Any units assigned to a dummy destination represent excess capacity Any units assigned to a dummy source represent unmet demand

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