It is used to save the time over stepping stone method
It provides a new means of finding the unused route with the largest negative improvement index.
Once largest index identified ,we are required to trace only one path, just as with the stepping stone approach, this helps to determine the maximum number of unit that can be shipped by the best unused route.
1.Construct a transportation table with the given cost of transportation and rim requirement.
3.For current basic feasible solution check degeneracy and non-degeneracy.
rim requirement=stone square(non-degeneracy)
rim requirement != stone square(degeneracy)
4.Find occupied matrix.
5.Find unoccupied matrix.
6.Find opportunity cost of unoccupied cells using formula:
opportunity cost =actual cost-implied cost
dij= cij - (ri+kj)
7.Unoccupied cell evaluation:
(a) if dij>0 then cost of transportation unchanged.
(b) if dij=0 then cost of transportation unchanged.
(c) if dij<0 then improved solution can be obtain and go to next step.
8.Select an unoccupied cell with largest –ve opportunity cost among all unoccupied cell.
9.Construct closed path for the occupied cells determined in step 8.
10.Assign as many as units as possible to the unoccupied cell satisfying rim conditions.
11.Go to step 4 and repeat procedure until
All dij>=0 i.e reached to the optimal solution.
Degeneracy :occurs in two cases
Degeneracy occurs in initial basic solution.
Degeneracy occurs in during the test of optimality.
. PROBLEM1:Shipping costs are Rs. 10 per kilometer.. What shipping schedule should be used. if the matrix given below the kilometers from source to destination. Requirement 3 3 2 8 x 50 30 220 1 y 90 45 170 3 z 50 200 50 4 destination a b c availability
A B C X Y Z 30 220 50 90 45 170 50 200 50 30 220 1 2 3 2 2 2 2 3 P1 P2 P3 P4 20 20 20 20 45 45 45 _ 150 150 _ _ P1 40 15 120 P2 40 15 _ P3 40 15 _ IBFS BY USING VAM METHOD
30 30 220 90 45 170 50 200 50 1 E 3 2 2 STONE SEQUARE =4 RIM REQUIREMENT =M+N-1=3+3-1=5 DEGENERACY NOW STONE SEQUARE =5 RIM REQUIREMENT =5 NON –DEGENERACY