Question

1. Find an initial solution and do one transpor tation iteration, then STOP Give the reason why your second tableau is optimal or no Supply D4 D3 D2 D1 20 12 16 14 100 S1 10 12 14 40 S2 15 16 10 60 S3 55 65 50 Demand 30 D4 Supply D3 D2 D1 20 16 12 14 100 s1 10 14 12 40 S2 15 16 10 60 S3 55 65 50 30 Demand ea Are there alternate opt imal routes for the transpor tation problem in # 1? Why or why not? What are the alternate routes, if any?

Answer 1

Final tableau is following:

v2 5 Supply D1 D2 D3 D4 u1 14 16 12 20 S1 100 50 14 u2 10 8 S2 40 0 0 40 10 u3 16 15 S3 0 45 15 Demand 30 50 65 55 0 Total Cost 2365 8 CO 20 12 30

Calculate reduced cost (Wij) for Non-Basic Variables Basic Variables ui vi -cij Wij X14 u1+y4- x11 u1+v1= 14 ul 0 v1 14 20 19 -20 -1 - - X12 u1+v2= X21 u2+v1- 16 u1 v2 = 16 12 -11 14 -12 -9 X22 u2+v2- X13 u1+v3= 12 ul= v3 12 14 -11 16 -14 -9 X23 u2+y3- X33 u3+v3 8 u3 -4 v3 12 10 -11 12 -10 -9 X34 u3+v4= X31 u3+v1- 15 u3 -4 v4 19 10 -4 14 -10 + X24 u2+v4= u 2= X32 u3+v2- 8 -11 v4 19 16 -4 16 -16 -4 = Wij corresponding to the non-basic variables is <= 0, therefore we have reached optimal solution O O O

The final tableau is optimal, because Wij value of all non-basic variables is <= 0

Yes, there are alternatie optimal routes, because reduced cost of non-basic variable X31 is 0.

This indicates that there is an alternative optimal solution for this problem.

The optimal solution using alternative routes is following:

v2 5 Supply D1 D2 D3 D4 u1 20 14 16 12. S1 100 0 50 50 0 u2 12 14 10 8 S2 40 0 0 0 40 u3 10 16 15 S3 60 0 30 15 15 Demand 30 50 65 55 0 Total Cost 2365 SST co