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Transportation Method

Transportation Method. Lecture 20 By Dr. Arshad Zaheer. RECAP. T ransportation model (Minimization) Illustration (Demand < Supply) Optimal Solution Modi Method. Maximization Total Demand exceeds Total Capacity (Supply). Maximization.

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Transportation Method

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  1. Transportation Method Lecture 20 By Dr. ArshadZaheer

  2. RECAP Transportation model (Minimization) Illustration (Demand < Supply) Optimal Solution Modi Method

  3. MaximizationTotal Demand exceeds Total Capacity (Supply)

  4. Maximization Maximization problem may be solved by the use of following method • Multiply the given pay off matrix of profits or gain by -1. Then use the transportation technique for minimization to obtain optimal solution. • To calculate the total profit or gain multiply the total cost by -1

  5. Illustration • Maximize the profit for this problem

  6. Introduce the fictitious supply to balance at zero profit

  7. Initial Solution by North West Corner Rule

  8. For maximization we multiply all the profits or gains by -1.

  9. Total Profit Total Cost =15*-10 + 15*-9 + 10*-8 + 10*-8 + 15* -20 = -745 Total Profit=-1*- 745 = 745

  10. No of Basic Variables= m+n-1 =4+3-1 =6 m= No of sources n= No of destinations

  11. For calculating shadow cost we need to find the values of U and V variables

  12. Equations U1+V1=-10 let U2=0 U2+V1=-9 U1=-1 V1=-9 U2+V2=-8 U2=O V2=-8 U3+V2=-8 U3=0 V3=-20 U3+V3=-20 U4=20 U4+V3=0

  13. We can calculate all the shadow cost in the same way for others Shadow cost of S1, D3 Vij = (Ui + Vj) –Cij V13 = (U1 + V3) –C13 =(-1-20) -12 =-9

  14. We add θ in maximum positive shadow cost to proceed further because our optimal condition is not yet satisfied

  15. Maximum θ= Min (10,15) ` = 10

  16. Total Cost= 15*-10 +15*-9 + 10*-8 + 25*-20 = -865 Total Profit/Gain = -1 * -865 = 865

  17. Equations U1+V1=-10 let U2=0 U2+V1=-9 U1=-1 V1=-9 U2+V2=-8 U2= 0 V2=-8 U3+V3=-20 U3=-12 V3=-8 U4+V2=0 U4=8 U4+V3=0

  18. Now we can calculate the shadow costs for all cells

  19. shadow costs are still positive so we use θ to proceed further

  20. Maximum θ= Min (10, 15) ` = 10

  21. Total Cost = 5*-10 + 10*-15 + 25*-9 + 25*-20 = - 925 Total Gain/Profit= = -1 * -925 = 925

  22. Equations U1+V1=-10 let U2=0 U1+V2=-15 U1=-1 V1=-9 U2+V1=-9 U2= 0 V2=-14 U3+V3=-20 U3=-6 V3=-14 U4+V2=0 U4=14 U4+V3=0

  23. Now the shadow cost for each cell can be calculated easily

  24. Criteria for optimality is not satisfied so we will proceed further with use of θ

  25. Maximum θ= Min (5, 10) ` = 5

  26. Total Cost =15*-15 + 25*-9 + 25*-20 = -950 Total Profit/Gain= = -1 * - 950 = 950

  27. Equations U1+V2=-15 let U2=0 U2+V1=-9 U1=-6 V1=-9 U3+V3=-20 U2= 0 V2=-9 U4+V1=0 U3=-11 V3=-9 U4+V2=0 U4=9 U4+V3=0

  28. Now calculate the shadow costs for non basic cells

  29. Criteria for optimality has been satisfied as all the shadow costs are non- positive

  30. Optimal Distribution • S1 ─ ─ ─ ─ > D2 = 15 • S2 ─ ─ ─ ─ > D1 = 25 • S3 ─ ─ ─ ─ > D3 = 25 • Sf ─ ─ ─ ─ > D1 = 5 • Sf ─ ─ ─ ─ > D2 = 5 • Sf ─ ─ ─ ─ > D3 = 5 Total = 80 Total Gain = 950

  31. Thank You

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