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Transportation Problem LECTURE 18 By Dr. Arshad Zaheer.

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Presentation on theme: "Transportation Problem LECTURE 18 By Dr. Arshad Zaheer."— Presentation transcript:

1 Transportation Problem LECTURE 18 By Dr. Arshad Zaheer

2 RECAP  Transportation model  Purpose  Initial Feasible Solution  North West Corner Method  Least Cost Method  Optimal Solution  Stepping Stone Method  Modi Method

3 Illustration 1 Minimize the transportation cost Sources D1D1 D2D2 D3D3 Capacity S1S1 24315 S2S2 53725 S3S3 87330 Demand102040 70

4 Initial Solution by North West Corner Rule SourcesDestinationCapacity D1D2D3 S1 2 Xij 4 Xij 3 Xij 15 S2 5 Xij 3 Xij 7 Xij 25 S3 8 Xij 7 Xij 3 Xij 30 Demand102040 70 S1D1 box is the most north column so we fill this first. See its column total which is 10 and row total which is 15. we place the less no in the box so is 10. Now the column total has been exhausted/completely filled and we mark zero to below box and the row total decrease to 5.

5 SourcesDestinationCapacity D1D2D3 S1 2 10 4 Xij 3 Xij 5 S2 5050 3 Xij 7 Xij 25 S3 8080 7 Xij 3 Xij 30 Demand02040 70 S1D2 box is the most adjacent west column so we fill this first now. See its column total which is 20 and row total which is 5. we place the less no in the box so is 5. Now the row total has been completely filled and we give zero to respective row total and the column total decreases to 15.

6 SourcesDestinationCapacity D1D2D3 S1 2 10 4545 3030 0 S2 5050 3 Xij 7 Xij 25 S3 8080 7 Xij 3 Xij 30 Demand01540 70 S2D2 box is the most adjacent north column to the previous cell just filled. So we fill this first. See its column total which is 15 and row total which is 25. We place the less no in the box so is 15. Now the column total has been exhausted and we assign zero to relevant column total and the row total decreases to 10.

7 SourcesDestinationCapacityy D1D2D3 S1 2 10 4545 3030 0 S2 5050 3 15 7 Xij 10 S3 8080 7070 3 Xij 30 Demand0040 70 S2D3 box is the most adjacent west column so we fill this first. See its column total which is 40 and row total which is 10. we place the less no in the box so is 10. Now the row total has been totally filled and the column total decrease to 30.

8 SourcesDestinationCapacity D1D2D3 S1 2 10 4545 3030 0 S2 5050 3 15 7 10 0 S3 8080 7070 3 Xij 30 Demand0030 70 S3D3 box is the last column so we fill this first. See its column and row total which is 30. So we place 30 in the box and both column and row balance will be utilized fully. Every time while finding the initial solution the column and row total of last box will be equal as in the given case 30.

9 SourcesDestinationCapacity D1D2D3 S1 2 10 4545 3030 0 S2 5050 3 15 7 10 0 S3 8080 7070 3 30 0 Demand000 70

10 Initial solution SourcesDestinationCapacity D1D2D3 S1 2 10 4545 3030 15 S2 5050 3 15 7 10 25 S3 8080 7070 3 30 Demand102040 70

11 No of Basic Variables= m+n-1 = 3+3-1 =5 m= No of sources n= No of destinations

12 SourcesDestinationCapacity D1D2D3 S1 2 10 4545 3030 15 S2 5050 3 15 7 10 25 S3 8080 7070 3 30 Demand102040 70 Basic variablesNon basic variables

13 Total cost Total cost = Xij * Cij Total is calculated on the basis of all the basic variables because non basic variables are zero and their cost will also becomes zero when multiplied with zero units. = 10 x 2 +5 x 4 +15 x 3 + 10 x 7+30 x 3 =245

14 Criteria for Optimality All the shadow cost, which must be non negative Shadow cost, for basic variables is always zero so there is no need to find their shadow cost Shadow cost: Vij = (Ui + Vj) –Cij So we need to find the U and V.

15 Shadow cost SourcesDestinationCapacity D1D2D3 S1 2 10 4545 3030 15 U1 S2 5050 3 15 7 10 25 U2 S3 8080 7070 3 30 U3 Demand10 V1 20 V2 40 V3 70

16 If we find the values of U1,U2,U3, and V1,V2 and V3 than we can easily find the shadow cost Write the equations for the basic variables U1+V1=2 U1+V2=4 U2+V2=3 U2+V3=7 U3+V3=3

17 There are 6 basic variables and 5 equations so we will give one variable arbitrary value which is equal to zero Let U2=0 By putting this U2 zero we can easily find the value of all other variables which will be as follows U1=1V1=1 U2=0V2=3 U3=-4V3=7

18 SourcesDestinationCapacity D1D2D3 S1 0 2 10 4545 5 3 0 15 U1=1 S2 5050 3 15 7 10 25 U2=0 S3 8080 7070 3 30 U3=-4 Demand10 V1=1 20 V2=3 40 V3=7 70 Shadow cost of S1D1 Vij = (Ui + Vj) –Cij V11 = (U1 + V1) –C11 =(1+1)-2 =0 So save time and do not calculate the shadow cost for basic variables. Write the shadow cost on opposite side of cost in the column Shadow cost of S1D3 Vij = (Ui + Vj) –Cij V13 = (U1 + V3) –C13 =(1+7)-3 =5

19 SourcesDestinationCapacity D1D2D3 S1 0 2 10 4545 5 3 0 15 U1=1 S2 -4 5 0 3 15 7 10 25 U2=0 S3 -11 8 0 -8 7 0 3 30 U3=-4 Demand10 V1=1 20 V2=3 40 V3=7 70 Our criteria of optimality is non positivity of shadow cost which is not satisfied as we have S1D3 box with positive shadow cost so we use θ

20 SourcesDestinationCapacity D1D2D3 S1 0 2 10 45-θ45-θ 5 3 0+θ 15 U1=1 S2 -4 5 0 3 15+θ 7 10-θ 25 U2=0 S3 -11 8 0 -8 7 0 3 30 U3=-4 Demand10 V1=1 20 V2=3 40 V3=7 70 Add θ in cell having largest positive shadow cost this will disturb the row and column balances for which you need to add and subtract θ from some other cells

21 Max θ = Min of (10,5) =5 The boxes from which the θ has been subtracted Replace the θ with 5 and make new tableau. ‘

22 SourcesDestinationCapacity D1D2D3 S1 2 10 4040 3535 15 U1 S2 5 0 3 20 7575 25 U2 S3 8 0 7070 3 30 U3 Demand10 V1 20 V2 40 V3 70 Again find the shadow cost by finding the same rules as discussed above

23 Total Cost: =10*2+5*3+20*3+5*7+30*3 =220 (cost decreased as compared to previous tableau) Equations: U1+V1=2 U1+V3=3 U2+V2=3 U2+V3=7 U3+V3=3

24 Let U2=0 By putting this U2 zero we can easily find the value of all other variables which will be as follows U1=-4V1=6 U2=0V2=3 U3=-4V3=7

25 SourcesDestinationCapacity D1D2D3 S1 2 10 -5 4 0 3535 15 U1= - 4 S2 1 5 0 3 20 7575 25 U2=0 S3 -6 8 0 -8 7 0 3 30 U3= - 4 Demand10 V1= 6 20 V2=3 40 V3=7 70 Our criteria of optimality is non positivity of shadow cost which is not satisfied as we have S2D1 box with positive shadow cost so we use θ

26 SourcesDestinationCapacity D1D2D3 S1 2 10-θ -5 4 0 35+θ35+θ 15 U1= - 4 S2 1 5 0+θ 3 20 7 5-θ 25 U2=0 S3 -6 8 0 -8 7 0 3 30 U3= - 4 Demand10 V1= 6 20 V2=3 40 V3=7 70 We can no add θ in S3D2 because its above and below boxes have zero units and we can not balance in zero.

27 Max θ = Min of (10,5) =5 The boxes from which the θ has been subtracted Replace the θ with 5 and make new tableau.

28 SourcesDestinationCapacity D1D2D3 S1 2 5 4040 3 10 15 U1= S2 5555 3 20 7070 25 U2= S3 8080 7070 3 30 U3= Demand10 V1= 20 V2= 40 V3= 70

29 Total Cost: =5*2+10*3+5*5+20*3+30*3 =215 (The cost further reduced) Shadow cost Equations: U1+V1=2 U1+V3=3 U2+V1=5 U2+V2=3 U3+V3=3

30 Let U2=0 By putting this U2 zero we can easily find the value of all other variables which will be as follows U1=-3V1=5 U2=0V2=3 U3=3V3=6

31 SourcesDestinationCapacity D1D2D3 S1 2 5 -4 4 0 3 10 15 U1= -3 S2 5555 3 20 -1 7 0 25 U2=0 S3 0 8 0 -7 7 0 3 30 U3= 3 Demand10 V1=5 20 V2=3 40 V3=6 70 All the shadow costs are non positive (negative) which is the indication of optimal solution

32 Optimal Distribution S1 ─ ─ ─ ─ > D1 = 5 S1 ─ ─ ─ ─ > D3 = 10 S2 ─ ─ ─ ─ > D1 = 5 S2 ─ ─ ─ ─ > D2 = 20 S3 ─ ─ ─ ─ > D3 = 30 Total = 70 Total Cost= 215

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