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Transportation Problem LECTURE 18 By Dr. Arshad Zaheer

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

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Illustration 1 Minimize the transportation cost Sources D1D1 D2D2 D3D3 Capacity S1S S2S S3S Demand

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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 Demand 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.

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SourcesDestinationCapacity D1D2D3 S Xij 3 Xij 5 S Xij 7 Xij 25 S Xij 3 Xij 30 Demand 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.

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SourcesDestinationCapacity D1D2D3 S S Xij 7 Xij 25 S Xij 3 Xij 30 Demand 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.

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SourcesDestinationCapacityy D1D2D3 S S Xij 10 S Xij 30 Demand 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.

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SourcesDestinationCapacity D1D2D3 S S S Xij 30 Demand 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.

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SourcesDestinationCapacity D1D2D3 S S S Demand000 70

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Initial solution SourcesDestinationCapacity D1D2D3 S S S Demand

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

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SourcesDestinationCapacity D1D2D3 S S S Demand Basic variablesNon basic variables

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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 x x 7+30 x 3 =245

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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.

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Shadow cost SourcesDestinationCapacity D1D2D3 S U1 S U2 S U3 Demand10 V1 20 V2 40 V3 70

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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

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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

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SourcesDestinationCapacity D1D2D3 S U1=1 S U2=0 S 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

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SourcesDestinationCapacity D1D2D3 S U1=1 S U2=0 S 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

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SourcesDestinationCapacity D1D2D3 S θ45-θ θ 15 U1=1 S θ 7 10-θ 25 U2=0 S 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

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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. ‘

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SourcesDestinationCapacity D1D2D3 S U1 S U2 S U3 Demand10 V1 20 V2 40 V3 70 Again find the shadow cost by finding the same rules as discussed above

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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

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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

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SourcesDestinationCapacity D1D2D3 S U1= - 4 S U2=0 S 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

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SourcesDestinationCapacity D1D2D3 S θ θ35+θ 15 U1= - 4 S θ θ 25 U2=0 S 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.

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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.

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SourcesDestinationCapacity D1D2D3 S U1= S U2= S U3= Demand10 V1= 20 V2= 40 V3= 70

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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

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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

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SourcesDestinationCapacity D1D2D3 S U1= -3 S U2=0 S 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

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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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