Transportation Method Lecture 20 By Dr. Arshad Zaheer

RECAP  Transportation model (Minimization)  Illustration (Demand < Supply)  Optimal Solution  Modi Method

Maximization Total Demand exceeds Total Capacity (Supply)

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

Illustration Maximize the profit for this problem Sources D1D1 D2D2 D3D3 Capacity S1S S2S S3S Demand

Introduce the fictitious supply to balance at zero profit Sources D1D1 D2D2 D3D3 Capacity S1S S2S S3S SfSf Demand

Sources DestinationCapacity D1D2D3 S1 10 Xij 15 Xij 12 Xij 15 S2 9 Xij 8 Xij 3 Xij 25 S3 12 Xij 8 Xij 20 Xij 25 SfSf 0 Xij 0 Xij 0 Xij 15 Demand

Initial Solution by North West Corner Rule

Sources DestinationCapacity D1D2D3 S S S SfSf Demand

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

Sources DestinationCapacity D1D2D3 S S S SfSf Demand

Total Profit Total Cost =15* * * * * -20 = -745 Total Profit=-1*- 745 = 745

No of Basic Variables= m+n-1 = =6 m= No of sources n= No of destinations

Sources DestinationCapacity D1D2D3 S U1= S U2= S U3= SfSf U4= Demand 30 V1= 20 V2= 30 V3= 80 For calculating shadow cost we need to find the values of U and V variables

Equations U1+V1=-10let U2=0 U2+V1=-9U1=-1V1=-9 U2+V2=-8U2=OV2=-8 U3+V2=-8U3=0V3=-20 U3+V3=-20U4=20 U4+V3=0

Sources DestinationCapacity D1D2D3 S U1=-1 S U2=0 S U3=0 SfSf U4=20 Demand 30 V1=-9 20 V2=-8 30 V3= Shadow cost of S1, D3 Vij = (Ui + Vj) –Cij V13 = (U1 + V3) –C13 =(-1-20)-12 =-9 We can calculate all the shadow cost in the same way for others

Sources DestinationCapacity D1D2D3 S U1=-1 S U2=0 S U3=0 SfSf U4=20 Demand 30 V1=-9 20 V2=-8 30 V3= We add θ in maximum positive shadow cost to proceed further because our optimal condition is not yet satisfied

Sources DestinationCapacity D1D2D3 S U1=-1 S U2=0 S θ θ 25 U3=0 SfSf θ 0 15-θ 15 U4=20 Demand 30 V1=-9 20 V2=-8 30 V3=-20 80

Maximum θ = Min (10,15) ` = 10

Sources DestinationCapacity D1D2D3 S U1= S U2= S U3= SfSf U4= Demand 30 V1= 20 V2= 30 V3= 80

Total Cost=15* * * *-20 =-865 Total Profit/Gain = -1 * -865 =865

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

Sources DestinationCapacity D1D2D3 S U1=-1 S U2=0 S U3=-12 SfSf U4=8 Demand 30 V1=-9 20 V2=-8 30 V3=-8 80 Now we can calculate the shadow costs for all cells

Sources DestinationCapacity D1D2D3 S U1=-1 S U2=0 S U3=-12 SfSf U4=8 Demand 30 V1=-9 20 V2=-8 30 V3=-8 80 shadow costs are still positive so we use θ to proceed further

Sources DestinationCapacity D1D2D3 S θ θ U1=-1 S θ θ U2=0 S U3=-12 SfSf U4=8 Demand 30 V1=-9 20 V2=-8 30 V3=-8 80

Maximum θ = Min (10, 15) ` = 10

Sources DestinationCapacity D1D2D3 S U1= S U2= S U3= SfSf U4= Demand 30 V1= 20 V2= 30 V3= 80

Total Cost = 5* * * *-20 =- 925 Total Gain/Profit= = -1 * -925 = 925

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

Sources DestinationCapacity D1D2D3 S U1=-1 S U2=0 S U3=-6 SfSf U4=14 Demand 30 V1=-9 20 V2= V3= Now the shadow cost for each cell can be calculated easily

Sources DestinationCapacity D1D2D3 S U1=-1 S U2=0 S U3=-6 SfSf U4=14 Demand 30 V1=-9 20 V2= V3= Criteria for optimality is not satisfied so we will proceed further with use of θ

Sources DestinationCapacity D1D2D3 S θ θ U1=-1 S U2=0 S U3=-6 SfSf θ 0 10-θ U4=14 Demand 30 V1=-9 20 V2= V3=-14 80

Maximum θ = Min (5, 10) ` = 5

Sources DestinationCapacity D1D2D3 S U1= S U2= S U3= SfSf U4= Demand 30 V1= 20 V2= 30 V3= 80

Total Cost =15* * *-20 =-950 Total Profit/Gain= =-1 * =950

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

Sources DestinationCapacity D1D2D3 S U1=-6 S U2=0 S U3=-11 SfSf U4=9 Demand 30 V1=-9 20 V2=-9 30 V3=-9 80 Now calculate the shadow costs for non basic cells

Sources DestinationCapacity D1D2D3 S U1=-6 S U2=0 S U3=-11 SfSf U4=9 Demand 30 V1=-9 20 V2=-9 30 V3=-9 80 Criteria for optimality has been satisfied as all the shadow costs are non- positive

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