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Transportation Method
Lecture 20
By
Dr. Arshad Zaheer
RECAP
Transportation model (Minimization) Illustration (Demand < Supply) Optimal Solution Modi Method
MaximizationTotal 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 D1 D2 D3 Capacity
S1 10 15 12 15
S2 9 8 3 25
S3 12 8 20 25
Demand 30 20 3065
80
Introduce the fictitious supply to balance at zero profit
Sources D1 D2 D3 Capacity
S1 10 15 12 15
S2 9 8 3 25
S3 12 8 20 25
Sf 0 0 0 15
Demand 30 20 30 80
Sources Destination Capacity
D1 D2 D3
S110
Xij
15
Xij
12
Xij15
S29
Xij
8
Xij
3
Xij25
S312
Xij
8
Xij
20
Xij25
Sf0
Xij
0
Xij
0
Xij15
Demand 30 20 30 80
Initial Solution by North West Corner Rule
Sources Destination CapacityD1 D2 D3
S110
1515
012
0 15
S29
158
103
0 25
S312
08
1020
15 25
Sf
00
00
015
15
Demand 30 20 30 80
For maximization we multiply all the profits or gains by -1.
Sources Destination CapacityD1 D2 D3
S1-10
15-15
0-12
015
S2-9
15-8
10-3
025
S3-12
0-8
10-20
1525
Sf
00
00
015
15
Demand 30 20 30 80
Total Profit
Total Cost =15*-10 + 15*-9 + 10*-8 + 10*-8 + 15* -20 = -745
Total Profit=-1*- 745= 745
No of Basic Variables= m+n-1=4+3-1=6
m= No of sourcesn= No of destinations
Sources Destination Capacity
D1 D2 D3
S1-10
15-15
0-12
015
U1=
S2-9
15-8
10-3
025
U2=
S3-12
0-8
10-20
1525
U3=
Sf
0
0
0
0
0
15
15
U4=
Demand 30V1=
20V2=
30V3=
80
For calculating shadow cost we need to find the values of U and V variables
Equations
U1+V1=-10 let U2=0U2+V1=-9 U1=-1 V1=-9U2+V2=-8 U2=O V2=-8U3+V2=-8 U3=0 V3=-20U3+V3=-20 U4=20U4+V3=0
Sources Destination Capacity
D1 D2 D3
S1-10
15-15
0-9 -12
015
U1=-1
S2-9
15-8
10-3
025
U2=0
S3-12
0-8
10-20
1525
U3=0
Sf
0
0
0
0
0
15
15
U4=20
Demand 30V1=-9
20V2=-8
30V3=-20
80
Shadow cost of S1, D3Vij = (Ui + Vj) –CijV13 = (U1 + V3) –C13
=(-1-20) -12=-9
We can calculate all the shadow cost in the same way for others
Sources Destination Capacity
D1 D2 D3
S1-10
156 -15
0-9 -12
015
U1=-1
S2-9
15-8
10-17 -3
025
U2=0
S33 -12
0-8
10-20
1525
U3=0
Sf
11 0
0
12 0
0
0
15
15
U4=20
Demand 30V1=-9
20V2=-8
30V3=-20
80
We add θ in maximum positive shadow cost to proceed further because our optimal condition is not yet satisfied
Sources Destination Capacity
D1 D2 D3
S1-10
156 -15
0-9 -12
015
U1=-1
S2-9
15-8
10-17 -3
025
U2=0
S33 -12
0-8
10-θ-20
15+θ25
U3=0
Sf
11 0
0
12 0
0+θ
0
15-θ
15
U4=20
Demand 30V1=-9
20V2=-8
30V3=-20
80
Maximum θ = Min (10,15)` = 10
Sources Destination Capacity
D1 D2 D3
S1-10
15 -15
0 -12
015
U1=
S2-9
15-8
10 -3
025
U2=
S3 -12
0-8
0-20
2525
U3=
Sf
0
0
0
10
0
5
15
U4=
Demand 30V1=
20V2=
30V3=
80
Total Cost= 15*-10 +15*-9 + 10*-8 + 25*-20 = -865
Total Profit/Gain= -1 * -865= 865
Equations
U1+V1=-10 let U2=0U2+V1=-9 U1=-1 V1=-9U2+V2=-8 U2= 0 V2=-8U3+V3=-20 U3=-12 V3=-8U4+V2=0 U4=8U4+V3=0
Sources Destination Capacity
D1 D2 D3
S1-10
15 -15
0 -12
015
U1=-1
S2-9
15-8
10 -3
025
U2=0
S3 -12
0-8
0-20
2525
U3=-12
Sf
0
0
0
10
0
5
15
U4=8
Demand 30V1=-9
20V2=-8
30V3=-8
80
Now we can calculate the shadow costs for all cells
Sources Destination Capacity
D1 D2 D3
S1-10
156 -150
3 -120
15U1=-1
S2-9
15-8
10-5 -3
025
U2=0
S3-9 -12
0-12 -8
0-20
2525
U3=-12
Sf
-1 0
0
0
10
0
5
15
U4=8
Demand 30V1=-9
20V2=-8
30V3=-8
80
shadow costs are still positive so we use θ to proceed further
Sources Destination Capacity
D1 D2 D3
S1-10
15-θ6 150+θ
3 -120
15U1=-1
S2-9
15+θ-8
10-θ-5 -3
025
U2=0
S3-9 -12
0-12 -8
0-20
2525
U3=-12
Sf
-1 0
0
0
10
0
5
15
U4=8
Demand 30V1=-9
20V2=-8
30V3=-8
80
Maximum θ = Min (10, 15)` = 10
Sources Destination Capacity
D1 D2 D3
S1-10
5 -15
10 -12
015
U1=
S2-9
25-8
0 -3
025
U2=
S3 -12
0 -8
0-20
2525
U3=
Sf
0
0
0
10
0
5
15
U4=
Demand 30V1=
20V2=
30V3=
80
Total Cost = 5*-10 + 10*-15 + 25*-9 + 25*-20 = - 925
Total Gain/Profit= = -1 * -925 = 925
Equations
U1+V1=-10 let U2=0U1+V2=-15 U1=-1 V1=-9U2+V1=-9 U2= 0 V2=-14U3+V3=-20 U3=-6 V3=-14U4+V2=0 U4=14U4+V3=0
Sources Destination Capacity
D1 D2 D3
S1-10
5 -15
10 -12
015
U1=-1
S2-9
25-8
0 -3
025
U2=0
S3 -12
0 -8
0-20
2525
U3=-6
Sf
0
0
0
10
0
5
15
U4=14
Demand 30V1=-9
20V2=-14
30V3=-14
80
Now the shadow cost for each cell can be calculated easily
Sources Destination Capacity
D1 D2 D3
S1-10
5 -15
10-3 -12
015
U1=-1
S2-9
25-6 -8
0-11 -3
025
U2=0
S3-3 -12
0-12 -8
0-20
2525
U3=-6
Sf
5 0
0
0
10
0
5
15
U4=14
Demand 30V1=-9
20V2=-14
30V3=-14
80
Criteria for optimality is not satisfied so we will proceed further with use of θ
Sources Destination Capacity
D1 D2 D3
S1-10
5-θ -15
10+θ-3 -12
015
U1=-1
S2-9
25-6 -8
0-11 -3
025
U2=0
S3-3 -12
0-12 -8
0-20
2525
U3=-6
Sf
5 0
0+θ
0
10-θ
0
5
15
U4=14
Demand 30V1=-9
20V2=-14
30V3=-14
80
Maximum θ = Min (5, 10)` = 5
Sources Destination Capacity
D1 D2 D3
S1-10
0 -15
15 -12
015
U1=
S2-9
25 -8
0 -3
025
U2=
S3 -12
0 -8
0-20
2525
U3=
Sf
0
5
0
5
0
5
15
U4=
Demand 30V1=
20V2=
30V3=
80
Total Cost =15*-15 + 25*-9 + 25*-20 = -950
Total Profit/Gain= = -1 * - 950 = 950
Equations
U1+V2=-15 let U2=0U2+V1=-9 U1=-6 V1=-9U3+V3=-20 U2= 0 V2=-9U4+V1=0 U3=-11 V3=-9U4+V2=0 U4=9U4+V3=0
Sources Destination Capacity
D1 D2 D3
S1-10
0 -15
15 -12
015
U1=-6
S2-9
25 -8
0 -3
025
U2=0
S3 -12
0 -8
0-20
2525
U3=-11
Sf
0
5
0
5
0
5
15
U4=9
Demand 30V1=-9
20V2=-9
30V3=-9
80
Now calculate the shadow costs for non basic cells
Sources Destination Capacity
D1 D2 D3
S1-5 -10
0 15
15-3 -12
015
U1=-6
S2-9
25-1 -8
0-6 -3
025
U2=0
S3-8 -12
0-12 -8
0-20
2525
U3=-11
Sf
0
5
0
5
0
5
15
U4=9
Demand 30V1=-9
20V2=-9
30V3=-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
Thank You
