Transportation Problem

LECTURE 18By

Dr. Arshad Zaheer

RECAP

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

Illustration 1

• Minimize the transportation cost

Sources D1 D2 D3 Capacity

S1 2 4 3 15

S2 5 3 7 25

S3 8 7 3 30

Demand 10 20 4070

70

Initial Solution by North West Corner RuleSources Destination Capacity

D1 D2 D3

S12

Xij

4

Xij

3

Xij15

S25

Xij

3

Xij

7

Xij25

S38

Xij

7

Xij

3

Xij30

Demand 10 20 40 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.

Sources Destination Capacity

D1 D2 D3

S12

10

4

Xij

3

Xij5

S25

0

3

Xij

7

Xij25

S38

0

7

Xij

3

Xij30

Demand 0 20 40 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.

Sources Destination Capacity

D1 D2 D3

S12

10

4

5

3

00

S25

0

3

Xij

7

Xij25

S38

0

7

Xij

3

Xij30

Demand 0 15 40 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.

Sources Destination Capacityy

D1 D2 D3

S12

10

4

5

3

00

S25

0

3

15

7

Xij10

S38

0

7

0

3

Xij30

Demand 0 0 40 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.

Sources Destination Capacity

D1 D2 D3

S12

10

4

5

3

00

S25

0

3

15

7

100

S38

0

7

0

3

Xij30

Demand 0 0 30 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.

Sources Destination Capacity

D1 D2 D3

S12

10

4

5

3

00

S25

0

3

15

7

100

S38

0

7

0

3

300

Demand 0 0 0 70

Initial solutionSources Destination Capacity

D1 D2 D3

S12

10

4

5

3

015

S25

0

3

15

7

1025

S38

0

7

0

3

3030

Demand 10 20 40 70

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

m= No of sourcesn= No of destinations

Sources Destination Capacity

D1 D2 D3

S12

10

4

5

3

015

S25

0

3

15

7

1025

S38

0

7

0

3

3030

Demand 10 20 40 70

Basic variables Non basic variables

Total cost

Total cost = Xij * CijTotal 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

Criteria for Optimality

• All the shadow cost, which must be non negativeShadow cost, for basic variables is always zero so there

is no need to find their shadow cost

Shadow cost:Vij = (Ui + Vj) –CijSo we need to find the U and V.

Shadow costSources Destination Capacity

D1 D2 D3

S12

10

4

5

3

0

15

U1

S25

0

3

15

7

10

25

U2

S38

0

7

0

3

30

30

U3

Demand 10

V1

20

V2

40

V3

70

• 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

• There are 6 basic variables and 5 equations so we will give one variable arbitrary value which is equal to zero

Let U2=0By putting this U2 zero we can easily find the value

of all other variables which will be as follows• U1=1 V1=1• U2=0 V2=3• U3=-4 V3=7

Sources Destination Capacity

D1 D2 D3

S10 2

10

4

5

5 3

0

15

U1=1

S25

0

3

15

7

10

25

U2=0

S38

0

7

0

3

30

30

U3=-4

Demand 10

V1=1

20

V2=3

40

V3=7

70

Shadow cost of S1D1Vij = (Ui + Vj) –CijV11 = (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 S1D3Vij = (Ui + Vj) –CijV13 = (U1 + V3) –C13

=(1+7) -3=5

Sources Destination Capacity

D1 D2 D3

S10 2

10

4

5

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

30

U3=-4

Demand 10

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

Sources Destination Capacity

D1 D2 D3

S10 2

10

4

5-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

30

U3=-4

Demand 10

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

• Max θ = Min of (10,5) =5

The boxes from which the θ has been subtractedReplace the θ with 5 and make new tableau.‘

Sources Destination Capacity

D1 D2 D3

S1 2

10

4

0

3

5

15

U1

S2 5

0

3

20

7

5

25

U2

S3 8

0

7

0

3

30

30

U3

Demand 10

V1

20

V2

40

V3

70

Again find the shadow cost by finding the same rules as discussed above

Total Cost:=10*2+5*3+20*3+5*7+30*3=220 (cost decreased as compared to previous tableau)

Equations:U1+V1=2U1+V3=3U2+V2=3U2+V3=7U3+V3=3

• Let U2=0By putting this U2 zero we can easily find the

value of all other variables which will be as follows

• U1=-4 V1=6• U2=0 V2=3• U3=-4 V3=7

Sources Destination Capacity

D1 D2 D3

S1 2

10

-5 4

0

3

5

15

U1= - 4

S21 5

0

3

20

7

5

25

U2=0

S3-6 8

0

-8 7

0

3

30

30

U3= - 4

Demand 10

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

Sources Destination Capacity

D1 D2 D3

S1 2

10-θ

-5 4

0

3

5+θ

15

U1= - 4

S21 5

0+θ

3

20

7

5-θ

25

U2=0

S3-6 8

0

-8 7

0

3

30

30

U3= - 4

Demand 10

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 .

• Max θ = Min of (10,5) =5

The boxes from which the θ has been subtractedReplace the θ with 5 and make new tableau.

Sources Destination Capacity

D1 D2 D3

S1 2

5

4

0

3

10

15

U1=

S25

5

3

20

7

0

25

U2=

S38

0

7

0

3

30

30

U3=

Demand 10

V1=

20

V2=

40

V3=

70

Total Cost:=5*2+10*3+5*5+20*3+30*3=215 (The cost further reduced)

Shadow cost Equations:

U1+V1=2U1+V3=3U2+V1=5U2+V2=3U3+V3=3

• Let U2=0By putting this U2 zero we can easily find the

value of all other variables which will be as follows

• U1=-3 V1=5• U2=0 V2=3• U3=3 V3=6

Sources Destination Capacity

D1 D2 D3

S1 2

5

-4 4

0

3

10

15

U1= -3

S25

5

3

20

-1 7

0

25

U2=0

S30 8

0

-7 7

0

3

30

30

U3= 3

Demand 10

V1=5

20

V2=3

40

V3=6

70All the shadow costs are non positive (negative) which is the indication of optimal solution

Optimal Distribution

• S1 ─ ─ ─ ─ > D1 = 5• S1 ─ ─ ─ ─ > D3 = 10• S2 ─ ─ ─ ─ > D1 = 5• S2 ─ ─ ─ ─ > D2 = 20• S3 ─ ─ ─ ─ > D3 = 30

Total = 70

Total Cost= 215

Thank You