Transportation Problem 1d

Solution by Vogel’s Approximation Method- First calculate the

Penalty of each row & each column. Penalty is the difference

of two least values. Parul Agrawal, BVIMR 9811880805

D1 D2 D3 D4 Supply

P1 20 30 50 17 7

P2 70 35 40 60 10

P3 40 12 60 25 18

Demand 5 8 7 15

Transportation Problem

Q-1 Following matrix represents the cost per unit of transportation from

three Places to 4 destinations. The capacity of each place & demand at

each destination is also given. Give the optimum assignments so as to

minimize the total transportation cost.

Parul Agrawal, BVIMR 9811880805

D1 D2 D3 D4 Supply Penalty

P1

20

30

50

17

7

3

P2

70

35

40

60

10

5

P3

40

12

60

25

18

13

Demand

5

8

7

15

Penalty

20

18

10

8

5

18 is the highest penalty in Column D2 and 12 is the least cost in this column, which is assigned 8 units. Penalties are calculated again.


D1 D2 D3 D4 Supply Penalties

P1

20

30

50

17

7

3

13

P2

70

35

40

60

10

5

5

P3

40

12

60

25

18

13

13

Demand

5

8

7

15

Penalties

20

18

10

8

18

10

8

8

5

35 is the highest penalty in row P3 and 25 is the least cost in this row, which is assigned 10 units. Calculate Penalties again.



P1

20

30

50

17

7

3

13

33

P2

70

35

40

60

10

5

5

20

P3

40

12

60

25

18

13

13

35

Demand

5

8

7

15

Penalties

20

18

10

8

18

10

8

10

8

5

10 8



P1

20

30

50

17

7

3

13

33

33

P2

70

35

40

60

10

5

5

20

20

P3

40

12

60

25

18

13

13

35

Demand

5

8

7

15

Penalties

20

18

10

8

18

10

8

10

8

10

43

5

10 8

2


P1

20

30

50

17

7

3

13

33

33

P2

70

35

40

60

10

5

5

20

20

20

P3

40

12

60

25

18

13

13

35

Demand

5

8

7

15

Penalties

20

18

10

8

18

10

8

10

8

10

43

40

60

5

3

8 10

2

60 is the highest penalty in column D4 & least cost is also 60 assign

3 units. Calculate penalties again Parul Agrawal, BVIMR 9811880805


P1

20

30

50

17

7 3 13 33 33

P2

70

35

40

60

10 5 5 20 20 20 40

P3

40

12

60

25

18 13 13 35

Demand

5

8

7

15

Penalties 20 18 10 8

18 10 8

10 8

10 43

40 60

40

5

7

8

3

10

2

40 is the last penalty which is assigned 7 units.

Pa

rul

Ag

raw

al,

BV

IMR

98

11

88

08

05

Feasibility Test- m+n-1 = No of Occupied Cells

Where m=No. of Rows & n= No. of columns

3+ 4 -1=6 = no of occupied cells.

Solution is Feasible.

Apply optimality test.



D1 D2 D3 D4 Supply Ui

P1

20

30

50

17

7

U1=17

P2

70

35

40

60

10

U2=60

P3

40

12

60

25

18

U3=25

Demand

5

8

7

15

Vj

V1= 3

V2= -13

V3= -20

V4 = 0

7

10

2 5

8

3

Optimality Test Calculate the opportunity cost of unoccupied cells.

Opportunity Cost = Cij-(Ui+Vj) P1-D2 30-(17-13)=26 P1-D3 50-(17-20)=53 P2-D1 70-(60+3) =7 P2-D2 35-(60-13)=-12 P3-D1 40-(25+3) =12 P3-D3 60-(25-20)=55


A solution to be optimal all the opportunity costs should

be zero or positive but here one cell (P2-D2) has negative

opportunity cost hence solution is not optimal.

It implies that this cell needs assignment.

To give assignment to cell P2-D2, find a closed loop with

other occupied cells.

There are three conditions of the loop.

1. it should be a closed loop.

2. loop should be formed with horizontal and vertical

lines, the no of lines should be even i.e 4, 6, 8.

3. all the corner points of that loop should be occupied

cells.



D1 D2 D3 D4 Supply

P1

20

30

50

17

7

U1=

P2

70

35 +

40

60 -

10

U2=

P3

40

12 -

60

25 +

18

U3=

Demand

5

8

7

15

V1= V2= V3= V4 =

5

7

8

3

10

2

Again Ui & Vj are calculated to apply optimality test.


D1 D2 D3 D4 Supply

P1

20

30

50

17

7

U1=17

P2

70

35

40

60

10

U2=48

P3

40

12

60

25

18

U3=25

Demand

5

8

7

15

V1= 3 V2= -13 V3= -8 V4 =0

5

3

2

5

7

13

All opportunity costs are positive hence the

solution is optimal & the Total cost is


Transportation Problem 1d

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