Presentation on Transportation

Problem

Presented By:- Group A1

Nikita Bali (11004)Neethi Nair (11044)Ranjini Nair (11045)Chandan Pahelwani (11047)Himani Parihar (11049)Sonia Dadlani (10022)

• QUESTION:A company has three production facilities P1, P2 and P3 with production capacity of 7, 10 and 18 units per week of a product, respectively. These units are to be shipped to four warehouses D1, D2, D3 and D4 with requirement of 5, 8, 7 and 15 units per week, respectively. The transportation cost per unit between factories to warehouses are given in the table below.

D1 D2 D3 D4 Capacity

P1 20 30 50 15 7

P2 70 35 40 60 10

P3 40 12 60 25 18

Demand 5 8 7 15 35

Step 1: Using Vogel’s Approximation Method (VAM) method to find out the initial basic feasible solution.

D2

20)5

7040

3035

D3

5040)7

D4

15)2

60)3

P1

P2

P312)8 60 25)10

Supply

7/2/0

10/7/3

18/10/0

5/0demand 8/0 7/0 15/13/3 35

20

Pena

lty

18 10 101010

Penalties

5513

15513

352035

D1

-

2035

18 1010

20 35

So the feasible solution is:

D1 D2 D3 D4

P1 20)5 - - 15) 2

P2 - - 40)7 60) 3

P3 - 12) 8 - 25) 10

Test for Optimality

Step 2:MODI Method

D1 D2 D3 D4

P1 U1=0

P2 U2=-45

P3 U3=-10

V1=-20

V2=-2V3=

5V4=-15

20

40

12

15

60

25

D1 D2 D3 D4

P1 2830

5550

U1=0

P2 570

-1235

U2=-45

P3 1040

5560

U3=-10

V1=-20

V2=-2

V3=5

V4=-15

20

12

40

25

60

1530-2 50+5

35-45-270-20-45

60-10+540-20-10

D1 D2 D3 D4

P1

P2

P3

0+X

8-X 10+X

3-X

• Improvisations of Solution

D1 D2 D3 D4

P1 20)5 - - 15)2

P2 - 35)3 40)7 -

P3 - 12)5 - 25)13

Take x = min {nos. with –x} = min {3,8} = 3

D1 D2 D3 D4

P1 U1 = 0

P2

U2 = -33

P3

U3 = -10

V1 = -20 V2 = -2 V3 = -7 V4 = -15

20 15

25

40

12

35

D1 D2 D3 D4

P1 28) 30 [30-2]

43) 50 [50-7]

U1 = 0

P2 17) 70[70-20-33]

12) 60[60-15-33]

U2 = -33

P3 10) 40[40-20-10]

43) 60[60-7-10]

U3=-10

V1 = -20 V2 = -2 V3 = -7 V4 = -15

20 15

25

40

12

35

Total transportation cost

20 * 5 = 100 35 * 3 = 10512 * 5 = 6040 * 7 = 280 15 * 2 = 30 25 * 13 = 325Total = Rs.900