This document presents a transportation problem involving 3 production facilities and 4 warehouses. The facilities have weekly production capacities of 7, 10, and 18 units. The warehouses have weekly demands of 5, 8, 7, and 15 units. The transportation costs between each facility-warehouse pair are given. Using the Vogel's Approximation Method, an initial basic feasible solution is found allocating specific facilities to meet warehouse demands. The MODI method is then used to test for optimality. Some reallocations are made to improve the solution. The optimal solution allocates production from the facilities to warehouses to meet demands at a total transportation cost of Rs. 900.

1. Presentation on
Transportation
Problem

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

3. • 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.

4. 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

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

6. D1 D2 D3 D4 Supply Penalties
P1 20)5 30 50 15)2 7/2/0 5 15 35 -
P2 70 35 40)7 60)3 10/7/3 5 5 20 20
P3 40 12)8 60 25)10 18/10/0 13 13 35 35
demand 5/0 8/0 7/0 15/13/3 35
20 18 10 10
Penalty
18 10 10
10 10
20 35

7. So the feasible solution is:
D1 D2 D3 D4
P1 20)5 - - 15) 2
P2 - - 40)7 60) 3
P3 - 12) 8 - 25) 10

8. Step 2:MODI Method
Test for
Optimality

9. D1 D2 D3 D4
P1 U1=
20 15
0
P2 60
U2=
40
-45
P3 12 25 U3=
-10
V1= V2= V3= V4=
-20 -2 5 -15

10. D1 D2 D3 D4
P1 28 55
50+5
U1=
20 30-2 15
30 50 0
P2 5 70-20-45 -12
35-45-2 40 60
U2=
70 35 -45
P3 1040-20-10 12
55
60-10+5 25 U3=
40 60 -10
V1= V2= V3= V4=
-20 -2 5 -15

11. • Improvisations of Solution
D1 D2 D3 D4
P1
0+X 3-X
P2
P3
8-X 10+X

12. Take x = min {nos. with –x}
= min {3,8}
=3
D1 D2 D3 D4
P1 20)5 - - 15)2
P2 - 35)3 40)7 -
P3 - 12)5 - 25)13

13. D1 D2 D3 D4
P1 U1 = 0
20 15
U2 = -33
35 40
P2
U3 = -10
P3
12 25
V1 = -20 V2 = -2 V3 = -7 V4 = -15

14. D1 D2 D3 D4
P1 28) 43) U1 = 0
30 50
20 [30-2] [50-7] 15
P2 17) 12) U2 = -33
70 60
[70-20-33] 35 40 [60-15-33]
P3 10) 43)
40 12 60 25 U3=-10
[40-20-10] [60-7-10]
V1 = -20 V2 = -2 V3 = -7 V4 = -15

15. Total transportation cost
20 * 5 = 100
35 * 3 = 105
12 * 5 = 60
40 * 7 = 280
15 * 2 = 30
25 * 13 = 325
Total = Rs.900