1. SEMINAR ON TRANSPORTATION PROBLEM
By Susant Kumar Indrajitsingha M.Sc. Part-1

2. CONTENTS Introduction Origin & Development of O.R
L.P formulation of the transportation problem
Definition of Transportation Problem
Algorithm of Transportation Problem
Example of a Transportation Problem & it’s solution

3. INTRODUCTION
The Transportation Problem is one of the subclasses of L.P.Ps in which the objective is to transport various quantities of a single homogeneous commodity, that are initially stored at various sources , to different destinations in such a way that the total transport cost is minimum. To achieve this objective we must know the amount and location of available supplies and the quantities demanded. In addition , we must know the costs that result from transporting one unit of commodity from various origins to various destinations.

4. Origin & Development of O.R
1917, A.K. Erlang,Danish Mathematician.
1940,McClosky & Trefthen ( World War II).
1949,R.R.L. Hyderabad.
Prof. Mahalonobis,Second Five Years plan in India.

5. LP FORMULATION OF TRANSPORTATION PROBLEM
A company manufacturing air-coolers has two plants located at Hyderabad and Mumbai with a capacity of 300 units and 100 units per week respectively.The company supplies the air-coolers to it’s four showrooms situated at Bangalore, Chennai, Delhi and Ernakulum which have a maximum demand of 85, 150, 150 & 55 units respectively. Due to the difference in raw material cost and transportation cost , the profit per unit per rupees differs which is shown in the table below :
City
Bangalore
Chennai
Delhi
Ernakulum
Hyderabad
110 90 75 55
Mumbai 65 50 80 45

6. Mathematical Formulation
Maximize z=110x11 +90x12+75x13+55x14+65x21+50x22+80x23+45x24
The constraints are :
X11+X12+X13+X14=300 ,
X21+X22+X23+X24=100 ,
X11+X21=85 ,
X12+X22=150 ,
X13+X23=120 ,
X14+X24=55,
Xij ≥ 0 , i=1,2 & j=1,2,3,4 .

7. DEFINITION C11 C12 . . . . . C1n a1 C21 C22 C2n a2 Cm1 Cm2 Cmn am
Destination 1 2 n
Availability
Origin
C11
C12
C1n a1
C21
C22
C2n a2 . m
Cm1
Cm2
Cmn am
Receiving Capacity b1 b2 bn

8. Solution of a Transportation problem
Formulate the given problem as a LPP.
Set up the given LPP in transportation table.
Examine whether total supply equals total demand. If not, introduce a dummy row/column.
Find an I.B.F.S.
Examine the solution obtained in step-4.
If the solution is not optimum, modify the shipping schedule by including that unoccupied cell whose inclusion may result in an improved solution.
Repeat step-4 until no further improvement is possible.

9. Finding an Initial Basic Feasible Solution
North-West Corner Method.
Least-Cost Method (Row/Column/Matrix minimum method), and
Vogel’s Approximation Method ( or Penalty Method)

10. Q. Find the initial basic feasible solution to the following Transportation Problem by North-West Corner Rule
Destination
Supply
Origin 1 2 3 4 6 8 10
Demand

11. Solution : 1 4 2 3 8 10

12. 3 4 2 8 10 1

13. 3 4 2 10 1

14. 2 4 10 1

15. 2 4 1 6

16. 1 4 2 3 8 10 6

17. Example : Obtain the optimal solution of the following transportation problem applying Vogel's approximation method for finding out the initial basic feasible solution . D1 D2 D3 D4
Supply S1 3 7 6 4 5 S2 2 8
Demand

18. Total demand = 3+3+2+2 = 10 Solution: Total supply = 5+2+3 = 10
Therefore it is a symmetrical transportation problem. Let us consider the transportation table . 3 7 6 4 5
4-3=1 2
3-2=1 8
6-3=3
4-2=2
Penalty↑ ←

19. 3 7 4 5
4-3=1 2
4-2=2
3-2=1
Penalty↑ ←

20. 3 7 4 5
4-3=1 2
7-3=4
5-4=1
Penalty↑ ←

21. 3 7 4 2 5

22. 3 7 6 4 2 5 8
u1+v1=3 ,u1+v2=7 ,u1+v4=4 u2+v3=3 ,u2+v4=2 ,u3+v2=3 By taking u1=0 & solving the above equations ,we get u1=0, u2=-2 , u3=-4, v1=3, v2=7 , v3=5 , v4=4.

23. ↓u →v 3 7 5 4 -2 1 2 -4 -1

24. zij - cj -1 -3 1 -5 -7 3
0-Ѳ
2+Ѳ 5 Ѳ 2

25. Choose Ѳ u1+v1=3 , u1+v4=4 , u2+v2=4 u2+v3=3 , u2+v4=2 , u3+v2=3
u1+v1=3 , u1+v4=4 , u2+v2=4
u2+v3=3 , u2+v4=2 , u3+v2=3
Taking u1=0 & solving the above equations, we get
u1=0 , u2=-2 , u3=-3 ,
v1=3 , v2=6 , v3=5 , v4=4 .

26. ↓u →v 3 6 5 4 -2 1 2 -3

27. zij - cj -1 -4 -6
Here zij-cj ≤ 0 so we have X11=3, X14=4 , X22=0 , X23=2 , X24=0 , X32=3 Minimum transportation cost is 3*3+2*7+2*3+3*3=38Unit . (Answer)

28. Impossible is nothing !
Question section ?