Transportation Problem

Case Study

The Transportation problem is one of the mot frequently encountered applications in real life situations and is a special type of linear programming problem.

The transportation problem has a number of origins and a number of destinations.

The problem indicates the amount of consignment to be transported from various origins to different destinations, so as to minimize the total transportation cost.

Defination:

Balanced Transportation Problem:o A problem in which the total supply

available at all the origins exactly satisfies the total demand required at all the destinations.

Unbalanced Transportation Problem:o A transportation problem where the total

availability at the origins is different from the total requirement at the destinations.

An initial feasible solution with an allocation of (m + n – 1) number of variables, xij ; i = 1,2,….,m;

j =1,2,….,n, is called a basic feasible solution.

A number of techniques are available for computing an initial basic feasible solution of a transportation problem. These are

a.North-West Corner Methodb.Matrix Minimum Methodc.Row Minima Methodd.Column Minima Methode.Vogel’s Approximation Method [VAM].

Feasible Solutions

Degeneracy

o A condition that occurs when the number of occupied cells in any solution is less than the number of rows plus the number of columns minus 1 in a transportation problem, i.e., (m + n – 1).

Test of Degeneracy

An optimal solution is one in which there is no opportunity cost, i.e., there is no other set of transportation routes (allocations) that will reduce the total transportation cost. The following two methods are widely used for testing the optimality:

1.Stepping Stone Methodo An iterative technique for moving from an initial

feasible solution to an optimal solution in transportation problems.

2.Modified Distribution Method (MODI)o Another algorithm for finding the optimal

solution to a transportation problem. It can be used in place of the stepping stone method.

Test of Optimality

The net cost of shipping one unit on a route not used in the current transportation problem solution.

Improvement Index

Reliance Equity Opportunities Fund had worked for many securities. The primary investment objective of the scheme is to seek to generate capital appreciation & provide long-term growth opportunities by investing in a portfolio constituted of equity securities & equity related securities. The company had estimated amount to get from the investors for 2005 till 2008 were,

Case Study

YEAR

Estimated Amount (Rs. In Crores)

2005 70

2006 40

2007 90

2008 30

After keeping the updates, the company selected four major sectors to invest in. These were, with the limit to invest in them:

SectorAmount

Invested (Rs. In Crores)

Automobile Sector 40

Banking Sector 50

Information Technology Sector

60

Metal Sector 60

The estimated net return on investment of one rupee till the end of the planning horizon is given below

Year

Net Return Data(in paisa) of selected Investments Amou

nt Availa

bleAuto

Sector

Banking

Sector

IT Sector

Metal Sector

2005 95 80 70 60 70

2006 75 65 60 50 40

2007 70 45 50 40 90

2008 60 40 40 30 30

Maximum

Investment

40 50 60 60210

230 Using the data given about the Reliance Equity Opportunities Fund, determine the optimal investment strategy.

The objective is to get the optimal investment strategy.

The problem is unbalanced which is to be further converted to a balanced one by adding a dummy investment and formulated as transportation problem to maximize return.

Solution:

Year

Net Return Data(in paisa) of selected Investments Amou

nt Availa

bleAuto

Sector

Banking

Sector

IT Sector

Metal Sector

Dummy

2005 95 80 70 60 0 70

2006 75 65 60 50 0 40

2007 70 45 50 40 0 90

2008 60 40 40 30 0 30

Maximum

Investment

40 50 60 60 20 230

Now, converting the maximization problem into a minimization problem by subtracting all the elements of the above pay-off matrix from the highest pay-off.

Year

Net Return Data(in paisa) of selected Investments Amou

nt Availa

bleAuto

Sector

Banking

Sector

IT Sector

Metal Sector

Dummy

2005 0 15 25 35 95 70

2006 20 30 35 45 95 40

2007 25 50 45 55 95 90

2008 35 55 55 65 95 30

Maximum

Investment

40 50 60 60 20 230

To get the initial feasible solution, we have applied the Vogel’s Approximation Method (VAM).

An algorithm used to find a relatively efficient initial feasible solution to a transportation problem by considering the ‘penalty cost’ of not using the cheapest available route.

It is preferred to the other methods because it gives an initial solution which is nearer to an optimal solution or is optimal solution itself.

Initial Feasible Solution

Year

Net Return Data(in paisa) of selected Investments Amo

unt Available

Auto

Sector

Banking

Sector

IT Sector

Metal

Sector

Dummy

2005 40 3070 30

0

2006 20 2040 20

0

2007 40 5090 50

0

2008 10 2030 20

0

Maximum

Investment

40 0

50 20 0

60 40 0

60 10 0

20 230

0 15 25 35 95

2 30

25 50

35 45

45 55

35 55 55 65

95

95

95

Row Penalty

I II

III

IV V

15

10 - - -

10 5 5 1

0 -

20

10 5 1

010

20 0 0 1

010

Column Penalty

I 20 15 10 10 0

II - 15 10 10 0

III - 20 10 10 0

IV - - 10 10 0

V - - 20 10 0

Initial Solution Obtained= (40×.95) + (30×.80) + (20×.65) + (20×.60) + (40×.50) + (50×.40) + (10×.30) + (20×0)

= Rs. 130 Crores

After getting the initial feasible solution, the next step is test of degeneracy. The total number of allocated cells in the table are 8. So, the allocated cells must be equal to (m +n-1) to get a non-degenerated.

‘m’ is number of origins and ‘n’ is number of destinations, i.e., m= 4 and n= 5

So, as per the formula, (m +n-1)= (4+5-1)= 8.

Therefore, the problem has a non-degenerated solution.

Test of Degeneracy

Since there are 8 allocations, the solution is tested straightway for optimality. We have tested optimality by Modified Distribution Method.

Test of Optimality

Year

Net Return Data(in paisa) of selected Investments Amo

unt Available

Auto

Sector

Banking

Sector

IT Sector

Metal

Sector

Dummy

2005 40 30 70

2006 20 20 40

2007 40 50 90

2008 10 20 30

Maximum

Investment

40 50 60 60 20 230

0 15 25 35 95

2 30

25 50

35 45

45 55

35 55 55 65

95

95

95

The row numbers (ri) and column numbers (kj) are added to such that it makes net return on investment of one rupee. We have assumed ‘r3= 0’.

r1 + k1 = 0 r1 + k2 = 15

r2 + k2 = 30r2 + k3 = 35

r3 + k3 = 45 r3 + k4 = 55

r4 + k4 = 65 r4 + k5 = 95

Assuming ‘r3= 0’, we have calculated all other values of ri and kj.

Year

Net Return Data(in paisa) of selected Investments Amo

unt Available

Auto

Sector

Banking

Sector

IT Sector

Metal

Sector

Dummy

2005 40 30 70

2006 20 20 40

2007 40 50 90

2008 10 20 30

Maximum

Investment

40 50 60 60 20 230

0 15 25 35 95

2 30

25 50

35 45

45 55

35 55 55 65

95

95

95

Row Number (ri)

r1 = -25

r2 = -10

r3 = 0

r4 = 10

Column Number

(kj)

k1 = 25

k2 = 40

k3 = 45

k4 = 55

k5 = 85

Unoccupied Cell

Expected Return

Imputed Return

Opportunity Cost

r1 k3 25 20 05

r1k4 35 30 05

r1k5 95 50 35

r2k1 20 15 05

r2k4 45 45 00

r2k5 95 75 20

r3k1 25 25 00

r3k2 50 40 10

r3k5 95 85 10

r4k1 35 35 00

r4k2 55 50 05

r4k3 55 55 00

Improvement Index

Since all the opportunity cost of non- allocated cells are positive, the initial solution obtained above is optimal.

Optimal Transport Schedule

Year Investment

Net Return

(Rs. Crores)

2005

Rs. 40 Crores to Automobile Sector

0.95× 40 = 38

Rs. 30 Crores to Banking Sector

0.80 × 30 = 24

2006

Rs. 20 Crores to Banking Sector

0.65 × 20 = 13

Rs. 20 Crores to IT Sector

0.60 × 20 = 12

2007

Rs. 40 Crores to IT Sector

0.50 × 40 = 20

Rs. 50 Crores to Metal Sector

0.40 × 50 = 20

2008Rs. 50 Crores to Metal

Sector0.30 × 10 = 3

Thank YouNitish Garg

023Sandeep R. Sharma

005Saket Maheshwari

006Rahul Sharma

043Surya Rathi

058Sumit Kumar

039Purav Riat 055