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Operations Research MBA-024
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Unit-II-2-TP

May 30, 2018

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Page 1: Unit-II-2-TP

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Operations Research

MBA-024

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TRANSPORTATION PROBLEM

UNIT II

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Transportation Model/Problem

TP is essentially an LPP.

But because of certain special characteristics

associated with the problem, we have aspecial algorithm to solve a TP.

It is not solved by applying the simplex

method.

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Transportation Model/Problem

Although in principle it can be solved by the

simplex method but its application would

make the problem very unmanageable.

The TP, along with the assignment problem is

a special type of LPP, therefore we have

special purpose algorithms for them.

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TP: Important Features

TP normally used for minimising

transportation cost, although the algorithm

can be applied on some other types of 

problems also.

A TP deals with problems related to

minimisation of cost when goods have to be

transported from different points of supply todifferent points of demand.

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TP: Important Features

The amounts available at different points of 

supply and amounts required at different

points of demand are known to us.

The unit cost of transportation is also known

to us.

In a general TP we may assume m points of 

supply and n points of demand.

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TP: Important Features

We have to design a transportation plan which

would minimise the total cost of 

transportation.

Total number of variables = m x n.

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TP: Formulation

General notation of variable: xij (amount of goodstransported from ith point of supply to jth point of demand).

Cij = cost coefficient attached with xij. Total number of constraints = m+n.

There would be m supply and n demandconstraints.

Availability at supply points: b1, b2, b3, , bm.

Requirement at demand points: d1, d2, d3, , dn.

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TP: Formulation

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TP: Formulation

This is a balanced TP because total supply is equalto total demand.

Because total supply = total demand, thereforeone constraint becomes redundant.

T

otal number of effective constraints = m+n-1. Number of basic variables = number of constraints = m+n-1

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There are 3 coalieries which can supply coal in

the following quantities:

a: 14 tonnes, b: 12 tonnes, c: 5 tonnes.

There are 3 consumers who require coal in the

following quantities:

A: 6 tonnes, B: 10 tonnes, C: 15 tonnes.

The cost of transporting 1 tonne of coal from

any one point a, b or c to any other point A, B orC are as follows:

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The problem is to minimise total cost of 

transportation.

6 4 1 a8 9 2 b

4 3 6 c

A B C

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North West Corner Rule

1. Select the north-west (upper left) corner cell.

2. Allocate as many units as possible.

3. Adjust supply and demand figures in therespective row and column.

4. If row supply is exhausted, move down one row.

5. If column demand is exhausted, move right onecolumn.

6. Repeat steps 2-5 till total available quantity isfully allocated.

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Least Cost Method

1.Select the cell with the lowest transportation

cost.

2.Allocate as many units as possible.3.Adjust supply and demand figures in the

respective row and column.

4.Repeat steps 1-3 till total available quantity is

fully allocated.

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Vogels Approximation Method

Best method of obtaining initial solution.

Consists of determining (row and column)penalties.

Row penalty is calculated by subtracting theminimum cost element from the second leastcost element in the row.

Column penalty is calculated by subtractingthe minimum cost element from the secondleast cost element in the column.

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Stepping Stone Method

Determine an initial basic feasible solution

using any of the above three methods.

Make sure that the number of occupied cellsis m+n-1.

Select an unoccupied cell.

Beginning at this cell, trace a closed path.

Move only horizontally or vertically.

Turn only at occupied cells.

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Stepping Stone Method

The cells at the turning points are called the

stepping stones.

Assign plus and minus signs alternatively toeach stepping stone, starting with a plus sign

at the unoccupied cell to be evaluated.

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Stepping Stone Method

Compute the net change in cost along theclosed path by finding the algebraic sum of the unit costs of the stepping stones,

considering the plus/minus signs assigned inthe previous step.

A negative sum indicates a decrease intransportation cost.

Repeat these steps for all the unoccupiedcells.

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Modified Distribution Method

Determine an initial basic feasible solution.

Compute the values of the row and column

variables (ui and v j). Compute the improvement index for each

unoccupied cell,

If all improvement indices are positive or zero,

the optimum solution has been reached.

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Factory Warehouse Capacity

(Supply)W1 W2 W3 W4

F

1 21 16 25 13 11F2 17 18 14 23 13

F3 32 27 18 41 19

Requirement

(Demand)

6 10 12 15 43

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W1 W2 W3 W4

F121 16 25 13

11

F2 17 18 14 23 13

F3

32 27 18 41

19

6 10 12 15 43

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N

orth West Corner Rule

NWCR

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W1 W2 W3 W4

F121 16 25 13

11

F2 17 18 14 23 13

F3

32 27 18 41

19

6 10 12 15

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W1 W2 W3 W4

F121 16 25 13

116

F2 17 18 14 23 13

F3

32 27 18 41

19

6 10 12 15

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W1 W2 W3 W4

F121 16 25 13

56

F2 17 18 14 23 13X

F3

32 27 18 41

19X

0 10 12 15

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W1 W2 W3 W4

F121 16 25 13

56 5

F2 17 18 14 23 13X

F3

32 27 18 41

19X

0 10 12 15

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W1 W2 W3 W4

F121 16 25 13

06 5 X X

F2 17 18 14 23 13X

F3

32 27 18 41

19X

0 5 12 15

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W1 W2 W3 W4

F121 16 25 13

06 5 X X

F2 17 18 14 23 13X 5

F3

32 27 18 41

19X

0 5 12 15

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W1 W2 W3 W4

F121 16 25 13

06 5 X X

F2 17 18 14 23 8X 5

F3

32 27 18 41

19X X

0 0 12 15

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W1 W2 W3 W4

F121 16 25 13

06 5 X X

F2 17 18 14 23 8X 5 8

F3

32 27 18 41

19X X

0 0 12 15

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W1 W2 W3 W4

F121 16 25 13

06 5 X X

F2 17 18 14 23 0X 5 8 X

F3

32 27 18 41

19X X

0 0 4 15

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W1 W2 W3 W4

F121 16 25 13

06 5 X X

F2 17 18 14 23 0X 5 8 X

F3

32 27 18 41

19X X 4

0 0 4 15

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W1 W2 W3 W4

F121 16 25 13

06 5 X X

F2 17 18 14 23 0X 5 8 X

F3

32 27 18 41

15X X 4

0 0 0 15

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W1 W2 W3 W4

F121 16 25 13

06 5 X X

F2 17 18 14 23 0X 5 8 X

F3

32 27 18 41

15X X 4 15

0 0 0 15

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W1 W2 W3 W4

F121 16 25 13

06 5 X X

F2 17 18 14 23 0X 5 8 X

F3

32 27 18 41

0X X 4 15

0 0 0 0

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Least Cost Method

LCM

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W1 W2 W3 W4

F121 16 25 13

11

F2 17 18 14 23 13

F3

32 27 18 41

19

6 10 12 15

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W1 W2 W3 W4

F121 16 25 13

1111

F2 17 18 14 23 13

F3

32 27 18 41

19

6 10 12 15

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W1 W2 W3 W4

F121 16 25 13

0X X X 11

F2 17 18 14 23 13

F3

32 27 18 41

19

6 10 12 4

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W1 W2 W3 W4

F121 16 25 13

0X X X 11

F2 17 18 14 23 1312

F3

32 27 18 41

19

6 10 12 4

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W1 W2 W3 W4

F121 16 25 13

0X X X 11

F2 17 18 14 23 112

F3

32 27 18 41

19X

6 10 0 4

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W1 W2 W3 W4

F121 16 25 13

0X X X 11

F2 17 18 14 23 11 12

F3

32 27 18 41

19X

6 10 0 4

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W1 W2 W3 W4

F121 16 25 13

0X X X 11

F2 17 18 14 23 01 X 12 X

F3

32 27 18 41

19X

5 10 0 4

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W1 W2 W3 W4

F121 16 25 13

0X X X 11

F2 17 18 14 23 01 X 12 X

F3

32 27 18 41

1910 X

5 10 0 4

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W1 W2 W3 W4

F121 16 25 13

0X X X 11

F2 17 18 14 23 01 X 12 X

F3

32 27 18 41

910 X

5 0 0 4

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W1 W2 W3 W4

F121 16 25 13

0X X X 11

F2 17 18 14 23 01 X 12 X

F3

32 27 18 41

95 10 X

5 0 0 4

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W1 W2 W3 W4

F121 16 25 13

0X X X 11

F2 17 18 14 23 01 X 12 X

F3

32 27 18 41

45 10 X

0 0 0 4

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W1 W2 W3 W4

F121 16 25 13

0X X X 11

F2 17 18 14 23 01 X 12 X

F3

32 27 18 41

45 10 X 4

0 0 0 4

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W1 W2 W3 W4

F121 16 25 13

0X X X 11

F2 17 18 14 23 01 X 12 X

F3

32 27 18 41

05 10 X 4

0 0 0 0

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Vogels Approximation Method

VAM

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W1 W2 W3 W4

F121 16 25 13

11

F2 17 18 14 23 13

F3

32 27 18 41

19

6 10 12 15

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0 -6 3 X 4

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0 0 0 0 43

Column

Penalty - - - -

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1111

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F3

32 27 18 41

197 12

6 10 12 15

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Stepping Stone Method

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F1W1F1W4F3W4F3W1 21-13+41-32=17

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F1W1F1W4F3W4F3W1 21-13+41-32=17

F1W2F1W4F3W4F3W2 16-13+41-27=17

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F1W1F1W4F3W4F3W1 21-13+41-32=17

F1W2F1W4F3W4F3W2 16-13+41-27=17

F

1W3F

1W4F

3W4F

3W1F

2W1F

2W3 25-13+41-32+17-14=24

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F1W1F1W4F3W4F3W1 21-13+41-32=17

F1W2F1W4F3W4F3W2 16-13+41-27=17

F1W3F1W4F3W4F3W1F2W1F2W3 25-13+41-32+17-14=24

F2W2F3W2F3W1F2W1 18-27+32-17=6

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11

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F1W1F1W4F3W4F3W1 21-13+41-32=17

F1W2F1W4F3W4F3W2 16-13+41-27=17

F1W3F1W4F3W4F3W1F2W1F2W3 25-13+41-32+17-14=24

F2W2F3W2F3W1F2W1 18-27+32-17=6

F2W4F3W4F3W1F2W1 23-41+32-17=-3

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F1W1F1W4F3W4F3W1 21-13+41-32=17

F1W2F1W4F3W4F3W2 16-13+41-27=17

F1W3F1W4F3W4F3W1F2W1F2W3 25-13+41-32+17-14=24

F2W2F3W2F3W1F2W1 18-27+32-17=6

F2W4F3W4F3W1F2W1 23-41+32-17=-3

F3W2F3W1F2W1F2W3 18-32+17-14=-11

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5- 10 4

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11

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1+ 12

F332 27 18 41

5- 10 4

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10 5 4

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F121 16 25 13

11

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6 7

F332 27 18 41

10 5 4

F1W1F1W4F3W4F3W3F2W3F2W1 21-13+41-18+14-17=28

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F1W1F1W4F3W4F3W3F2W3F2W1 21-13+41-18+14-17=28

F1W2F1W4F3W4F3W2 16-13+41-27=17

F1W3F1W4F3W4F3W3 25-13+41-18=35

F2W2F2W3F3W3F3W2 18-14+18-27=-5

F2W4F3W4F3W3F2W3 23-41+18-14=-14

F3W1F2W1F2W3F3W3 32-17+14-18=11

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10 9

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10 9

W1 W2 W3 W4

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11

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6 3 4

F3 32 27 18 41

10 9

W1 W2 W3 W4

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F121 16 25 13

11

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6 3 4

F3 32 27 18 41

10 9

F1W1F1W4F2W4F2W1 21-13+41-32=17

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F1W2F1W4F2W4F2W3F3W3F3W2 16-13+23-14+18-27=3

F1W3F1W4F2W4F2W3 25-13+23-14=21

F2W2F2W3F3W3F3W2 18-14+18-27=-5

F3W1F2W1F2W3F3W3 32-17+14-18=11

F3W4F3W3F2W3F2W4 41-18+14-23=14

W1 W2 W3 W4

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F121 16 25 13

11

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6 3- 4

F3 32 27 18 41

10- 9+

W1 W2 W3 W4

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W1 W2 W3 W4

F121 16 25 13

11

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6 3 4

F3 32 27 18 41

7 12

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11

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6 3 4

F3 32 27 18 41

7 12

W1 W2 W3 W4

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F3 32 27 18 41

7 12

W1 W2 W3 W4

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6 3 4

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7 12

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F121 16 25 13

11

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6 3 4

F3 32 27 18 41

7 12

F1W1F1W4F2W4F2W1 21-13+41-32=17

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F1W2F1W4F2W4F2W2 16-13+23-18=8

F1W3F1W4F2W4F2W2F3W2F3W3 25-13+23-18+27-18=26

F2W3F3W3F3W2F2W2 14-18+27-18=5

F3W1F2W1F2W2F3W2 32-17+18-27=6

F3W4F3W2F2W2F2W4 41-27+18-23=9

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5 10 4

W1 W2 W3 W4

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F121 16 25 13

u111

F217 18 14 23

u21 12

F332 27 18 41

u3

5 10 4

v1 v2 v3 v4

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Degeneracy

A manufacturing company has three plants X Y

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A manufacturing company has three plants X, Y

and Z, which supply to the distributors locatedat A, B, C, D and E. Monthly plant capacities are

80, 50 and 90 units respectively. Monthly

requirements of distributors are 40, 40, 50, 40

and 80 units respectively. Unit transportationcosts are given below in rupees:

F T

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Determine an optimal

distribution for the

company in order tominimise the total

From To

A B C D E

X 5 8 6 6 3

Y 4 7 7 6 6

Z 8 4 6 6 3