Transportation Problems 5

8/4/2019 Transportation Problems 5

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Transportation Problems


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Consider a commodity which is produced at various centers

called SOURCES and is demanded at various otherDESTINATIONS.

The production capacity of each source (availability) and the

requirement of each destination are known and fixed.

The cost of transporting one unit of the commodity from eachsource to each destination is also known.

The commodity is to be transported from various sources to

different destinations in such a way that the requirement of

each destination is satisfied and at the same time, the totalcost of transportation is minimized.

This optimum allocation of the commodity from various

sources to different destinations is called TRANSPORTATION

PROBLEM.

Transportation Problems


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A transportation problem is a special type of

linear programming problem and hence can

be formulated and solved as such.

Generally, transportation costs are involved

in such problems but the scope of problems

extends well beyond to cover situations

which have nothing to do with these costs

like scheduling production problem,

controlling inventory and management of

funds over different time periods.

Transportation Problems..


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The method allows the manager to seek answers to the

questions like the following: What is the optimal way of shipping goods from various

sources (warehouses) to different markets so as to

minimize the total cost involved in the shipping?

How to handle a situation when some routes are notavailable or when some units have to be necessarily

transported from a particular source to a particular

market?

If an item can be produced at different locations at

varying costs and sold in different markets at different

prices, then what production and shipping plan will yield

maximum profit?

Transportation Problems..


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A transportation problem can be stated mathematically as follows:

Let there be m SOURCES and n DESTINATIONS.

Let ai : the availability at the ith source

bj : the requirement of the jth destination.

Cij : the cost of transporting one unit of commodity

from the ith source to the jth destination

xij

: the quantity of the commodity transported from

ith source to the jth destination

Where i = 1, 2, m

j = 1, 2, ...... n

Problem Statement


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A transportation problem can be stated as a LPP as :

Problem Statement

)(1 1

MinimizeXcZ ij

m

i

n

j

ij

tosubject

i

n

j

ij aX 1

mi ,....,2,1

j

m

i

ij bX 1

nj ,.....,2,1

njandmiforXij ,...2,1,...2,10

for

for


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Origin (i)

Destination (j)

Supply, ai

1 2 .. n

1 C11 C12 .. C1n a1

2 C21 C22 .. C2n a2

.. .. .. .. .. ..

m Cm1 Cm2 Cm3 Cmn am

Demand, bj b1 b2 b3 Bn ai =bj

Transportation Tableau

x11 x12 x1n

x2nx22x21

xm1 xm2 xmn


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Transportation Tableau When a transportation problem is solved, some of the xijs would assume

positive values indicating utilized routes. The cells containing such values are

called occupied or filled cells and each of them represents the presence of abasic variable.

For the remaining cells, called the empty cells, xijs would be zero. These are

the routes that are not utilized by the transportation pattern and their

corresponding variables (xijs) are regarded to be non-basic.

In general, a transportation problem with m-sources and n-destinations, with

matching aggregate supply and demand, may be expressed as an LPP with m *

n decision variables and m + n - 1 constraints.

The number of variables required for forming a basis in one less, i.e. m +

n 1. This is so, because there are only m + n 1 independent variables

in the solution basis.

A basic feasible solution of a transportation problem has exactly m + n 1

positive components in comparison to the m + n positive components

required for a basic feasible solution in respect of LPP in which there are m +

n structural constraints to satisfy.


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Solution to The Transportation Problem

A firm owns facilities at seven places. It has manufacturing plants at places

A, B and C with daily output of 500, 300 and 200 units of an item

respectively. It has warehouses at places P, Q, R and S with daily

requirements of 180, 150, 350 and 320 units respectively. Per unit

shipping charges on different routes are given below.

To: P Q R S

From A: 12 10 12 13

From B: 7 11 8 14

From C: 6 16 11 7

The firm wants to send the output from various plants to warehouses

involving minimum transportation cost. How should it route the product

so as to achieve its objective?


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Solution to The Transportation Problem

A transportation problem can be solved by

two methods:

a) Simplex Method

b) Transportation Method

P Q R S Supply

A 12 10 12 13 500

B 7 11 8 14 300

C 6 16 11 7 200

Demand 180 150 350 320 1000

From

To


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The Transportation Method

Three Steps Involved:

Obtaining an initial solution, that is to say making an

initial assignment in such a way that a basic feasible

solution is obtained.

Ascertaining whether it is optimal or not, by

determining opportunity costs associated with the

empty cells. If the solution is optimal then exit and if it

is not optimal, proceed to next step.

Revise the solution until an optimal solution is reached.


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Step 1: Initial Feasible Solution

The first step in using the transportation method is to obtain a feasible

solution, namely, the one that satisfies the rim requirements (i.e. the

requirements of demand and supply).

The commonly used methods to obtain the initial feasible solution areas follows:

1. North-West Corner (NWC) Rule

2. Least Cost Method (LCM)

3. VogelsApproximation Method (VAM)


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North-West Corner Rule

P Q R S Supply

A 12 10 12 13 500

B 7 11 8 14 300

C 6 16 11 7 200

Demand 180 150 350 320 1000

From

To

180

0

320150

0

170170

180

0

180

0

120120

200

0

200

0

0

Total Cost = 12 x 180 + 10 x 150 + 12 x 170 + 8 x 180 + 14 x 120 + 7 x 200

= Rs. 10,220This routing of the units meets all the rim requirements and involves a

total of 6 (3+4-1) shipments as there are six occupied cells.

This method ignores the cost factor which is sought to be minimized.


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Least Cost Method (LCM)

P Q R S Supply

A 12 10 12 13 500

B 7 11 8 14 300

C 6 16 11 7 200

Demand 180 150 350 320 1000

From

To

180

0

Total Cost = 10 x 150 + 12 x 50 + 13 x 300 + 8 x 300 + 6 x 180 + 7 x 20= Rs. 9,620

2020

300

0

300

50

0

150

0

35050

0

300300

0

0

If there is a tie in the minimum cost, so that two or more routes have

the same least cost of shipping, then, conceptually, either of them may

be selected.

However a better initial solution is obtained if the route chosen is theone where largest quantity can be assigned.


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Vogels Approximation Method (VAM)

P Q R S Supply

A 12 10 12 13 500

B 7 11 8 14 300

C 6 16 11 7 200

Demand 180 150 350 320 1000

From

To

I 1 1 3 6

II 5 1 4 1

III - 1 4 1

I

2

1

1

II

2

1

-

III

2

3

-

6

200

120

0

5

180

0

120

4

120

230

0

150

0

350230

0

120120

0

0



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Vogels Approximation Method (VAM)

Important Points

The VAM is also called the Penalty Method because the cost

differences that it uses are nothing but the penalties of not using

the least-cost routes. Since the objective function is the

minimization of the transportation cost, in each iteration thatroute is selected which involves the maximum penalty of not

being used.

The initial solution obtained by VAM is found to be the best of all

as it involves the lowest total cost among all the three initialsolutions. Of course, it does not come as a rule, but usually VAM

provides the best initial solution. Therefore this method is

generally preferred in solving the problems.


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Step 2: Testing the Optimality

There are two methods of testing the optimality of a basic feasible

solution i.e.

1. Stepping-stone Method

2. Modified Distribution Method (MODI)

Both the methods involve determining opportunity costs of empty

cells, the methodology employed by them differs. The opportunity

cost values indicate whether the given solution is optimal or not. Both the methods can be used only when the solution is a basic

feasible solution, so that it has m + n 1 basic variables.


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Step 3: Improving the Solution

By applying either of the methods, if the solution is found to be

optimal, then the process terminates as the problem is solved.

If solution is not seen to be optimal, then a revised and improved

basic feasible solution is obtained. This is done by exchanging a non-

basic variable for a basic variable.

For this, re-arrangement is made by transferring units from an

occupied cells to an empty cell that has the largest opportunity cost

and then adjusting the units in other related cells in a way that all the

rim requirements are satisfied. This is done by first tracing a closed

loop.

The solution obtained is again tested for optimality and revised, if

necessary.

We continue this process until an optimal solution is finally obtained.


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Stepping-stone Method (Using NW Corner Rule)

P Q R S Supply

A 12 10 12 13 500

B 7 11 8 14 300

C 6 16 11 7 200

Demand 180 150 350 320 1000

From

To

180 150 170

180 120

200


Initial Feasible Solution: Testing for Optimality

+

_

_

+


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Stepping-stone Method..

A shipment of one item on the route AS would cause an

increase of Rs. 13 but it will also mean a reduction of one

unit from AR and thereby, a reduction of Rs. 12.

Further, an increase of a unit in BR would raise the cost by

Rs. 8 while a unit lesser in BS would save Rs. 14.

The net effect of the operation would be a saving of Rs. 5 (13

12 + 8 - 14).

This implies that a reduction of Rs. 5 can be effected by

adopting the route AS.

The opportunity cost of this route is Rs. 5.

Since the opportunity cost is positive, it means that it is

worth considering making cell AS a basic variable.


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In the same manner, evaluate each of the remaining empty

cells as follows:

CR CR - CS - BS - BR 11 - 7 + 14 - 8 = 10 -10

Cell Closed Loop Net Cost Change Opportunity Cost

AS AS - AR + BR - BS 13- 12 + 8 -14 = -5 5

BP BP - BR - AR - AP 7 - 8 + 12 -12 = -1 1

BQ BQ - BR - AR - AQ 11 - 8 + 12 - 10 = 5 -5

CP CP - CS - BS - BR - AR - AP 6 - 7 + 14 - 8 + 12 - 12 = 5 -5

CQ CQ - CS - BS - BR - AR - AQ 16 - 7 + 14 - 8 + 12 - 10 = 17 -17

S i h d


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Tracing A Closed Loop To draw a close loop, always begin with an empty cell and move alternatively

horizontally and vertically, through occupied cell only, until reaching back

to the starting point.

In the process of moving from one occupied cell to another

a) Move only horizontally or vertically, but never diagonally;

b) Step over empty and if the need be, over occupied cells withoutchanging them.

A closed loop would always have corners only on the occupied cells.

Having traced the path, place plus and minus signs alternately in the cells

on each turn of the loop, beginning with a plus (+) sign in the empty cell.

An important restriction is that there must be exactly one cell with a plus

sign and one cell with a minus sign in any row or column in which the loop

takes a turn.

This restriction ensures that the rim requirements would not be violated

when units are shifted among cell.


S i M h d


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An even number of at least four cells must participate in a

closed loop and an occupied cell can be considered only once

and not more.

If there exists a basic feasible solution with m + n 1

positive variables, then there would be one and only one

closed loop for each cell. This is irrespective of the size of

the matrix given.

All cells that receive a plus or a minus sign, except the

starting empty cell, must be the occupied cells.

Closed loop may or may not be square or rectangular in

shape.

Tracing A Closed Loop



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CR CR - CS - BS - BR 11 - 7 + 14 - 8 = 10 -10


AS AS - AR + BR - BS 13- 12 + 8 -14 = -5 5

BP BP - BR - AR - AP 7 - 8 + 12 -12 = -1 1

BQ BQ - BR - AR - AQ 11 - 8 + 12 - 10 = 5 -5

CP CP - CS - BS - BR - AR - AP 6 - 7 + 14 - 8 + 12 - 12 = 5 -5

CQ CQ - CS - BS - BR - AR - AQ 16 - 7 + 14 - 8 + 12 - 10 = 17 -17

Continuing with testing optimality, the rule is: if none of the emptycells has a positive opportunity cost, the solution is optimal.

The solution in question, therefore, is not optimal.

Here, the most favorable cell is AS for it has the largest opportunity

cost equal to 5. So we will include AS as a basic variable in the solution.


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P Q R S Supply

A 12 10 12 13 500

B 7 11 8 14 300

C 6 16 11 7 200

Demand 180 150 350 320 1000

From

To

180 150 170

180 120

200


Improved Solution: Non-Optimal


50

300


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This solution involves a total cost of Rs. 9,620 which is lower by Rs.

600 (5 x 120) in comparison to the initial solution.

We again apply step 2 to determine optimality.

CR CR - CS - AS AR 11 - 7 + 13 - 12 = 5 -5


BS BS - AS - AR - BR 14- 13 + 12 - 8 = 5 -5

BP BP - BR - AR AP 7 - 8 + 12 -12 = -1 1

BQ BQ - BR - AR - AQ 11 - 8 + 12 - 10 = 5 -5

CP CP - CS - AS - AP 6 - 7 + 13 - 12 = 0 0

CQ CQ - CS

AS - AQ 16 - 7 + 13 - 10 = 12 -12

The solution is also not optimal here as Cell BP has the positive

opportunity cost equal to 1.

So we will include BP as a basic variable in the solution.


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P Q R S Supply

A 12 10 12 13 500

B 7 11 8 14 300

C 6 16 11 7 200

Demand 180 150 350 320 1000

From

To

180 150 50

300

120

200


Improved Solution: Optimal


230

120


28/50


This solution involves a total cost of Rs. 9,440.

We again apply step 2 to determine optimality.

CR CR - CS - AS AR 11 - 7 + 13 - 12 = 5 -5


BS BS - AS - AR - BR 14- 13 + 12 - 8 = 5 -5

AP AP - AR - BR BP 12 - 12 + 8 - 7 = 1 -1

BQ BQ - BR - AR - AQ 11 - 8 + 12 - 10 = 5 -5

CP CP - CS - AS AR BR BP 6 - 7 + 13 12 + 8 - 7 = 1 -1

CQ CQ - CS

AS - AQ 16 - 7 + 13 - 10 = 12 -12

Since the opportunity cost of all the empty cells are negative, the

solution obtained is optimal.

A carefully look at the solution reveals that it is identical to theinitial solution obtained by VAM.

d f d b h d( )


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Modified Distribution Method(MODI)

It is an efficient method of testing the optimality of a

transportation solution.

It avoids the kind of extensive scanning and reduces the

number of steps required in the evaluation of the empty cells

in case of stepping-stone method.

It gives a straightforward computational scheme whereby we

can determine the opportunity cost of each of the empty

cells.

difi d i ib i h d ( )


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Modified Distribution Method (MODI)

P Q R S Supply

A 12 10 12 13 500

B 7 11 8 14 300

C 6 16 11 7 200

Demand 180 150 350 320 1000

From

To

180



ui

0

vj

150 170

180 120

200

12 10 12

- 4

18

- 11

Having determined all ui and vj values, calculate for each unoccupied cell ij = ui +

vj cij. The ijs represent the opportunity costs of various cells. After obtaining theopportunity costs, proceed in the same way as in the stepping stone method.

+5

+1 -5

-5 -17 -10

difi d i ib i h d ( O I)


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P Q R S Supply

A 12 10 12 13 500

B 7 11 8 14 300

C 6 16 11 7 200

Demand 180 150 350 320 1000

From

To

180

Here the cell AS has the largest positive opportunity cost. Select AS for inclusion asa basic variable.


+

_

_

+

ui

0

vj

150 170

180 120

200

12 10 12

- 4

18

- 11

+5

+1 -5

-5 -17 -10

M difi d Di ib i M h d (MODI)


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P Q R S Supply

A 12 10 12 13 500

B 7 11 8 14 300

C 6 16 11 7 200

Demand 180 150 350 320 1000

From

To

180


ui

0

vj

150 170

180 120

200

12 10 12

- 4

18

- 11

50

300


M difi d Di ib i M h d (MODI)


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P Q R S Supply

A 12 10 12 13 500

B 7 11 8 14 300

C 6 16 11 7 200

Demand 180 150 350 320 1000

From

To

180


ui

0

vj

150 50

300

120

200

12 10 12

- 4

13

- 6

+1 -5 -5

0 -12 -5

Here the cell BP has the largest positive opportunity cost. Select BP for inclusion asa basic variable.

+

_

_

+

M difi d Di t ib ti M th d (MODI)


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P Q R S Supply

A 12 10 12 13 500

B 7 11 8 14 300

C 6 16 11 7 200

Demand 180 150 350 320 1000

From

To

180


ui

0

vj

150 50

300

120

200

12 10 12

- 4

13

- 6

230

120


M difi d Di t ib ti M th d (MODI)


35/50


P Q R S Supply

A 12 10 12 13 500

B 7 11 8 14 300

C 6 16 11 7 200

Demand 180 150 350 320 1000

From

To

180

Improved Solution: Optimal

ui

0

vj

150 120

200

11 10 12

- 4

13

- 6

230

120

-1

-5 -5

-1 -12 -5

Here all empty cells are less than or equal to zero. So the solution is found to beoptimal. Since all ij values are negative, the solution is also unique.


= Rs. 9,440

U b l d T t ti P bl


36/50

Unbalanced Transportation Problems

Sometimes aggregate supply does not match the aggregate demand.

Such problems are called unbalanced transportation problems.

Balancing must be done before they can be solved.

When the aggregate supply exceeds the aggregate demand, a column

of slack variables is added to the transportation tableau which

represents a dummy destination with a requirement equal to theamount of excess supply and the transportation costs equal to zero.

On the other hand, when the aggregate demand exceeds the

aggregate supply, balance is restored by adding a dummy origin, The

row representing it is added with an assumed total availability equal

to the difference between the total demand and supply and witheach of the cells having a zero unit cost.

In some cases, however, when the penalty of not satisfying the

demand at a particular destination(s) is given, then such penalty

value should be considered and not zero.

P hibit d R t


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Prohibited Routes

Sometimes in a given transportation problem some route(s)may not be available due to unfavorable weather conditions

on a particular route, strike on a particular route etc.

To handle a situation of this type, we assign a very large cost

represented by M to each of such routes which are not

available.

The effect of adding a large cost element would be that such

routes would automatically be eliminated in the final

solution.

U i M lti l O ti l S l ti


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Unique vs Multiple Optimal Solutions

The solution in case of transportation problem will be

optimal if all the ij

values are less than or equal to zero.

The solution will be unique if all the ij values are negative.

If some cell (or cells) has ij = 0, then multiple optimal

solutions are indicated so that there exist transportation

pattern(s) other than the one obtained which can satisfy allthe rim requirements for the same cost.

To obtain an alternate optimal solution, traced a closed loop

beginning with a cell having ij = 0 and get the revised

solution in the same way as a solution is improved.

This revised solution would also entail the same total cost as

before.

Transportation Problem


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WarehouseMarket

SupplyA B C

1 10 12 7 180

2 14 11 6 100

3 9 5 13 160

4 11 7 9 120

Demand 240 200 220

Transportation Problem

It is known that currently nothing can be sent from

warehouse 1 to market A and from warehouse 3 to market

C. Solve the problem and determine the least cost

transportation schedule. Is the optimal solution obtained by

you unique? If not, what is/are the other optimalsolution/s?

Maximization Problem


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Maximization Problem

A transportation tableau may contain unit profits instead of unit costs and the

objective function be maximization of total profits.

In this case the given problem is first converted into an equivalentminimization problem by subtracting each element of the given matrix from a

constant value k, which can be any number. Generally we take the largest one.

The procedure yields what is termed as the opportunity loss matrix to which

the transportation method is applied.

The entries in the opportunity loss matrix indicate how much each of the

values is away from that constant. The minimization of the opportunity loss

automatically leads to maximization of total profit.

The total profit value is obtained by using the transportation pattern of the

solution on the unit profit matrix and not through the opportunity loss matrix.

If a maximization type of transportation problem is unbalanced, then it should

be balanced by introducing the necessary dummy row/column for a

source/destination, before converting it into a minimization problem.

Similarly, if such a problem has a prohibited route, then the payoff element for

such a route should be substituted by M before proceeding to convert to

minimization type.

Transportation Maximization Problem


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Transportation Maximization Problem

Solve the problem for maximum profit.

Per Unit Profit (Rs.)

Market

A B C D

Warehouse

X 12 18 6 25

Y 8 7 10 18

Z 14 3 11 20

Available at warehouses:

X: 200 units

Y: 500 unitsZ: 300 units

Demand in the markets:

A: 180 units

B: 320 unitsC: 100 units

D: 400 units

Transhipment Problems


42/50

Transhipment Problems

Sometimes multi-plant firms find it necessary to send some

goods from one plant to another in order to meet the

substantial increase in the demand in the second market. Thesecond plant here would act both as a source and a destination

and there is no real distinction between source and

destination.

A transportation problem is regarded as a transhipmentproblem when shipment of goods is allowed from one source to

another and from one destination to another.

A transportation problem with m-origins and n-destinations

becomes a transhipment problem with m + n sources and an

equal number of destinations.

Fortunately, the optimal solution to a transhipment problem

can be found by solving a transportation problem.

Transhipment Problem


43/50

Transhipment ProblemA firm owns facilities at seven places. It has manufacturing plants at places A, B

and C with daily output of 500, 300 and 200 units of an item respectively. It has

warehouses at places P, Q, R and S with daily requirements of 180, 150, 350 and

320 units respectively. Per unit shipping charges on different routes are givenbelow.

To: P Q R S

From A: 12 10 12 13

From B: 7 11 8 14

From C: 6 16 11 7

How should the firm route the product to achieve the target of minimum cost?

From

Plant

To Plant

A B CA 0 2 8

B 5 0 7

C 10 9 0

Additional cost data:

From

Warehouse

To Warehouse

P Q R SP 0 6 5 8

Q 8 0 3 2

R 5 4 0 10

S 9 4 8 0



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P Q R S Supply

A 12 10 12 13 500

B 7 11 8 14 300

C 6 16 11 7 200

Demand 180 150 350 320 1000

From

To

180

Optimal Solution (Transportation Model)

150 120

200

230

120




45/50


A B C P Q R S Supply

A 12 10 12 13 500

B 7 11 8 14 300

C 6 16 11 7 200

P

Q

R

S

Demand 180 150 350 320 1000

Arranging the tableau

150 230 120

180 120

200



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A 0 2 8 12 10 12 13 500

B 5 0 7 7 11 8 14 300

C 10 9 0 6 16 11 7 200

P 12 7 6 0 6 5 8 0

Q 10 11 16 8 0 3 2 0

R 12 8 11 5 4 0 10 0

S 13 14 7 9 4 8 0 0

Demand 0 0 0 180 150 350 320 1000

Initial Feasible Solution: Non-optimal

0150 230 120180 120

200

ui

vj

- 0

- 0

- 0

- 0

- 0

- 0

- 0

10 12 13

-4-6

11

-11-10-12-13

640Find out the opportunity cost of each unoccupied cell i.e.

ij = ui + vj

cij .For example opportunity cost of cell AB = 0 + 4 2 = 2

This method suggests that the route AB be brought into the solution. Trace aclosed loop and find out the revised solution.

+

_ +

_



47/50



A 0 2 8 12 10 12 13 500

B 5 0 7 7 11 8 14 300

C 10 9 0 6 16 11 7 200

P 12 7 6 0 6 5 8 0

Q 10 11 16 8 0 3 2 0

R 12 8 11 5 4 0 10 0

S 13 14 7 9 4 8 0 0

Demand 0 0 0 180 150 350 320 1000

Revised Solution: Non-optimal

150 230 120

180 120

200

- 0

- 0

- 0

- 0

- 0

- 0

- 0

350-230


= Rs. 8,980



48/50



A 0 2 8 12 10 12 13 500

B 5 0 7 7 11 8 14 300

C 10 9 0 6 16 11 7 200

P 12 7 6 0 6 5 8 0

Q 10 11 16 8 0 3 2 0

R 12 8 11 5 4 0 10 0

S 13 14 7 9 4 8 0 0

Demand 0 0 0 180 150 350 320 1000

Revised Solution: Non-optimal

0150230 120180

200

ui

vj

- 0

- 0

- 0

- 0

- 0

- 0

10 10 13

-2-6

9

-9-10-10-13

620

350-230

Find out the opportunity cost of each unoccupied cell i.e. ij = ui + vj cij .

For example opportunity cost of cell AC = 0 + 6 8 = -2

This method suggests that the route QS be brought into the solution. Trace a

closed loop and find out the revised solution.

+

_ +

_



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A 0 2 8 12 10 12 13 500

B 5 0 7 7 11 8 14 300

C 10 9 0 6 16 11 7 200

P 12 7 6 0 6 5 8 0

Q 10 11 16 8 0 3 2 0

R 12 8 11 5 4 0 10 0

S 13 14 7 9 4 8 0 0

Demand 0 0 0 180 150 350 320 1000

Revised Solution: optimal

150230 120

180

200

- 0

- 0

- 0

- 0

- 0

- 0

350-230

270

- 120

0ui

-2-5-9-10-10-12

vj10 10 129520

Find out the opportunity cost of each unoccupied cell. Now here all these

cells have negative opportunity costs. So the solution is optimal here.



50/50


Optimal Solution

Send from Plant A: 230 units to Plant B and 270 units

to Warehouse Q,

Send from Plant B: 180 units to warehouse P and 350

units to Warehouse R,

Send from Plant C: 200 units to Warehouse S and

Send from warehouse Q: 150 units to Warehouse S.

Total Cost = 2 x 230 + 10 x 270 + 7 x 180 + 8 x 350 + 7 x200 + 2 x 120 = Rs. 8,860

The transportation pattern would involve a total cost of Rs.

8860 resulting in a saving of Rs. 9440 8860 = Rs. 580 onaccount of the possibility of transhipment.

Documents

Transportation Problems 5