MANEESH P

DEPT. OF APPLIED ECONOMICS

The basic transportation problem was developed in 1941 by F.I.

Hitchaxic. However it could be solved for optimally as an answer to

complex business problem only in 1951,when Geroge B. Dantzig

applied the concept of Linear Programming in solving the

Transportation models.

Transportation problems are primarily concerned with the optimal

(best possible) way in which a product produced at different factories

or plants (called supply origins) can be transported to a number of

warehouses (called demand destinations).

INTRODUCTION

Transportation problem is a special kind of LPproblem in which goods are transported from a setof sources to a set of destinations subject to thesupply and demand of the source and thedestination respectively, such that the total cost oftransportation is minimized.

The objective in a transportation problem is:-To

fully satisfy the destination requirements within

the operating production capacity constraints at

the minimum possible cost.

Whenever there is a physical movement of goods from

the point of manufacture to the final consumers through

a variety of channels of distribution (wholesalers,

retailers, distributors etc.), there is a need to minimize

the cost of transportation so as to increase the profit on

sales. Transportation problems arise in all such cases.

.

It aims at providing assistance to the topmanagement in ascertaining how many units of aparticular product should be transported from eachsupply origin to each demand destinations to thatthe total prevailing demand for the company’sproduct is satisfied, while at the same time the

total transportation costs are minimized

i.e.; The total supply available at the plants exactly matches

the total demand at the destinations. Hence, there is neither

excess supply nor excess demand.

Such type of problems where supply and demand are

exactly equal are known as Balanced Transportation

Problem.

Supply (from various sources) are written in the rows,

while a column is an expression for the demand of different

warehouses. In general, if a transportation problem has m

rows an n columns, then the problem is solvable if there are

exactly (m + n –1) basic variables

.UNBALANCED TRANSPORTATION PROBLEM :

A transportation problem is said to be unbalanced if the

supply and demand are not equal.

Two situations are possible:-

1. If Supply < demand, a dummy supply variable is

introduced in the equation to make it equal to demand.

2. If demand < supply, a dummy demand variable is

introduced in the equation to make it equal to supply.

Then before solving we must balance the demand & supply.

When supply exceeds demand, the excess supply is assumed to go to

inventory. A column of slack variables is added to the transportation

table which represents dummy destination with a requirement equal to

the amount of excess supply and the transportation cost equal to zero.

When demand exceeds 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 total demand & supply and

with each cell having a zero unit cost.

Demand Less Than Supply

Suppose that a plywood factory increases its rate of production from 100 to 250 desks

The firm is now able to supply a total of 850 desks each period

Warehouse requirements remain the same (700) so the row and column totals do not balance

We add a dummy column that will represent a fake warehouse requiring 150 desks

This is somewhat analogous to adding a slack variable

We use the northwest corner rule and either vogel’s approximation method or MODI to find the optimal solution

FROM

TO

A B C TOTAL AVAILABLE

I 5 4 3 250

II 8 4 3 300

III 9 7 5 300

WAREHOUSE REQUIREMENTS

300 200 200 850

700

FROM

TO

A B C D TOTAL AVAILABLE

I 5 4 3 0 250

II 8 4 3 0 300

III 9 7 5 0 300

WAREHOUSE REQUIREMENTS

300 200 200 150850= 850

FROM

TO

A B C D TOTAL AVAILABLE

I5 4 3 0

250

II8 4 3 0

300

III9 7 5 0

300

WAREHOUSE REQUIREMENTS

300 200 200 150

(1)

(1)

(2)

(3) (0) (0) (0)

250

300-250=50

FROM

A B C D

TOTAL AVAILABLE

II8 4 3 0 300

III 9 7 5 0 300

WAREHOUSE

REQUIREMENTS

50 200 200 150

TO

(1)

(2)

(1) (3) (2) (0)

200

300-200=100

FROM

A C D

TOTAL AVAILABLE

II 8 3 0 100

III 9 5 0 300

WAREHOUSE

REQUIREMENTS

50 200 150

TO

100 (5)

(4)

(1) (2) (0)

200-100=100

FROMA C D

TOTAL AVAILABLE

III 9 5 0 300

WAREHOUSEREQUIREMENTS 50 100 150

TO

(4)

(9) (5) (0)

50

300-50=250

FROMC D

TOTAL AVAILABLE

III 5 0 250

WAREHOUSEREQUIREMENTS 100 150

TO

(5)

(5) (0)

150

250-150=100

FROM

C

TOTAL AVAILABLE

III 5 100

WAREHOUSEREQUIREMENTS

100

TO

100

100-100=0

FROMA B C D

I 5 4 3 0

II 8 4 3 0

III 9 7 5 0

TO

WAREHOUSEREQUIREMENTS

TOTAL AVAILABLE

300 200 200 150

250

300

300

250

200 100

50 100 150

m+n-1 = 3+4-1= 6

FROMA B C D

I 5 4 3 0

II 8 4 3 0

III 9 7 5 0

TO

WAREHOUSEREQUIREMENTS

TOTAL AVAILABLE

300 200 200 150

250

300

300

250

200 100

50 100 150

u1

u2

U3

v1 v2 v3 v4

U1+V1=5

U2+V2=4

U2+V3=3

U3+V1=9

U3+V3=5

U3+V4=0

ASSUMING U3=0

0+V1=9V1=9

0+V3=5V3=5

0+V4=0V4=0

U1+V1=5U1+9=5U1= 5-9 = -4

U2+V3=3U2+5=3U2= 3-5= -2

(V3=5)

U3+V4=0U3+0=0U3=0

U2+V2=4-2+V2=4V2= 4-(-2) = 6

(V4=0)

(U2= -2)

CIJ TABLE

4 3 0

8 0

7

V1 V2 V3 V4

U1

U2

U3

-4

-2

0

0569

uI-VJ TABLE

2 1 -4

7 -2

7

V1 V2 V3 V4

U1

U2

U3

DIJ TABLE

2 2 4

1 2

0

V1 V2 V3 V4

U1

U2

U3

Total cost = 250*5+200*4+100*3+50*9+100*5+150*0

1250+800+300+450+500+0 = 3,300

Initial basic feasible solution

Here all the dij values are positive, therefore the solution is optimal.

Demand Greater than Supply

The second type of unbalanced condition occurs when total demand is greater than total supply

In this case we need to add a dummy row representing a fake factory

The new factory will have a supply exactly equal to the difference between total demand and total real supply

FROM

A B CPLANTSUPPLY

W 6 4 9 200

X 10 5 8 175

Y 12 7 6 75

WAREHOUSEDEMAND 250 100 150

450

TO

500

Totals do notbalance

FROM

A B CPLANTSUPPLY

W 6 4 9 200

X 10 5 8 175

Y 12 7 6 75

Z 0 0 0 50

WAREHOUSEDEMAND 250 100 150

500

TO

500

FROM

A B CPLANTSUPPLY

W 6 4 9 200

X 10 5 8 175

Y 12 7 6 75

Z 0 0 0 50

WAREHOUSEDEMAND 250 100 150

TO

200

0 100 75

50

75

CIJ TABLE

4 9

12 7

0 0

V1 V2 V3

U1

U2

U3

U4

-4

0

-2

-10

10 5 8

UI+VJ TABLE

1 4

8 3

-5 -2

V1 V2 V3

U1

U2

U3

U4

DIJ TABLE

3 5

4 4

5 2

V1 V2 V3

U1

U2

U3

U4

DIJ= CIJ-(UI+VJ)

Here, all the dij values are positive, therefore the solution is optimal.

TC= 200*6+0*10+100*5+75*8+75*6+50*0

1200+500+600+450= 2750

CONCLUSION

In real-life problems, total demand is frequently not equal to total supplyThese unbalanced problems can be handled easily by introducing dummy sources or dummy destinationsIf total supply is greater than total demand, a dummy destination (warehouse), with demand exactly equal to the surplus, is created If total demand is greater than total supply, we introduce a dummy source (factory) with a supply equal to the excess of demand over supply

Any units assigned to a dummy destination represent excess capacity

Any units assigned to a dummy source represent unmet demand

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