transportation model pgdm

8/6/2019 Transportation Model Pgdm

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THE TRANSPORTATIONTHE TRANSPORTATION

MODELMODEL


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MODULE IVMODULE IV

Transportation Model - Vogelsapproximation method - MODImethod - Minimization case -

Maximization case Unbalancedproblem


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REFERENCE BOOKS

1. An Introduction to Management Science:

Quantitative Approaches to decision making

- Anderson, Sweeney, Williams2. Operations Research - Kanti Swaroop

3. Operations Research - Hamdy A Taha

4. OperationsR

esearch - S.Kalavathy


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THE TRANSPORTATION MODEL

The transportation model is a special class ofLPPs that deals with transporting(shipping) a

commodity from sources (e.g. factories) to

destinations(e.g. warehouses). The objectiveis to determine the transportation schedule

that minimizes the total transportation

cost while satisfying supply and dem

andlimits.


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We assume that there are m sources 1,2, , m and n

destinations 1, 2, , n. The cost of transporting one

unit from Source i to Destination j is cij.We assume that the availability(supply) at source i is

ai (i=1, 2, , m) and the demandat the destination j is

bj (j=1, 2, , n). We make an important assumption:the problem is a balanced one.That is

!!

!n

j

j

m

i

i ba

11

That is, total availability equals total demand.


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Letx

ij be the amount of commodity tobe shipped from the source i to the

destination j.

We present the data in an mvn table

called the transportation matrix or the

cost effectiveness matrix shown below.


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xx1111 xx1212 xx1n1n a1

xx2121 xx2222 xx2n2n a2

xxm1m1

xxm2m2

xxmnmn am

b1

b2

bn

S

o

u

r

c

e

1

2

.

.

m

Destination

1 2 . . n Supply

Demand

cc1212

CC2121

CCm2m2

CC2222CC2n2n

CC1n1n

CCm1m1 CCmnmn

CC1111


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Thus the problem becomes the LPP

!!

!

n

j

ijij

m

i

xcz11

Minimize

subject to

),...,2,1(

),...,2,1(

1

1

njbx

miax

j

m

i

ij

i

n

j

ij

!!

!!

!

!

0uijx


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OBJECTIVE OF THE TRANSPORTATION MODEL

To determine the amount of the commodityto be shifted from each source to eachdestination such that the total

transportation cost is minimized and thedemand at each destination (requirementcentre) is met.


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DEFINTIONS

A set of non-negative values

(allocations) xij that satisfies theconstraints (rim conditions) and also thenon - negativity restrictions is called a

feasible solution to the transportationproblem.

A feasible solution to an m x n

transportation problem that has notmore than (m + n 1) non negativeallocations is called a basic feasiblesolution to the transportation problem.


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DEFINTIONS (contd.)

A basic feasible solution to an m x n

transportation problem is said to be non degenerate if it contains exactly (m + n 1)non negative allocations and degenerate if

it contains less than (m + n 1) non negative allocations in independentpositions.

A feasible solution (not necessarily basic) issaid to be an optimal solution if it minimizesthe total transportation cost (or maximisesthe profit)


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CONDITION FORA FEASIBLE SOLUTION OF THE

TRANSPORTATION PROBLEM

The necessary and sufficient condition forthe transportation problem to have afeasible solution is that the

Total supply = Total demand

That is, the problem must be balanced.


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OPTIMAL SOLUTION OF THE TRANSPORTATION

PROBLEM PROCEDURAL STEPS

1. Find the initial basic feasible solution using

a) North West Corner Rule

b) Least Cost Matrix

c) Vogels Approximation Method/ PenaltyMethod

2. Find an optimal solution by makingsuccessive improvements using MODIMethod.


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NORTH WEST CORNER RULENORTH WEST CORNER RULE

1. Starting with the cell at the upper left

(north-west) corner of the transportationmatrix, allocate as much as possible. Thatis, x11= min (a1, b1).

2. (i) If min (a1, b1) = a1 put x11= a1, decreaseb

1

by a1

, and move vertically to the secondrow and make the second allocation x21=min (a2, b1 - x11) in the cell (2,1) . Crossout the first row.(ii)If min (a1, b1) = b1 put x11= b1,

decrease a1 by b1, and move horizontally tothe second row and make the secondallocation x12= min (a1 - x11, b2) in the cell(1,2) . Cross out the first column.


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NORTH WEST CORNER RULE (contd.)NORTH WEST CORNER RULE (contd.)

(iii) If a1 = b1 then put x11= a1 = b1.Cross out the first row and the firstcolumn and move diagonally .

3.Repeat steps 1 and 2 until rim

requirements are met.


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PROBLEM 1PROBLEM 1

1. MG Auto has three plants in Chennai, Fatehpur and

Kolkata, and five distribution centers in Coimbatore,

Hyderabad, Mumbai, Delhi and Chandigarh. The

capacities of the plants (each quarter), the quarterly

demands at the distribution centers and the

transportation cost per car from the plants to the

distribution centers are given below. Find the initial basic

feasible solution to this problem using North West

Corner Rule.

Distribution Centers Supply

Plants

Chennai 2 11 10 3 7 4

Fatehpur 1 4 7 2 1 8

Kolkata 3 9 4 8 12 9

Demand 3 3 4 5 6


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PROBLEM 2PROBLEM 2Three refineries with daily capacities of 7, 12 and 11

million gallons of petrol, respectively, supply three

distribution areas with daily demands of 10 million

gallons each. Petrol is distributed to the three

distribution areas through a network of pipelines. The

table below gives the transportation cost oftransportation (in thousands of rupees) between the

refineries and the distribution areas. Using the North

West Corner Rule approximation find the initial

solution. Distribution Centres Supply

Refineries

1 2 6 7

0 4 2 12

3 1 5 11


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PROBLEM 3PROBLEM 3

Obtain an initial basic feasible solution to thefollowing Transportation problem using the

North West corner Rule.


Supply

Centers

1 2 3 4

1 2 3 11 7 6

2 1 0 6 1 13 5 8 15 9 10

Demand 7 5 3 2


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HOMEWORKHOMEWORK

1. Three orchards supply crates of oranges to four

retailers. The daily demand at the four retailers is 200,225, 275, and 250 crates, respectively. Supply at the

three orchards is estimated at 250, 300, and 400 crates

daily. The transportation costs (in rupees) per crate

from the orchards to the retailers are given in Tablebelow. Find the initial basic feasible solution to this

problem using the North West Corner Rule.

R

etailers Supply

Orchards

11 13 17 14 250

16 18 14 10 300

21 24 13 10 400

Demand 200 225 275 250


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LEAST COST METHODLEAST COST METHOD

1. Identify the cell with the smallest cost

and allocate as much as possible. Thatis, xij= min (ai, bj).

2. (i) If min (ai, bj) = ai put xij= ai,

decrease bj by ai. Cross out the ith row.(ii)If min (ai, bj) = bj put xij= bj,decrease ai by bj. Cross out the j

th

column.(iii)If min (ai, bj) = ai = bj then put

xij = ai = bj. Cross outeither the ith row

or the jth

column (but not both).


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LEAST COST METHOD (contd.)LEAST COST METHOD (contd.)

3. Repeat step 1 for the resulting reducedtransportation matrix until all rimrequirements are met.


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PROBLEM 1PROBLEM 1

Mangoes have to be transported from farms in Tamil

Nadu, Andhra Pradesh and Maharashtra to 4 factories of

Tasty Squash Ltd. located elsewhere in India. Determine

using the Least Cost Method, how many tons of mangoes

must be transported from each farm to each factory so

that the total cost of transportation is minimized giventhe following data pertaining to the transportation cost

(in thousands of rupees).FACTORIES

F

A

R

M

S

1 2 3 4 Supply

TAMIL NADU 1 2 1 4 30

ANDHRA PRADESH 3 3 2 1 50

MAHARASHTRA 4 2 5 9 20

Demand 20 40 30 10


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PROBLEM 2PROBLEM 2

The following table gives the cost of transportation

per ton (in thousands of rupees) of wheat fromthe distribution centers to the supply centers of a

Public Distribution System. Obtain an initial basic

feasible solution to the following Transportation

problem using the Least Cost Method.


Supply

Centers

1 2 3 4

1 5 3 7 2 302 8 2 1 5 70

3 6 2 3 2 50

Demand 20 40 40 50


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PROBLEM 3PROBLEM 3

Obtain an initial basic feasible solution to the

following Transportation problem using the

Least Cost Method.


Supply

Centers

1 2 3 4

1 2 3 11 7 6

2 1 0 6 1 13 5 8 15 9 10

Demand 7 5 3 2


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HOMEWORKHOMEWORK



Least Cost Method.


Supply

Centers

1 2 3

1 1 2 6 7

2 0 4 2 123 3 1 5 11

Demand 10 10 10


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VOGELS APPROXIMATION METHODVOGELS APPROXIMATION METHOD

1. Find the difference (penalty) between the

smallest and the next smallest element in eachrow(column) and write them in brackets besideeach row(column).

2. Identify the row/column with largest penalty (iftie occurs break it arbitrarily). Choose the cellwith the smallest cost in the selectedrow/column and allocate as much as possible to

this cell. Cross out the satisfied row/column.3. Compute the row and column penalties for the

reduced transportation matrix and go to step 2.Repeat till rim requirements are satisfied.


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PROBLEM 1PROBLEM 1

1. Three orchards supply crates of oranges to four

retailers. The daily demand at the four retailers is 200,225, 275, and 250 crates, respectively. Supply at the

three orchards is estimated at 250, 300, and 400 crates

daily. The transportation costs (in rupees) per crate

from the orchards to the retailers are given in Tablebelow. Find the initial basic feasible solution using Vogels

approximation method.

R

etailers Supply

Orchards

11 13 17 14 250

16 18 14 10 300

21 24 13 10 400

Demand 200 225 275 250


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PROBLEM 2PROBLEM 2Three refineries with daily capacities of 7, 12 and 11

million gallons of petrol, respectively, supply three

distribution areas with daily demands of 10 milliongallons each. Petrol is distributed to the three

distribution areas through a network of pipelines. The

table below gives the transportation cost oftransportation (in thousands of rupees) between the

refineries and the distribution areas. Using Vogels

approximation find the initial solution.

Distribution Centres Supply

Refineries

1 2 6 7

0 4 2 12

3 1 5 11

Demand 10 10 10


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PROBLEM 3PROBLEM 3

Determine how many tons of wheat must be transported from each

grain elevator to each mill on a monthly basis in order to minimize the

total cost of transportation given the following data?

Grain Elevator Supply Mill Demand

1. Amritsar 150 A. Jaipur 200

2. Coimbatore 175 B. Mysore 100

3. Kalahandi 275 C. Puri 300

Jaipur Mysore Puri Supply

Amritsar 6 8 10 150

Coimbatore 7 11 11 175

Kalahanadi 4 5 12 275

Demand 200 100 300


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HOMEWORKHOMEWORK



Vogels approximation Method.


Supply

Centers

1 2 3 4

1 2 3 11 7 6

2 1 0 6 1 1

3 5 8 15 9 10

Demand 7 5 3 2

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