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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.
Distribution Centers Supply
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.
Distribution Centers Supply
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.
Distribution Centers Supply
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
Obtain an initial basic feasible solution to the
following Transportation problem using the
Least Cost Method.
Distribution Centers Supply
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
Obtain an initial basic feasible solution to the
following Transportation problem using the
Vogels approximation Method.
Distribution Centers Supply
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