Operations Research Unit 5 Sikkim Manipal University Page No. 93 Unit 5 Transportation Problem Structure: 5.1 Introduction Learning objectives 5.2 Formulation of Transportation Problem (TP) 5.3 Transportation Algorithm (MODI Method) 5.4 The Initial Basic Feasible Solution North west corner rule Matrix minimum method Vogel‟s approximation method 5.5 Moving Towards Optimality Improving the solution Modified distribution method / MODI method / U – V Method Degeneracy in transportation problem 5.6 Summary 5.7 Terminal Questions 5.8 Answers to SAQs and TQs Answers to Self Assessment Questions Answers to Terminal Questions 5.9 References 5.1 Introduction Welcome to the unit on transportation model in Operations Research Management. Transportation model is an important class of linear programs. For a given supply at each source and a given demand at each destination, the model studies the minimisation of the cost of transporting a commodity from a number of sources to several destinations. The transportation problem involves m sources, each of which has available a i (i = 1, 2… m) units of homogeneous product and n destinations, each of which requires b j (j = 1, 2…., n) units of products. Here a i and b j are positive integers. The cost c ij of transporting one unit of the product from the i th source to the j th destination is given for each i and j. The objective is to develop an integral transportation schedule that meets all demands from the inventory at a minimum total transportation cost.
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Operations Research Unit 5
Sikkim Manipal University Page No. 93
Unit 5 Transportation Problem
Structure:
5.1 Introduction
Learning objectives
5.2 Formulation of Transportation Problem (TP)
5.3 Transportation Algorithm (MODI Method)
5.4 The Initial Basic Feasible Solution
North west corner rule
Matrix minimum method
Vogel‟s approximation method
5.5 Moving Towards Optimality
Improving the solution
Modified distribution method / MODI method / U – V Method
Degeneracy in transportation problem
5.6 Summary
5.7 Terminal Questions
5.8 Answers to SAQs and TQs
Answers to Self Assessment Questions
Answers to Terminal Questions
5.9 References
5.1 Introduction
Welcome to the unit on transportation model in Operations Research
Management. Transportation model is an important class of linear
programs. For a given supply at each source and a given demand at each
destination, the model studies the minimisation of the cost of transporting a
commodity from a number of sources to several destinations.
The transportation problem involves m sources, each of which has available
ai (i = 1, 2… m) units of homogeneous product and n destinations, each of
which requires bj (j = 1, 2…., n) units of products. Here ai and bj are positive
integers. The cost cij of transporting one unit of the product from the ith
source to the jth destination is given for each i and j. The objective is to
develop an integral transportation schedule that meets all demands from the
inventory at a minimum total transportation cost.
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Sikkim Manipal University Page No. 94
It is assumed that the total supply and the total demand are equal.
n
1j
m
1i
i bja (1)
The condition (1) is guaranteed by creating either a fictitious destination with
a demand equal to the surplus if total demand is less than the total supply or
a (dummy) source with a supply equal to the shortage if total demand
exceeds total supply. The cost of transportation from the fictitious
destination to all sources and from all destinations to the fictitious sources
are assumed to be zero so that total cost of transportation will remain the
same.
Learning objectives
By the end of this unit, you should be able to:
Formulate the transportation problem
Find the initial basic feasible solution
Compare the advantages of various methods of finding initial basic
feasible solution
Solve the degeneracy in the transportation problem
Apply the model to minimise the cost of transporting a commodity
5.2 Formulation of Transportation Problem
The standard mathematical model for the transportation problem is as
follows.
Let xij be number of units of the homogenous product to be transported from
source i to the destination j
Then objective is to
Minimise z = ij
m
1i
n
1j
ij xC
Subject to
n.,..........,2,1ij;bjx
m,......,2,1i,ax
m
1i
ij
n
1j
iij (2)
(2) (2)
With all xij 0 and integrals
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Theorem: A necessary and sufficient condition for the existence of a
feasible solution to the transportation problem (2) is:
n
1j
m
1i
i bja
Self Assessment Questions
Fill in the blanks
1. Transportation problems are a special type of ___________.
2. The number of rows and columns need not always be ___________.
3. Transportation problem develops a schedule at _______ and ________.
5.3 Transportation Algorithm (MODI Method)
The first approximation to (2) is integral. Therefore, you always need to find
a feasible solution. Rather than determining a first approximation by a direct
application of the simplex method, it is more efficient to work with the
transportation table given below. The transportation algorithm is the simplex
method specialised to the format of table involving the following steps:
i) Finding an integral basic feasible solution
ii) Testing the solution for optimality
iii) Improving the solution, when it is not optimal
iv) Repeating steps (ii) and (iii) until the optimal solution is obtained
The solution to TP is obtained in two stages.
In the first stage, you find the basic feasible solution using any of the
following methods a) North-west corner rule b) Matrix Minima Method or
least cost method c) Vogel‟s approximation method. In the second stage,
you test the basic feasible solution for its optimality either by MODI method
or by stepping stone method.
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Table 5.1: Transportation Table
D1 D2 Dn Supply ui
S1
x11
x12
x1n a1 u1
S2
x21
x22
x2n a2 u2
S3
x31
x32
x3n a3 u3
Sm
xm1
xm2
xmn am um
Demand b1 b2 bn ai = bi
vj v1 V2 vm
Self Assessment Questions
State Yes or No
4. In transportation problems, ai = bj is a sufficient and necessary
condition for getting a feasible solution.
5. Transportation problems can also be solved by simplex method.
6. Matrix-minima method gives optimum solution.
5.4 The Initial Basic Feasible Solution
Let us consider a TP involving m-origins and n-destinations. Since the sum
of origin capacities equals the sum of destination requirements, a feasible
solution always exists. Any feasible solution satisfying m + n – 1 of the
m + n constraints is a redundant one and hence it can be deleted. This also
means that a feasible solution to a TP can have only m + n – 1 positive
component; otherwise the solution will degenerate.
It is always possible to assign an initial feasible solution to a TP, satisfying
all the rim requirements. This can be achieved either by inspection or by
following some simple rules. You can begin by imagining that the
transportation table is blank that is initial xij = 0. The simplest procedures for
initial allocation are discussed in the following section.
C11
C21
C31
Cm
1
C12
C22
C32
Cm
2
C1n
C2n
C3n
Cm
n
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5.4.1 North West Corner rule
Step1: The first assignment is made in the cell occupying the upper left
hand (north-west) corner of the transportation table. The maximum feasible
amount is allocated here is
x11 = min (a1, b1)
Either the capacity of origin O1 is used up or the requirement at destination
D1 is satisfied or both. This value of x11 is entered in the upper left hand
corner (small square) of cell (1, 1) in the transportation table.
Step 2: If b1 > a1, the capacity of origin O is exhausted and the requirement
at destination D1 is still not satisfied. Then at least one variable in the first
column will have to take on a positive value. Move down vertically to the
second row and make the second allocation of magnitude:
x21 = min (a2, b1 – x21) in the cell (2, 1)
This either exhausts the capacity of origin O2 or satisfies the remaining
demand at destination D1.
If a1 > b1 ,the requirement at destination D1 is satisfied, but the capacity of
origin O1 is not completely exhausted. Move to the right in a horizontal
position to the second column to make the second allocation of magnitude:
x12 = min (a1 – x11, b2) in the cell (1, 2)
This either exhausts the remaining capacity of origin O1 or satisfies the
demand at destination D2.
If b1 = a1, the origin capacity of O1 is completely exhausted as well as the
requirement at destination is completely satisfied, then there is a tie at the
second allocation. An arbitrary tie breaking choice is made. Make the
second allocation of magnitude
x12 = min (a1 – a1, b2) = 0 in the cell (1, 2)
OR
x21 = min (a2, b1 – b2) = 0 in the cell (2, 1)
Step 3: Start from the new north-west corner of the transportation table
satisfying the destination requirements and exhausting the origin capacities
one at a time, moving down towards the lower right corner of the
transportation table until all the rim requirements are satisfied.
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Solved Problem 1
Determine an initial basic feasible solution to the following transportation
problem using the north west corner rule:
D1 D2 D3 D4
Availability
01 6 4 1 5 14
02 8 9 2 7 16
03 4 3 6 2 5
6 10 15 4 35
Requirements
Where Oi and Dj represent the ith origin and the jth destination respectively.
Solution: The transportation table of the given T.P. has 12 cells.
14
16
5
6 10 15 4
Following north west corner rule, the first allocation is made in the cell
(1,1), the magnitude being x11 = min (14, 6) = 6
The second allocation is made in the 6 10 15 4 cell (1, 2) and the
magnitude of allocation is given by x12 = min (14 – 6, 10) = 8
The third allocation is made in the cell (2, 2), the magnitude being
x22 = min (16, 10 – 8) = 2.
The magnitude of fourth allocation, in the cell (2, 3) is given by x23 = min
(16 – 2, 15) = 14.
The fifth allocation is made in the cell (3, 3), the magnitude being x33 = min
(5, 15 –14) =1.
The sixth allocation in the cell (3, 4) is given by x34 = min (5 –1, 4) = 4.
Now all the rim requirements have been satisfied and hence an initial
feasible solution to the TP has been obtained. The solution is displayed as
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Table 5.2: Initial feasible solution to the TP
D1 D2 D3 D4
01
02
03
6
6
8
4
4
5 14
8
2
9
14
2
16
4
3
1
6
4
2 5
6 10 15 4
Clearly, this feasible solution is non-degenerate basic feasible solution as
the allocated cells do not form a loop. The transportation cost according to