SVKM’S NMIMS ASMSOC
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Chapter 2
Transportation Problem
Introduction
The distribution of goods produced by a factory from various warehouses (or
sources) to different markets (or destinations) where they are required, causes
problems to almost every business. The transportation method is developed to
deal with the transportation of goods from different destinations, given the
relevant data like available quantities at various sources, demand at each of the
destinations, the cost of shipping along each route and non-availability of certain
routes; if any.
A typical transportation problem is like this. A matrix is given where sources are
given row-wise, destinations are indicated column-wise and unit cost of
transportation from each source to each destination is provided. Also indicated is
the supply at each source and demand at every market.
Plant
Market
A
Supply
X
B
Y
C
Z
Demand
Balanced and Unbalanced Problems
If aggregate demand (AD) is equal to aggregate supply (AS), the problem is called
balanced transportation problem and if the two do not match, it is called
unbalanced. An unbalanced problem is balanced first by introducing a dummy
source (if AS<AD) or a dummy destination (if AD>AS). The cost elements of the
dummy row/column are taken to be zero. If, however, penalties for not satisfying
demand are given, they should be taken instead of zeros. In any case, the solution
to a problem begins only when it is a balanced one.
Problem Statement
The mathematical formulation of Transportation Problem is :
Let = quantity of product available at the origin i
= Quantity of the goods required at destination j
= Transportation cost per unit from origin i to destination j.
= Quantity of the goods transported from origin i to destination j.
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The problem is “balanced” if
∑
=∑
With these, the problem can be stated as a linear programming problem as:
Minimise
Total Cost Z = ∑ ∑
∑
Subject to
=
for i = 1,2,3,…..,m
∑
And
=
for j = 1,2,3,…..,n
>0
for all i = 1,2,3,…..,m and j = 1,2,3,…..,n
The transportation model can be portrayed in a tabular form:
Origin (i)
Destinations
1
Source
1
2
……
m
2
Supply
n
c11
c12
……
c1n
c21
c22
……
c2n
……
……
……
……
cm1
cm2
……
cmn
Demand
……
……
∑
=∑
Solution to Transportation Problem
Three steps are involved in obtaining solution to a transportation problem. They
are:
1. Find initial solution to the problem.
2. Test whether the solution is optimal. Stop if it is optimal. If not, go to step 3.
3. Find an improved solution. Go to step 2.
1. Finding Initial Solution There are three methods for finding initial solution to
the problem.
(a) North-west Corner Rule: Begin from the north-west corner of the table.
Consider first plant supply and first market demand. The lower of the two shall be
assigned to the first-plant first-market route. If supply is more than demand, shift
to the next the market and if supply is less than demand then move to the next
plant. In case the supply and demand are equal, move diagonally to the next plant
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and next market route. In any case, again consider the demand and supply and
make the allocation by taking lower of the two. Move in a similar manner until all
allocations are made. This gives initial feasible solution to the problem.
(b) Least Cost Method: In this method, the route with the minimum cost (that is to
say, the cell with the least unit cost) is selected and the supply and the demand at
the plant and market involved are considered. The lower of these two is allocated
in the cell chosen and the plant or market whichever is satisfied is deleted. Both
are deleted if they are both satisfied due to equal demand and supply. If the plant
is satisfied in the allocation, then the demand at the market is adjusted and if the
market is satisfied, then the plant supply is adjusted. In case there is more than
one cell with the same minimum cost, the one where larger number of units can
be allocated is selected. Again the remaining plants and markets are considered
and the same steps are taken. The process is continued until all allocations are
made.
(c) Vogel’s Approximation Method (VAM) or Penalty Method: This method works
as follows.
(i) Obtain the difference between pair of minimum cost values for each of the
rows and columns.
(ii) Select the largest of the cost differences and choose the least-cost cell in that
row/column. Consider the supply and the demand at the plant and market
involved. The lower of these two is allocated in the cell chosen and the plant or
market whichever is satisfied is deleted. Both are deleted if they are both satisfied
due to equal demand and supply. If the plant is satisfied in the allocation, then the
demand at the market is adjusted and if the market is satisfied, then the plant
supply is adjusted. In case there is a tie in the largest cost difference values, the
one corresponding to which larger number of units can be allocated is selected.
(iii) Calculate the cost differences again for the reduced problem and proceed in
the same manner as above. Repeat until all allocations are made.
In a given problem, one of these methods is used to find the initial solution.
Usually, the north-west corner rule is not used because it does not take cost into
consideration while making allocations. The other two methods tend to provide
initial solution with comparatively lower cost. Even in these two, the penalty
method usually gives better results.
2. Testing Optimality of the Solution Once the initial solution is obtained, it is
tested for optimality. For testing the solution for optimality, it is necessary that it
should have
, where m is the number of rows and n is the number of
columns, number of occupied cells. If the number is less, the solution is termed as
degenerate. Degeneracy is discussed later.
For all rows and columns,
and
values are obtained. Each row and each
column is assigned one value in such a way that, corresponding to every occupied
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SVKM’S NMIMS ASMSOC
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cell, the row and the column
adds up to the cost value,
For this, one of the
rows/columns is assigned a value arbitrarily. Usually, the first row value u1 value
is set equal to zero and other values are determined one by one, using this and the
other successively derived values. Once all these values are obtained, calculate
= +
–
. Now, if all
≤0, then the solution is optimal, otherwise not. A
positive
in a cell indicates that allocating goods in that cell can reduce cost.
Thus, if a cell 2,3 in the matrix has = 3, it means that every unit allocated to this
route, that is, every unit sent from plant 2 to market 3 would save cost at the rate
of Rs 3 per unit.
3. Improving a Non-optimal Solution If a solution is found to be non-optimal, it is
improved as follows. Beginning with the cell that has the largest
value, draw a
closed path by moving alternately horizontally and vertically, taking turns at right
angles, and halting in occupied cells only, in such a way as to reach the same cell
from where it began. It may be noted that irrespective of the size of the given
matrix; one and only closed path can be traced starting from a given cell. The path
can take any shape and the movements can be clock-wise or anti clock-wise. Once
the closed path is drawn, plus (+) and minus (–) signs are placed alternately on
the cells lying at the path, the first movement being assigned a negative sign. After
this, cells with the negative signs are considered and the minimum quantity in
them is determined. This is the maximum quantity that can be transferred over
the path for obtaining revised solution. Finally, the quantity is added to the cells
with (+) sign and subtracted from each cell with a (–) sign and the revised
solution is obtained. This solution is tested for optimality and improved in the
same manner, if necessary. The process is repeated until an optimal solution is
found.
Practical Applications Of Transportation Problems
The use of transportation method is not limited to solution of the transportation
problems alone. Problems of scheduling production, controlling inventory and
management of funds over different time periods illustrate some other areas
which can use transportation method.
Journal Problems - I
1.
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 0f 180, 150, 350 and
320 units respectively. Per unit shipping charges on different routes are:
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
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SVKM’S NMIMS ASMSOC
a.
b.
c.
2.
a.
b.
c.
d.
3.
a.
b.
c.
d.
OPERATIONS RESEARCH
SYBBA/SYBSc (FINANCE)
minimum transportation cost.
Is the given transportation problem balance?
Initial basic feasible solution using NWC, LCE and VAM. Compare and analyze the
solutions.
Test the optimality using MODI method.
A company has three plants at locations A, B and C which produce the same
product. It has to supply this to buyer located at D, E and F. The weekly plant
capacities for A, B and C are 100, 800 and 150 units respectively, while the buyer
requirements are 750, 200 and 500 units respectively for D, E and F. The unit
shipping costs (in Rs) are given here
Buyer
Plant
D
E
F
A
8
4
10
B
9
7
9
C
6
5
8
Assume that the penalty for failing to supply buyer requirement is Rs. 4, Rs. 3 and
Rs. 3 per unit in respect of D, E and F respectively. Determine the optimal
distribution for the company so as to minimize the cost of transportation and
penalty payable. Use VAM and MODI methods.
Balance the problem by using penalty.
Find the IBFS by VAM
Test the optimality of the solution by MODI method.
Give the transportation schedule and total transportation cost. How much is the
amount of penalty to be paid for the unsatisfied demand?
Consider the following data for the transportation problem:
Destination
Factory
Supply
1
2
3
A
8
4
10
10
B
9
7
9
80
C
6
5
8
15
Demand
75
20
50
Since there is not enough supply, some of the demands at the three destinations
may not be satisfied. For the unsatisfied demands let the penalty costs be rupees
4, 5 and 6 for destinations (1), (2) and (3), respectively. Find the optimal
allocation that minimizes the transportation and penalty costs.
Balance the problem by using penalty.
Find the IBFS by VAM.
Test the optimality of the solution by MODI method.
Give the transportation schedule and total transportation cost. How much is the
amount of penalty to be paid for the unsatisfied demand?
5
7
3
10
2
4
2
2
2
4
2
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OPERATIONS RESEARCH
SYBBA/SYBSc (FINANCE)
A company is spending Rs 1200 on transportation of its units from three plants to 10
four distribution centers. The supply and demand of units with unit cost of
transportation are given as under:
Distribution Centers
Plants
Supply
1
2
3
4
P1
20
30
50
17
7
P2
70
35
40
60
10
P3
40
12
60
25
18
Demand
5
8
7
15
What can be the maximum saving by optimal scheduling?
Unique vs. Multiple Optimal Solutions
In the optimal solution, if all ∆ij < 0, then the solution is unique optimal solution.
However, if some ∆ij = 0, then the problem has multiple optimal solutions. If an
alternative optimal solution is desired, then draw the closed path beginning with
the cell having ∆ij = 0 and find improved solution in same manner as discussed
earlier.
Prohibited Routes
Prohibited Routes If some route(s) is known to be prohibited, so that currently it
is not possible to send goods through that, then the cost element for that cell
should be replaced by a very prohibitive cost value, equal to M. After this, the
problem is solved in ordinary way.
Journal Problems - II
5.
a.
b.
c.
The table below records transportation costs per unit of a product from origins
O1, O2, O3 and O4 to destinations D1, D2, D3, D4 and D5. The capacities and
requirements are mentioned in the table.
Destination
Origin
Capacity
D1
D2
D3
D4
D5
O1
12
4
9
5
9
55
O2
8
1
6
6
7
45
O3
1
12
4
7
7
30
O4
10
15
6
9
1
50
Requirement
40
20
50
30
40
Employing VAM, make the initial allocations to the allocations to the origins to
satisfy the requirements of the destinations and test the optimality of these
allocations.
Use MODI method for obtaining the optimal solution that would minimize the
total cost of transportation.
Find the total cost implied by the solution.
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4
3
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SYBBA/SYBSc (FINANCE)
d.
Is the solution unique? If not, find an alternative solution as well.
2
a.
A cement company has three factories which manufacture cement which is then
transported to four distribution centers. The quantity of monthly production of
each factory, the demand of each distribution center and the associated
transportation cost per quintal are given as follows:
Distribution Centers
Monthly
Factories
Production (in
W
X
Y
Z
quintals)
A
10
8
5
4
7,000
B
7
9
15
8
8,000
C
6
10
14
8
10,000
Monthly Demand(in
6,000
6,000
8,000
5,000
quintals)
Suggest the optimal transportation schedule.
8
6.
b.
c.
d.
Is there any other transportation schedule which is equally attractive? If so, write
that.
If the company wants at least 5,000 quintals of cement are transported from
factory C to distribution center Y, will the transportation schedule be different? If
so, what will be the new optimal schedule and the effect on cost?
Suppose the company desires to send at most 500 quintals of cement from factory
C to distribution center Y, what will the optimal transportation schedule be? Also,
obtain the total transportation cost in such case.
Degeneracy
Degeneracy As indicated, a transportation problem solution is said to be
degenerate if the number of occupied cells is less than r + c – 1. In such a solution,
optimality cannot be tested since all ui and vj values cannot be determined. To
remove degeneracy, an infinitesimally small value ε is placed in each of the
required number of unoccupied cells.
The quantity ε is defined such that it satisfies the following conditions:
k + ε = k;
k – ε = k;
k X ε = 0;
and ε - ε = 0.
The cell/s in which ε is placed must be independent so that it is not possible to
draw a closed path starting with that particular cell. Among independent cells,
preference should be given to a cell that has the lowest cost.
7
2
5
5
SVKM’S NMIMS ASMSOC
7.
OPERATIONS RESEARCH
SYBBA/SYBSc (FINANCE)
A company has three cement plants from which cement has to be transported to
four distribution centers. With identical costs of production at the three plants,
the only variable costs involved are transportation costs. The monthly demand at
the four distribution centers and the unit costs the plants to the distribution
centers (in Rs.) are given below:
Distribution Centers
Monthly
Plants
Production
W
X
Y
Z
(tonnes)
A
500
1000
150
800
10000
B
200
700
500
100
12000
C
600
400
100
900
8000
Monthly
Demand
9000
9000
10000
4000
(tonnes)
a.
b.
8.
a.
b.
Suggest optimal transportation schedule and indicate the minimum 12
transportation cost.
If for certain reasons, route from Plant C to distribution center X is closed down, 8
will the transportation change? If so, suggest the new schedule and effect on total
cost.
A company has four factories manufacturing the same commodity, which are
required to meet the demands at four warehouses. The supplies and demands as
also the cost per transportation from factory to warehouse in rupees per unit of
product are given in the following table:
Distribution Centers
Supply
Factory
(units)
X
Y
Z
W
A
25
55
40
60
60
B
35
30
50
40
140
C
36
45
26
66
150
D
35
30
41
50
50
Demand
90
100
120
140
(units)
Derive an optimal strategy of transportation of goods from factories to the 9
warehouses and assess the optimal cost.
If a new transporter agrees to transport goods from factory C to warehouse W at 1
a unit cost of Rs.50, analyze the impact of this on your current optimal solution.
8
SVKM’S NMIMS ASMSOC
9.
a.
b.
OPERATIONS RESEARCH
SYBBA/SYBSc (FINANCE)
Given the following transportation problem:
Market
Warehouse
Supply
A
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
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.
8
Is the optimal solution obtained by you unique? If not, what is/are the other 2
optimal solution/s.
10.
Determine optimal solution to the problem given below. Obtain the initial solution 10
by VAM.
To Markets
Plants
Supply
M1
M2
M3
M4
P1
4
2
9
1
40
P2
20
6
11
3
40
P3
7
1
0
14
50
P4
7
1
12
6
90
Demand
90
30
50
30
11.
The table given below has been taken from the solution procedure of a
transportation problem, involving minimization of cost (in rupees).
Stockists
Monthly
Factories
Capacity
X
Y
Z
(units)
A
31
25
56
4
B
8
41
41
16
24
16
77
8
a.
b.
82
77
C
Monthly
Demand
(units)
8
72
16
102
Show that the above solution is not optimal.
Find an optimal solution.
9
25
41
SVKM’S NMIMS ASMSOC
c.
d.
e.
f.
OPERATIONS RESEARCH
Does the problem have multiple solutions? Give reasons. If so, find one more
optimal solution.
Comment upon the managerial significance of multiple optimal solutions.
If it is considered necessary to transport 20 units from factory A to stockist Z,
what will be the least-cost distribution schedule and the effect on cost?
In the alternate solution, If the transport cost from factory A to stockist Z is
increased by Rs. 5 per unit, will the solution change? If so, find the new solution.
The following table gives one of the possible solutions to a transportation problem
involving three sources and four destinations:
12.
D1
D2
10
S2
D3
200
S1
150
9
23
13
10
350
50
7
d.
18
50
S3
a.
b.
c.
D4
100
6
4
13.
SYBBA/SYBSc (FINANCE)
13
15
5
Is the above solution degenerate?
1
Is the above solution optimal? Is it a unique solution? If yes, why?
2
What is the opportunity cost of transporting one unit from source S1 to 1
destination D4?
How would the cost be affected if it is decided to transport one unit from S2 to 1
D2?
A company has four warehouses and five stores. The warehouses have total 10
surplus of 430 units of a given commodity that is divided among them as follows:
Warehouse:
W1
W2
W3
W4
Surplus:
150
30
120
130
The five stores have, in all, a requirement of 450 units of the commodity.
Individual requirements are:
Store:
Requirement:
S1
80
S2
60
S3
20
S4
210
S5
80
Cost of shipping one unit from the ith warehouse to the jth store is as displayed in
the following table: M in this table indicates that the route is not available.
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SVKM’S NMIMS ASMSOC
Warehouse
W1
W2
W3
W4
OPERATIONS RESEARCH
S1
9
5
10
5
Store
S3
10
12
7
2
S2
12
18
M
6
SYBBA/SYBSc (FINANCE)
S4
10
11
3
M
S5
6
2
20
8
How should the company arrange to transport the units so that the transportation
cost is minimized?
The following transportation table shows all necessary information on the
availability of supply to each warehouse, the requirement of each market and the
unit transportation cost from each warehouse to each market:
14.
Markets
Supply
P
Q
R
S
A
6
3
5
4
22
B
5
9
2
7
15
C
5
7
8
6
8
Demand
7
12
17
9
The shipping clerk has worked out the following schedule from experience: 12
units from A to Q, 1 unit from A to R, 9 units from A to S, 15 units from B to R, 7
units from C to P and 1 unit from C to R.
Check and see if the clerk has the minimum total transport cost.
Find the optimal schedule and minimum total transportation cost.
Is there any alternate solution? Give reason.
If the clerk is approached by a courier to route C to Q, who offers to reduce his
rate in the hope of getting some business, by how much the rate should be
reduced that the clerk will offer him the business?
Warehouse
a.
b.
c.
c.
Maximization Problems
Maximization Transportation Problems Sometimes, a problem may involve
transportation of goods from various plants to different markets so that profits
may be maximized. In such a case, unit profit matrix may be given along with
demand and supply values and it may be sought to determine how the allocations
are done so that total profit may be maximized. To solve such a problem, it is
converted into a minimization problem by subtracting all the profit values from a
constant value, usually the largest value in the table, to get opportunity loss
matrix. After this, the problem is solved in usual manner.
11
3
4
1
1
SVKM’S NMIMS ASMSOC
15.
OPERATIONS RESEARCH
Solve the following Transportation problem for maximum profit:
Warehouses
A
X
12
Y
8
Z
14
Availability at warehouses
X : 200 units
Y : 500 units
Z : 300 units
16.
SYBBA/SYBSc (FINANCE)
10
Per Unit Profit
Markets
B
C
D
18
6
25
7
10
18
3
11
20
Demand in the markets
A : 180 units
B : 320 units
C : 100 units
D : 400 units
A company has four manufacturing plants and five warehouses. Each plant 10
manufactures the same product which is sold at different prices at each
warehouse area. The cost of manufacturing and cost of raw materials are different
in each plant due to various factors. The capacities of the plants are also different.
The data given in the following table:
Plant
Item
1
2
3
4
Manufacturing Cost (Rs.) per unit
12
10
8
7
Raw material cost (Rs.) per unit
8
7
7
5
Capacity per unit time
100
200
120
80
The company has five warehouses. The sales price, transportation costs and
demand are given in the following table:
Transportation Cost (Rs) per unit
Sale price
(Rs) per unit
1
2
3
4
A
4
7
4
3
30
B
8
9
7
8
32
C
2
7
6
10
28
D
10
7
5
8
34
E
2
5
8
9
30
Formulate and solve this transportation problem to maximize profit.
Warehouse
12
Demand
80
120
150
70
90
SVKM’S NMIMS ASMSOC
17.
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SYBBA/SYBSc (FINANCE)
A company has three plants in which it produces a standard product. It has four
agencies in different parts of the country where this product is sold. The
production cost varies from factory to factory and the selling price from market to
market. The shipping cost per unit of the product from each plant to each of the
agencies is known and is stable. The relevant data are given in the following tale:
Weekly production capacity
(units)
1
400
2
300
3
800
Shipping cost (in Rs) per unit:
Plant
Unit production cost (Rs)
18
24
20
Agency
Plant
1
2
3
1
2
5
7
2
8
4
6
3
3
4
4
Demand (units)
300
400
300
Selling Price (Rs)
32
35
31
Determine the optimal plan so as to maximize the profits.
4
3
2
5
500
36
18.
Describe transportation problem and give its mathematical model.
2
19.
Discuss and compare the various methods of finding IBFS to a TP.
2
20.
What is meant by optimality test? How do you determine whether a given solution 2
is optimal or not?
21.
What is unbalanced TP? How will you convert it into a balanced one?
22.
How do you know that a transportation problem has:
Alternate solution
Feasible solution
a.
b.
23.
What is degeneracy in transportation problem? How can we deal with this
problem?
24.
In the case of TP, why is it necessary to resolve degeneracy before testing any
basic feasible solution for optimality?
25.
How can the transportation method be applied to a TP where objective function is
maximization type?
26.
How do you deal with the problem of prohibited route in transportation method?
13
2