Transportation problems
Operational Research Level 4
Prepared by T.M.J.A.Cooray
Department of Mathematics
MA 4020-Transportation problems
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Introduction
 Transportation
problem is a special
kind of LP problem in which goods
are transported from a set of
sources to a set of destinations
subject to the supply and demand
of the source and the destination
respectively, such that the total cost
of transportation is minimized.
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Examples:

Sources
factories,
finished goods
warehouses ,
raw
materials
houses,
suppliers etc.

Destinations
Markets
Finished goods ware
house
ware raw
materials
ware
houses,
factories,
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A schematic representation of a transportation problem is shown below
a1
S1
a2
ai
am
D1
b1
D2
b2
Si
Dj
Dn
Sm
MA 4020-Transportation problems
bj
bn
4
 m-
number of sources
 n- number of destinations
 ai- supply at source I
 bj – demand at destination j
cij – cost of transportation per unit from
source i to destination j
Xij – number of units to be transported
from the source i to destination j
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Destination j

1
2
j
n
S 1 c11
O 2
U
R
C
E
c12
c1j
c1n
i ci1
ci2
i
m
Demand
a2
cij
cm1 cm2
b1
Supply
a1
b2
MA 4020-Transportation problems
bj
cin
ai
cmn
am
bn
6
Transportation problem: represented as
a
LP model
m
n
Minimize : Z   cij X ij
i 1 j 1
n
subject to
X
j 1
m
X
i 1
ij
ij
 ai i  1,2,...., m
 bj
j  1,2,....., n
X ij  0 for i  1,...m and j  1,..n
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The ideal situation is shown below.,with equalities
instead of inequalities. There are “mn” unknown
variables and m+n-1 independent equations.
m
n
Minimize : Z   cij X ij
i 1 j 1
n
subject to
X
j 1
m
X
i 1
ij
ij
 ai i  1,2,...., m
 bj
j  1,2,....., n
X ij  0 for i  1,...m and j  1,..n
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When solving the transportation problem ,the
number of possible routes should be  m+n1.
If it is <m+n-1, it is called a degenerate
solution.
In such a case evaluation of the solution
will not be possible.
In order to evaluate the cells /routes
(using the u-v method or the stepping
stone method ) we need to
imagine/introduce
some
used
cells/routes carrying / transporting a
very small quantity, say . That cell
should be selected
at the correct place.
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Example: Consider a transportation problem
involving 3 sources and 3 destinations.
Sourc
e
Destination
1
2
3
20
10
15
Supply
200
1
10
12
9
2
25
30
18
300
500
3
200
400
400
1000
Deman
d
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Types of transportation
problems
 Balanced
transportation problems
m
a
 Unbalanced
i 1
i

b
transportation problems
m
a
i 1
i

n
j 1
j
n
b
j 1
j
Include a dummy source or a dummy destination
having a supply “d” or demand “d” to convert it to a
balanced transportation problem.
n
bj
Where d= 
j 1
m
  ai
i 1
m
or
a
i 1
i
n
  b j respectively.
MA 4020-Transportation problems
j 1
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Example
Plant
W
A1
R
E2
H
O3
U
S4
E
Supply
1
2
3
4
5
Deman
d
10
2
3
15
9
25
5
10
15
2
4
30
15
5
14
7
15
20
20
15
13
-
8
30
20
20
30
10
25
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Solution of transportation
problems
 Two
phases:
 First phase:
 Find an initial feasible solution
 2nd phase:
 Check for optimality and improve the
solution
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Find an initial feasible solution
 North
west corner method
 Least cost method
 Vogel’s approximation method
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Checking for optimality
 U-V
method
 Stepping-Stone method
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Example-( having a degenerate solution)
Introduce  to for phase 2..
1 Destinations
2
Sources
3
Supply
3
2
3
25
5
6
5
15
1
3
4
20
2
5
7
10
20
20
30
S1
S2
S3
S4
Demand
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Transshipment models.
 In
transportation problems ,shipments are
sent directly from a particular source to a
particular destination to minimize the total
cost of shipments.
 It is sometimes economical if the shipment
passes through some transient nodes in
between the sources and destinations.
 In transshipment models it is possible for
a shipment to pass through one or more
intermediate nodes before it reaches its
destination.
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Transshipment problem with sources and
destinations acting as transient nodes
 Number
of starting nodes as well as the
number of ending nodes is the sum of number
of sources and the number of destinations of
the original
m problem.
n
 a  b
 Let
B=
i 1
i
j 1
j
 be
the buffer stock and it is added to all the
starting nodes and all the ending nodes.
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a1+B
..
aj+B
am+B
B
B
S1
S1
B
..
Sj
B
Sj
Sm
Sm B
D1
…
D1 b1+B
…
Dn
Dn bn+B
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 Destinations
D1,D2,….Dn are included as
additional starting nodes mainly to act as
transient nodes.they don’t have any original
supply and the supply of these nodes
should be at least B.
 The sources S1,S2,….Sm are included as
additional ending nodes mainly to act as
transient nodes.these nodes are not having
any original demand.But each of these
transient nodes is assigned with B units as
the demand value.
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 We
need to know the transshipment cost
between the sources ,between the
destinations and between sources and
destinations .
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Example

Supplies at the sources are 100,200,150 and 350 and
Demand at the destinations are 350 and 450
respectively.
S1
S2
S3
S4
D1
D2
S1
0
4
20
5
25
12
S2
10
0
6
10
5
20
S3
15
20
0
8
45
7
S4
20
25
10
0
30
6
D1
20
18
60
15
0
10
D2
10
25
30
23
4
0
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S1
S2
S3
S4
D1
D2
S1 0
4
20
5
25
12
800+100=900
S2
S3
S4
D1
D2
10
15
20
20
10
0
20
25
18
25
6
0
10
60
30
10
8
0
15
23
5
45
30
0
4
20
7
6
10
0
800+200=1000
800
800
800
800
800+35
0=1150
800+45
0=1250
800+150=950
800+350=1150
800
800
Same algorithms can be used to solve this transshipment
problem.
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Transportation problem with some transient
nodes between sources and destination.
 Consider
the case where the shipping items
are first sent to intermediate finished goods
ware houses from the supply points/factories
and then to the destinations.
 To solve these problems the capacity at each
transient node is made equal to B.
 Where
m
n
i 1
j 1
B =  ai   b j
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Example
 Multi
plant organization has 3 plants and
three market places.
 The goods from the plants are sent to market
places through two intermediate finished
goods warehouses.
 Cost of transportation per unit between plants
and warehouses and warehouses to market
places and also supply values of plants and
demand values of the markets are shown in
the table.
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M1
M2
M3 W1 W2
SUPPLY
P1



15
30
200
P2



28
10
300
P3



30
15
400
W1
10
40
30 0
20
900
W2
25
15
35 25
0
900
DEMAND
100 400 40
0
900
900
900
Solution of the problem is same as Ordinary transportation Problems.
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