Post on 07-Apr-2018
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Transportation Problems
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Consider a commodity which is produced at various centers
called SOURCES and is demanded at various otherDESTINATIONS.
The production capacity of each source (availability) and the
requirement of each destination are known and fixed.
The cost of transporting one unit of the commodity from eachsource to each destination is also known.
The commodity is to be transported from various sources to
different destinations in such a way that the requirement of
each destination is satisfied and at the same time, the totalcost of transportation is minimized.
This optimum allocation of the commodity from various
sources to different destinations is called TRANSPORTATION
PROBLEM.
Transportation Problems
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A transportation problem is a special type of
linear programming problem and hence can
be formulated and solved as such.
Generally, transportation costs are involved
in such problems but the scope of problems
extends well beyond to cover situations
which have nothing to do with these costs
like scheduling production problem,
controlling inventory and management of
funds over different time periods.
Transportation Problems..
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The method allows the manager to seek answers to the
questions like the following: What is the optimal way of shipping goods from various
sources (warehouses) to different markets so as to
minimize the total cost involved in the shipping?
How to handle a situation when some routes are notavailable or when some units have to be necessarily
transported from a particular source to a particular
market?
If an item can be produced at different locations at
varying costs and sold in different markets at different
prices, then what production and shipping plan will yield
maximum profit?
Transportation Problems..
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A transportation problem can be stated mathematically as follows:
Let there be m SOURCES and n DESTINATIONS.
Let ai : the availability at the ith source
bj : the requirement of the jth destination.
Cij : the cost of transporting one unit of commodity
from the ith source to the jth destination
xij
: the quantity of the commodity transported from
ith source to the jth destination
Where i = 1, 2, m
j = 1, 2, ...... n
Problem Statement
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A transportation problem can be stated as a LPP as :
Problem Statement
)(1 1
MinimizeXcZ ij
m
i
n
j
ij
tosubject
i
n
j
ij aX 1
mi ,....,2,1
j
m
i
ij bX 1
nj ,.....,2,1
njandmiforXij ,...2,1,...2,10
for
for
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Origin (i)
Destination (j)
Supply, ai
1 2 .. n
1 C11 C12 .. C1n a1
2 C21 C22 .. C2n a2
.. .. .. .. .. ..
m Cm1 Cm2 Cm3 Cmn am
Demand, bj b1 b2 b3 Bn ai =bj
Transportation Tableau
x11 x12 x1n
x2nx22x21
xm1 xm2 xmn
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Transportation Tableau When a transportation problem is solved, some of the xijs would assume
positive values indicating utilized routes. The cells containing such values are
called occupied or filled cells and each of them represents the presence of abasic variable.
For the remaining cells, called the empty cells, xijs would be zero. These are
the routes that are not utilized by the transportation pattern and their
corresponding variables (xijs) are regarded to be non-basic.
In general, a transportation problem with m-sources and n-destinations, with
matching aggregate supply and demand, may be expressed as an LPP with m *
n decision variables and m + n - 1 constraints.
The number of variables required for forming a basis in one less, i.e. m +
n 1. This is so, because there are only m + n 1 independent variables
in the solution basis.
A basic feasible solution of a transportation problem has exactly m + n 1
positive components in comparison to the m + n positive components
required for a basic feasible solution in respect of LPP in which there are m +
n structural constraints to satisfy.
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Solution to The Transportation Problem
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 of 180, 150, 350 and 320 units respectively. Per unit
shipping charges on different routes are given below.
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 minimum transportation cost. How should it route the product
so as to achieve its objective?
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Solution to The Transportation Problem
A transportation problem can be solved by
two methods:
a) Simplex Method
b) Transportation Method
P Q R S Supply
A 12 10 12 13 500
B 7 11 8 14 300
C 6 16 11 7 200
Demand 180 150 350 320 1000
From
To
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The Transportation Method
Three Steps Involved:
Obtaining an initial solution, that is to say making an
initial assignment in such a way that a basic feasible
solution is obtained.
Ascertaining whether it is optimal or not, by
determining opportunity costs associated with the
empty cells. If the solution is optimal then exit and if it
is not optimal, proceed to next step.
Revise the solution until an optimal solution is reached.
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Step 1: Initial Feasible Solution
The first step in using the transportation method is to obtain a feasible
solution, namely, the one that satisfies the rim requirements (i.e. the
requirements of demand and supply).
The commonly used methods to obtain the initial feasible solution areas follows:
1. North-West Corner (NWC) Rule
2. Least Cost Method (LCM)
3. VogelsApproximation Method (VAM)
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North-West Corner Rule
P Q R S Supply
A 12 10 12 13 500
B 7 11 8 14 300
C 6 16 11 7 200
Demand 180 150 350 320 1000
From
To
180
0
320150
0
170170
180
0
180
0
120120
200
0
200
0
0
Total Cost = 12 x 180 + 10 x 150 + 12 x 170 + 8 x 180 + 14 x 120 + 7 x 200
= Rs. 10,220This routing of the units meets all the rim requirements and involves a
total of 6 (3+4-1) shipments as there are six occupied cells.
This method ignores the cost factor which is sought to be minimized.
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Least Cost Method (LCM)
P Q R S Supply
A 12 10 12 13 500
B 7 11 8 14 300
C 6 16 11 7 200
Demand 180 150 350 320 1000
From
To
180
0
Total Cost = 10 x 150 + 12 x 50 + 13 x 300 + 8 x 300 + 6 x 180 + 7 x 20= Rs. 9,620
2020
300
0
300
50
0
150
0
35050
0
300300
0
0
If there is a tie in the minimum cost, so that two or more routes have
the same least cost of shipping, then, conceptually, either of them may
be selected.
However a better initial solution is obtained if the route chosen is theone where largest quantity can be assigned.
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Vogels Approximation Method (VAM)
P Q R S Supply
A 12 10 12 13 500
B 7 11 8 14 300
C 6 16 11 7 200
Demand 180 150 350 320 1000
From
To
I 1 1 3 6
II 5 1 4 1
III - 1 4 1
I
2
1
1
II
2
1
-
III
2
3
-
6
200
120
0
5
180
0
120
4
120
230
0
150
0
350230
0
120120
0
0
Total Cost = 10 x 150 + 12 x 230 + 13 x 120 + 7 x 180 + 8 x 120 + 7 x 200= Rs. 9,440
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Vogels Approximation Method (VAM)
Important Points
The VAM is also called the Penalty Method because the cost
differences that it uses are nothing but the penalties of not using
the least-cost routes. Since the objective function is the
minimization of the transportation cost, in each iteration thatroute is selected which involves the maximum penalty of not
being used.
The initial solution obtained by VAM is found to be the best of all
as it involves the lowest total cost among all the three initialsolutions. Of course, it does not come as a rule, but usually VAM
provides the best initial solution. Therefore this method is
generally preferred in solving the problems.
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Step 2: Testing the Optimality
There are two methods of testing the optimality of a basic feasible
solution i.e.
1. Stepping-stone Method
2. Modified Distribution Method (MODI)
Both the methods involve determining opportunity costs of empty
cells, the methodology employed by them differs. The opportunity
cost values indicate whether the given solution is optimal or not. Both the methods can be used only when the solution is a basic
feasible solution, so that it has m + n 1 basic variables.
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Step 3: Improving the Solution
By applying either of the methods, if the solution is found to be
optimal, then the process terminates as the problem is solved.
If solution is not seen to be optimal, then a revised and improved
basic feasible solution is obtained. This is done by exchanging a non-
basic variable for a basic variable.
For this, re-arrangement is made by transferring units from an
occupied cells to an empty cell that has the largest opportunity cost
and then adjusting the units in other related cells in a way that all the
rim requirements are satisfied. This is done by first tracing a closed
loop.
The solution obtained is again tested for optimality and revised, if
necessary.
We continue this process until an optimal solution is finally obtained.
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Stepping-stone Method (Using NW Corner Rule)
P Q R S Supply
A 12 10 12 13 500
B 7 11 8 14 300
C 6 16 11 7 200
Demand 180 150 350 320 1000
From
To
180 150 170
180 120
200
Total Cost = 12 x 180 + 10 x 150 + 12 x 170 + 8 x 180 + 14 x 120 + 7 x 200= Rs. 10,220
Initial Feasible Solution: Testing for Optimality
+
_
_
+
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Stepping-stone Method..
A shipment of one item on the route AS would cause an
increase of Rs. 13 but it will also mean a reduction of one
unit from AR and thereby, a reduction of Rs. 12.
Further, an increase of a unit in BR would raise the cost by
Rs. 8 while a unit lesser in BS would save Rs. 14.
The net effect of the operation would be a saving of Rs. 5 (13
12 + 8 - 14).
This implies that a reduction of Rs. 5 can be effected by
adopting the route AS.
The opportunity cost of this route is Rs. 5.
Since the opportunity cost is positive, it means that it is
worth considering making cell AS a basic variable.
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Stepping-stone Method..
In the same manner, evaluate each of the remaining empty
cells as follows:
CR CR - CS - BS - BR 11 - 7 + 14 - 8 = 10 -10
Cell Closed Loop Net Cost Change Opportunity Cost
AS AS - AR + BR - BS 13- 12 + 8 -14 = -5 5
BP BP - BR - AR - AP 7 - 8 + 12 -12 = -1 1
BQ BQ - BR - AR - AQ 11 - 8 + 12 - 10 = 5 -5
CP CP - CS - BS - BR - AR - AP 6 - 7 + 14 - 8 + 12 - 12 = 5 -5
CQ CQ - CS - BS - BR - AR - AQ 16 - 7 + 14 - 8 + 12 - 10 = 17 -17
S i h d
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Tracing A Closed Loop To draw a close loop, always begin with an empty cell and move alternatively
horizontally and vertically, through occupied cell only, until reaching back
to the starting point.
In the process of moving from one occupied cell to another
a) Move only horizontally or vertically, but never diagonally;
b) Step over empty and if the need be, over occupied cells withoutchanging them.
A closed loop would always have corners only on the occupied cells.
Having traced the path, place plus and minus signs alternately in the cells
on each turn of the loop, beginning with a plus (+) sign in the empty cell.
An important restriction is that there must be exactly one cell with a plus
sign and one cell with a minus sign in any row or column in which the loop
takes a turn.
This restriction ensures that the rim requirements would not be violated
when units are shifted among cell.
Stepping-stone Method..
S i M h d
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An even number of at least four cells must participate in a
closed loop and an occupied cell can be considered only once
and not more.
If there exists a basic feasible solution with m + n 1
positive variables, then there would be one and only one
closed loop for each cell. This is irrespective of the size of
the matrix given.
All cells that receive a plus or a minus sign, except the
starting empty cell, must be the occupied cells.
Closed loop may or may not be square or rectangular in
shape.
Tracing A Closed Loop
Stepping-stone Method..
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Stepping-stone Method..
CR CR - CS - BS - BR 11 - 7 + 14 - 8 = 10 -10
Cell Closed Loop Net Cost Change Opportunity Cost
AS AS - AR + BR - BS 13- 12 + 8 -14 = -5 5
BP BP - BR - AR - AP 7 - 8 + 12 -12 = -1 1
BQ BQ - BR - AR - AQ 11 - 8 + 12 - 10 = 5 -5
CP CP - CS - BS - BR - AR - AP 6 - 7 + 14 - 8 + 12 - 12 = 5 -5
CQ CQ - CS - BS - BR - AR - AQ 16 - 7 + 14 - 8 + 12 - 10 = 17 -17
Continuing with testing optimality, the rule is: if none of the emptycells has a positive opportunity cost, the solution is optimal.
The solution in question, therefore, is not optimal.
Here, the most favorable cell is AS for it has the largest opportunity
cost equal to 5. So we will include AS as a basic variable in the solution.
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P Q R S Supply
A 12 10 12 13 500
B 7 11 8 14 300
C 6 16 11 7 200
Demand 180 150 350 320 1000
From
To
180 150 170
180 120
200
Total Cost = 12 x 180 + 10 x 150 + 12 x 50 + 13 x 120 + 8 x 300 + 7 x 200= Rs. 9,620
Improved Solution: Non-Optimal
Stepping-stone Method..
50
300
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Stepping-stone Method..
This solution involves a total cost of Rs. 9,620 which is lower by Rs.
600 (5 x 120) in comparison to the initial solution.
We again apply step 2 to determine optimality.
CR CR - CS - AS AR 11 - 7 + 13 - 12 = 5 -5
Cell Closed Loop Net Cost Change Opportunity Cost
BS BS - AS - AR - BR 14- 13 + 12 - 8 = 5 -5
BP BP - BR - AR AP 7 - 8 + 12 -12 = -1 1
BQ BQ - BR - AR - AQ 11 - 8 + 12 - 10 = 5 -5
CP CP - CS - AS - AP 6 - 7 + 13 - 12 = 0 0
CQ CQ - CS
AS - AQ 16 - 7 + 13 - 10 = 12 -12
The solution is also not optimal here as Cell BP has the positive
opportunity cost equal to 1.
So we will include BP as a basic variable in the solution.
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P Q R S Supply
A 12 10 12 13 500
B 7 11 8 14 300
C 6 16 11 7 200
Demand 180 150 350 320 1000
From
To
180 150 50
300
120
200
Total Cost = 10 x 150 + 12 x 230 + 13 x 120 + 7 x 180 + 8 x 120 + 7 x 200= Rs. 9,440
Improved Solution: Optimal
Stepping-stone Method..
230
120
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Stepping-stone Method..
This solution involves a total cost of Rs. 9,440.
We again apply step 2 to determine optimality.
CR CR - CS - AS AR 11 - 7 + 13 - 12 = 5 -5
Cell Closed Loop Net Cost Change Opportunity Cost
BS BS - AS - AR - BR 14- 13 + 12 - 8 = 5 -5
AP AP - AR - BR BP 12 - 12 + 8 - 7 = 1 -1
BQ BQ - BR - AR - AQ 11 - 8 + 12 - 10 = 5 -5
CP CP - CS - AS AR BR BP 6 - 7 + 13 12 + 8 - 7 = 1 -1
CQ CQ - CS
AS - AQ 16 - 7 + 13 - 10 = 12 -12
Since the opportunity cost of all the empty cells are negative, the
solution obtained is optimal.
A carefully look at the solution reveals that it is identical to theinitial solution obtained by VAM.
d f d b h d( )
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Modified Distribution Method(MODI)
It is an efficient method of testing the optimality of a
transportation solution.
It avoids the kind of extensive scanning and reduces the
number of steps required in the evaluation of the empty cells
in case of stepping-stone method.
It gives a straightforward computational scheme whereby we
can determine the opportunity cost of each of the empty
cells.
difi d i ib i h d ( )
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Modified Distribution Method (MODI)
P Q R S Supply
A 12 10 12 13 500
B 7 11 8 14 300
C 6 16 11 7 200
Demand 180 150 350 320 1000
From
To
180
Total Cost = 12 x 180 + 10 x 150 + 12 x 170 + 8 x 180 + 14 x 120 + 7 x 200= Rs. 10,220
Initial Feasible Solution: Testing for Optimality
ui
0
vj
150 170
180 120
200
12 10 12
- 4
18
- 11
Having determined all ui and vj values, calculate for each unoccupied cell ij = ui +
vj cij. The ijs represent the opportunity costs of various cells. After obtaining theopportunity costs, proceed in the same way as in the stepping stone method.
+5
+1 -5
-5 -17 -10
difi d i ib i h d ( O I)
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Modified Distribution Method (MODI)
P Q R S Supply
A 12 10 12 13 500
B 7 11 8 14 300
C 6 16 11 7 200
Demand 180 150 350 320 1000
From
To
180
Here the cell AS has the largest positive opportunity cost. Select AS for inclusion asa basic variable.
Initial Feasible Solution: Testing for Optimality
+
_
_
+
ui
0
vj
150 170
180 120
200
12 10 12
- 4
18
- 11
+5
+1 -5
-5 -17 -10
M difi d Di ib i M h d (MODI)
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Modified Distribution Method (MODI)
P Q R S Supply
A 12 10 12 13 500
B 7 11 8 14 300
C 6 16 11 7 200
Demand 180 150 350 320 1000
From
To
180
Initial Feasible Solution: Testing for Optimality
ui
0
vj
150 170
180 120
200
12 10 12
- 4
18
- 11
50
300
Total Cost = 10 x 150 + 12 x 180 + 12 x 50 + 13 x 120 + 7 x 200 + 8 x 300= Rs. 9,620
M difi d Di ib i M h d (MODI)
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Modified Distribution Method (MODI)
P Q R S Supply
A 12 10 12 13 500
B 7 11 8 14 300
C 6 16 11 7 200
Demand 180 150 350 320 1000
From
To
180
Improved Solution: Non-Optimal
ui
0
vj
150 50
300
120
200
12 10 12
- 4
13
- 6
+1 -5 -5
0 -12 -5
Here the cell BP has the largest positive opportunity cost. Select BP for inclusion asa basic variable.
+
_
_
+
M difi d Di t ib ti M th d (MODI)
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Modified Distribution Method (MODI)
P Q R S Supply
A 12 10 12 13 500
B 7 11 8 14 300
C 6 16 11 7 200
Demand 180 150 350 320 1000
From
To
180
Improved Solution: Non-Optimal
ui
0
vj
150 50
300
120
200
12 10 12
- 4
13
- 6
230
120
Total Cost = 10 x 150 + 12 x 230 + 13 x 120 + 7 x 180 + 7 x 200 + 8 x 120= Rs. 9,440
M difi d Di t ib ti M th d (MODI)
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Modified Distribution Method (MODI)
P Q R S Supply
A 12 10 12 13 500
B 7 11 8 14 300
C 6 16 11 7 200
Demand 180 150 350 320 1000
From
To
180
Improved Solution: Optimal
ui
0
vj
150 120
200
11 10 12
- 4
13
- 6
230
120
-1
-5 -5
-1 -12 -5
Here all empty cells are less than or equal to zero. So the solution is found to beoptimal. Since all ij values are negative, the solution is also unique.
Total Cost = 10 x 150 + 12 x 230 + 13 x 120 + 7 x 180 + 7 x 200 + 8 x 120
= Rs. 9,440
U b l d T t ti P bl
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Unbalanced Transportation Problems
Sometimes aggregate supply does not match the aggregate demand.
Such problems are called unbalanced transportation problems.
Balancing must be done before they can be solved.
When the aggregate supply exceeds the aggregate demand, a column
of slack variables is added to the transportation tableau which
represents a dummy destination with a requirement equal to theamount of excess supply and the transportation costs equal to zero.
On the other hand, when the aggregate demand exceeds the
aggregate supply, balance is restored by adding a dummy origin, The
row representing it is added with an assumed total availability equal
to the difference between the total demand and supply and witheach of the cells having a zero unit cost.
In some cases, however, when the penalty of not satisfying the
demand at a particular destination(s) is given, then such penalty
value should be considered and not zero.
P hibit d R t
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Prohibited Routes
Sometimes in a given transportation problem some route(s)may not be available due to unfavorable weather conditions
on a particular route, strike on a particular route etc.
To handle a situation of this type, we assign a very large cost
represented by M to each of such routes which are not
available.
The effect of adding a large cost element would be that such
routes would automatically be eliminated in the final
solution.
U i M lti l O ti l S l ti
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Unique vs Multiple Optimal Solutions
The solution in case of transportation problem will be
optimal if all the ij
values are less than or equal to zero.
The solution will be unique if all the ij values are negative.
If some cell (or cells) has ij = 0, then multiple optimal
solutions are indicated so that there exist transportation
pattern(s) other than the one obtained which can satisfy allthe rim requirements for the same cost.
To obtain an alternate optimal solution, traced a closed loop
beginning with a cell having ij = 0 and get the revised
solution in the same way as a solution is improved.
This revised solution would also entail the same total cost as
before.
Transportation Problem
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WarehouseMarket
SupplyA 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
Transportation Problem
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. Is the optimal solution obtained by
you unique? If not, what is/are the other optimalsolution/s?
Maximization Problem
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Maximization Problem
A transportation tableau may contain unit profits instead of unit costs and the
objective function be maximization of total profits.
In this case the given problem is first converted into an equivalentminimization problem by subtracting each element of the given matrix from a
constant value k, which can be any number. Generally we take the largest one.
The procedure yields what is termed as the opportunity loss matrix to which
the transportation method is applied.
The entries in the opportunity loss matrix indicate how much each of the
values is away from that constant. The minimization of the opportunity loss
automatically leads to maximization of total profit.
The total profit value is obtained by using the transportation pattern of the
solution on the unit profit matrix and not through the opportunity loss matrix.
If a maximization type of transportation problem is unbalanced, then it should
be balanced by introducing the necessary dummy row/column for a
source/destination, before converting it into a minimization problem.
Similarly, if such a problem has a prohibited route, then the payoff element for
such a route should be substituted by M before proceeding to convert to
minimization type.
Transportation Maximization Problem
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Transportation Maximization Problem
Solve the problem for maximum profit.
Per Unit Profit (Rs.)
Market
A B C D
Warehouse
X 12 18 6 25
Y 8 7 10 18
Z 14 3 11 20
Available at warehouses:
X: 200 units
Y: 500 unitsZ: 300 units
Demand in the markets:
A: 180 units
B: 320 unitsC: 100 units
D: 400 units
Transhipment Problems
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Transhipment Problems
Sometimes multi-plant firms find it necessary to send some
goods from one plant to another in order to meet the
substantial increase in the demand in the second market. Thesecond plant here would act both as a source and a destination
and there is no real distinction between source and
destination.
A transportation problem is regarded as a transhipmentproblem when shipment of goods is allowed from one source to
another and from one destination to another.
A transportation problem with m-origins and n-destinations
becomes a transhipment problem with m + n sources and an
equal number of destinations.
Fortunately, the optimal solution to a transhipment problem
can be found by solving a transportation problem.
Transhipment Problem
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Transhipment ProblemA 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 of 180, 150, 350 and
320 units respectively. Per unit shipping charges on different routes are givenbelow.
To: P Q R S
From A: 12 10 12 13
From B: 7 11 8 14
From C: 6 16 11 7
How should the firm route the product to achieve the target of minimum cost?
From
Plant
To Plant
A B CA 0 2 8
B 5 0 7
C 10 9 0
Additional cost data:
From
Warehouse
To Warehouse
P Q R SP 0 6 5 8
Q 8 0 3 2
R 5 4 0 10
S 9 4 8 0
Modified Distribution Method (MODI)
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Modified Distribution Method (MODI)
P Q R S Supply
A 12 10 12 13 500
B 7 11 8 14 300
C 6 16 11 7 200
Demand 180 150 350 320 1000
From
To
180
Optimal Solution (Transportation Model)
150 120
200
230
120
Total Cost = 10 x 150 + 12 x 230 + 13 x 120 + 7 x 180 + 7 x 200 + 8 x 120= Rs. 9,440
Transhipment Problem
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Transhipment Problem
A B C P Q R S Supply
A 12 10 12 13 500
B 7 11 8 14 300
C 6 16 11 7 200
P
Q
R
S
Demand 180 150 350 320 1000
Arranging the tableau
150 230 120
180 120
200
Transhipment Problem
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Transhipment Problem
A B C P Q R S Supply
A 0 2 8 12 10 12 13 500
B 5 0 7 7 11 8 14 300
C 10 9 0 6 16 11 7 200
P 12 7 6 0 6 5 8 0
Q 10 11 16 8 0 3 2 0
R 12 8 11 5 4 0 10 0
S 13 14 7 9 4 8 0 0
Demand 0 0 0 180 150 350 320 1000
Initial Feasible Solution: Non-optimal
0150 230 120180 120
200
ui
vj
- 0
- 0
- 0
- 0
- 0
- 0
- 0
10 12 13
-4-6
11
-11-10-12-13
640Find out the opportunity cost of each unoccupied cell i.e.
ij = ui + vj
cij .For example opportunity cost of cell AB = 0 + 4 2 = 2
This method suggests that the route AB be brought into the solution. Trace aclosed loop and find out the revised solution.
+
_ +
_
Transhipment Problem
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Transhipment Problem
A B C P Q R S Supply
A 0 2 8 12 10 12 13 500
B 5 0 7 7 11 8 14 300
C 10 9 0 6 16 11 7 200
P 12 7 6 0 6 5 8 0
Q 10 11 16 8 0 3 2 0
R 12 8 11 5 4 0 10 0
S 13 14 7 9 4 8 0 0
Demand 0 0 0 180 150 350 320 1000
Revised Solution: Non-optimal
150 230 120
180 120
200
- 0
- 0
- 0
- 0
- 0
- 0
- 0
350-230
Total Cost = 2 x 230 + 10 x 150 + 13 x 120 + 7 x 180 + 7 x 200 + 8 x 350
= Rs. 8,980
Transhipment Problem
8/4/2019 Transportation Problems 5
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Transhipment Problem
A B C P Q R S Supply
A 0 2 8 12 10 12 13 500
B 5 0 7 7 11 8 14 300
C 10 9 0 6 16 11 7 200
P 12 7 6 0 6 5 8 0
Q 10 11 16 8 0 3 2 0
R 12 8 11 5 4 0 10 0
S 13 14 7 9 4 8 0 0
Demand 0 0 0 180 150 350 320 1000
Revised Solution: Non-optimal
0150230 120180
200
ui
vj
- 0
- 0
- 0
- 0
- 0
- 0
10 10 13
-2-6
9
-9-10-10-13
620
350-230
Find out the opportunity cost of each unoccupied cell i.e. ij = ui + vj cij .
For example opportunity cost of cell AC = 0 + 6 8 = -2
This method suggests that the route QS be brought into the solution. Trace a
closed loop and find out the revised solution.
+
_ +
_
Transhipment Problem
8/4/2019 Transportation Problems 5
49/50
Transhipment Problem
A B C P Q R S Supply
A 0 2 8 12 10 12 13 500
B 5 0 7 7 11 8 14 300
C 10 9 0 6 16 11 7 200
P 12 7 6 0 6 5 8 0
Q 10 11 16 8 0 3 2 0
R 12 8 11 5 4 0 10 0
S 13 14 7 9 4 8 0 0
Demand 0 0 0 180 150 350 320 1000
Revised Solution: optimal
150230 120
180
200
- 0
- 0
- 0
- 0
- 0
- 0
350-230
270
- 120
0ui
-2-5-9-10-10-12
vj10 10 129520
Find out the opportunity cost of each unoccupied cell. Now here all these
cells have negative opportunity costs. So the solution is optimal here.
Transhipment Problem
8/4/2019 Transportation Problems 5
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Transhipment Problem
Optimal Solution
Send from Plant A: 230 units to Plant B and 270 units
to Warehouse Q,
Send from Plant B: 180 units to warehouse P and 350
units to Warehouse R,
Send from Plant C: 200 units to Warehouse S and
Send from warehouse Q: 150 units to Warehouse S.
Total Cost = 2 x 230 + 10 x 270 + 7 x 180 + 8 x 350 + 7 x200 + 2 x 120 = Rs. 8,860
The transportation pattern would involve a total cost of Rs.
8860 resulting in a saving of Rs. 9440 8860 = Rs. 580 onaccount of the possibility of transhipment.