technical analysis
TRANSCRIPT
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UN BALANCED TRANSPORTATION
PROBLEMMANEESH P
DEPT. OF APPLIED ECONOMICS
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The basic transportation problem was developed in 1941 by F.I.
Hitchaxic. However it could be solved for optimally as an answer to
complex business problem only in 1951,when Geroge B. Dantzig
applied the concept of Linear Programming in solving the
Transportation models.
Transportation problems are primarily concerned with the optimal
(best possible) way in which a product produced at different factories
or plants (called supply origins) can be transported to a number of
warehouses (called demand destinations).
INTRODUCTION
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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.
The objective in a transportation problem is:-To fully satisfy the destination requirements within the operating production capacity constraints at the minimum possible cost.
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Whenever there is a physical
movement of goods from the point
of manufacture to the final
consumers through a variety of
channels of distribution
(wholesalers, retailers, distributors
etc.), there is a need to minimize the
cost of transportation so as to
increase the profit on sales.
Transportation problems arise in all
such cases.
.
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It aims at providing assistance to the top management in ascertaining how many units of a particular product should be transported from each supply origin to each demand destinations to that the total prevailing demand for the company’s product is satisfied, while at the same time the total transportation costs are minimized
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i.e.; The total supply available at the plants exactly matches
the total demand at the destinations. Hence, there is neither
excess supply nor excess demand.
Such type of problems where supply and demand are
exactly equal are known as Balanced Transportation
Problem.
Supply (from various sources) are written in the rows,
while a column is an expression for the demand of different
warehouses. In general, if a transportation problem has m
rows an n columns, then the problem is solvable if there are
exactly (m + n –1) basic variables
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.UNBALANCED TRANSPORTATION PROBLEM :
A transportation problem is said to be unbalanced if the
supply and demand are not equal.
Two situations are possible:-
1. If Supply < demand, a dummy supply variable is
introduced in the equation to make it equal to demand.
2. If demand < supply, a dummy demand variable is
introduced in the equation to make it equal to supply.
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Then before solving we must balance the demand & supply.
When supply exceeds demand, the excess supply is assumed to go to
inventory. A column of slack variables is added to the transportation
table which represents dummy destination with a requirement equal to
the amount of excess supply and the transportation cost equal to zero.
When demand exceeds 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 total demand & supply and
with each cell having a zero unit cost.
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Demand Less Than Supply
Suppose that a plywood factory increases its rate of production from 100 to 250 desks
The firm is now able to supply a total of 850 desks each period
Warehouse requirements remain the same (700) so the row and column totals do not balance
We add a dummy column that will represent a fake warehouse requiring 150 desks
This is somewhat analogous to adding a slack variable We use the northwest corner rule and either vogel’s
approximation method or MODI to find the optimal solution
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FROM
TO
A B C TOTAL AVAILABLE
I 5 4 3 250
II 8 4 3 300
III 9 7 5 300
WAREHOUSE REQUIREMENTS
300 200 200 850
700
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FROM
TO
A B C D TOTAL AVAILABLE
I 5 4 3 0 250
II 8 4 3 0 300
III 9 7 5 0 300
WAREHOUSE REQUIREMENTS
300 200 200 150 850= 850
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FROM
TO
A B C D TOTAL AVAILABLE
I5 4 3 0
250
II8 4 3 0
300
III9 7 5 0
300
WAREHOUSE REQUIREMENTS
300 200 200 150
(1)
(1)
(2)
(3) (0) (0) (0)
250
300-250=50
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FROMA B C D
TOTAL AVAILABLE
II 8 4 3 0 300
III 9 7 5 0 300
WAREHOUSE
REQUIREMENTS
50 200 200 150
TO
(1)
(2)
(1) (3) (2) (0)
200
300-200=100
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FROM
A C DTOTAL
AVAILABLE
II 8 3 0 100
III 9 5 0 300
WAREHOUSE
REQUIREMENTS
50 200 150
TO
100 (5)
(4)
(1) (2) (0)
200-100=100
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FROMA C D
TOTAL AVAILABLE
III 9 5 0 300
WAREHOUSE REQUIREMENTS 50 100 150
TO
(4)
(9) (5) (0)
50
300-50=250
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FROMC D
TOTAL AVAILABLE
III 5 0 250
WAREHOUSE REQUIREMENTS 100 150
TO
(5)
(5) (0)
150
250-150=100
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FROM
C
TOTAL AVAILABLE
III 5 100
WAREHOUSE REQUIREMENTS
100
TO
100
100-100=0
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FROMA B C D
I 5 4 3 0
II 8 4 3 0
III 9 7 5 0
TO
WAREHOUSE REQUIREMENTS
TOTAL AVAILABLE
300 200 200 150
250
300
300
250
200 100
50 100 150
m+n-1 = 3+4-1= 6
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FROMA B C D
I 5 4 3 0
II 8 4 3 0
III 9 7 5 0
TO
WAREHOUSE REQUIREMENTS
TOTAL AVAILABLE
300 200 200 150
250
300
300
250
200 100
50 100 150
u1
u2
U3
v1 v2 v3 v4
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U1+V1=5
U2+V2=4
U2+V3=3
U3+V1=9
U3+V3=5
U3+V4=0
ASSUMING U3=0
0+V1=9V1=9
0+V3=5V3=5
0+V4=0V4=0
U1+V1=5U1+9=5U1= 5-9 = -4
U2+V3=3U2+5=3U2= 3-5= -2
(V3=5)
U3+V4=0U3+0=0U3=0
U2+V2=4-2+V2=4V2= 4-(-2) = 6
(V4=0)
(U2= -2)
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CIJ TABLE
4 3 0
8 0
7
V1 V2 V3 V4
U1
U2
U3
-4
-2
0
0569
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uI-VJ TABLE
2 1 -4
7 -2
7
V1 V2 V3 V4
U1
U2
U3
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DIJ TABLE
2 2 4
1 2
0
V1 V2 V3 V4
U1
U2
U3
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Total cost = 250*5+200*4+100*3+50*9+100*5+150*0
1250+800+300+450+500+0 = 3,300
Initial basic feasible solution
Here all the dij values are positive, therefore the solution is optimal.
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Demand Greater than SupplyThe second type of unbalanced condition
occurs when total demand is greater than total supply
In this case we need to add a dummy row representing a fake factory
The new factory will have a supply exactly equal to the difference between total demand and total real supply
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FROM
A B CPLANTSUPPLY
W 6 4 9 200
X 10 5 8 175
Y 12 7 6 75
WAREHOUSEDEMAND 250 100 150
450
TO
500
Totals do not balance
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FROM
A B CPLANTSUPPLY
W 6 4 9 200
X 10 5 8 175
Y 12 7 6 75
Z 0 0 0 50
WAREHOUSEDEMAND 250 100 150
500
TO
500
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FROM
A B CPLANTSUPPLY
W 6 4 9 200
X 10 5 8 175
Y 12 7 6 75
Z 0 0 0 50
WAREHOUSEDEMAND 250 100 150
TO
200
0 100 75
50
75
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CIJ TABLE
4 9
12 7
0 0
V1 V2 V3
U1
U2
U3
U4
-4
0
-2
-10
10 5 8
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UI+VJ TABLE
1 4
8 3
-5 -2
V1 V2 V3
U1
U2
U3
U4
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DIJ TABLE
3 5
4 4
5 2
V1 V2 V3
U1
U2
U3
U4
DIJ= CIJ-(UI+VJ)
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Here, all the dij values are positive, therefore the solution is optimal.
TC= 200*6+0*10+100*5+75*8+75*6+50*0
1200+500+600+450= 2750
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CONCLUSION
In real-life problems, total demand is frequently not equal to total supplyThese unbalanced problems can be handled easily by introducing dummy sources or dummy destinationsIf total supply is greater than total demand, a dummy destination (warehouse), with demand exactly equal to the surplus, is created If total demand is greater than total supply, we introduce a dummy source (factory) with a supply equal to the excess of demand over supply
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Any units assigned to a dummy destination represent excess capacity
Any units assigned to a dummy source represent unmet demand
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THANK YOU FOR YOUR
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