unbalanced transportation problem

1. MANEESH P
DEPT. OF APPLIED ECONOMICS

2. 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

3. 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.

4. 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.
.

5. 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

6. 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

7. .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.

8. 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.

9. 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

10. 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

11. 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

12. 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
(1)
(1)
(2)
(3) (0) (0) (0)
250
300-250=50

13. FROM
A 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

14. FROM
A C D
TOTAL
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

15. FROM
A C D
TOTAL
AVAILABLE
III 9 5 0 300
WAREHOUSE
REQUIREMENTS 50 100 150
TO
(4)
(9) (5) (0)
50
300-50=250

16. FROM
C D
TOTAL
AVAILABLE
III 5 0 250
WAREHOUSE
REQUIREMENTS 100 150
TO
(5)
(5) (0)
150
250-150=100

17. FROM
C
TOTAL
AVAILABLE
III 5 100
WAREHOUSE
REQUIREMENTS
100
TO
100
100-100=0

18. FROM
A 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

19. FROM
A 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

20. U1+V1=5
U2+V2=4
U2+V3=3
U3+V1=9
U3+V3=5
U3+V4=0
ASSUMING U3=0
0+V1=9
V1=9
0+V3=5
V3=5
0+V4=0
V4=0
U1+V1=5
U1+9=5
U1= 5-9 = -4
U2+V3=3
U2+5=3
U2= 3-5= -2
(V3=5)
U3+V4=0
U3+0=0
U3=0
U2+V2=4
-2+V2=4
V2= 4-(-2) = 6
(V4=0)
(U2= -2)

21. CIJTABLE
4 3 0
8 0
7
V1 V2 V3 V4
U1
U2
U3
-4
-2
0
0569

22. uI-VJ TABLE
2 1 -4
7 -2
7
V1 V2 V3 V4
U1
U2
U3

23. DIJ TABLE
2 2 4
1 2
0
V1 V2 V3 V4
U1
U2
U3

24. 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.

25. Demand Greater than Supply
 The 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

26. FROM
A B C
PLANT
SUPPLY
W 6 4 9 200
X 10 5 8 175
Y 12 7 6 75
WAREHOUSE
DEMAND 250 100 150
450
TO
500
Totals do not
balance

27. FROM
A B C
PLANT
SUPPLY
W 6 4 9 200
X 10 5 8 175
Y 12 7 6 75
Z 0 0 0 50
WAREHOUSE
DEMAND 250 100 150
500
TO
500

28. FROM
A B C
PLANT
SUPPLY
W 6 4 9 200
X 10 5 8 175
Y 12 7 6 75
Z 0 0 0 50
WAREHOUSE
DEMAND 250 100 150
TO
200
0 100 75
50
75

29. CIJ TABLE
4 9
12 7
0 0
V1 V2 V3
U1
U2
U3
U4
-4
0
-2
-10
10 5 8

30. UI+VJ TABLE
1 4
8 3
-5 -2
V1 V2 V3
U1
U2
U3
U4

31. DIJ TABLE
3 5
4 4
5 2
V1 V2 V3
U1
U2
U3
U4
DIJ= CIJ-(UI+VJ)

32. 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

33. CONCLUSION
In real-life problems, total demand is frequently not
equal to total supply
These unbalanced problems can be handled easily by
introducing dummy sources or dummy destinations
If 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

34. Any units assigned to a dummy destination represent
excess capacity
Any units assigned to a dummy source represent unmet
demand

35. THANK YOU FOR YOUR
ATTENTION!!!