Since a feasible solution exists only for balanced problem, it is necessary that the total availability be made equal to the total demand. To solve an unbalanced transportation problem, first of all it is converted into balance transportation problem by introducing a fictitious origin or destination which will provide the surplus supply or demand

1. Unbalanced Transportation Problem
Dr. Deepa Chauhan
Unbalanced
Transportation Problems
By
Dr. Deepa Chauhan
Associate Professor,
Applied Science & Humanities Department,
Axis Institute of Technology, Kanpur

2. Unbalanced Transportation Problem
Dr. Deepa Chauhan
The Unbalanced Transportation Problem
If the total availability from all origins is equal to the total demand at all the destination i.e.,
, then this problem is called balanced transportation problem. In many real life
situations, however, the total availability may not be equal to total demand i.e.,
, such problems are called unbalanced transportation problems. In these problems
either some available resources will remain unused or some requirements will remain unfilled.
Since a feasible solution exists only for balanced problem, it is necessary that the total availability
be made equal to the total demand. To solve an unbalanced transportation problem, first of all it is
converted into balance transportation problem by introducing a fictitious origin or destination
which will provide the surplus supply or demand. The following two cases do arise
 If the total capacity or availability is more than the total demand i.e. , we add a
dummy destination to take up the excess capacity and the costs of transportation to this
destination are set equal to zero. The zero cost cells are treated the same way as real cost cells
and the problem is solved as a balanced problem.
 In case the total demand is more than availability i.e. , we add a dummy origin
(source) to fill the balance requirement and the transportation costs are again set equal to zero.
However, in real life, the cost of unfilled demand is seldom zero since it may lose sales, lesser
profits, possibility of losing customers or even business or the use of more costly substitute.
Thus an unbalanced transportation problem is converted into balanced transportation problem
that can be solved by the method described earlier.

3. Unbalanced Transportation Problem
Dr. Deepa Chauhan
Example 1: Determine the optimal transportation plan from the following table giving the plant to
market shipping costs and quantities required at each market and available at each plant:
Plant ↓ Market Availability
W1 W2 W3 W4
F1 11 20 7 8 50
F2 21 16 10 12 40
F3 8 12 18 9 70
Requirement 30 25 35 40
Solution: We have
Total availability, =50+40+70=160
Total requirement, =30+25+35+40=130
Since , the given problem is unbalanced. To convert it into a balanced transportation
problem, we introduce a dummy market W5 with demand 160-130=30, such that the unit
transportation cost from each plant to market W5 is equal to zero.
Thus we obtain the following balanced TP:
Plant ↓ Market Availability
W1 W2 W3 W4 W5
F1 11 20 7 8 0 50
F2 21 16 10 12 0 40
F3 8 12 18 9 0 70
Requirement 30 25 35 40 30 Total=160

4. Unbalanced Transportation Problem
Dr. Deepa Chauhan
Using Vogel’s method, an initial basic feasible solution is obtained as
Table-1
Plant ↓ Market Availability Row
Penalties
W1 W2 W3 W4 W5
F1 11 20 7 8 0 50 7-0=7
F2 21 16 10 12 0 (30) 40 10-0=10
F3 8 12 18 9 0 70 8-0=8
Requirement 30 25 35 40 30 Total=160
Column
Penalties
11-8=3 16-12=4 10-7=3 9-8=1 0-0=0
Table-2
Plant ↓ Market Availability Row
Penalties
W1 W2 W3 W4
F1 11 20 7 8 50 1
F2 21 16 10 12 10 2
F3 8 12 (25) 18 9 70 1
Requirement 30 25 35 40 Total=160
Column
Penalties
3 4 3 1
Table-3
Plant ↓ Market Availability Row
Penalties
W1 W3 W4
F1 11 7 8 50 1
F2 21 10 12 10 2
F3 8 (30) 18 9 45 1
Requirement 30 35 40 Total=160
Column
Penalties
3 3 1

5. Unbalanced Transportation Problem
Dr. Deepa Chauhan
Table-4
Table-5
Plant ↓ Market Availability Row
Penalties
W3 W4
F1 7 8 (25) 50 1
F2 10 12 10 2
Requirement 35 25 Total=160
Column
Penalties
3 4
Table-6
Plant ↓ Market Availability Row
Penalties
W3
F1 7 (25) 25 1
F2 10 (10) 10 2
Requirement 35 Total=160
Column
Penalties
3
Plant ↓ Market Availability Row
Penalties
W3 W4
F1 7 8 50 1
F2 10 12 10 2
F3 18 9 (15) 15 9
Requirement 35 40 Total=160
Column
Penalties
3 1

6. Unbalanced Transportation Problem
Dr. Deepa Chauhan
All allocations made during the above procedure are shown in below in the allocation matrix
Plant ↓ Market Availabilit
y
W1 W2 W3 W4 W5
F1 11 20 7 (25) 8 (25) 0 50
F2 21 16 10 (10) 12 0 (30) 40
F3 8 (30) 12 (25) 18 9 (15) 0 70
Requirement 30 25 35 40 30 Total=160
The total transportation cost corresponding to this BFS is given by
Z=7x25+8x25+10x10+0x30+8x30+12x25+9x15
₹1150
Here m+n-1=3+5-1=7
No. of allocations=7
Since m+n-1= No. of allocations, this BFS is non-degenerated.
Hence the optimality test can be performed.
Plant ↓ Market Availabilit
y
W1 W2 W3 W4 W5
F1 11 20 7 (25) 8 (25) 0 50
F2 21 16 10 (10) 12 0 (30) 40
F3 8 (30) 12 (25) 18 9 (15) 0 70
Requirement 30 25 35 40 30 Total=160
Now we determine a set of values ui and vj such that cij=ui+vj for each occupied cell
u1+v3=7
u1+v4=8
u2+v3=10
u2+v5=0
u3+v1=8
u3+v2=12

7. Unbalanced Transportation Problem
Dr. Deepa Chauhan
u3+v4=9
Let u3=0 (max allocation), v1=8, v2=12, v4=9, u1= - 1, v3=8, u2=2, v5= -2
Fill the vacant cells with dij=cij-(ui+vj)
Plant ↓ Market ui
W1 W2 W3 W4 W5
F1 11-(-1+8) 20-(-1+12) 0-(-1-2) -1
F2 21-(2+8) 16-(2+12) 12-(2+9) 2
F3 18-(0+8) 0-(0-2) 0
vj 8 12 8 9 -2
Plant ↓ Market Availability
W1 W2 W3 W4 W5
F1 4 9 3 50
F2 11 2 1 40
F3 10 2 70
Requirement 30 25 35 40 30 Total=160
Since all dij>0, the BFS under test is a unique optimal solution.
Zmin=₹ 1150 Answer

8. Unbalanced Transportation Problem
Dr. Deepa Chauhan
Example 2: Solve the following transportation problem
A B C Supply
X 11 21 16 14
Y 7 17 13 26
Z 11 23 21 36
Demand 18 28 25
Solution: We have
Total supply, 1 =76
Total demand, 1 5 =71
Since , the given problem is unbalanced. To convert it into a balanced transportation
problem, we introduce a dummy column D with demand 76-71=5, such that the unit transportation
cost is equal to zero.
Thus we obtain the following balanced TP:
A B C D Supply
X 11 21 16 0 14
Y 7 17 13 0 26
Z 11 23 21 0 36
Demand 18 28 25 5 Total=75
Using Vogel’s method, an initial basic feasible solution is obtained as
A B C D Supply Row
penalties
X 11 21 16 0 (5) 14 11
Y 7 17 13 0 26 7
Z 11 23 21 0 36 11
Demand 18 28 25 5 Total=75
Column
penalties
4 4 3 0

9. Unbalanced Transportation Problem
Dr. Deepa Chauhan
Reduced table is
A B C Supply Row
penalties
X 11 21 16 9 5
Y 7 17 13 26 6
Z 11 (18) 23 21 36 10
Demand 18 28 25 Total=75
Column
penalties
4 4 3
Reduced table is
B C Supply Row
penalties
X 21 16 (9) 9 5
Y 17 13 26 4
Z 23 21 18 2
Demand 28 25 Total=75
Column
penalties
4 3
Reduced table is
B C Supply Row
penalties
Y 17 13 (16) 26 4
Z 23 21 18 2
Demand 28 16 Total=75
Column
penalties
6 8

10. Unbalanced Transportation Problem
Dr. Deepa Chauhan
Reduced table is
B Supply
Y 17 (10) 10
Z 23 (18) 18
Demand 28 Total=75
All allocations made during the above procedure are shown in below in the allocation matrix
A B C D Supply
X 11 21 16 (9) 0 (5) 14
Y 7 17 (10) 13 (16) 0 26
Z 11 (18) 23 (18) 21 0 36
Demand 18 28 25 5 Total=75
Total transportation cost=144+0+170+208+198+414 ₹1134
Here m+n-1=3+4-1=6
No. of allocations=6
Since m+n-1= No. of allocations, this BFS is non-degenerated.
Hence the optimality test can be performed.
Now we determine a set of values ui and vj such that cij=ui+vj for each occupied cell
u1+v3=16
u1+v4=0
u2+v2=17
u2+v3=13
u3+v1=11
u3+v2=23
Let v1=0, then

11. Unbalanced Transportation Problem
Dr. Deepa Chauhan
u3=11, v2=12, u2=5, v3=8, u1=8, v4= - 8
Fill the vacant cells with dij=cij-(ui+vj)
A B C D Supply
X 11-(8) 21-(20) 0 (5) 14
Y 7-(5) 0-(-3) 26
Z 21-(19) 0-(3) 36
Demand 18 28 25 5 Total=75
A B C D Supply
X 3 1 16 (9) 0 (5) 14
Y 2 17 (10) 13 (16) 3 26
Z 11 (18) 23 (18) 2 -3* 36
Demand 18 28 25 5 Total=75
This matrix is called cell evaluation matrix
Since one cell value is –ve (d34<0), the basic feasible solution is not optimal.
In the cell evaluation matrix, identify the cell with most negative entry. It is the cell (3,4)
Write down again the feasible solution
A B C D Supply
X 11 21 16 (9) 0 (5) 14
Y 7 17 (10) 13 (16) 0 26
Z 11 (18) 23 (18) 21 0* 36
Demand 18 28 25 5 Total=75

12. Unbalanced Transportation Problem
Dr. Deepa Chauhan
Trace a closed path in matrix. Mark the identified cell as positive and each occupied cell at corners
of the path alternately-ve, +ve, -ve and so on
A B C D Supply
X
11 21
16 (9)
+
0 (5)
-
14
Y
7
17 (10)
+
- 13
(16)
0
26
Z
11 (18)
23 (18)
-
21 +*
36
Demand 18 28 25 5 Total=75
A B C D Supply
X
11 21
16 (9+5) 0 (5-5)
14
Y
7
17 (10+5)
13 (16-5) 0
26
Z
11 (18)
23 (18-5)
21 0 (+5)
36
Demand 18 28 25 5 Total=75

13. Unbalanced Transportation Problem
Dr. Deepa Chauhan
A B C D Supply
X
11 21
16 (14) 0
14
Y
7
17 (15)
13 (11) 0
26
Z
11 (18)
23 (13)
21 0 (5)
36
Demand 18 28 25 5 Total=75
Now transportation cost of this second feasible solution
Z= 55 1 19 99 ₹ 1119
Now we determine a set of values ui and vj such that cij=ui+vj for each occupied cell
u1+v3=16
u2+v2=17
u2+v3=13
u3+v1=11
u3+v2=23
u3+v4=0
Let v1=0, then
u3=11, v2=12, u2=6, v3=7, u1=9, v4= -11

14. Unbalanced Transportation Problem
Dr. Deepa Chauhan
Fill the vacant cells with dij=cij-(ui+vj)
A B C D Supply
X
11-(9) 21-(21)
16 (14) 0-(-2)
14
Y
7 –(6)
17 (15)
13 (11) 0-(-9)
26
Z
11 (18)
23 (13)
21-(18) 0 (5)
36
Demand 18 28 25 5 Total=75
A B C D Supply
X
2 0
2
14
Y
1 9
26
Z
3
36
Demand 18 28 25 5 Total=75
Since all dij≥0, second feasible solution is optimal solution of this transportation problem.
Zmin ₹ 1119