Max (–10,–4, 0) = 0. Since the maximum payoff is 0, the alternative Z is selected by the businessman

Since the number of columns is less than the number of rows, given assignment problem is unbalanced one. To balance it , introduce a dummy column with all the entries zero. The revised assignment problem is

Here only 3 tasks can be assigned to 3 men.
Step 1: Its not necessary, since each row contains zero entry. Go to Step 2.
Step 2:

Step 3 (Assignment) :

Since each row and each columncontains exactly one assignment,all the three men have been assigned a task. But task S is not assigned to any Man. The optimal assignment schedule and total cost is

The optimal assignment (minimum) cost = Rs. 35

Given transportation table is

Total Availability = Total Requirement 
Therefore the given problem is balanced transportation problem.
Hence there exists a feasible solution to the given problem.
First allocation :

Second allocation :

Third Allocation :

Fourth Allocation :

Fifth allocation :

Final allocation :

Transportation schedule : O₁⟶D₁, O₁⟶D₂, O₂⟶D₂, O₂⟶D₃, O₃⟶D₃,O₃⟶D₃.
The transportation cost
= (6\(\times\)6)+(8\(\times\)4)+(2\(\times\)9)+(14\(\times\)2)+(1\(\times\)6)+(4\(\times\)2)
= Rs.128

Here the number of rows and columns are equal.
\(\therefore\) The given assignment problem is balanced.
Now let us find the solution.
Step 1: Select a smallest element in each row and subtract this from all the elements in its row.
The cost matrix of the given assignment problem is

Column 3 contains no zero. Go to Step 2.
Step 2: Select the smallest element in each column and subtract this from all the elements in its column.

Since each row and column contains atleast one zero, assignments can be made.
Step 3: (Assignment):
Examine the rows with exactly one zero. Row B contains exactly one zero. Mark that zero by \(\square\) (i.e) Person B is assigned to Job 1. Mark other zeros in its column by ×.

Now, Row C contains exactly one zero. Mark that zero by \(\square\). Mark other zeros in its column by × .

Now, Row D contains exactly one zero. Mark that zero by \(\square\) . Mark other zeros in its column by × .

Row E contains more than one zero, now proceed column wise. In column 1, there is an assignment. Go to column 2. There is exactly one zero. Mark that zero by \(\square\) . Mark other zeros in its row by × .

There is an assignment in Column 3 and column 4. Go to Column 5. There is exactly one zero. Mark that zero by \(\square\) . Mark other zeros in its row by × .

Thus all the five assignments have been made. The Optimal assignment schedule and total cost is

The optimal assignment (minimum) cost = Rs. 9

Here \(\sum { { a }_{ i }=\sum { { b }_{ j } } =950 } \)
(i.e) Total Availability = Total Requirement
\(\therefore\) The given problem is balanced transportation problem.
Hence there exists a feasible solution to the given problem.
First let us find the difference (penalty) between the first two smallest costs in each row and column and write them in brackets against the respective rows and columns

Choose the largest difference. Here the difference is 5 which corresponds to column D₁ and D₂. Choose either D₁ or D₂ arbitrarily.
Here we take the column D₁. In this column choose the least cost. Here the least cost corresponds to (S₁, D₁). Allocate min (250, 200) = 200 units to this Cell.
The reduced transportation table is 

Choose the largest difference. Here the difference is 5 whichcorresponds to column D₂. In this column choose the least cost. Here the least cost corresponds to (S₁, D₂) . Allocate min(50,175) = 50 units to this Cell.
The reduced transportation table is

Choose the largest difference. Here the difference is 6 which corresponds to column
D₂. In this column choose the least cost. Here the least cost corresponds to (S₂, D₂). Allocate min(300,175) =175 units to this cell.
The reduced transportation table is

Choose the largest difference. Here the difference is 4 corresponds to row S2. In this row choose the least cost. Here the least cost corresponds to (S2, D4). Allocate min(125, 250) = 125 units to this Cell.
The reduced transportation table is

The Allocation is

Thus we have the following allocations:

Transportation schedule :
S₁⟶D₁,S₁⟶D₂,S₂⟶D₂,S₂⟶D₄,S₃⟶D₃,S₃⟶D₄
This initial transportation cost
= (200 x 11)+(500 x 13)+(175 x 18)+(125 x 10)+(275 x 12)+(1255 x 10)
= Rs.12,075

If the total supply = total demand, then the given problem is a balanced transportation problem.

A feasible solution to a transportation problem is a set of non negative values x_ij (i = 1, 2, m, j = 1, 2, ....... n) that satisfies the constraints.
If a basic feasible solution to a transportation problem contains exactly m + n - l allocations. in independent positions, it is called a non degenerate basic feasible solution. Here m is the number of rows and n is the number of columns in a transportation problem.

The objective function is minimize Z = \(\overset { m }{ \underset { i=1 }{ \Sigma } } \overset { n }{ \underset { j=1 }{ \Sigma } } { { C }_{ ij } }{ x }_{ ij }\) subject to the constraints
\(\overset { n }{ \underset { j=1 }{ \Sigma } }{ x }_{ ij } = a_i, i=1,2,.....m\) (Supply constraints)
\(\overset { m }{ \underset { i=1 }{ \Sigma } }{ x }_{ ij } = b_j, j=1,2,.....n\) (demand constraints)
x_ij ≥, 0 for all i, j (non-negative restrictions)