First check whether the number of rows in equal to the numbers of columns, if it is so, the assignment problem is said to be balanced.
Step 1: Choose the least element in,each row and subtract it from all the elements of that row.
Step 2: Choose the least element in each column and subtract it from all the elements of that column. Step 2 has to be performed from the table olblained in step 1. 
Step 3: Check whether there is atleast o ir results in maximunm volume of sales. each row and each column and make an assignment as follows.
(i) Examine the rows successively until a row with exactly one zero is found. Mark that zero by \(\square\), that means an assignment is made there. Cross (x) all other zeros in its column. Continue this until all the rows have been examined.
(ii) Examine the columns successively until a columns with exactly one zero is found. Mark that zero by \(\square\), that means an assignment is made there. Cross (x) all other zeros in its row. Continue this until all the columns have been examined.
Step 4: If each row and each column contains exactly one assignment, then the solution is optimal.

The optimal solution is

Number of rows \(\neq\) number of columns.
Hence we add 2 dummy columns E and F.


The optimum allocation is

Total failure 118

Number of lines = 4 <5 (no. of rows)
 

Total allocation cost = 23. 

Here the number of rows and columns are equal.
∴ The given assignment problem is balanced.
Step 1 :
Select a minimum element in each row and subtract this from all the elements in its row.

Here IV column has no zero. Go to step 2.
Step 2:
Select the minimum 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:
Examine the rows with only one zero. Mark that. zero by ׈ and draw a vertical line.

Thus, all the assignments have been made.
The optimal assignment schedule and total cost is
 

Σa_i = 12 + 14 + 4 = 30
Σb_j = 9 + 10 + 11 = 30
Σa_i =  Σb_j
∴ The given problem is a balanced transportation problem.
Hence, there exists a feasible solution to the given problem.
I - allocation :

[∵ the highest penalty is 7, In C, least cost is 0 & min (11, 14) = 11] 
II - allocation :

[∵ the highest penalty is 4, In S1'least cost is 1& min (10, 12) = 10]
III - allocation :

[∵ In A, least cost is 2 & min (9, 3) = 3]
IV - allocation :

[∵ In A, least cost is 3 & min (6,4) = 4]
V - allocation :

[∵ min (2, 2) = 2]
Thus, the allocations are

∴ The transportation schedule is
S₁ → A, S₁ → B, S₂ → A, S₂ → C  S₃ → A
Hence, the total transportation cost is
= 2(5) + 10(1) + 3(2) + 11(0) + 4(3)
= 10 + 10 + 6 + 0 + 12 = Rs. 38

Here, total capacity = 7+ 9 + 18 = 34
total demand = 5 + 8 + 7 + 14 = 34
Total capacity = Total demand
∴ The given problem is a balanced transportation problem.
Hence, there exists a feasible solution to the given problem,
I - allocation:

[∵ The least cost is 8 & min (8, 18) = 8]
II - allocation:

[∵ The least cost is 7 & min (7,14) = 7]
III - allocation:

[∵ The least cost is 20 & min (5, 9) = 5]
IV - allocation:

[∵ The least cost is 40 & min (4, 7) = 4]
VI - allocation:

[∵  min (3, 3) = 1]
Thus, the allocations are

∴ The transportation schedule is
O₁ → D₁, O₂ → A₁, O₂ → C₁, O₃ → B_{1 ,} O₃ → C_1, O₃ → D₁,
Hence, the total transportation cost is
= 7(10) + 5(20) + 4(40) + 8(8) + 3(70) + 7(20)
= 70+ 100+ 160+64+210+ 140 = Rs. 744 

Here total supply = 300 + 400 + 500 = 1200
Total demand = 250 + 350 + 400 + 200 =1200
∴ Total supply = total demand
The given problem is a balanced transportation problem.
Hence, there exists a feasible solution to the given problem
I-allocation:

[∵ Min (250, 300) = 250]
II-allocation:

[∵ Min (50,350) = 50]
III-allocation:

[∵ Min (300,400) = 300]
IV-allocation:

[∵ Min (400,100) = 100]
V-allocation:

[∵ Min (300, 500) = 300]
VI-allocation:

[∵ Min (200, 200) = 200]
Thus, the allocations are

∴ The transportation schedule is
A → P, A → Q, B → Q, B → R, C → R, C → S
Hence, the total transportation cost is
= 250 (3) + 50 (1) + 300 (6) + 100 (5) + 300 (3) + 200 (2)
= 750 + 50 + 1800 + 500 + 900 + 400
= Rs. 4400

(i) Max (50, 10, 5) = 50
∴ A is the best alternative under maximin principle
(ii) Min (100, 80,40) = 40
∴ C is the best alternative under minimax principle