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