min( 27, 25, 23, 32) = 23. Since the minimum cost is 23, the best alternative is E₃ according to minimax principle.

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

Max (5,7,12,15) = 15 ஃ Action A4 is the best

Total Capacity = Total Demand
\(\therefore\) The given problem is balanced transportation problem.
Hence there exists a feasible solution to the given problem.
Given Transportation Problem is

First Allocation:

Second Allocation:

Third Allocation:

Fourth Allocation:

Fifth Allocation:

Sixth Allocation:

Transportation schedule :
T⟶H,T⟶P,B⟶C,B⟶H,M⟶H,M⟶K
The total Transportation cost = ( 5×8) + (25×5)+ (35×5) + (5×11)+ (18×9) + (32×7)
= 40+125+175+55+162+224
= Rs.781

Total Supply = Total Demand = 24
\(\therefore\)The given problem is a balanced transportation problem.
Hence there exists a feasible solution to the given problem.
Given Transportation Problem is:

The least cost is 1 corresponds to the cells (O₁, D₁) and (O3, D₄)
Take the Cell (O₁, D₁) arbitrarily.
Allocatemin (6,4) = 4 units to this cell.

The reduced table is

The least cost corresponds to the cell (O₃, D₄). Allocate min (10,6) = 6 units to this cell.

The reduced table is

The least costis 2 corresponds to the cells (O₁, D₂), (O₂, D₃), (O₃, D₂), (O₃, D₃)
Allocate min (2,6) = 2 units to this cell.

The reduced table is

The least cost is 2 corresponds to the cells (O₂, D₃), (O₃, D₂), (O₃, D₃)
Allocate min ( 8,8) = 8 units to this cell.

The reduced table is

Here allocate 4 units in the cell (O₃, D₂)

Thus we have the following allocations:

Transportation schedule :
O₁⟶D₁, O₁⟶D₂,O₂⟶D₃,O₃⟶D₂,O₃⟶D₄
Total transportation cost
= (4×1)+ (2×2)+(8×2)+(4×2)+(6×1)
= 4+4+16+8+6
= Rs. 38.

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 total supply = 5 + 8 + 7 + 14 = 34, Total demand = 7 + 9 + 18 = 34
(i.e) Total supply =Total demand 
Therefore The given problem is balanced transportation problem.
\(\therefore\) we can findan initial basic feasible solution to the given problem.
From the above table we can choose the cell in the North West Corner. Here the cell is (1, A)
Allocate as much as possible in this cell so that either the capacity of first row is exhausted or the destination requirement of the first column is exhausted.
i.e. x₁₁ = min (5, 7) = 5

Reduced transportation table is

Now the cell in the North west corner is (2,  A)
Allocate as much as possible in the first cell so that either the capacity of second row is exhausted or the destination requirement of the first column is exhausted.
i.e. x₁₂ = min (2, 8) = 2

Reduced transportation table is

Here north west corner cell is (2, B) Allocate as much as possible in the first cell so that either the capacity of second row is exhausted or the destination requirement of the second column is exhausted.
i.e. x₂₂ = min (6, 9) = 6

Reduced transportation table is

Here north west corner cell is (3,B).
Allocate as much as possible in the first cell so that either the capacity of third row is exhausted or the destination requirement of the second column is exhausted.
i.e. x₃₂ = min (7, 3) = 3

Reduced transportation table is

Here north west corner cell is (3,C) Allocate as much as possible in the first cell so that either the capacity of third row is exhausted or the destination requirement of the third column is exhausted.
i.e. x₃₃ = min (4, 18) = 4

Reduced transportation table and final allocation is x₄₄ = 14

Thus we have the following allocations

Transportation schedule : 1⟶A, 2⟶B, 3⟶B, 3⟶C, 4⟶C
The total transportation cost.
= (5 \(\times\) 2) + (2 x\(\times\) 3) + (6 \(\times\) 3) + (3 \(\times\) 4) + (4 \(\times\) 7) + (14 \(\times\) 2)
= Rs. 102

Here \(\sum { { a }_{ i }=\sum { { b }_{ j } } =80 } \)
(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 Allocation:

Second Allocation:

Third Allocation:

Fourth Allocation:

Fifth Allocation:

Sixth Allocation:

Thus we have the following allocations:


Transportation schedule :
A⟶I, A⟶II, A⟶III, A⟶IV, B⟶I, C⟶IV, D⟶II
Total transportation cost:
= (6×5)+(6+1)+(17×3)+(5×3)+(15×3)+(12×3)+(19×1)
= 30+6+51+15+45+36+19
= Rs.202

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