First allocation:

Second allocation:

Third allocation:

Fourth allocation:

Fifth allocation:

Final allocation:

The transportation cost is
\( = (30 \times 6)+(5 \times 5)+(28 \times 11)+ (7 \times 9)+(25 \times 7)+(25 \times 13) \)
= 180 + 25 + 308 + 63 + 175 + 325  
= Rs. 1076

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

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

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