Minimum (8, 3, 5,7) = 3
Since minimum cost is 3, the best alternative is E₂

Max (2, 4, 0, -2) = 4 Since maximum pay off is 4, the alternative 2 is selected

Max (9, 8, 5) = 9
∴ d₁ is the best strategy using maximin principle.

Min (45, 70, 53) = 45
∴ Choosing Bonds is the best decision under minimax principle.

Max (6, 2, 4) = 6
∴ Action Al is the best according to maximin principle

Here total availability = 150 + 100 + 250 = 500
total requirement = 50 + 150 + 300 = 500
∴ Total availability = total requirement
∴ The given problem is a balanced transportation problem
Hence, there exists a feasible solution to the given problem
I - allocation:

[∵ least cost is 4 & min (50,150) = 50]
II - allocation:

[∵ least cost is 6 & min (150, 250) = 150]
III - allocation:

[∵ least cost is 8 & min (300, 100) = 100]
IV - allocation:

[∵ least cost is 9 & min (200,100) = 100]
V - allocation:

[∵ min (100, 100) = 100]
Thus, the allocations are

∴ The transportation schedule is
O₁ → D₁, O₁ → D₃, O₂ → D₃, O₃ → D₂, O₃ → D₃
Hence, the total transportation cost is
= 50(4) + 100(8) + 100(11) + 150(6) + 100(9)
= 200 + 800 + 1100 + 900 + 900
= Rs. 3900

Here, total supply = 10 + 5 + 3 = 18
total demand = 5 + 4 + 6 + 3 = 18
∴ Total supply = total demand
∴ The given problem is a balanced transportation problem.
∴ We can find an initial basic feasible solution to the given problem.
I - allocation:

[∵ min (5, 10) = 5]
II - allocation:

[∵ min (4, 5) = 4]
III - allocation:

[∵ min (1, 6) = 1]
IV - allocation:

[∵ min (5, 5) = 5]
V - allocation:

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

∴ The transportation schedule is
1 → A, 1 → B, 1 → C, 2 → C, 3 → D
Hence, the total transportation cost
= 5(3) + 4(1) + 1(7) + 5(5) + 3(2)
= 15 + 4 + 7 + 25 + 6 = Rs. 57