Question

Consider the following transportation problem

Determine an initial basic feasible solution using 

(a) Least cost method 

(b) Vogel’s approximation method

Answer 1

Let ‘a_i ‘ denote the availability and ‘b_j ‘ denote the requirement. Then, 

Σa_i = 30 + 50 + 20 = 100 and Σb_j = 30 + 40 + 20 + 10 = 100 

Since Σa_i = Σb_j . the given problem is a balanced transportation problem and we can get an initial basic feasible solution. 

(a) Least Cost Method (LCM) : 

First allocation:

Second allocation:

Third allocation:

Fourth allocation:

Fifth allocation:​​​​​​​

We first allocate 10 units to cell (O₂ , D₂ ). Since it has a minimum cost. Then we allocate the balance 10 units to cell (O₁ , D₂ ). Thus we get the final allocation table as given below.

Transportation schedule: 

O₁ → D₂ , O₁ → D₃ , O₂ → D₁ , O₂ → D₂ , O₂ → D₄ , O₃ → D₂ 

(i.e) x₁₂ = 10, x₁₃ = 20, x₂₁ = 30, x₂₂ = 10, x₂₄ = 10, x₃₂ = 20 

The total cost is = (10 × 8) + (20 × 3) + (30 × 4) + (10 × 5) + (10 × 4) + (20 × 2)

= 80 + 60+ 120 + 50 + 40 + 40 

= 390 

Thus the LCM, we get the minimum cost for the transportation problem as Rs. 390.

(b) Vogel’s approximation method: 

First allocation:

The largest penalty = 3. So allocate 20 units to the cell (O₃ , D₂ ) which has the least cost in column D₂ 

Second allocation:

Largest penalty = 4. So allocate min (20, 30) units to the cell (O₁ , D₃ ) which has the least cost in column D₃ 

Third allocation:

The largest penalty is 3. So allocate min (20, 50) to the cell (O₂ , D₂ ) which has the least cost in column D₂ . 

Fourth allocation:

Largest penalty = 2. So allocate min (10, 30) to cell (O₂ , D₄ ) which has the least cost in column D₄ 

Fifth allocation:

Largest penalty = 1. Allocate min (30, 20) to cell (O₂ , D₁ ) which has the least cost in column D₁ . Finally allot the balance 10 units to cell (O₁ , D₁ ) 

We get the final allocation table as follows.

Transportation schedule: 

O₁ → D₁ , O₁ → D₃ , O₂ → D₁ , O₂ → D₂ , O₂ → D₄ , O₃ → D₂ 

(i.e) x₁₁ = 10, x₁₃ = 20, x₂₁ = 20, x₂₂ = 20, x₂₄ = 10, x₃₂

= 20 

Total cost is given by = (10 × 5) + (20 × 3) + (20 × 4) + (20 × 5) + (10 × 4) + (20 × 2) 

= 50 + 60 + 80+ 100 + 40 + 40 

= 370 

Hence the minimum cost by YAM is Rs. 370