Question

Consider the following transportation problem

	Detination				Availabiity
	D1	D2	D3	D4
O1	5	8	3	6	30
O2	4	5	7	4	50
O3	6	2	4	6	20
Requirement	30	40	20	10

Determine an initial basic feasible solution using Least cost method

Answer 1

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

`sum"a"_"i"` = 30 + 50 + 20 = 100 and `sum"b"_"j"` = 30 + 40 + 20 + 10 = 100

Since `sum"a"_"i" = sum"b"_"j"`

The given problem is a balanced transportation problem and we can get an initial basic feasible solution.

Least Cost Method (LCM)

First allocation:

Total Availability = Total Requirement = 100

∴ The given problem is balanced transformation problem.

Hence there exists a feasible solution to the given problem

First allocation:

	D1	D2	D3	D4	(ai)
O1	5	8	3	6	30
O2	4	5	7	4	50
O3	6	(20)2	4	6	20/0
(bj)	30	40/20	20	10

Second allocation:

	D1	D2	D3	D4	(ai)
O1	5	8	(20)3	6	30/10
O2	4	5	7	4	50
(bj)	30	20	20/0	10

Third allocation:

	D1	D2	D4	(ai)
O1	5	8	6	10
O2	(30)4	5	4	50/20
(bj)	30/0	20	10

Fourth allocation:

	D2	D4	(ai)
O1	8	6	10
O2	5	(10)4	20/10
(bj)	20	10/0

Fifth allocation:

	D2	(ai)
O1	(10)8	10/0
O2	(10)5	10/00
(bj)	20/10/0

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.

	D1	D2	D3	D4	Availability
O1	5	(10)8	(20)3	6	30
O2	(30)4	(10)5	7	(10)4	50
O3	6	(20)2	4	6	20
Requirement	30	40	20	10

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 ₹ 390.