Tamil Nadu Board of Secondary EducationHSC Commerce Class 12th

# Consider the following transportation problem Destination Availability 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 - Business Mathematics and Statistics

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Sum

Consider the following transportation problem

 Destination Availability 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 Vogel’s approximation method

#### Solution

Let ‘ai‘ denote the availability and ‘bj‘ 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.

Vogel’s approximation method:

First allocation:

 D1 D2 D3 D4 (ai) penalty O1 5 8 3 6 30 (2) O2 4 5 7 4 50 (1) O3 6 (20)2 4 6 20/0 (2) (bj) 30 40/20 20 10 penalty (1) (3) (1) (2)

The largest penalty = 3.

So allocate 20 units to the cell (O3, D2) which has the least cost in column D2

Second allocation:

 D1 D2 D3 D4 (ai) penalty O1 5 8 (20)3 6 30/10 (2) O2 4 5 7 4 50 (1) (bj) 30 20 20/0 10 penalty (1) (3) (4) (2)

Largest penalty = 4

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

Third allocation:

 D1 D2 D4 (ai) penalty O1 5 8 6 10 (1) O2 4 (20)5 4 50/30 (1) (bj) 30 20/10 10 penalty (1) (3) (2)

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:

 D1 D4 (ai) penalty O1 5 6 10 (1) O2 4 (10)4 30/20 (0) (bj) 30 10/0 penalty (1) (2)

Largest penalty = 2.

So allocate min (10, 30) to cell (O2, D4) which has the least cost in column D4

Fifth allocation:

 D1 (ai) penalty O1 (10)5 10/0 – O2 (20)4 20/0 – (bj) 30/10/0 penalty (1)

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.

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

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 ₹ 370.

Concept: Transportation Problem
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