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Tamil Nadu Board of Secondary EducationHSC Commerce Class 12

Determine basic feasible solution to the following transportation problem using North west Corner rule. Sinks A B C D E Supply P 2 11 10 3 7 4 Origins Q 1 4 7 2 1 8 R 3 9 4 8 12 9 Demand 3 3 4 5 6

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Question

Determine basic feasible solution to the following transportation problem using North west Corner rule.

    Sinks  
    A B C D E Supply
  P 2 11 10 3 7 4
Origins Q 1 4 7 2 1 8
  R 3 9 4 8 12 9
Demand   3 3 4 5 6  
Chart
Sum
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Solution

For the given problem, total supply is 4 + 8 + 9 = 21 and total demand is 3 + 3 + 4 + 5 + 6 = 21.

Since the total supply equals total demand, it is a balanced problem and we can find a feasible solution by the North West Comer rule.

First allocation:

  A B C D E (ai)
P (3)2 11 10 3 7 4/1
Q 1 4 7 2 1 8
R 3 9 4 8 12 9
(bj) 3/0 3 4 5 6  

Second allocation:

  B C D E (ai)
P (1)11 10 3 7 1/0
Q 4 7 2 1 8
R 9 4 8 12 9
(bj) 3/2 4 5 6  

Third allocation:

  B C D E (ai)
Q (2)4 7 2 1 8/6
R 9 4 8 12 9
(bj) 2/0 4 5 6  

Fourth allocation:

  C D E (ai)
Q (4)7 2 1 6/2
R 4 8 12 9
(bj) 4/0 5 6  

Fifth allocation:

  D E (ai)
Q (2)2 1 2/0
R 8 12 9
(bj) 5/3 6  

Final allocation:

  D E (ai)
R (3)8 (6)12 9/6/0
(bj) 3/0 6/0  

First, we allow 3 units to (R, D) cell.

Then balance 6 to (R, E) cell.

Thus we have the following allocations:

  A B C D E Supply
P (3)2 (1)11 10 3 7 4/1
Q 1 (2)4 (4)7 (2)2 1 8
R 3 9 4 (3)8 (6)12 9
Demand 3 3 4 5 6  

Transportation schedule:

P → A

P → B

Q → B

Q → C

Q → D

R → D

R → E

i.e x11 = 3

x12 = 1

x22 = 2

x23 = 4

x24 = 2

x34 = 3

x35 = 6

Total cost = (3 × 2) + (1 × 11) + (2 × 4) + (4 × 7) + (2 × 2) + (3 × 8) + (6 × 12)

= 6 + 11 + 8 + 28 + 4 + 24 + 72

= 153

Thus the minimum cost of the transportation problem by Northwest Comer rule is ₹ 153.

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Transportation Problem
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Chapter 10: Operations Research - Exercise 10.1 [Page 251]

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Samacheer Kalvi Business Mathematics and Statistics [English] Class 12 TN Board
Chapter 10 Operations Research
Exercise 10.1 | Q 10 | Page 251

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