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1 vote
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The following transportation problem

  A B C Supply
I 50 30 220 1
II 90 45 170 3
III 250 200 50 4
Demand 4 2 2 0

 

$\begin{array}{|l|l|l|l|l|} \hline & \text{A} & \text{B} & \text{C} & \text{Supply} \\\hline \text{I} & \text{50} & \text{30} & \text{220} & \text{1} \\\hline \text{II} & \text{90} & \text{45} & \text{170} & \text{3} \\\hline \text{III} & \text{250} & \text{200} & \text{50} & \text{4} \\\hline \text{Demand} & \text{4} & \text{2} & \text{2} & \text{0} \\\hline  \end{array}$

Has a solution

  A B C
I 1    
II 3 0  
III   2 2

 

$\begin{array}{|l|l|l|}\hline \text{} & \text{A} & \text{B} & \text{C} \\\hline \text{I} & \text{1} & \text{} & \text{} \\\hline\text{II} & \text{3} & \text{0} & \text{} \\\hline\text{III} & \text{} & \text{2} & \text{2} \\\hline\end{array}$ 

The above solution of a given transportation problem is

  1. Infeasible solution
  2. optimum solution
  3. non-optimum solution
  4. unbounded solution
in Others
edited by
2.1k views

2 Answers

2 votes
 
Best answer

Ans is B optimum solution

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1
Hi sir,

your answer is correct but small correction for 1st row 2nd column dij=+25

d12=c12-(u1+v2)

=30-(-195+200)=25
0
yes it will be  25
0
The number of allocations are 4 only (0 allocation is by default for every cell), so shouldn't it be (A) Infeasible?
0
that is the condition for degeneracy and not of infeasibility
0
In your answer to UGCNET-Dec2014-III-68 you stated that occurrence of degeracy while solving transportation problem means the solution obtained is not feasible.
0
to resolve degeneracy allocation of 0 is done
0
Can you please clear how u1 = -195. Aren't we to suppose that u1 = 0 in the beginning?

Sanjay Sir's solution is not showing up.
0 votes
Answer should be (A) Infeasible.

The number of allocations are 4 only (0 allocation is by default for every cell) and as they're less than m+n-1 that is 3+3-1=5, the solution is Infeasible.
0
yeah , But condition is that number of allocation should be less then m+n-1

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