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The following transportation problem :

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

Has a solution

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

The above solution of a given transportation problem is

  1. Infeasible solution
  2. optimum solution
  3. non-optimum solution
  4. unbounded solution
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Ans is B optimum solution

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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.
Answer:

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