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The difference between maximum possible profit for $0/1$ Knapsack and fractional Knapsack problem with capacity $(W)=20$.

$\begin{array}{|c|c|c|c|c|c|} \hline \text{Item} &  a & b & c & d & e & f & g & h & i & j\\  \hline \text{Weight} &3 & 5 & 2 & 1 & 12 & 10 & 9 & 9 & 4 &1  \\  \hline\text{Profit} & 7 & 10 & 3 & 3 & 26 & 19 & 18 & 17 & 5 & 4 \\\hline \end{array}$

For 0/1 knapsack, we have to make the whole table or there is any direct method to this?
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so difference=6-3=3

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Ans :- 3.4

Common part for both (j+d+e)

For fractional knap sack, common part + 6/10 * f

For 0/1 knapsack , common part + c + i

X = difference betn profits = 3.4

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