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Fractional Knapsack Problem exhibits Greedy Choice Property

Proof:

First show that as much as possible of the highest value/pound item must be included in the optimal solution.

Let wh, vh be the weight available and value of the item with the highest value/pound ratio (item h).

Let L(i) be the weight of item i contained in the thief's loot L.

Total Value V = \(\sum_{i=1}^n L(i) \frac{v_i}{w_i}\)

If some of item h is left, and L(j) $\neq$ 0 for some j $\neq$ h, then replacing j with h will yield a higher value.

L(j) \(\frac{v_j}{w_j} \leq L(j) \frac{v_h}{w_h}\)

\(\frac{v_j}{w_j} \;\leq\; \frac{v_h}{w_h}\) true by definition of h



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