Fractional Knapsack Problem

Now we need to show that an initial greedy choice of item h leads to an optimal solution. We can make the greedy choice first.

Case I:

If W < w_h, then L(h) = W and for all j $\neq$ h L(j) = 0 is the only take.

We can fill the knapsack with the highest value-per-pound item.

Case II:

There is room left after greedy choice. L(i), L(j) > 0, i $\neq$ j. Assume item i was the first choice and j = h a subsequent choice.

Thus, V = L(i) + L(h) .

If we choose the same amount L(h) of item h first (we cannot choose more or we would have by the above result), then the value will be

V' = L(h) + L(i)

V' = V

Thus we can make the greedy choice first.