My questions is as follows:
Does the fractional knapsack (multiple knapsack case) also exhibit the
same greedy solution or is it NP-Hard.
I have two versions of this problem that i am considering:
Case 1: you can use the remaining fraction of an item in another
knapsack
Case 2: you cannot use the remaining fraction of an item in another
knapsack
Could someone direct me if the complexity of this problem is known.
Thanks
AG
This looks like homework...
Anyway, Case 1 is polynomially solvable; you should try some LP-formulation.
Case 2 is NP-hard, and this really is just a two-line exercise once you
know the NP-hardness proof for subset-sum or knapsack.
--Gerhard
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Gerhard J. Woeginger http://www.win.tue.nl/~gwoegi/