# How to solve a knapsack problem with increased weight limit?

Let us consider the knapsack problem. Given a set $P$ of $n$ items where each item has weight $w_i$ and value $v_i$ for all $i=1,2,\ldots,n$. We have two bins, one has a weight limit of $W$ and the other has a weight limit of $W'=2W$. Assume that we have an optimal algorithm $\texttt{OPT}$ that solves the knapsack problem.

Let $O$ denotes the value of an optimal solution when we apply $\texttt{OPT}$ to the first bin (that has a weight limit of $W$) and to the set of items $P$. Also, let $O'$ denotes the value of an optimal solution when we apply $\texttt{OPT}$ to the second bin (that has a weight limit of $W'=2W$) and to the same set of items $P$.

Can we compare $O$ and $O'$?

I guessed that $$O'\leq 2O.$$

If all weights are W < $w_i$ ≤ 2W, then O = 0 and O' = highest value of a single item.
There is very little that you can say except the obvious O ≤ O'. You might have "valuable" items with a total weight ≤ W, and all other items are worthless. Then O' is not or only very little greater than O. Or all items with weight ≤ W might be worthless with some items with weight W < $w_i$ ≤ 2W very valuable, then the ratio O' / O can be arbitrarily large.
• Thank you. Can we compare $O$ and $O'$ in general? I mean maybe there exists some constant $c$ such that $O'\leq 2O+c$ or something? – Ribz Jun 17 '17 at 15:28