Shrinking Item 0-1 Knapsack problem
Maybe I misunderstood you, but we can definitely find a dynamic programming solution, a very similar one to the one for knapsack itself. Let $T[i, b]$ denote the maximum profit that can be made for the first $1 \leq i \leq n$ items, and $b$ is the remaining weight of the knapsack. Here is the recurrence.

$$T[i, b] = \begin{cases} 0 &amp; \text{if } i =0 \vee b = 0 \\ T[i - 1, b] &amp; \text{if } w(i)  &gt; b \\ \max \{T[i - 1, b], T[i-1, b-V(i)] + V(i)\} &amp; \text{if } w(i) \le b \end{cases}$$

For the base case, the 0th item doesn't exist so we have no profit. Similar for having run out of space. 

Next, if $w(i) := W(i) + V(i) &gt; b$, then we obviously cannot add item $i$ to our knapsack, and must continue on. 

In the last case, we have two choices, either not include the item, or include it. For the latter, we then have $b - V(i)$ weight remaining, and $V(i)$ extra profit.

This dynamic programming algorithm has a running time of $O(nB)$, which is pseudopolynomial. Returning $T[n, B]$ is your maximum profit for all your items.