Unbounded Knapsack - Transition

For each capacity $w$ and each item $i$: $dp[w] = \max(dp[w], dp[w-w_i] + v_i)$ if $w_i \le w$ Notice the difference from 0/1: you use $dp[w-w_i]$, not $dp_{prev}[w-w_i]$.

This allows taking the same item multiple times. Process weights from $0$ to $W$ (not reversed like 0/1). When computing $dp[w]$, you want $dp[w-w_i]$ to already include items from this round. This formula captures the relationship between smaller problems.