The profile sweep of Section 5 as a word RAM program. Read the word, sort the jobs by start time ($O(n log n)$), compute $q_max$ and a sentinel, then sweep the jobs in that order, carrying for every profile $x$ (how many selected jobs are due at each of the next $q_max$ instants, a base $m+1$ number of $q_max$ digits) and every weight $c ≤ W$ the least preprocessing load of a feasible selection of weight at least $c$. One job is one marginalisation pass (the shift of the profile is a division by a power of $m+1$, the low digits being minimised away) and one take pass, each $O(1)$ per table cell, so a job costs $O((m+1)^q_max (W+1))$ and the whole run is within $c · (W+1) · (m+1)^q_max · (n+1) + c · sortCost$. The answer is read off the profiles at weight $W$. The correctness is that of recursion (5) in its repaired form ($Q3Model$); jobs may tie in start time ($δ = 0$), no rescaling is needed. The domain restricts to positive processing times, the paper's standing assumption for Lemma 3, exactly as the concept's statement now says.