This is a homework problem for my introduction to algorithms course.

Recall the scheduling problem from Section 4.2 in which we sought to minimize the maximum lateness. There are $n$ jobs, each with a deadline $d_i$ and a required processing time $t_i$, and all jobs are available to be scheduled starting at time $s$. For a job $i$ to be done, it needs to be assigned a period from $s_i \geq s$ to $f_i$ = $s_i + t_i$, and different jobs should be assigned nonoverlapping intervals. As usual, such an assignment of times will be called a schedule.

In this problem, we consider the same setup, but want to optimize a different objective. In particular, we consider the case in which each job must either be done by its deadline or not at all. We’ll say that a subset $J$ of the jobs is schedulable if there is a schedule for the jobs in $J$ so that each of them finishes by its deadline. Your problem is to select a schedulable subset of maximum possible size and give a schedule for this subset that allows each job to finish by its deadline.

(a) Prove that there is an optimal solution $J$ (i.e., a schedulable set of maximum size) in which the jobs in $J$ are scheduled in increasing order of their deadlines.

(b) Assume that all deadlines $d_i$ and required times $t_i$ are integers. Give an algorithm to find an optimal solution. Your algorithm should run in time polynomial in the number of jobs $n$, and the maximum deadline $D = \max_i d_i$.

I've solved the problem as worded with the recurrence

$Opt(i, d) = \max\left \{ \begin{array} \\ Opt(i-1, d-t_i) + 1 \hspace{20 mm} d\leq d_i \\ Opt(i-1, d) \end{array} \right \}$

but our instructor added a new requirement that our algorithm must not be dependent on D. This recurrence seems like it would produce an $O(nD)$ running time if implemented with dynamic programming.

I can't figure out how to reduce its running time from $O(nD)$ to $O(n^k)$. To me it seems like it's a variation on the knapsack problem with all values equal to 1. In which case it seems like this is the best that can be done.

If I'm doing something wrong could someone point me in the right direction, or if I've done everything right so far, could someone at least give me a hint as to how I can make an $O(n^k)$ recurrence or algorithm.

  • 1
    $\begingroup$ "Recall the scheduling problem from Section 4.2" -- what was that? Can you summarise the problem in a meaningful way, and focus on the parts essential to your question? $\endgroup$
    – Raphael
    Oct 31, 2012 at 9:58

1 Answer 1


Use a dynamic programming algorithm to compute an $n \times n$ table $T$, where the entry $T(j,k)$ answers the question: suppose you wish to schedule $j$ out of the first $k$ jobs. What is the earliest time you can complete processing these?

How do we compute $T(j,k+1)$? Either the job $k+1$ is included in the best set of $j$ out of $k+1$ jobs, so $$T(j,k+1) = T(j-1,k) + t_{k+1},$$ or job $k$ is not included in this set, so $$T(j,k+1) = T(j,k).$$ We also have to worry about the deadline. We can do this by making the entries of $T$ where the task is impossible equal to $\infty$, and checking whether we have exceeded the deadline at every step. So the pseudocode for computing the $T(j,k+1)$ entry of the table is

if T[j-1,k] + t[k+1] > d[k+1]
    then T[j,k+1] = T[j,k]
    else T[j,k+1] = min ( T[j,k-1]+t[k+1], T[j,k] )

We initialize by setting $T(j,k) = \infty$ if $j > k$, and $T(1,1) = \infty$ if $t_1 > d_1$, and $T(1,1) = t_1$ otherwise.

  • $\begingroup$ I already used dynamic programming. If you compute the values to sub problems in an efficient order (i.e. dynamic programming), you still end up with $O(nD)$. This homework is done, and if I remember correctly I came up with an answer that worked, so I'll post that later if I have time. $\endgroup$
    – Manual5355
    Oct 31, 2012 at 14:34
  • $\begingroup$ @Joseph: There can be different algorithms that use dynamic programming to solve the same problem, and have different running times. This hint is for a dynamic programming algorithm that runs in $O(n^2)$ time. I'll elaborate later if I have time. $\endgroup$
    – Peter Shor
    Oct 31, 2012 at 15:46

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