MINIMUM SEQUENCING WITH RELEASE TIMES

• INSTANCE: Set T of tasks, for each task $t\in T$ a release time $r(t)\in Z^+$, a length $l(t)\in Z^+$, and a weight $w(t)\in Z^+$.
• SOLUTION: A one-processor schedule for T that obeys the release times, i.e. a function $f : T \rightarrow N$ such that, for all $u \ge 0$, if S(u) is the set of tasks t for which $f(t)\le u <f(t)+l(t)$, then $\vert S(u)\vert = 1$ and for each task t, $f(t)\ge r(t)$.
• MEASURE: The weighted sum of completion times, i.e. $\sum_{t\in T} w(t)(f(t)+l(t))$.

  
  

• Good News: Approximable within 1.6853 [196].

• Comment: Approximable within $e/(e-1)\approx 1.58$ if all weights w(t)=1 [101]. Variation in which there are precedence constraints on T instead of release times is approximable within 2 [216]. The preemptive case is approximable within 4/3 [431]. Variation in which there are precedence constraints and release times is approximable within e [433]. If all weights w(t)=1 and the objective is to minimize the max-stretch $\max_{t\in T} (f(t)+l(t)-r(t))/l(t)$ the nonpreemptive problem is not approximable within $O(\vert T\vert^{1-\varepsilon })$ for any $\varepsilon >0$, but the preemptive problem admits a PTAS [70].