Consider a directed acyclic graph $G=(V,E)$ of $n$ tasks where every node is a task and every edge is a dependency. If there is an edge from a task $v$ to another task $w$, that means we need to finish $v$ before starting $w$. We have $p$ processors of the same capability and every processor can finish any task in one unit of time.

We define a schedule function $s$ to map a task $v$ to a positive integer time step $s(v)$. Since we have only $p$ processors, so the system can only process at most $p$ tasks in one time step. Also, a schedule needs to respect the dependency of edges.