给定任务序列J1J2 ...Jn，假定只有一台处理机为这批作业服务。每件任务Ji给定一个时限di和对应利润Pi。只有在规定时限之内完成作业，才会得到利润Pi。通过设计合适的算法选择和安排任务子集Ji，使得J中的每个作业都能在各自的时限内完工，且使获得的利润∑Pi总和最大。

令d=max{di} 1≤i≤n，b=min{n,d}

任何最大利润的可完工子序列中的作业个数必不大于b。

-Given task sequence J1J2 ... Jn, the assumption that only one service processor for these operations. Ji each task given a time limit and the corresponding di profits Pi. Only the stipulated time-frame to complete operations, profit would have been Pi. By designing a suitable algorithm for the selection and arrangement task subset Ji, makes J each operation can be in their respective time-frame for completion, and so was the largest sum of profits ΣPi. Order d = max (di) 1 ≤ i ≤ n, b = min (n, d) the highest profits of any sequence can be completed in the number of operations will not be greater than b.