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<b><font face="宋体" size="4" color="#FF0033">最优调度方案:</font><font face="宋体" size="4" color="#000000">
<font color="#0033CC">1.</font>将{a1,a2,a3,…an, b1,b2,…bn}排成非递减序列
<font color="#0033CC">2.</font>依次从序列中抽出最小元素 m,如果m = aj。且作业j还设有排入调
度表,则把作业安排在调度表可达的最左面一项定位上。(n个作业
的调度有n项。开始全部为空。)如果m = bj,且作业j未排入,则把
作业j排在最右项上。如果作业j 已排入,则取序列下一个最小元素。
<font face="Arial, Helvetica, sans-serif" color="#0033CC">3.</font>继续1、2,直到元素取完为止。
<font color="#FF0033">例3.2 <font color="#000000">n=4</font></font><font color="#FF0033"> </font>(n个作业,每个作业有<font color="#FF0000">2</font>个任务<font color="#FF0000">a</font>、<font color="#FF0000">b</font>)<font color="#FF0033">
</font><img src="IMAGE/ta-13.gif" width="149" height="90" align="middle" hspace="200"><font color="#FF0033">
</font>(a1, a2, a3, a4 ) = (3, 4, 8, 10)
(b1, b2, b3, b4) = (6, 2, 9, 15)
排序后为<font color="#0033CC"><i><font face="Arial, Helvetica, sans-serif">(b2, a1, a2, b1, a3, b3, a4, b4)= (2, 3, 4, 6, 8, 9, 10, 15</font></i></font>)
设<font color="#FF0000"> <img src="IMAGE/a.gif" width="13" height="11" align="middle">1 <img src="IMAGE/a.gif" width="13" height="11" align="middle">2 <img src="IMAGE/a.gif" width="13" height="11" align="middle">3 <img src="IMAGE/a.gif" width="13" height="11" align="middle">4</font> 为最优调度。
M = b2 = 2 -> <img src="IMAGE/a.gif" width="13" height="11" align="middle"><font color="#FF0000">4</font> = T2 (最右)
a1 = 3 -> <img src="IMAGE/a.gif" width="13" height="11" align="middle"><font color="#FF0000">1</font> = T1 (最左)
a2 = 4 作业2已被调度
b1 = 6 作业1已被调度
a3 = 8 -> <img src="IMAGE/a.gif" width="13" height="11" align="middle"><font color="#FF0000">2</font> = T3 (最左)
<img src="IMAGE/a.gif" width="13" height="11" align="middle"><font color="#FF0000">3</font> = T4
<img src="IMAGE/text10-f.gif" width="505" height="242">
任务个数 = 2时,动态规则法能极大减低复杂度,<font color="#FF0033">(O(nlogn))</font>
任务个数 > 2时,已证为NP难题。<font color="#FF0033">(O(ni))
</font></font></b></pre>
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