希望的人能够帮助我了解如何在下文中计算服务时间。 教科书中的每一例都使用8毫秒,虽然这是固定的,但在解决问题时,计算为5002微秒。
考虑采用一种需求量计算系统,并配有一个纸质,平均可进入和转移20毫米秒。 地址通过一个主要记忆页表翻译,每个记忆存取时间为1秒。 因此,每一页表中的记忆参考资料有两条。 为了改进这一时间,我们增加了一个联名记忆,如果页码的条目是联名记忆的话,就会减少查阅一个记忆参考的时间。
假设80%的通道属于附属记忆,其余80%(占总数的2%)造成页面故障。 有效的记忆存取时间是什么?
答复:
effective access time = (0.8) ×(1 μsec) + (0.1) ×(2 μsec) + (0.1) ×(5002 μsec) = 501.2 μsec = 0.5 millisec
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