Page replacement and Demand Paging(需求分頁問題)

ckjboy2003

我有几个关于”需求分页”的问题要请问前辈们,希望可以得到解答

Consider a system that uses demand paging to bring pages of a process into main memory. Suppose normal memory access time are 10 naoseconds and page faults are serviced in a total of 5 milliseconds.

A. If one memory access in 10000 causes a page fault, what is the effective access time(EAT)?
我的计算方法如下:
EAT=(1-P)*ma+P*处理分页错误的时间

=(1-P)*10ns+P*5ms
=(1-P)*10+p*5000000
=10+4999990P
这一题这样子解对吗?
我不了解的地方是题目里指出的10000是什么?


B.By what factors does demand paging slow the system down, compared to a system in which pages are always in main memory?


請前輩們指導,用中文就可以了,非常感謝

qtdszws

A.
EAT = (1–p) x memory access+ p x page fault
p=1/10000

B.内存大小,磁盘速度

scutan

第一道题实际上就是一个概率中的求数学期望的问题。求解方法2楼已经给了。

ckjboy2003

謝謝前輩們指導,我了解了