Consider a demand-paging system with a paging disk that has an average
access and transfer time of 20 milliseconds.Addresses are translated
through a page table in main memory, with an access time of 1 microsecond
per memory access. Thus, each memory reference through the page
table takes two accesses. To improve this time, we have added an
associative memory that reduces access time to one memory reference,
if the page-table entry is in the associative memory.
Assume that 80 percent of the accesses are in the associative memory
and that, of the remaining, 10 percent (or 2 percent of the total) cause
page faults.What is the effective memory access time?
恐龍本習題 跟95政大的考題類似 但給的答案
EAT = (0.8) × (1 us) + (0.1) × (2 us) + (0.1) × (5002 us)
= 501.2 us
= 0.5 ms
其中 5002us 滿怪的 我算的是 應該2*20 ms + 2*1 us =40002us
有高手可以解釋整個架構嗎?
謝謝!
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※ 編輯: yesa315 來自: 140.127.208.96 (10/20 10:24)