Re: [理工] [OS]-E.A.T
: :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?
經過了一個月後 我有點感覺了 綜合各位的想法
1ns + 0.2*(1us) + 0.02 * (20ms) = 401.2us
^^^^^^^^^^^^
^^^^ ^^^^^^^^^ |
| TLB miss時 要存取 |
mem存取 page table |
(每次存取都要做的) |
都前面都miss時 則要去存取disk
這是計組方面的想法
話說答案差那0.xxus 教授就不會給分嗎? 有時候算EAT時 說什麼men存取時間太小
所以忽略 但我習慣算不忽略的 答案就差那麼一點點 到底該怎麼辦?
有錯請指教!
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