: :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? 经过了一个月後 我有点感觉了 综合各位的想法 1ns + 0.2*(1us) + 0.02 * (20ms) = 401.2us ^^^^^^^^^^^^ ^^^^ ^^^^^^^^^ | | TLB miss时 要存取 | mem存取 page table | (每次存取都要做的) | 都前面都miss时 则要去存取disk 这是计组方面的想法 话说答案差那0.xxus 教授就不会给分吗? 有时候算EAT时 说什麽men存取时间太小 所以忽略 但我习惯算不忽略的 答案就差那麽一点点 到底该怎麽办? 有错请指教! --

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