※ 引述《yesa315 (XD)》之铭言： : 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? : 恐龙本习题 跟95政大的考题类似 但给的答案 : EAT = (0.8) × (1 us) + (0.1) × (2 us) + (0.1) × (5002 us) : = 501.2 us : = 0.5 ms : 其中 5002us 满怪的 我算的是 应该2*20 ms + 2*1 us =40002us : 有高手可以解释整个架构吗? : 谢谢! EAT = (0.8)*(1us) + (0.2)*(1us + (1-0.1)*(1us) + (0.1)*(20ms)) = 0.8us + 0.2us + 0.18us + 400us = 401.18us = 0.4ms 想法： EAT = (TLB hit)+(TLB miss) = (1次 memory access) + (1次memory access + 资料存在时1次 + 资料不存在时) 读取page table资料 memory access page fault 这样对吗?? 我不是很确定 --

※ 发信站: 批踢踢实业坊(ptt.cc) ◆ From: 122.125.50.152