Re: [理工] [ OS ]-中山95

看板Grad-ProbAsk作者 (who I am)時間15年前 (2010/03/27 14:07), 編輯推噓4(400)
留言4則, 3人參與, 最新討論串2/4 (看更多)
※ 引述《KarmaPolice (Karma Police)》之銘言: : 4. Assume we have a demand-paged memory. The page table is held in registers. : It takes 8 milliseconds to service a page fault if an empty page is available : or the replaced page is not modified, and 20 milliseconds if the replaced page : is modified. Memory access time is 100 nanoseconds. Assume that the page to be : replaced is modified 70 percent of the time. What is the maximum acceptable : page-fault rate for an effective access time of no more than 200 nanoseconds? : 這題想請教一下該如何解? : 我解出來的答案 讓我覺得有點離譜 想問問大家都是算多少? EAT=(1-p)*100ns + p * (0.3*8ms+0.7*20ms) =0.1ms-0.1ms*p+p*16.4ms =0.1ms+(16.3ms)*p and EAT<200ns 故 16.3ms*p<200ns-100ns=0.1ms => p<0.1ms/16.3ms =0.6% -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 124.12.49.53

03/27 14:12, , 1F
謝謝~~
03/27 14:12, 1F

03/27 18:13, , 2F
1ms不是等於10^6ns嗎??
03/27 18:13, 2F

03/27 18:41, , 3F
所以我才覺得這題很怪....答案很離譜
03/27 18:41, 3F

03/27 20:11, , 4F
所以100ns是0.0001ms吧= =
03/27 20:11, 4F
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