Re: [理工][計系]台大電機99

看板Grad-ProbAsk作者ai305428d (可愛小小羅)時間15年前 (2011/02/08 19:26)推噓1(1推 0噓 4→)

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※ 引述《charliejack (charliejack)》之銘言： : 8. A hard disk has a track seek time of 10ms. : The disk rotation speed is 9000 rpm. : Each track on the disk jas 600 sectors. : Each sector has total 512 bytes data. : What is the average time it takes to read 1024 bytes data? : (A) 10.5 ms : (B) 13.355 ms : (C) 14.55 ms : (D) 15.333 ms : (E) None of the above : 我的式子 : 10ms + ( 1s / (9000rpm/60s) / 2 )*1000 = 13.333 ms : 這裡我將 trasfer time 給忽略?! (題目沒給冏) : 但還差一點點不知道是不是哪裡列錯 : 請大家指教一下我是參考某年交大寫的這裡transfer time 要自己導首先計算rotation time: 9000/60=150 rps rotation time = 1000 * 1/150 = 6.67 ms/轉 transfer time = 6.67 / 600 = 0.011 ms/sector Effect access time = seek + avg.rotation + transfer = 10 ms + 1/2 * 6.67 ms +0.011 ms = 13.346 ms 題目是要讀取1024 bytes 的data 1024 / 512 = 2 sector 所以最後total accsee time = 13.346 * 2 =26.692 ms ......所以答案是(e) ? -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 111.255.10.122

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