[問題] 91NCCU-CS 磁碟空間+資料讀取速度計算

看板TransCSI作者aweila75 (David)時間17年前 (2007/05/20 09:20)推噓3(3推 0噓 1→)

留言4則, 3人參與討論串1/1

3. (a) A 2HD floppy disk has a storage capacity of 1440KB. Given that there are 18 sectors per track and each sector contains 512 Bytes of data, what is the number of tracks on this 2HD disk? (b) Suppose the floppy disk drive has a rotation speed of 300 rpm. Assume the arm movement time = 10 ms fixed startup time + 1 ms for each track crossed. Compute the best-case, worst-cast and average-case (assume, on average, the read-write head moves 20 tracks) access times for this disk drive. (a)我的計算是 18*512bytes = 9216 byte (1440*1024bytes)/9216byte = 160 track (b) 300 rpm 所以 60/300 = 0.2秒 0.2*(1/2) = 100 ms 平均旋轉時間接下來的best-case 100 + 10 + 1*1(track) = 111 ms average-case 100 + 10 + 1*20(track) = 130 ms worst-cast 100 + 10 + 1*160(track) = 270 ms 不知道這樣算是否正確？請指教。 -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 218.162.129.136

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zelkova

05/20 10:34, , 1^F

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aweila75

05/20 21:40, , 2^F

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MrAshan

05/27 23:59, , 3^F

05/27 23:59, 3^F