作者aweila75 (David)
看板TransCSI
標題[問題] 91NCCU-CS 磁碟空間+資料讀取速度計算
時間Sun May 20 09:20:10 2007
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。
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1F：推 zelkova:(a)第一式還要乘以2 因為是2HD 05/20 10:34
2F：→ aweila75:原來2HD是兩顆的意思喔，我以為是形容詞或是品牌 05/20 21:40
3F：推 MrAshan:可是題目一開始說2HD能儲存的容量是1440KB呀為何還乘2 05/27 23:59
4F：推 aweila75:樓上的說的好像也對...結論是不需要乘嗎？ 06/03 09:35