作者aweila75 (David)
看板TransCSI
标题[问题] 91NCCU-CS 磁碟空间+资料读取速度计算
时间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)我的计算是 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 平均旋转时间
接下来的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
不知道这样算是否正确？请指教。
--
※ 发信站: 批踢踢实业坊(ptt.cc)
◆ From: 218.162.129.136
1F：推 zelkova:(a)第一式还要乘以2 因为是2HD 05/20 10:34
2F：→ aweila75:原来2HD是两颗的意思喔，我以为是形容词或是品牌 05/20 21:40
3F：推 MrAshan:可是题目一开始说2HD能储存的容量是1440KB呀为何还乘2 05/27 23:59
4F：推 aweila75:楼上的说的好像也对...结论是不需要乘吗？ 06/03 09:35