Re: [問題] 89NCCU-CS 磁碟存取時間與空間計算
※ 引述《just1016 (誰陪我玩五子棋?)》之銘言：
: ※ 引述《aweila75 (David)》之銘言：
: : 2.A multiplattered hard disk is divided into 40 sectors and 400 cylinders.
: : There are four platter surfaces. The total capacity of the disk is 128MB.
: : A cluster consists of 4 sectors. The disk is rotating at a rate of 5400rpm.
: : The disk has an average seek time of 12 msec.
: : (a) what is the capacity of a cluster for this disk?
: : (b) what is the average access time before starting data transfer?
: : (c) what is the maximal disk transfer rate in bytes per second?
: : 想了很久，算不出來XD，高手請幫忙，感謝您。
: (a)4*400*40 = 64000 sectors
: 一sector的capacity：128MB / 64000 = 2000 bytes
: 一cluster的capacity：2000*4 = 8000 bytes
: (b)（60*1000ms）/ 5400 = 11.11ms
: 11.11ms / 2＝5.55ms
: 12ms＋5.55ms = 17.55ms
: (c)迴轉一圈所須時間：（60*1000ms）/ 5400 = 11.11ms = 0.01111second
: 資料量:2000 bytes*40 = 80000 bytes
^^^^^^^^^^^^^^^
為什麼資料量要用sector*40 ，又為什麼資料量是這樣算來的？
看不太懂，麻煩高手解釋一下，謝謝。
: maximal disk transfer rate：
: 80000 / 0.01111 = 7200720 bytes/second
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 218.162.129.136
推
05/20 10:08, , 1F
05/20 10:08, 1F
→
05/20 10:10, , 2F
05/20 10:10, 2F
推
05/20 10:15, , 3F
05/20 10:15, 3F
→
05/20 10:21, , 4F
05/20 10:21, 4F
→
05/23 08:40, , 5F
05/23 08:40, 5F
討論串 (同標題文章)
本文引述了以下文章的的內容：
完整討論串 (本文為第 3 之 3 篇)：