作者aweila75 (David)
看板TransCSI
标题Re: [问题] 89NCCU-CS 磁碟存取时间与空间计算
时间Sun May 20 10:07:28 2007
※ 引述《just1016 (谁陪我玩五子棋?)》之铭言:
: ※ 引述《aweila75 (David)》之铭言:
: : 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,算不出来XD,高手请帮忙,感谢您。
: (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)回转一圈所须时间:(60*1000ms)/ 5400 = 11.11ms = 0.01111second
: 资料量:2000 bytes*40 = 80000 bytes
^^^^^^^^^^^^^^^
为什麽资料量要用sector*40 ,又为什麽资料量是这样算来的?
看不太懂,麻烦高手解释一下,谢谢。
: maximal disk transfer rate:
: 80000 / 0.01111 = 7200720 bytes/second
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1F:推 zelkova:sector/份=2000bytes 一圈的sector=40份 05/20 10:08
2F:→ aweila75:此外,为什麽资料量/旋转一圈时间就是资料传输时间阿? 05/20 10:10
3F:推 zelkova:相当於这些 资料量 在 一圈时间内 传出去的速率 05/20 10:15
4F:→ zelkova:就好像瞬时速度 v =(m)/(sec) 05/20 10:21
5F:→ aweila75:了解,谢谢你的讲解 05/23 08:40