作者gnqwertyuiop (中华炒面)
看板TransCSI
标题[问题] 96北大计概第一题
时间Sun Jul 1 21:29:04 2007
我忘了题目
所以拿东华91的来冒充
What is the average time to read or white
a 512 byte sector for a typical disk rotating
at 4500 RPM?The advertised average seek time is
20ms, the transfer rate is 2MB/sec, and the
controller overhead is 2ms. Assume that the disk
is idle so that there is no waiting time.
看解答:
seek time: 20ms
rotating time:(60*1000)/4500=13.33ms
13.33/2=6.66ms
data transfer time:[(1*1000)/2000000]*512=0.256ms
access time=20+6.66+0.256+2=28.916ms
我看讲义上写
Access time=Seek time+Rotational Delay Time+Data Transfer Time
为啥旋转延迟时间要除以2
我看它中文解释为 每旋转一圈的时间
然後第二个问题是
seek time跟Controller overhead有啥不同呢?
感谢大家^^
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◆ From: 59.115.54.125
1F：→ XrGodz:除以2...是取平均时间 (当资料在第一笔或是最後一笔的时候) 07/01 22:02
2F：→ gnqwertyuiop:感谢^^ 07/01 22:25
※ 编辑: gnqwertyuiop 来自: 59.115.54.125 (07/01 22:25)
※ 编辑: gnqwertyuiop 来自: 59.115.54.125 (07/01 22:27)
3F：推 queenisi:有没有人可以说明一下为什麽这样算吗?我看好久看不懂@@ 07/13 20:31