[問題] 96北大計概第一題

[gnqwertyuiop (中華炒麵)]

我忘了題目
所以拿東華91的來冒充
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
我看講義上寫
Access time=Seek time+Rotational Delay Time+Data Transfer Time
為啥旋轉延遲時間要除以2
我看它中文解釋為  每旋轉一圈的時間

然後第二個問題是
seek time跟Controller overhead有啥不同呢?
感謝大家^^

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◆ From: 59.115.54.125
→ XrGodz:除以2...是取平均時間 (當資料在第一筆或是最後一筆的時候)
→ gnqwertyuiop:感謝^^
※ 編輯: gnqwertyuiop    來自: 59.115.54.125        (07/01 22:25)
※ 編輯: gnqwertyuiop    來自: 59.115.54.125        (07/01 22:27)
推 queenisi:有沒有人可以說明一下為什麼這樣算嗎?我看好久看不懂@@