計概---硬碟讀取時間相關 [96關務] - 考試
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By Donna
at 2013-06-19T11:27
at 2013-06-19T11:27
Table of Contents
對於一般轉速5400RPM的磁碟,假設
平均尋找時間為12ms,
傳送速率5MB/sec,
控制時間 2ms,
且磁碟是閒置的 沒有任何等待時間
試問 寫入512 位元組,
需要的平均時間是多少 ?
A)21.5ms B)19.7ms C)16.5ms D)14.1ms [96關務]
什麼是控制時間哩?
書上寫的只有 ACCESS TIME = seek time + rotational latency + transfer time
而且只給轉速 但沒給需要幾轉
那該怎麼算出旋轉時間呢??
Ans: B).19.7 ms
平均尋找時間為12ms,
傳送速率5MB/sec,
控制時間 2ms,
且磁碟是閒置的 沒有任何等待時間
試問 寫入512 位元組,
需要的平均時間是多少 ?
A)21.5ms B)19.7ms C)16.5ms D)14.1ms [96關務]
什麼是控制時間哩?
書上寫的只有 ACCESS TIME = seek time + rotational latency + transfer time
而且只給轉速 但沒給需要幾轉
那該怎麼算出旋轉時間呢??
Ans: B).19.7 ms
→ carterdunk:Access Time = Seek + Control + Latency + Transfer 06/19 13:36
→ carterdunk:背公式吧...研究所考試的東西幾乎都忘了 06/19 13:36
※ 編輯: beatfuture 來自: 42.72.151.221 (06/19 15:58) 推 crave:把控制時間另外加進公式內即可 旋轉時間就是轉速的一半 06/19 17:06
→ crave:自己用我這個算法下去算 答案沒錯 它自動進位了 應該是19.66 06/19 17:07
→ nalum219:平均時間可以想成[最快+最慢]/2 06/19 17:41
→ nalum219:所以旋轉來說 最快就是0 最慢就是轉一圈 06/19 17:41
→ nalum219:所以平均轉速用轉半圈的時間來算 06/19 17:42
推 ABJones:關鍵確實在樓上所言,使用平均來破題 06/20 11:38
※ 編輯: beatfuture 來自: 42.72.169.131 (06/20 14:05) Tags:
考試
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at 2013-06-21T06:06
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