作者boy5548 (小YO)
看板Grad-ProbAsk
標題[理工] 一題OS
時間Fri Jan 28 13:27:25 2011
Consider a system running ten I/O-bound tasks and one CPU-bound task.
Assume that the I/O-bound tasks issue an I/O operation once for every
millisecond of CPU computing and that each I/O operation takes 10 millisecond
to compute.Also assume that the context-switching overhead is 0.1 millisecond
and that all processes are long-running tasks.
Describe the CPU utilization for a round-robin scheduler when:
(a)The time quantum is 1 millisecond
(b)The time quantum is 10 milliseconds
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sol:
(a) 1.0/(1.0+0.1) = 1/1.1 = 91%
(b) 20/((10*1.1)+10.1) = 20/21.1 = 94%
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這題考滿多次的，可是就是沒有很了解怎麼解這題...
可以請高手詳細說明一下嗎?(尤其是(b)小題)
謝謝!!
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◆ From: 114.39.13.219
※ 編輯: boy5548 來自: 114.39.13.219 (01/28 13:29)
推 cakeboy:io bound 每做1 time unit 就會一次context switch so 01/28 20:09
→ cakeboy:10個IO bound 10*1.1=11 cpu bound 剛好10 time unit 01/28 20:11
→ cakeboy:做完也要一次context switch so 10+0.1=10.1 所以 實際用 01/28 20:12
→ cakeboy:在execute 的時間/總時間 20/21.1 成立 01/28 20:12
→ boy5548:請問為什麼cpu bound要做10 time unit? 01/28 23:26
→ boy5548:而且為什麼剛好10 time unit做完也要一次context-switch? 01/28 23:27
→ dy957:因為time quantum 是10 01/28 23:34
推 cakeboy:因為他寫 long-time 表示這些行程一直在跑 01/29 19:26
→ aoqq12:有個更簡單的想法你試試。。你就畫個 ready queue 01/29 23:27
→ aoqq12:然後cpu bound一開始就在裡面 你每隔1m就抓一個i/o bound 01/29 23:28
→ aoqq12:放進去ready queue 01/29 23:28
→ aoqq12:在照順序去給他cpu就好了 01/29 23:29