[理工] [OS] scheduling

看板Grad-ProbAsk作者 (Drinking)時間14年前 (2011/10/14 22:22), 編輯推噓4(402)
留言6則, 2人參與, 最新討論串3/5 (看更多)
恐龍的習題 Consider a system running ten IO-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 tasks 10 milliseconds to complete.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. .. .. .. .. 10 millisecond 不太懂這題的意思 請問怎麼算.. -- ※ 發信站 :批踢踢實業坊(ptt.cc) ◆ From: 114.37.178.177 ※ 編輯: ceo890710 來自: 114.37.178.177 (10/14 22:24)

10/15 22:00, , 1F
CPU 1ms->I/O->CPU 1ms->I/O.....
10/15 22:00, 1F

10/15 22:03, , 2F
context switch的時間不算在time quantan內?
10/15 22:03, 2F

10/15 22:42, , 3F
不算吧,那不是process的cpu burst time,是額外負擔
10/15 22:42, 3F

10/16 02:13, , 4F
抱歉我錯了 應該要算....
10/16 02:13, 4F

10/16 02:20, , 5F
不對應該不用算,上面精神錯亂了= = context switching
10/16 02:20, 5F

10/16 02:21, , 6F
應該當作OS處理的latency,而不算在process的time quantu
10/16 02:21, 6F
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文章代碼(AID): #1Ec4MMFR (Grad-ProbAsk)