[理工] [OS] scheduling

[ceo890710 (Drinking)]

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)
推 da0910cc:CPU 1ms->I/O->CPU 1ms->I/O.....
推 da0910cc:context switch的時間不算在time quantan內?
推 genius945:不算吧,那不是process的cpu burst time,是額外負擔
→ genius945:抱歉我錯了  應該要算....
推 genius945:不對應該不用算,上面精神錯亂了= = context switching
→ genius945:應該當作OS處理的latency,而不算在process的time quantu