. Consider a system running ten I/O-bound tasks and one CPU-boundtask. Assume th

Question

. Consider a system running ten I/O-bound tasks and one CPU-boundtask. Assume that the I/O-
bound tasks issue an I/O operation once for every millisecond ofCPU computing and that each
I/O operation takes 10 milliseconds to complete. Also assume thatthe context-switching
overhead is 0.1 millisecond and that all processes are long-runingtasks. What is the CPU
utilization for a round-robin scheduler when: The timequantum is 10 milliseconds


I am not really understand that how to get answer

I was assuming that total time will be 11 sec because (1 +.1)* 10 which is cpu time add by context switch however, Total time ( 1ms + .1ms )*10 + 10+.1 = 21.1

therefore, i am getting confused to get answer....

help me ..

Explanation / Answer

Your assumption is right . But you did nt add the time takenby the I/O bound task which is 10+0.1 ( 10 ms for exec + .1ms for contextswitch ) = 10.1 now add this result to one which you have got ( ie time taketo cycle thru all processes) ( round robin fashion) 10.1+11 = 21.1 cpu util : (20/21.1 )*100 = 94.786 hope this helps ram

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