Consider a system running 2 (two) CPU-bound tasks and 4 (four) I/O-bound tasks.
ID: 3871571 • Letter: C
Question
Consider a system running 2 (two) CPU-bound tasks and 4 (four) I/O-bound tasks. Each I/O bound task issues an I/O operation once every 1 millisecond of CPU. Each I/O operation takes 4 milliseconds. Assume that there is only one I/O device (so multiple I/O requests may have to queue). Assume that the context switch takes 0.2 milliseconds. Assume that each CPU-bound requires 10 milliseconds of CPU to complete and each I/O-bound task requires 2 milliseconds of CPU time. Show through a Ganttt chart how the I/O and CPU are allocated and compute the average turnaround times for the CPU-bound and I/O bound tasks.Explanation / Answer
cpu bound process p1 and p2,io bound process p3,p4,p5,p6
all arrive at t=0 so we use fcfs and use them by their number
process no arrival time cpu time io1 io2 completion time tat
p1 0 10 0 0 10.2 10.2
p2 0 10 0 0 20.4 20.4
p3 0 2 4 4 31 31
p4 0 2 4 4 35 35
p5 0 2 4 4 39 39
p6 0 2 4 4 43 43
average turn around time=10.2+20.4+31+35+39+43/6
178.6/6=29.75
i=overhead
0.2
gantt chart
time 0.2 10.2 10.4 20.4 20.6 21.6 21.8 22.8 23 24 24.2 25.2 25.6 25.8 26.8 29.6 29.8 30.8 31 33.6 33.8 34.8 35 37.6 37.8 38.8 39 42.8 43
i p1 i p2 i p3 i p4 i p5 i p6 idle i p3 idle i p4 i p3 idle i p5 i p4 idle i p6 i p5 idle i p6