(a) Utilization of 1/0 controller (U) *\^v)" (b) Total response time (R) t” (c) Average Queue size. (d) Total time spent by a request in the queue +- Considering the same disk IiO system and calculate the above, g$ah^ur.s of the disk performance if the disk service time is halved i.e. R.:4ms.

Solution to 1st given problem

Given:100 IOPS

Service time (RS):- 8ms

Arrival time (Ra)=1/IOPS=1/100=0.01

1) Utilization (u)=RS/RA=8/0.01=0.8 or 80%

2) Total response time(R)= Rs/(1-U)= 8/(1-80)=0.10 ms

3) Average queue time=U*2/(1-U)=80/79=2.02 ms

4) Total time spent by a request in the queue= UR=0.80.10=0.08 ms

Solution to 2nd given problem

Service time (RS)=4ms

Arrival time (Ra)=1/100=0.01

1) Utilization (u)=RS/RA = 4/0.01=40%

2) Total Response time(R)=Rs/(1-U)=4/(39)=0.10 ms

3) Average Queue time = U*2/(1-U)=80/39=2.05ms

4) Total time spent by a request in the queue=UR=0.40.10=0.04 ms