A) 20× 1000 × 16 × 1KB=3,20,000 × 1KB=3,20,000KB
B)
Step1:- find the rotational latency:-
Rotational speed=3000 rpm
3000 rpm=60 sec
1 rpm= 60/3000 sec= rotational delay=20 msec
Rotational latency (Ravg) = 1/2 x 1 rpm= 1/2 x 20 =10 msec
Step2:- To find transfer rate
Transfer rate = bytes on track/rotational delay= (16 x 1 KB)/20 ms=800 KB/sec
C) Data is transferred Byte wise; given in the question.
CPU read/write time for a Byte = 0.1 μs
Interrupt overhead (counted in CPU utilization time only) = 0.4 μs
Transfer time for 1 Byte data which took place at the rate of 800 KB/sec = 1/800KB=1.25 μs
Percentage of CPU time required for this job =
Total overhead =0.1+0.4=0.5 μs
Total time = 0.5 + 1.22 = 1.72 μs
Percentage of CPU time required for this job = 0.5/1.72 x 100 =29%
D) Percentage of CPU time held up for disk I/O for cycle stealing DMA transfer =0.1/1.25 x100 =8 %