Suppose you’ve been asked to evaluate the performance of a mechanical disk drive
ID: 3739142 • Letter: S
Question
Suppose you’ve been asked to evaluate the performance of a mechanical disk drive with the following specifications:
a. What is the average time to read or write a single sector?
b. What is the average time to read or write an entire track on the disk (128 consecutive sectors)?
c. Suppose an I/O system using a single disk of this specification receives 15 I/O requests per second such that each request reads or writes a single sector. What is the utilization of the I/O system? Assume the I/O system is in equilibrium.
Average Seek Time 8ms Rotation Rate 15000rpm Transfer Rate 10MB/s Controller Overhead 1ms # cylinders 16394 Sector Size 1KB Sectors / Track 128Explanation / Answer
Solution:
a. What is the average time to read or write a single sector?
Average time to read or write a single sector
1 sector is of size 1KB
Rotational latency => 15000 rpm = 60/15000= 0.004/2= 0.002 sec
anf for 1 KB transfer it will take 1KB/1MB= 2^(-10)= 0.0009765625 seconds
Average time will be= seek time + rotational latency= 0.008+0.002 + 0.0009765625= 0.0109765625 sec
b. What is the average time to read or write an entire track on the disk (128 consecutive sectors)?
Size of the track= 128*1KB= 128KB
time taken = 8 + 128 * (0.002) + 128*0.0009765625 = 0.389 s
c. Suppose an I/O system using a single disk of this specification receives 15 I/O requests per second such that each request reads or writes a single sector. What is the utilization of the I/O system? Assume the I/O system is in equilibrium.
15 I/O for sectors means 15 KB is requested every second
We can provide 10MBPS
so utilization per second= 15KB/10MB= 0.146484375%
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