Suppose a given bus protocol requires 4 ns for devices to make requests, 8 ns fo

Question

Suppose a given bus protocol requires 4 ns for devices to make requests, 8 ns for arbitration, and 15 ns to complete each operation. How many operations are completed per second? a. A hard disk with 4 platters has 1024 tracks/platter, 1024 sectors/track, and 512 bytes/sector. If the platter rotation speed is 15,000 rpm with 1 ms seek time for every 100 tracks traversed. Calculate the following:i. Total capacity of this hard disk (3 marks)ii. The average seek time (4 marks)iii. The average rotational latency (3 marks)iv. The transfer time for a sector (3 marks)v. The average access time (3 marks)

Explanation / Answer

Request time use only ones because after request approved ,there is no need to requst

and arbitration time like a acknowledgement time, used for acknowledge the sender

one second = 1000 ns

time for operation = 1000-4 = 996

Operation per second= 996 / 23 = 43.30 = 43

2-------------------------

i

total capacity of disk = 4*1024*1024*512

= 22*210*210*29

= 231

= 2 GB

ii)----no of track per platter = 1024

1 ns seek time for 100 track

average seek time = 1024/100 * 1ms

= 10.24ms

iii)--- 15000 rotation = 60 second

1 rotation = 60/ 15000 = 1/250

Average rotational latency = 1/250 * 1/2

= 2 ms

iv)--- 1 track rotate = 2/1024 ms

1 sector rotate = 2/ 1024*1024

=1.9 * 10-6

v)------

Average access time = seek time + rotational time + transfer rate

= 10.24 ms + 2ms + 1.9*10-6

= 12.24 ms

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.