A multiplattered hard disk is divided into 1100 sectors and 40,000 cylinders. There are six platter surfaces. Each block holds 512 bytes. The disk is rotating at 4800 rpm and has an average seek time of 12 msec.please with the formulas in each question.1. What is the total capacity of this disk?2. What is the disk transfer rate in bytes per second?3. What are the minimum and maximum latency times for this disk?4. What is the average latency time of the disk?5. How does the average lattency time of this disk compare to today's average latency time for disks?

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dannwaeze45 dannwaeze45 · Answer 1 · 2019-12-23T23:20:41+01:00

Answer:

1. 135168000000 bytes

2. 270336000 /sec

3. min latency is 0, max latency is 12.5

4. 0.00625 sec

Explanation:

1.

capacity= number of sectors*number of cylinders* platter surfaces*size of block

1100*40000*6*512=135168000000 bytes

2.

transfer rate= number of sectors*platter surfaces*size of block*rpm/60

(since we are converting rpm to revolutions per second)

1100*6*512*4800/60=270336000 /sec

3.

minimum latency =0

maximum latency = 60*1000/4800=12.5

4.

average latency time = ( 60 / rpm) / 2

0.5*60/4800=0.00625 sec