I/O Disk Drive Calculations

Question

So I am studying for an up and coming exam, one of the questions involves calculating various disk drive properties. I have spent a fair while researching sample questions and formula but because I'm a bit unsure on what I have come up with I was wondering could you possibly help confirm my formulas / answers?

Information Provided:

Rotation Speed = 6000 RPM
Surfaces = 6
Sector Size = 512 bytes
Sectors / Track = 500 (average)
Tracks / Surfaces = 1,000
Average Seek Time = 8ms
One Track Seek Time = 0.4 ms
Maximum Seek Time = 10ms

Questions:

Calculate the following

(i) The capacity of the disk
(ii) The maximum transfer rate for a single track
(iii) Calculate the amount of cylinder skew needed (in sectors)
(iv) The Maximum transfer rate (in bytes) across cylinders (with cylinder skew)

My Answers:

(i) Sector Size x Sectors per Track x Tracks per Surface x No. of surfaces

512 x 500 x 1000 x 6 = 1,536,000,000 bytes

(ii) Sectors per Track x Sector Size x Rotation Speed per sec

500 x 512 x (6000/60) = 25,600,000 bytes per sec

(iii) (Track to Track seek time / Time for 1 Rotation) x Sectors per Track + 4

(0.4 / 0.1) x 500 + 4 = 24

(iv) Really unsure about this one to be honest, any tips or help would be much appreciated.

I fairly sure a similar question will appear on my paper so it really would be a great help if any of you guys could confirm my formulas and derived answers for this sample question. Also if anyone could provide a bit of help on that last question it would be great.

Thanks.

Answer 1

(iv) The Maximum transfer rate (in bytes) across cylinders (with cylinder skew)

500 s/t (1 rpm = 500 sectors) x 512 bytes/sector x 6 (reading across all 6 heads maximum) 1 rotation yields 1536000 bytes across 6 heads you are doing 6000 rpm so that is 6000/60 or 100 rotations per second so, 153,600,000 bytes per second (divide by 1 million is 153.6 megabytes per second)

takes 1/100th of a second or 10ms to read in a track then you need a .4ms shift of the heads to then read the next track. 10.0/10.4 gives you a 96.2 percent effective read rate moving the heads perfectly.

you would be able to read at 96% of the 153.6 or 147.5 Mb/s optimally after the first seek. where 1 Mb = 1,000,000 bytes