Question 5 A floppy disk rotates at a speed of 360 RPM. The disk is divided into

Question

Question 5

A floppy disk rotates at a speed

of 360 RPM. The disk is divided into

9 sectors per track and there are 80

tracks. The disk is double sided. Each

sector stores 512 bytes.

(a) What is the disk capacity in bytes?

(b) What is the minimum and maximum latency time for the disk?

(c) What is the transfer time for a single sector?

Question 6

A hard disk spins at a rate of 9600 RPM.

Each cylinder (a track than spans all the

platters) contains 64 sectors on each platter surface.

(a)

How much time does it take to read 16 consecutive sectors?

(b)

If there are four platters and both sides of the platters are used and there are 1200 cylinders and each sector stores 512 bytes, what is the drive capacity?

Explanation / Answer

Here are the answers:

a) The disk capacity is:

512 bytes/sector * 9 sectors= 4608 bytes per track
4608* 80 tracks = 368640 bytes per side
368640 * 2 sides = 737280 bytes total

b) Latency is the time it takes for a sector to pass under the read/write head of a disk after the head is positioned over the appropriate disk track. The maximum is the time it takes for a complete rotation of a disk. In this case 167 milliseconds. (1min / 360rpm * 60 sec.)

The minimum latency would be the time it takes if the read/write head is next to the desired sector. In this case 18,5 ms. (167 ms / 9 sectors)

c) The transfer time for a single sector is the same as the minimum latency: 18,5 ms.

6

a) Time for a complete spin: 1/9600*60 =6,25 ms.
1 cylinder contains of 64 sectors. 6,25ms / 64 =0,09765625 ms for 1 sector
0,09765625 * 16 = 1,5625ms for 16 sectors.

b) 512 bytes/sector * 64 = 32768 bytes per cylinder
32768 bytes * 1200 cylinders = 39321600 bytes per platter
39321600 bytes * 4 platters * 2 sides =NNN-NN-NNNNbytes totalNNN-NN-NNNNbytes / 1024 = 307200 kbytes / 1024 = 300 megabytes

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