1. Given a disk drive with the following characteristics: Number of surfaces = 2
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Question
1. Given a disk drive with the following characteristics:
Number of surfaces = 200
Number of tracks per surface = 100
Number of sectors per track = 50
Bytes per sector = 256
Speed = 2400 RPM
What is the total disk capacity?
What is the average rotational latency?
Solution:
2. A disk has 20 surfaces (i.e. 10 double sided platters). Each surface has 1000 tracks. Each tracks has 128 sectors. Each sector can hold 64 bytes. The disk space allocation policy allocates an integral number of contiguous cylinders to each file,
a.) How many bytes are contained in one cylinder?
b.) How many cylinders are needed to accommodate a 5-Megabyte file on the disk?
c.) How much space is wasted in allocating this 5-Megabyte file?
Solution:
3. A disk has the following configuration:
The disk has 310 MB
Track size: 4096 bytes
Sector Size: 64 bytes
A programmer has 96 objects each being 50 bytes in size. He decides to save each object as an individual file. How many bytes (in total) are actually written to disk?
Solution:
5. A disk drive has 3 double-sided platters. The drive has 300 cylinders. How many tracks are there?
Solution:
6. Given the following specifications for a disk drive:
- 256 bytes per sector
- 200 sectors per track
- 1000 tracks per surface
- 2 platters
- Rotational speed 7500 RPM
- Normal recording
What is the transfer rate of the disk?
Solution:
7. Given the following specifications for a disk drive:
- 256 bytes per sector
- 100 sectors per track
- 1000 tracks per surface
- 3 platters
- Average seek time of 8 ms
- Rotational speed 15000 RPM
- Normal recording
a) What would be the time to read 10 contiguous sectors from the same track?
b) What would be the time to read 10 sectors at random?
Solution:
Please answer all questions completely. No half answers. Thanks!
Explanation / Answer
1)
total disk capacity :- total number of sectors * each sector size
total number of sectors = 200*100*50 = 1000000 sectors
therefore, total disk capacity = 1000000 * 256 = 256000000 bytes
SECTOR SIZE = 50 * 256 = 12800 bytes can be retrieved by one rotation
Given rotation speed is 2400/60 in each second
therefore, rotational latency is 12800 * (2400/60) = 12800 * 40 = 512000 bytes
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