Question

Consider a magnetic disk consisting of 16 heads and 400 cylinders. This disk has four 100-cylinder zones with the cylinders in different zones containing 160, 200, 240. and 280 sectors, respectively. Assume that each sector contains 512 bytes, average seek time between adjacent cylinders is 1 msec, and the disk rotates at 7200 RPM. Calculate the (a) disk capacity, (b) optimal track skew, and (c) maximum data transfer rate.

Answer 1

Answer:

Disk Capacity  = 45056000 bytes

Optimal skew = 26.455 ≈ 27

Maximum Data transfer rate =  262.5 GB per second

Explanation:

given data

heads = 16

cylinders = 400

cylinder zones = 100

each sector contains = 512 bytes

average seek time = 1 msec

disk rotates = 7200 RPM

to find out

disk capacity, optimal track skew, and maximum data transfer rate

solution

first we get total number of sectors that is  

total number of sector = number of zones × (number of sectors in different zones

total number of sector = 100 × (160+200+240+280)

total number of sector = 88000

so

Disk Capacity = total number of sectors  ×  size of each sector

Disk Capacity =  88000 × 512

Disk Capacity  = 45056000 bytes

and

Rotation time = $\frac{60}{7200}$

Rotation time = 8.33 milli seconds

so Optimal number of sectors in a track = average of ( 160,200,240,280 )

Optimal number of sectors in a track  = 220

now New sector is read every  $\frac{8.33}{220}$ i.e = 0.0378 ms

here Optimal skew = seek time ÷ new sector read time

Optimal skew = $\frac{1}{0.0378}$

Optimal skew = 26.455 ≈ 27

and

here we know that for maximum transfer rate we will select cylinder with maximum number of sectors i.e here  280 sectors

so

capacity of one track with maximum = 280 × 512

capacity of one track with maximum =  = 143360 bytes

and Number of rotations in 1 second is = $\frac{7200}{60}$

Number of rotations in 1 second is = 120

so Data transfer rate = Number of heads × Capacity of one track × Number of rotations in one second

Data transfer rate =  16 × 143360 × 120

Data transfer rate = 275251200 bytes per second

Data transfer rate =  262.5 GB per second

Answer 2

Following are the calculation to the given points:

Calculating value:

For point a):

Calculating the total number of sectors = number of zones $\times$ (number of sectors in different zones)

$= 100 \times (160+200+240+280) \\\\= 100 \times (880) \\\\ = 100 \times 880 \\\\ =88000$

Calculating the capacity of disk = total number of sectors  $\times$  size of each sector

$= 88000 \times 512\\\\= 45056000\ \text{bytes}$

For point b):

Calculating the Rotation time:

$= \frac{60}{7200}\\\\= \frac{6}{720}\\\\= \frac{1}{120}\\\\= 8.33 \ \text{milli seconds}$

Calculating the Optimal number of sectors in a track = average of ( 160,200,240,280 )

$=\frac{ 160+200+240+280}{4}\\\\=\frac{880}{4}\\\\= 220$

Calculating the new sector of every read:

$=\frac{8.33 \ ms}{ 220} \\\\= 0.0378\ ms$

Calculating the Optimal skew = seek time / new sector read time

$=\frac{1\ ms}{0.0378\ ms}\\\\ =\frac{1}{0.0378 }\\\\ = 26.455 \approx 27$

For point c):

To achieve the highest transfer rate, we must choose the cylinder with the most sectors, 280.

length of a single track (track with the highest number of sectors):

$\to 280\times 512 = 143360\ bytes$

No of rotations in one second $\to \frac{7200}{60} \\\\\to \frac{720}{6}\\\\ \to 120$

Data transfer rate = Number of heads  $\times$  Capacity of one track  $\times$ Number of rotations in one second

$= 16\times 143360 \times 120\ \frac{bytes}{second}\\\\= 275251200 \ \frac{bytes}{second}\\\\= 262.5 \ \ GBps \\\\ OR \\\\ = \frac{275251200}{2^{21}}$

Find out more information about the magnetic disk here:

brainly.com/question/9480984