secondary memory

Question

consider a disk pack with the following:

16 surfaces

128 tracks/surface

256 sectors/track

512 bytes/sectors

(consider that each surface has 1 read/write head)

The disk is rotating at 3600 rpm. What is the data transfer rate ?

Answer 1

Concept

$\Rightarrow$ In this question we have a R/W head on each surface and all the heads Read/Write data from their respective surface simultaneously.

$\textrm{Data tranfer rate}$ = $\frac{\textrm{16 * Total data present on 1 surface of disk}}{\textrm{Time taken to transfer that data}}$

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$\textrm{Total data present on 1 surface of disk}$

=$128 \frac{tracks}{surface}256\frac{sector}{track}*512\frac{bytes}{sector}$

= $128* 256 * 512 \frac{bytes}{surface}$

= $2^{24}\frac{bytes}{surface}$

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$\textrm{Time taken to transfer that data}$

Disk is rotating at $3600$ revolutions per minute

$\Rightarrow$ $3600$ revolutions in $60$ seconds

$\Rightarrow$ The Disk performs $1$ revolution in $\frac{1}{60}$ seconds

Also, in $1$ revolution a R/W head can read data of $1$ track and a surface has $128$ tracks.

$\Rightarrow$ Time to R/W $1$ surface(time to read $128$ tracks) = $128* \frac{1}{60}$ seconds = $\frac{128}{60}$ seconds

$\Rightarrow$ Time to R/W $16$ surface = $\frac{128}{60}$ seconds ($\because$ $16$ R/W heads are reading data simultaneously from each surface)

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$\textrm{Data tranfer rate }$

=$\frac{\textrm{16 * Total data present on a surfaces of disk}}{\textrm{Time taken to transfer that data}}$

= $16 *2^{24} *\frac{60}{128} Bytes/second $

= $ 2^{21}* 60 Bytes/second$

=$2*60 MB/second$

=$120 MB/second$

Comments

@Satbir, by default we assume that all surfaces have a R/W head right?

https://gateoverflow.in/1393/gate2005-70 In this question, why is the number of surfaces not considered for taking the data transfer rate, ie. we should be able to read 16*512 sectors in 1 revolution. Other people have also commented this doubt on the question.

Answer 2

3600 Rotations in 60 secs
i.e. 1 Rotation in 1/60 sec

In one rotation we can read/Write 1 Track amount of data.
1 track= 256 sectors= 256 * 512B = 256 * 512 * 8 bits =2²⁰ Bits

since there is a Read/Write head for every surface i.e. 16 
therefore 16* 2²⁰ = 2²⁴ bits in One rotation
Now, One rotation takes 1/60 secs, so we can do 60 rotaions in 1 Sec

Therefore 60*2²⁴ bits per second wiz. approx 1006.63 MegaBits per Second

Comments

Thanks.

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So it is

(number of rotations in 1 sec)* (track capacity).

Right??

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getting answer 120 mbps

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Yes i am also getting 120 mbps