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The storage area of a disk has the innermost diameter of $10$ cm and outermost diameter of $20$ cm. The maximum storage density of the disk is $1400$ bits/cm. The disk rotates at a speed of $4200$ RPM. The main memory of a computer has $64$-bit word length and $1$µs cycle time. If cycle stealing is used for data transfer from the disk, the percentage of memory cycles stolen for transferring one word is

  1. $0.5 \%$
  2. $1 \%$
  3. $5\%$
  4. $10\%$
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This is an AMAZING question! It's got beautiful small details.


Data transfer rate.

To calculate that, we need track capacity.

The maximum storage density of the disk is 1400 bits/cm.

All the tracks have equal capacity. The outermost track is the largest, and the innermost track is the smallest. Still they got the same capacity, how? Because we vary the density (ie we pack the bits tightly/loosely)

The innermost track would be the most dense, so as to get the required capacity.

So, the given density is of the innermost track.

 

Now, what is its circumference?

$2\pi r=\pi d$

=>$10\pi $

=> $31.4cm$

 

1 cm of this track = 1400 bits.

31.4 cm of this track = 43960 bits. = 5495 bytes.

 

So, Data transfer rate = $\frac{4200(5495)}{60}$ bytes per second.

Therefore, to transfer 64-bits = 8 bytes, time required = $20.798\mu s$


Now, percent of cycles stolen:

$\frac{1 \mu s}{20.798 \mu s}$

=> $\frac{1}{21}$

=> $5\%$

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The capacity of track = л x d x Density = 3.14 x 10 x 1400 bits = 14000*pi bits

4200 rotation = 60 sec

So, 1 rotation time = 60/4200 =(1/70) sec

 DMA controller transfer 64 bits = 1 usec

In 1 sec = 64 × 10^6 bits.

In 1 sec the disk can read = 14000л /(1/70)
                                             = 3.08 x 10^6 bits

Total memory cycle stolen =

       (3.08*10^6) / (64*10^6 )%

       =5%
Answer:

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