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Consider the disk drive with the following specification:

$16$ surfaces, $1024$ tracks/surface, $1024$ sectors/track, $1KB/sector$, rotation speed is $3000 rpm$ and the disk is operated in burst Mode. The processor runs at $600 MHZ$ and takes$300$ & $900$ clock cycle to initiate & complete $DMA$ transfer respectively, if the size of transferred data is $20KB$.

 The percentage of processor time consumed for the transfer operation is ________.
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Initial time = 300 * (1/ 600 * 10^6 ) = 0.5 microsec
terminate time = 900 * (1/ 600 * 10^6) = 1.5 microsec
time for 1 revolution = 60/3000 sec
so in 1 revoluution data transfer is 1024 KB
in 1 sec data transfer is  (1024 * 3000 )/ 60 KB = 50 MB ,  now rate is 50 MB/sec
Data transfer time for 20kB = (20 KB/ 50MB) sec =   390.625 microsecs

% CPU consumed = [ ( Initial time + termination time)/ total time ] * 100

= % CPU consumed = [ ( 0.5 + 1.5) / 392.625 ] * 100 

=  0.509 %

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Processor runs at $600 MHz$

$1$ sec ---- $600*10^6$ cycle

1 cycle ---- $\frac{1}{600*10^6}$ -- $\frac{1}{600}$ micro sec

DMA will take $300+600=1200$ cycle

$1200$ cycle -- $\frac{1}{600} * 1200 =2$ micro sec

$3000$ rotation --- $60$ sec

$1$ rotation --- $20$ msec

$20$ msec -- $1024$ sector rotates-- $1024$  KB

$20$ KB -- $390.625$ micro sec

% processor time consumed = $\frac{2}{390.625} *100$ = $0.512$%
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