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Consider the disk drive with the following specification $16$ surface. $1024$ tracks per surface. $1024$ sectors/track, $1$ KB /sector  rotation speed is $3000$ $\text{rpm}$.The disk is operated in Burst Mode.The processor runs at $600$ $\text{MHZ}$ and takes $300$ & $900$ clock cycle to initiate & complete DMA transfer respectively. If the size of transfer is $20$$\text{KB}$

What is the percentage of processor time consumed for the transfer operation?

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total cycle which dma need to initiate and then complete the total cycles that will be required will be 1200 cycles. Total time by dma will be ( initiate + transfer + complete ) out of which only transfer time will be useful time. 

so wastage time = 1200 cycles * time of one cycle. 

1cycle = $\frac{1}{frequency}$

1 cycle= $\frac{1}{600Mhz}$

1200 cycles = 2 micro second time. 

now we have to calculate the time for transfer which can be easily calculated using the disk parameters given . assuming a movable read write head is available . if it is movable in one round i will be able to read one track only else if it is fixed . every surface will have a read write head and 16 tracks can be read in one rotation. 

3000 rounds = 60 seconds

1 round = $\frac{1}{50}$

total data on track = 1024 *1KB

1024*1KB data = $\frac{1}{50}$ seconds

20 KB data = $\frac{20}{50*1024}$= 390 micro second. 

so total time = 392 micro second. whle transfer time is 390. 

cpu will be involved in the transfer time only . so percentage will be $\frac{2*100}{392}$

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