A hard disk with a transfer rate of 10 Mbytes/ second is constantly transferring data to memory using DMA. The processor runs at 600 MHz, and takes 300 and 900 clock cycles to initiate and complete DMA transfer respectively. If the size of the transfer is 20 Kbytes, what is the percentage of processor time consumed for the transfer operation ?
(A) 5.0%
(B) 1.0%
(C) 0.5%
(D) 0.1%
Answer: (D)
Explanation: Transfer rate=10 MB per second
Data=20 KB=20* 2 10
So Time=(20 * 2 10)/(10 * 2 20)= 2* 10-3 =2 ms
Processor speed= 600 MHz=600 Cycles/sec
Cycles required by CPU=300+900 =1200
For DMA=1200
So time=1200/(600 *10 6)=.002 ms
In %=.002/2*100=.1%
So (D) is correct option
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