Total cycle required for initial set up of a DMA transfer and DMA completion
= (1000 + 700 ) clock cycles
=1700 clock cycles
Given clock rate= 2 GHz. So, clock cycle time=(1/clock rate)= (1/2 GHz) = (1/ 2*109 ) * 109 ns = 0.5 ns
No of cycles for Initialization and Completion of the block transfer is 1700 cycles,
so total time takes by processor =1700*0.5 ns=850 ns . let X = 850 ns , this is preparation time or total clock cycle time use by cpu .
Disk transfer rate is 4000 KB/sec. So, to transfer 16 KB, the time required is ( 16/4000 ) sec=4000000 ns.
Let Y= 40,00000 ns , this is transfer time.
% of time processor "involved" is = X/(X+Y) * 100
= {850/(850+4000000)} * 100 [ as question explicitly says it is burst mode, so we use this formula x/x+y where x =clock cycle time use by cpu and y = Transfer time ]
=0.02124 %
[ In burst mode all data is transferred in a single cycle . so we need to consider x+y in this formula x/x+y ]
Therefore, % of processor time is free= 100 - % of processor time involved
=(100- 0.02124)
=99.97876
=99.978