retagged by
3,425 views
8 votes
8 votes
Consider a system, that uses Interrupt Driven I/O for a particular device which has a data transfer rate of 10 KBPS . The processing of the interrupt ( Jump, Execute and Return to the main program ) takes 250 us . What percentage of CPU time is consumed by I/O device , if I/O device interrupts for every 2 bytes ?
retagged by

2 Answers

1 votes
1 votes
x = time taken by cpu in interrupt processing

 y = time taken by i/o device to prepare data.

y= 200

x=250

x/y =( 250 /200 ) * 100 = 1.25 *100 = 125%

it should be 200/250 = 0.8

% is  0.8 * 100 = 80%
edited by
0 votes
0 votes
kapil i m getting 125% .
and i think explanation given by debashish deka is absloutely right...
edited by
Answer:

Related questions

3 votes
3 votes
2 answers
2
1 votes
1 votes
1 answer
3
Ayush Upadhyaya asked Oct 30, 2018
1,175 views
I think percentage of processor time consumed should be $\frac{ \,ISR\,time}{ISR\,time+\,Data\,Transfer\,Time}=\frac{100}{100+122}$.Is it correct?
1 votes
1 votes
1 answer
4
Na462 asked Sep 24, 2018
803 views