polling_RC test series

Question

A keyboard must be interrogated at least 10 times per second to make sure that no keystroke by the user is missed. Assume that each such interrogation and data transfer takes 800 clock cycles in a processor with a 1 GHz clock. what % of CPU time is spent polling the keyboard? (answer up to 4 decimal places).

Comments

0.0008 %

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yes , its correct!

pls explain!

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It takes 800 cc for the entire process.

We have to interrogate 10 times per second.

So in 1 second we require 8000 cc of CPU

Freq is 1GHz which means there are $10^9$ cc per sec.

% of time wasted = $\frac{8*10^3}{10^9}*100$=$8*10^{-4}$ = 0.0008%

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@Gate Fever, 

Please add the question source in title.

Answer 1

CPU Clock Cycle Time= 10^-9 sec =1nsec

Each interrogation and transfer take =800*1ns=800nsec

In 1 sec keyboard is interrogated 10 times

So in 1 nsec it is interrogated =10 ×10^(-9) times

So, total time spent in polling=800*10*10^(-9) nsec

% of time spent in polling=((time spent on polling in a cycle)/(total time) )*100

=(((800*10^(-8))nsec)/(1nsec)) *100

=0.0008

Comments

yes , thanks!