#CO DOUBT_RC Test series

Question

pls explain!

Comments

Option B...?

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@Gate Fever, 

Please add the question source in title

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@Soumya Tiwari

yes, pls explain

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pls post the solution Soumya Tiwari 

Answer 1

Given,data rate of 3MB per sec and 4 Byte is transferred at a time

So no of times hard disk is accessed in a second = 3MB/4B=750000

Now, total time take to service 1 transfer=1200*1ns(Clock Time)

No of times Hard disk will be accessed in 1ns=750000*10^(-9)=75*10^(-5) times

Total time taken to service hard disk transfer=1200*75*10^(-5)nsec

It's also given that hard disk is only active for 5% of time =0.05*1200*75*10^(-5) =0.45nsec

% CPU time spent on hard disk=(0.045nsec/1nsec)*100=4.5%

Comments

My approach was as below:

Disk has data rate 3MB per sec

3MB-->1sec

4MB-->$(4 * 10^{-6})$$/3$ sec==> $1.667* 10^{-6}$

Disk is active 5% of the time ==> 0.05*$1.667* 10^{-6}$  => $83.35 ns$

Over head of CPU and Performing transfer = $1200*10^{-9}s$==> $1200ns$

% of CPU time spent in attending to the hard drive = $\frac{Disk active time }{Over head of CPU and performing transfer}$

=$\frac{83.35 }{1200}*100$

=$6.94$

@Soumya Tiwari could you help me out where I'm wrong.

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Soumya Tiwari 

is there any specific approach to solve these kind of questions!

bcoz am really bad in solving questions related to i/o interface!

 

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No of times Hard disk will be accessed in 1ns=750000*1ns=75*10^(-5)

dont u think its unit would be sec. 

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Thanks :) I have edited now you can check!
Regarding approch, whenever Question ask for % of time wasted in CPU spent on attending hard disk or key stroke like in other question you posted.
That question is based on polling and this question is based on interrupt I/O If not given then try to find the no of times the i/o is accessed in given period.
The no of cycle used to service interrupt or polling will be given in the question this 2 along with the clock cycle time of CPU help us to find the Total time spent on i/o operation in 1 clock cycle period
Thus helping us to get the % CPU invested for i/o.
Correct me if my approch is wrong!

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@Hemanth_13 

Don't you think ,The time spent in attending the hard disk once should be 0.05*1200nsec 

So we need to find total time spent on attending the disk in 1 clock cycle period which will be equal to no of times hard disk is accessed * time spend on attending hard disk once.

Correct me if I'm wrong!