[discussion_vu] Re: :::::::::VU-CS-Experts::::::::: Re: ●๋•νυ ¢αƒєтєяια●๋• ``**``.....COME n Answer CS - EXPERTS...``**``

Thanx but In my opinion ....10 processors per minute ko...

60/10 = 6  sec ayegaa....

And service time = 4 sec...

 

So the fraction of time CPU is busy = service time / staying time

                                                          = 4 / 6

          = 2 / 3

Taking percentage fraction               = 2 / 3 × 100

          = 66.6%

So the CPU is busy for 66.6% of the time.

On Mon, May 2, 2011 at 12:57 AM, ~M~~M~ <d.angel1112@gmail.com> wrote:

HERE WE GO CHECK THE SOLUTION...

Given that there are on an average 10 processes per minute.

So the arrival rate = 10 process/min.

Service time = 4 sec.

Hence the fraction of time CPU is busy = service time / staying time

= 4 / 10

=0.4

So the CPU is busy for 40% of the time

On 1 May 2011 16:02, ~☆~SHINING STAR~☆~ <bc080402322@vu.edu.pk> wrote:

YES ..

CS EXPERT ....

Answer this Question...of OPERATING SYSTEMS..CS604

 

Suppose there are number of processes running on a system. New process in this system arrives at an average of ten processes per minute. Each process requires an average of 4 seconds of service time. Estimate the fraction of CPU busy time in a system with a single processor.

 

Waiting..

Best Regards..

($$)

 

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