Scheduling

Question

A process spends 30% of its execution time waiting for completion of I/O operation. If there are 5 processes in memory at once, then the probability of CPU time utilized is _______ %. (Assume all I/O operations are overlapped). (Upto 2 decimal places)

Comments

Made Easy Answer Please Justify the Formula :- 

 

Answer 1

n = total number of processes

p = fraction of time a process is waiting for I/O

Probability (all processes waiting for I/O) = $p^n$

CPU utilization = 1 – $p^n$

Suppose that a process spends a fraction p of its time in I/O wait state. With n processes in memory at once, the probability that all processes are waiting for I/O is $p^n$ .

The CPU utilization then given by the formula CPU utilization = 1 - $p^n$ . where n is called the degree of multiprogramming.

 

Here we are talking in terms of fraction, if no process is waiting for I/O, then CPU Utilization=1 means 100% CPU is utilized

Comments

cant i do it like this;

if every process do 30% I/O and 70% CPU

then lets say each process 0.3 time for i/o and 0.7 time for cpu

then cpu utilization will be (0.35/0.38) * 100= 92.105

pls tell me why this is wrong??

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Probability (all processes waiting for I/O) = p^n

can you explain how  you derive this??

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let the probability of one process doing I/O = p.

if n processes are doing/waiting for I/O simultaneously then the probability of n processes doing I/O = p^n (Multiplication Rule Probability--->

Just multiply the probability of the first event by the second. For example, if the probability of event A is 1/9 and the probability of event B is 2/9 then the probability of both events happening at the same time is (1/9)*(2/9) = 2/81.)

Now, if there are n processes in the Main Memory, then the probability that all processes are waiting for I/O (and therefore the CPU is idle) is: p*p*p*p*p...(n times) = p^n

Therefore, Probability of CPU utilization is: 1 – p^n.

Answer 2

Let required burst time per process = x

Waiting time for a process = 30% of x = $\frac{30}{100} $ x= $\frac{3}{10}$ = 0.3 x

if all operations are not overlapped then

total waiting time = no.of processes * Waiting time for a process = 5 * 0.3x = 1.5 x

total required time (cpu usage + waiting time) = 5x + 1.3 x

if all operations are overlapped then (assuming running operations also overlapped)

total waiting time = Waiting time for a process = 0.3x

total required time (cpu usage + waiting time) = x + 0.3 x

probability of CPU time utilized is ( all operations are overlapped )= $\frac{x-0.3x}{x+0.3x} $ =  $\frac{0.7 x}{1.3x} $ =  $\frac{7}{13} $

Comments

can this be the solution

cpu utilisation is 1-p^5

where p is i/o burst time

so here it should be 1-.3^5

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I also think this would be the answer.