Log In
6 votes

     Measurements  of a certain  system  have shown that a process  runs, on the
     average,  for time T  before  blocking  for  Input/Output.  Process  switch
     requires time S which is overhead.  For round robin scheduling with quantum
     Q, give a formula for the cpu efficiency in each of the following cases:

     (a) Q=infinity  (b) Q>T  (c) S<Q<T  (d) Q=S  (e)  Q nearly 0
     cpu efficiency is useful cpu time/total cpu time, i.e.  sum of cpu
     useful time and cpu overhead in switching processes.

     For (a) and (b) a  process  runs  for T and a  switch  occurs  when it gets
     blocked.  Efficiency is T/(T+S)

     In (c), since Q<T, each run of T requires T/Q process  switches,  resulting
     in overhead of ST/Q and therefore, the efficiency is T/(T+(ST/Q))  which is

     For (d), take Q for S in (c) and we get Q/(Q+Q) which is 50%
     For (e), the efficiency goes to 0 as Q goes to 0

Can somebody please explain option c and hence option D with example ,Thanks and sorry if it is naive question :)

in Operating System 4.1k views

1 Answer

3 votes
Best answer

There is one more solution to this question:

 Assume T is the total processing time and S is the overhead for each 
     content switch. Then the CPU efficiency can be represented by the 
     equation: T / ([T/Q] - 1) * S + T 
               where [T/Q] = k is an integer and k >= T/Q and k < T/Q + 1.

     (a) Q = infinity. It becomes T / T = 1

     (b) Q > T. It becomes T / T = 1

     (c) S < Q < T. The efficieny is T / ([T/Q] - 1) * S + T. 

     (d) Q = S. 

         If T > Q,
         T / (T/Q * S - S) + T = T / (T - S) + T = T / 2T - S = T / 2T - Q.
         If T < Q, it becomes T / T = 1.

     (e) Q is nearly zero. It becomes T / infinity = 0.

Let us solve option c first:


here the closer the value of Q is from T then more the efficiency 

In option d if the value of T < S then the efficiency is 100% 

selected by
In your solution T is the processing time for each process and the second solution T is the total processing time
what i want : In efficiency formula how you write T/Q ,I want logic behind it or how you got entire formula,Please explain that.Thanks for reply.

Execution Time = T

[T/Q] * S since we need the total switching times, but now we are counting switch after the last process has finished(an extra count) so we subtract 1. 

In my example when i put Q = 2 we are switching 3 times and when we put Q = 3 we can do the execution in just 1 switch between each process P1 and P2 so that's that.

Efficiency = Execution Time/ Execution time + Switching time 

Efficieny = T / T + ([T/Q] - 1) * S  
what i understood is Q is less than T, so before process go for I/O i.e before T ,it is switching due to time quantum Q and now T=6 and Q=2 S=2 ,then there will be 3 switch and each switch require 2 unit as follows

S*(T/Q)=(6/2) *2 =6

efficiency =6/(6+6)  or from your formula 6/(6+4).

Thanks :)
Yeah but if T itself depends on the execution time of individual processes. like if there are 2 process with execution time 3 units then there will be one switch ((6/3)-1) with Q = 3

So if T is 6 and Q=2 no matter what processes you have there can never be 3 switches.

Related questions

5 votes
3 answers
If we have only one process in ready queue with burst time "m", then how many context switching will happen using round robing scheduling with time quantum q ,where q<m.Assume that dispatching the process first time is not counted as a context switch.
asked Aug 26, 2017 in Operating System rahul sharma 5 1.6k views
0 votes
1 answer
Choose correct answer from the below options: If the following jobs are to be executed on a single processor system The jobs are arrived at time 0 and in the order a, b, c, d, e. Calculate the departure time (completion time) for job ‘a’ if scheduling is round robin with time slice 1 15 5 9 11
asked Dec 10, 2017 in Operating System Parshu gate 595 views
3 votes
2 answers
Here what are the number of context switches ? Is it 5 or 6? Do we consider context switch before P1 (i.e during the start) ?
asked Aug 12, 2017 in Operating System Xylene 3.4k views
1 vote
1 answer
How many time context switch in Round Robin? ( if only one process remain in ready queue at end of scheduling does that count every time, when time slice over or only one time?)
asked Jan 7, 2017 in Operating System sanyam53 973 views