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+4 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

Answer:

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

Q/(Q+S).

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 :)**

+2 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%

0

In your solution T is the processing time for each process and the second solution T is the total processing time

0

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.

+1

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

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