Process Scheduling

Question

On a system using round-robin scheduling, let S represents the time needed to perform a switch, Q represents the round-robin time quantum, and R represents the average time a process runs before blocking on I/O.

the CPU efficiency when S=Q<R

a) R/(R+S)

b)Q/(Q+S)

c)1/2

d)zero

Answer 1

The answer is 1/2 i.e. option c. Here's the explanation : -

‘R’ is the average time a process runs before I/O block and ‘S’ is the time needed for switch. ‘Q’ represents the round-robin time quantum.

Efficiency of CPU = usefull time / total time

 Usefull time = R

Total time = usefull + switching time = R + (R/Q)*S     (when S < R and Q < R)

Efficiency = R/(R + (R/Q)*S) = Q/(Q+S)

When S = Q

Efficiency = Q/2Q = 1/2.

Comments

@ank73811

How switching time = (R/Q)*S

I couldn't understand this part.

Please suggest

Thanks

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given R is avg running time of process ,Q is time quantum since R > Q

So no of switches = R/Q

so total time in switching is S*(R/Q)

Answer 2

i think its zero...as every process will get blocked by R units . So no process will be completed.

hence eff. = 0

Comments

NO

ANS: 1/2 only