#self doubt #round robin #cpu scheduling

Question

in round robin algorithm if a process with burst time = 10 arrived at t=0 and time quantum is 2 units, context switch time is 2 units . Then completion time of the process will be?

Answer 1

 

Arrival time = 0, Burst time = 10 units, Time quantum = 2 units and Context switch time = 2 units.

Ready queue = P1 P1 P1 P1 P1

P1

P1

P1

P1

P1

0                       2 4                     6 8                       10 12           14 16                    18         

Number of context switch = 4

so CS time = 4*2 = 8

Completion time = 8 + 10 = 18 units.

 

Comments

if the same process again getting the turn, then it requires CONTEXT SWITCH ? - depends upon implementation of OS

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It do sir, but we can assume windows OS which implements RR and CS happens for the single process.

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implementation depends question - will not ask in GATE.

you can't give any assurance that, question based on Windows OS( as you said, Windows Os can do CS for one process.)

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Is context switching required for only one process in Round robin scheduling ? Yes, even if there is just 1 process, after using up its time slice, it would save the current execution state, go back to the ready queue and be scheduled to run again.  
if we have only one process of 10 time units. If the quantum is 12 time units, the process finishes in less than 1 time quantum, with no overhead. If the quantum is 6 time units, however, the process requires 2 quanta, resulting in a context switch. If the time quantum is 1 time unit, then nine context switches will occur, slowing the execution of the process accordingly