05/08/2015 RoundRobinCPUSchedulingGATEFundas

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ShortestRemainingtimefirst(SRTF)CPUscheduling

MiscellaneousexamplesonCPUschedulingwhereprocessesspendsometimedoingCPUandsometimedoingI/O

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RoundRobinCPUSchedulingOct112014

RoundRobinschedulingismostpracticalschedulingalgorithmsusedforschedulingtheprocessesontotheCPU.Inthisalgorithmeveryprocessisexecutedforatimequantum(timeslice)andthenpreemptedtogivechancetootherprocess.

Criteria:Timequantum(Timeslice),arrivaltimeMode:Preemptive

Flowchartofroundrobinscheduling

Example1:Findaverageresponsetime,averageturnaroundtimeandaveragewaitingtime

P.No. Arrivaltime Bursttime

1 0 4

2 1 5

3 2 2

4 3 1

5 4 6

6 6 3

Timequantum=2

Solution:Inroundrobinschedulingwehavetomaintainonereadyqueueinwhichprocessesarekeptandexecutedinthesameordertheyentered.

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05/08/2015 RoundRobinCPUSchedulingGATEFundas

http://www.gatefundas.com/roundrobincpuscheduling/ 2/5

Readyqueue:emptyP1arrivesatt=0Readyqueue:P1soaccordingtoflowchartB.T

05/08/2015 RoundRobinCPUSchedulingGATEFundas

http://www.gatefundas.com/roundrobincpuscheduling/ 3/5

2 4 6

3 3 7

4 1 9

5 2 2

6 6 3

Timequantum=3

Solution:FollowingtheexplanationofaboveexampleGANTTchartandreadyqueueofthisexampleareasfollows

ReadyQueue:P4,P5,P3,P2,P4,P1,P6,P3,P2,P4,P1,P3

P4 P5 P3 P2 P4 P1 P6 P3 P2 P4 P1 P3

0 1 4 6 9 12 15 18 21 24 27 303233

Responsetime=firstresponsetimearrivaltimeTurnaroundtime=completiontimearrivaltimeWaitingtime=turnaroundtimebursttime

P.No. Arrivaltime Bursttime Completiontime Turnaroundtime Waitingtime responsetime

1 5 5 32 27 22 10

2 4 6 27 23 17 5

3 3 7 33 30 23 3

4 1 9 30 29 20 0

5 2 2 6 4 2 2

6 6 3 21 15 12 12

Total=128 Total=96 Total=32

AverageResponsetime=32/6=5.3Averageturnaroundtime=128/6=21.3Averagewaitingtime=96/6=16

Example3:Consider4processesP1,P2,P3,P4allarrivinginthesameorder.Thebursttimerequirementsoftheprocessesare4,1,8,1respectivelythenwhatisthecompletiontimeofprocessP1.AssumingroundrobinschedulingwithTimeQuantum=1a)7b)8c)9d)10

Solution:Givenallprocessesarriveinsameordersotakearrivingtimeofallprocessesas0

P.No Arrivaltime Bursttime

1 0 4

2 0 1

3 0 8

4 0 1

AsarrivaltimesofalltheprocessesaresamesofirstplacetheprocessesinreadyqueuebasedonlowestprocessnumberfirstReadyQueue:=P1,P2,P3,P4Nowproceedinthesamewayasexplainedabove.

05/08/2015 RoundRobinCPUSchedulingGATEFundas

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P1 P2 P3 P4 P1 P3 P1 P3 P1 P3

0 1 2 3 4 5 6 7 8 9

CompletiontimeofP1is9

Example4:[GATE2012]Considerthreeprocessesshowninthetable

Process Arrivaltime Bursttime

P1 0 5

P2 1 7

P3 3 4

ThecompletionorderoftheprocessesunderthepoliciesFCFSandroundrobinwithtimequantumof2units(a)FCFS:P1,P2,P3RR:P1,P2,P3(b)FCFS:P1,P3,P2RR:P1,P3,P2(c)FCFS:P1,P2,P3RR:P1,P3,P2(d)FCFS:P1,P3,P2RR:P1,P2,P3

Solution:

P1arrivesatt=0ReadyQueue:P1forP1B.T

05/08/2015 RoundRobinCPUSchedulingGATEFundas

http://www.gatefundas.com/roundrobincpuscheduling/ 5/5

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Examplesonlosslessjoindecomposition2commentsayearago

nishantagrawal25AC+isacbdegfororiginaldependencysetbutwhenyouwilllookindecomposedrelationitisonly

MinimalcoverorCanonicalcoverofgivenFDset2commentsayearago

nishantagrawal25GladtoknowthattheseexampleshelpedyouThanksforpointingoutthetypo,It'scorrectednowPleasetellif

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meli 7monthsagoAreusurethat?MynoteshowTurnarround=completesubmited..notTurnarround=completearrival

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nishantagrawal25 7monthsagoMod >meliSubmittedandarrivalmeanssamethinghere...SubmittedmeanswhenjobissubmittedtotheCPUwhichissameasarrivaltime

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