For this example, how large of a quantum value will Round Robin need to be for i

Question

For this example, how large of a quantum value will Round Robin need to be for it to mimic FCFS?

EntriesAverage Time Name Key Time Processes Finished CPU Utilization Throughput CSTLA CPUIOCPUIO 00993611 0.00113.1611 229208|| 8.7 10.00 01000010-00110.431 229208|| 8.7 10.00 009950 0.001117.1512107 2089510.00 009970 0.001 16.35 518 208 3.86110.00 009936 0.001115.55 315 208 6.3510.00 1 FCFS 2012.91 myrun 3 RR 1 2009.99 myrun 4 RR 5 2005.95 myrun-5ER 1012012.9 201-201 myrun6ER132012.911-201 20 myrun RR 201201 2.911-201-201 myrun 8ER 2512012.912011-201 993 5900993610.00115 51 3151208][ 6.35][ 10.00! 009936 0.00 14.89 285 2087.0210.00 009936 0.00 14.16 258 2087.7510.00 009936 0.00 14.23 258 2087.7510.00 NameKevAverage Minimum MaximumSDAverage Minimum MaximumSD myrun 1|| FCFS| 152898|| 950.0 2012.91|443.37|1324.98|| 830.06|| 170291|18.14 1 . 31.54|1880.0 2847 myrun 3||RR1||1928.06|| 1847.5 2009.99||6437|| 1724.06|| 169776| 1754.58|| .97 myrun 4RR 5 1843.68 1627.562005.95141.89 1639.681507.561729.573.17 myrun 5]|RR101768. 55][ 1478.731 2012.91 20302|1564.55][ 1358.75]| 1702.916.17 myrun 6 RR 151702.661317.20 2012.91 269.621498.66|1197.201702.91 9.48 myrun_7||RR 2011 1629.21|| 1235.1 2012.91||339.86|| 1425.21|| 1115.12|| 1702.91||12.93 myrun-8IRR 251635.96| 270.1 2012.91][332.46 [143 1.96 1150. 1211702.911256| myrun_2|SJF 1246.93391.54 2000.00494.361042.93 RR 25 1635.96 1270.12 2012.91 332.46 1431 96 1150.12 30-5-7 33|3|3 || |

Explanation / Answer

3.) SJF- Shortest Job First algoithm executes the jobs with less burst time first .

It is the optimal algorithm ; it gives minimum average waiting time for a given set of processes becuase by moving a short process before a long one, the waiting time of short process decreases and the algorithm increases waiting time of the long process, so avg. waiting time decreases.

This is because wait time is calculated by C.T.- A.T.=B.T.+ W.T.(Completion time+ Waiting time = Burst time+ Arrival time)

For lesser burst time, if we keep completion time less , the wait time would be low.

While for the processes with higher burst time, higher completion time can also work so this overall produces a lesser average waiting time

4.)

The performance of Round Robin Scheduling depends on the size of time quantum. If the quantum value is too large, it behaves as if it is a FCFS scheduling algorithm. If the time quantum value is too low, context switchcs are too frequent causing additional burden of workload on the CPU. .

If both the quantum size and the context-switch overhead are close to zero, then RR is called processor sharing because it appears to the processes that they each have its own CPU. All ready processes have equal share of CPU time.

For Round Robin to behave as FCFS, the time quantum should equal to the maximum of Burst time among all the processes.

e.g. say there are three processes: P1, P2, P3 with burst time 3,4,7 respectively then any time quantum above 7ms would make round robin behave as FCFS

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