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This is an old revision of this page, as edited by 81.26.170.61 (talk) at 11:54, 6 June 2024 (Process scheduling: error on the example: Reply). The present address (URL) is a permanent link to this revision, which may differ significantly from the current revision.

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Someone needs to expand on this

- I completely agree with you. —Preceding unsigned comment added by 213.86.119.247 (talk) 13:33, 12 February 2008 (UTC)[reply]


"Round-robin scheduling results in max-min fairness if the data packets are equally sized, since the data flow that has waited longest time is given scheduling priority"

These two articles should not be merged, they are about two widely different subjects. /Marius —Preceding unsigned comment added by 83.108.127.10 (talk) 13:08, 20 May 2010 (UTC)[reply]

Q: How come the data flow that has waited the longest time is given scheduling priority if Round-Robin is based on forwarding the following flow if a current serviced flow is absent?

A: Perhaps because of the second clause is wrong. Please rephrase it. "... since all active flows will take turns in a fair manner, independently of how many packets that are in queue in the flow"? Mange01 (talk) 21:24, 6 May 2008 (UTC)[reply]

a — Preceding unsigned comment added by 124.106.71.44 (talk) 05:07, 3 September 2011 (UTC)[reply]

Fairness of round-robin scheduling

It was not immediately clear to me why round-robin scheduling would favor the biggest processes or data packet flows. I eventually figured out that this would only be true in the complete absence of quantum-slicing, or if the quanta were large relative to the size of the processes or data packets, or if the data packet size varied from one job to another. I made edits accordingly. However, it would be good if someone could confirm this, preferably with reference to a source. -AlanUS (talk) 15:20, 11 October 2011 (UTC)[reply]

Process scheduling: error on the example

The example provided in the section 'Process Scheduling' as an image is an excelent resource!

However after following the process I found in 'Excute time = 475' a little mistake. P3 cannot be completed at execute time 475 because the quanta ends at 400. Then, by 475 P1 should have been completed and P4 should be in processing state. Jcammmmm (talk) 12:49, 20 May 2024 (UTC)[reply]

It is a good idea to update the diagram 81.26.170.61 (talk) 11:54, 6 June 2024 (UTC)[reply]