基于PS排队模型的单台流媒体服务器性能分析
发布时间:2019-01-05 00:05
【摘要】:在有限容量的M/M/1/N处理机共享(processor-sharing,PS)模型基础上,引入等待队列,建立了单台服务器的性能评价模型.首先,通过求解系统微分方程,推导出稳态下用户平均服务响应时间和平均排队等待时间的表达式;然后,通过引入稳态时平均等待时间阈值和平均服务时间阈值,并按照服务质量好坏由用户评价的原则,提出了一种服务器性能评价方案,为服务器的设计提供了定量参考依据;最后,在并发用户数N=1和N→∞这两种情况下,计算出性能评价模型中用户平均服务响应时间和平均排队等待时间的值,并与典型的M/M/1排队模型以及无限容量的M/M/1-PS模型中的相应结果进行了比较.
[Abstract]:Based on the finite capacity M/1/N processor sharing (processor-sharing,PS) model, the performance evaluation model of a single server is established by introducing the waiting queue. First, by solving the differential equations of the system, the expressions of the average service response time and the average queue waiting time of the users under steady state are derived. Then, by introducing the mean waiting time threshold and the average service time threshold, and according to the principle that the quality of service is evaluated by the user, a server performance evaluation scheme is proposed. It provides a quantitative reference for the design of the server. Finally, the average service response time and queue waiting time of the users in the performance evaluation model are calculated under the two cases of concurrent user number Nu 1 and N N 鈭,
本文编号:2400985
[Abstract]:Based on the finite capacity M/1/N processor sharing (processor-sharing,PS) model, the performance evaluation model of a single server is established by introducing the waiting queue. First, by solving the differential equations of the system, the expressions of the average service response time and the average queue waiting time of the users under steady state are derived. Then, by introducing the mean waiting time threshold and the average service time threshold, and according to the principle that the quality of service is evaluated by the user, a server performance evaluation scheme is proposed. It provides a quantitative reference for the design of the server. Finally, the average service response time and queue waiting time of the users in the performance evaluation model are calculated under the two cases of concurrent user number Nu 1 and N N 鈭,
本文编号:2400985
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