以包为单位进行数据传输的网络模型中,一般会有这三个表示网络性能的指标:包时延,丢包率和网络吞吐量。
- 包时延(packet delay)
- 处理时延:误差检测,决定输出链路
- 排队时延:等待传输的输出链路,以来与路由器处理拥堵的水平(当路由缓冲区有包在排队(等待转发)时,当包到达率超过链路容量)
- 传输时延:将比特推送到链路的时间<-包长度/链路带宽(bps)
- 传播时延:信号到达终点的时间<-物理链路的长度/传输速度(传输媒介)
- 丢包率(packet loss rate)
包到达时缓冲队列已满,即丢包/由传输层协议决定丢包率
- 吞吐量(throughput)
成功传输的比特率;端到端路径会限制端到端吞吐量,从而产生瓶颈链路。
无线网络中为了方便研究,会经常使用无记忆性的指数分布和均匀概率的泊松分布来对随机事件进行建模。
指数分布https://en.wikipedia.org/wiki/Exponential_distribution
- If we think of X as being the lifetime of some instrument, then equation states that the probability (that the instrument lives for at least x+t hours given that it has survived t hours) is the same as the initial probability (that it lives for at least x hours).
- In other words, if the instrument is alive at time t , then the distribution of the remaining amount of time that it survives is the same as the original lifetime distribution; that is, the instrument does not remember that it has already been in use for a time t .
注:只有连续的指数分布才有无记忆的性质。
泊松分布https://en.wikipedia.org/wiki/Poisson_distribution
- Customers that arrive at a post office during a day
- Wrong phone calls received during a week
- Students that go to the instructor’s office during office hours
- … and packets that arrive at a network switch
- n traffic streams, with independent identically distributed (iid) interarrival times with PDF F(s) – not necessarily exponential
- Arrival rate of each stream l/n
- As n→∞, combined stream can be approximated by Poisson under mild conditions on F(s) – e.g., F(0)=0, F’(0)>0
小理论
- 系统的平均数量=每个顾客在系统花费的平均时间*顾客到达率
- 排队的顾客数量=每个顾客排队的平均时间*顾客到达率
概率论
1/conditional probabilities: P(E|F)=P(EF)|P(F)
2/Independent events.
Definition: Two events E and F are said to be independent if P(EF) = P(E)P(F)
The above equation implies that E and F are independent if P(E|F) = P(E)
[which also implies that P(F|E) = P(F)].
Remarks: E and F are independent if knowledge that F has occurred does not affect the probability that E occurs.
3/Law of total probability
Let E and F be events. We may express E as E = EF ∪ EFc
Because in order for a point to be in E, it must either be in both E and F, or it must be in E and not in F. Since EF and EFc are mutually exclusive, we have that
P(E) = P(EF)+P(EFc)
= P(E|F)P(F) +P(E|Fc)P (Fc)
= P(E|F)P(F) +P(E|Fc)(1 −P(F))