# 计算机代写|密码学与网络安全代写cryptography and network security代考|CS6260

## 计算机代写|密码学与网络安全代写cryptography and network security代考|Solution for a Stable Network

The formula that relates the probability of $k$ existence packets at a given time on a network is given by the solution of the following equations, which depends on the configuration assumed for the system (Kleinrock, 1975)
$$p_k=p_o \prod_{i=0}^{k-1} \frac{\lambda_i}{\mu_{i+1}},$$

$$\begin{gathered} p_o=\frac{1}{1+\sum_{k=1}^{\infty} \prod_{i=0}^{k-1} \frac{\lambda_i}{\mu_{i+1}}} \ N=\sum_{k=0}^{\infty} k p_k, \ \sigma_N^2=\sum_{k=0}^{\infty}(k-N)^2 p_k, \ N=\lambda T \end{gathered}$$
in which:

• $p_o$ – probability that there will be no packages in the system;
• $p_k$ – probability of having $k$ packages in the system;
• $\lambda_i$ – package arrival fee to state $i$;
• $\mu_{i+1}$ – departure fee for state packages $i+1$
• $N$ – average number of packages in the system;
• $\sigma_N^2$ – variance of the number of packages in the system;
• $T$ – average waiting time in line.
These equations serve as a starting point for calculating several queuing theory problems and for other solutions shown below in this chapter.

## 计算机代写|密码学与网络安全代写cryptography and network security代考|Constant Arrival Rate System

The results described do not depend on the parameters $\lambda$ and $\mu$ individually, but only on the quotient between them, the ratio between arrival, and processing fees.

The average waiting time in the queue is given by the quotient between the average number of packages and the average rate of arrival of packages to the system. Therefore,
$$T=\frac{N}{\lambda}=\frac{1 / \mu}{(1-\rho)} .$$
Figure $6.9$ illustrates the increase in latency in a network, as the $\rho$ coefficient increases.

When $\rho=0$, the $T$ latency corresponds to the expected service time for the first package, being equal to $1 / \mu$. This average time depends on the parameter $\mu$ individually, contrary to the previous ones.

Note that, as $\rho$ tends to the unit, that is, the rate of packets that arrives on the network approaches the processing rate of the network’s servers, the average number of packets in the system and the average waiting time in the queue tend to grow in an unlimited way. This type of behavior when $\rho$ tends to 1 is characteristic of almost all queuing systems cncountered.

In general, this feature is exploited by hackers for an attack known as denial of service (DoS), where the network receives a flood of packets so that the arrival rate approaches processing rate, and the number of packets on the network tends to infinity.

Another interesting amount to be analyzed is the probability that there must be at least $k$ packages in the system. This probability is given by
\begin{aligned} P[x&\geq k \text { in the system }]=\sum_{i=k}^{\infty} p_i \ &=\sum_{i=k}^{\infty}(1-\rho) \rho^i=\rho^k \end{aligned}
for $\rho$ less than 1 , since $\sum_{i=k}^{\infty} \rho^i$ only converges if $\rho<1$.

# 密码学与网络安全代考

## 计算机代写|密码学与网络安全代写cryptography and network security代考|稳定网络解决方案

$$p_k=p_o \prod_{i=0}^{k-1} \frac{\lambda_i}{\mu_{i+1}},$$

$$\begin{gathered} p_o=\frac{1}{1+\sum_{k=1}^{\infty} \prod_{i=0}^{k-1} \frac{\lambda_i}{\mu_{i+1}}} \ N=\sum_{k=0}^{\infty} k p_k, \ \sigma_N^2=\sum_{k=0}^{\infty}(k-N)^2 p_k, \ N=\lambda T \end{gathered}$$

• $p_o$ -系统中没有包的概率;
• .
• $\lambda_i$ -包裹到达州的费用 $i$
• .$\mu_{i+1}$ -国家包裹的离境费 $i+1$
• $N$ -系统中平均包数;
• $\sigma_N^2$ -系统中包数量的方差;
• $T$ -平均排队等候时间。
这些方程作为计算几个排队理论问题和本章中所示其他解的起点
计算机代写|密码学与网络安全代写密码学和网络安全代考|固定到达率系统
所描述的结果不单独依赖于参数$\lambda$和$\mu$，而只依赖于它们之间的商、到货和加工费之间的比率队列中的平均等待时间由包裹的平均数量与包裹到达系统的平均速率之间的商给出。因此，
$$T=\frac{N}{\lambda}=\frac{1 / \mu}{(1-\rho)} .$$
图$6.9$说明了随着$\rho$系数的增加，网络中延迟的增加当$\rho=0$时，$T$延迟对应第一个包的预期服务时间，等于$1 / \mu$。这个平均时间分别取决于参数$\mu$，与前面的参数相反注意，当$\rho$趋于单位时，即到达网络的包的速率接近网络服务器的处理速率，系统中的平均包数和队列中的平均等待时间趋于无限增长。当$\rho$趋向于1时，这种类型的行为是几乎所有遇到的排队系统的特征
通常情况下，黑客利用这一特性进行DoS (denial of service)攻击，网络收到大量报文，使到达速率接近处理速率，网络上的报文数量趋于无穷大 另一个要分析的有趣的量是系统中至少有$k$个包的概率。这个概率由$\rho$
\begin{aligned} P[x&\geq k \text { in the system }]=\sum_{i=k}^{\infty} p_i \ &=\sum_{i=k}^{\infty}(1-\rho) \rho^i=\rho^k \end{aligned}
给出，因为$\sum_{i=k}^{\infty} \rho^i$只在$\rho<1$ 时收敛

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