梅威瑟：
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\begin{enumerate}
\item Probability that more than 4 crews will be needed
\begin{align*}
&=\mbox{Probability that there are at least $5$ ships in the system}\\
&=1-(P_0+P_1+P_2+P_3+P_4)\\
&=1-\left[e^{-\rho}+\dfrac{\rho e^{-\rho}}{1!}+\dfrac{\rho^2 e^{-\rho}}{2!}+\dfrac{\rho^3 e^{-\rho}}{3!}+\dfrac{\rho^4 e^{-\rho}}{4!}\right]\quad\left[\because P_n=\dfrac{\rho^n e^{-\rho}}{n!}\right]\\
&=1-e^{-\rho}\left[1+\dfrac{\rho }{1!}+\dfrac{\rho^2 }{2!}+\dfrac{\rho^3 }{3!}+\dfrac{\rho^4 }{4!}\right]\\
&=1-e^{3/2}\left[1+\dfrac{{\left(3/2\right)} }{1!}+\dfrac{\left(3/2\right)^2 }{2!}+\dfrac{\left(3/2\right)^3 }{3!}+\dfrac{\left(3/2\right)^4 }{4!}\right]\\
&=0.019
\end{align*}
\item Problems arrive at a computer center in a Poisson fashion at an average rate of 5/day. The rules of the computer center are that any man waiting to get his problem solved must aid the man whose problem is currently being solved. If the time to solve a problem with one man has an exponential distribution with mean time $1/3$ day and if the average solving time is inversely proportional to the number of people working on the problem, find the expected system time for a person entering the line.\\
\textbf{Solution:}\\
Here $\lambda=5$ problems/day and $\mu=3$ problems/day.\\
Sine the expected time in a system is given by $W_s=\dfrac{L_s}{\lambda}$, therefore first we find the expected number of persons working at any instant, i.e. $L_s$. Now
\begin{equation*}
L_s =\sum_{n=0}^{\infty} nP_n=\sum_{n=0}^{\infty} n\dfrac{\rho^ne^{-\rho}}{n!}=\rho e^{-\rho}\sum_{n=1}^{\infty}\dfrac{\rho^{n-1}}{(n-1)!}=\rho e^{-\rho}\left[1+\dfrac{\rho}{1!}+\dfrac{\rho^2}{2!}
+\ldots\right]=\rho e^{-\rho}e^{\rho}=\rho=\dfrac{\lambda}{\mu}=\dfrac53
\end{equation*}
Therefore $W_s=\dfrac{5/3}{5}$ day $=\dfrac13$ day $=8$ hours.
\end{enumerate}
\subsection{Queuing Model-III: $(M/M/1):(N/FCFS)$}
This model is also based on the assumptions made for Model-I except a limit on the capacity of the system to accommodate only $N$ customers. Therefore in this model, when the queue length reaches to its maximum capacity $N$, no further arrival will be allowed to enter the system. Such a situation may arise in an emergency room in a hospital or at a barber shop with finite number of chairs for waiting customers etc.\\
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我遇到了一个重叠行的问题。每当我编译主要文章中的代码时，我都会发现标记的行重叠了。
为了重新检查，我仅编译了该段并创建了一个新文件，发现没有重叠的行。
在我的原始文章中，为了消除此错误，我必须\\在每个\end{align}或\end{enumerate}或之后添加\end{equation}。我仔细搜索了文章，发现问题始于第一幅图中标记的线之后，这意味着equation*我用于标记方程的环境可能存在一些问题。在标记线之前，无论我在哪里使用任何环境，都不会发生这种重叠。
我不知道这是否是编译器问题，或者是由于页边距缺少行高而导致的。任何形式的帮助都将不胜感激。
（我知道当你编译它时你不会得到任何重叠的线条，因为我单独编译它时没有得到重叠的线条。这就是为什么为了支持我的主张，我添加了重叠线条的图像（从我的原始文章中捕获的第一张图片）。我还提供了我用来制作这篇文章的每个序言）