194 lines
9.1 KiB
TeX
194 lines
9.1 KiB
TeX
\documentclass[全部作业]{subfiles}
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\input{mysubpreamble}
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\setcounter{chapter}{1}
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\begin{document}
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\chapter{应用层}
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\begin{enumerate}
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\cnitem[2] Consider an HTTP client that wants to retrieve a Web document at a given URL.
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The IP address of the HTTP server is initially unknown. What transport and
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application-layer protocols besides HTTP are needed in this scenario?
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\begin{zhongwen}
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首先，客户端需要根据URL中的域名找到IP地址，这需要DNS协议（这是应用层协议），而DNS协议是基于UDP协议的（这是传输层协议）。之后的HTTP协议如果不是HTTP3.0，则是基于TCP协议的（这是传输层协议），如果是HTTP3.0，则是基于UDP协议的（这是传输层协议）。
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总结：如果是HTTP3.0，则还需要应用层协议DNS，传输层协议UDP；
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如果不是HTTP3.0，则还需要应用层协议DNS，传输层协议UDP、TCP。
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\end{zhongwen}
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\cnitem[8] Suppose within your Web browser you click on a link to obtain a Web page. The
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IP address for the associated URL is not cached in your local host, so a DNS lookup is
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necessary to obtain the IP address. Suppose that $n$ DNS servers are visited before your
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host receives the IP address from DNS; the successive visits incur an RTT of
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RTT$_1$, $\cdots$, RTT$_n$. Further suppose that the Web page associated with the link contains
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exactly one object, consisting of a small amount of HTML text. Let RTT$_0$ denote the
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RTT between the local host and the server containing the object. Assuming zero
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transmission time of the object, how much time elapses from when the client clicks on
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the link until the client receives the object?
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\begin{zhongwen}
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首先，需要查询DNS，由于DNS基于UDP，因此需要$\sum_{i=1}^{n}RTT_i$的时间完成DNS查询。之后的网页应该是通过HTTP协议传输，假定这里使用的不是HTTP 3.0，那么这里是基于TCP协议的HTTP协议，因此需要$RTT_0$的时间建立TCP连接，之后请求和响应还需要一个$RTT_0$的时间。
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因此总的时间为：
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$$
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\sum_{i=1}^{n}RTT_{i} + 2RTT_0
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$$
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\end{zhongwen}
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\cnitem[9] Referring to Problem P8, suppose the HTML file references three very small
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objects on the same server. Neglecting transmission times, how much time elapses
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with
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\begin{enumerate}
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\item Non-persistent HTTP with parallel connections?
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\begin{zhongwen}
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每次请求和响应都需要开启一个新的TCP连接，但这些连接可以同时进行，因此需要的时间为：
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$$
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\sum_{i=1}^{n}RTT_{i}+2RTT_0+2RTT_0=\sum_{i=1}^{n}RTT_i+4RTT_0
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$$
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\end{zhongwen}
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\item Non-persistent HTTP with no parallel TCP connections?
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\begin{zhongwen}
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每次请求和响应都需要开启一个新的TCP连接，而且这些连接不能同时进行，因此需要的时间为：
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$$
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\sum_{i=1}^{n}RTT_{i}+2RTT_0+3\times 2RTT_0=\sum_{i=1}^{n}RTT_i+8RTT_0
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$$
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\end{zhongwen}
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\item Persistent HTTP?
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\begin{zhongwen}
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只开启一个TCP连接，并且在这个TCP连接中，HTTP请求互相不阻塞，也就是说不需要等到前一个HTTP请求收到响应后再发送后一个HTTP请求，由于忽略传输时间，因此可以认为这三个请求同时发出，同时得到响应。但这三个请求还是会依赖于HTML文件的响应，因此在DNS查询后，需要建立一次TCP连接，之后在这个连接上进行两批HTTP的请求和响应，因此总的时间为：
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$$
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\sum_{i=1}^{n}RTT_{i}+RTT_0+2RTT_0=\sum_{i=1}^{n}RTT_{i}+3RTT_0
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$$
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\end{zhongwen}
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\end{enumerate}
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\cnitem[11] What is the difference between \textit{MAIL FROM}: in SMTP and \textit{From}: in the mail
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message itself?
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\begin{zhongwen}
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SMTP中的\textit{MAIL FROM}表示发信的SMTP服务器的地址与发信的用户名，而邮件信息中的\textit{From}是用户自己填写的内容。
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也就是说，一封邮件可能经过很多次转发，而SMTP中的\textit{MAIL FROM}就是每次转发时直接发信的服务器和用户名，而邮件信息中的\textit{From}是写信的人。
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当然，由于\textit{From}是可以由用户自己编辑的，所以用户也可以填写一个虚假的发信的邮箱地址，而用户就无法在\textit{MAIL FROM}上造假了（除非用户用自己的SMTP服务器）。
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\end{zhongwen}
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\cnitem[15] Consider accessing your e-mail with POP3.
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\begin{enumerate}
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\item Suppose you have configured your POP mail client to operate in the
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\textit{download-and-keep} mode. Complete the following transaction:
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% 这种需要分栏的情况可以用minipage也可以用multicol，具体哪个更好呢？
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\begin{minipage}[H]{0.3\linewidth}
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\begin{minted}{text}
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C: list
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S: 1 498
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S: 2 912
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S: .
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C: retr 1
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S: blah blah ...
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S: ..........blah
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S: .
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?
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?
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\end{minted}
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\end{minipage}
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{\Huge$\longrightarrow$}
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\begin{minipage}[H]{0.3\linewidth}
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\begin{minted}{text}
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C: list
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S: 1 498
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S: 2 912
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S: .
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C: retr 1
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S: blah blah ...
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S: ..........blah
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S: .
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C: retr 2
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S: blah blah ...
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S: ..........blah
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S: .
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C: quit
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S: +OK POP3 server signing off
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\end{minted}
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\end{minipage}
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\item Suppose you have configured your POP mail client to operate in the
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\textit{download-and-delete} mode. Complete the following transaction:
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\begin{minipage}[H]{0.3\linewidth}
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\begin{minted}{text}
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S: ..........blah
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S: .
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?
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?
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\end{minted}
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\end{minipage}
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{\Huge$\longrightarrow$}
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\begin{minipage}[H]{0.3\linewidth}
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\begin{minted}{text}
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S: ..........blah
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S: .
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C: dele 1
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C: retr 2
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S: blah blah ...
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S: ..........blah
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S: .
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C: dele 2
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C: quit
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S: +OK POP3 server signing off
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\end{minted}
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\end{minipage}
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\item Suppose you have configured your POP mail client to operate in the
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\textit{download-and-keep} mode. Using your transcript in part (b), suppose you retrieve
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messages 1 and 2, exit POP, and then five minutes later you again access POP to
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retrieve new e-mail. Suppose that in the five-minute interval no new messages have
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been sent to you. Provide a transcript of this second POP session.
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\begin{minted}{text}
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C: list
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S: 1 498
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S: 2 912
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S: .
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C: retr 1
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S: blah blah ...
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S: ..........blah
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S: .
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C: retr 2
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S: blah blah ...
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S: ..........blah
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S: .
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C: quit
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S: +OK POP3 server signing off
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\end{minted}
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\end{enumerate}
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\item Consider query flooding, as discussed in Section 2.6. Suppose that each peer is
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connected to at most $N$ neighbors in the overlay network. Also suppose that the
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node-count field is initially set to $K$. Suppose Alice makes a query. Find an upper
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bound on the number of query messages that are sent into the overlay network.
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\begin{zhongwen}
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每个节点最多向$N$个邻居节点发送查询信息，但是不应向给自己发送查询信息的服务器发送查询信息，因此除了Alice外，每个邻居最多向$N-1$个邻居节点发送查询信息。而Alice因为是查询的发起者，所以最多向$N$个邻居节点发送查询信息。
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因此，设Alice发起的查询为第$0$次查询，邻居发送给邻居的邻居为第$1$次查询，依此类推。设$a_i$ $(i\in \mathbb{N})$表示第$i$次查询时新收到查询的邻居数，则$a_{i+1}=a_i\times (N-1)$，且$a_0=N$。这样的查询最多执行$K$轮，那么最多发送的查询信息数也就是所有收到查询的邻居数的总和，因此只需要将每次查询时新收到查询的邻居数求和即可。
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因此最多发送的查询信息数为：
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$$
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\underbrace{a_0+a_1+a_2+ \cdots}_{K个}=\underbrace{N+N(N-1)+N(N-1)^{2}+ \cdots }_{K个}
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$$
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可以看到$\{ a_n \}_{n=1}^{\infty}$构成等比数列，所求的值为它的前$K$项求和，因此上式可化为：
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$$
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\frac{N(1-(N-1)^{K})}{1-(N-1)}=\frac{N(1-(N-1)^{K})}{2-N}
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$$
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\end{zhongwen}
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\end{enumerate}
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\end{document} |