2. cnp3-ebook
3. English
4. Zen

3 ../../protocols/dns.rst:12
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — The last component of the Domain Name System is the DNS protocol. The original DNS protocol runs above both the datagram and the bytestream services. In practice, the datagram service is used when short queries and responses are exchanged, and the bytestream service is used when longer responses are expected. In this section, we first focus on the utilization of the DNS protocol above the datagram service. We will discuss later other recently proposed protocols to carry DNS information. 491/10000 · 491
3 ../../principles/transport.rst:12
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — There are two important models used to organize a networked application. The first and oldest model is the client-server model. In this model, a server provides services to clients that exchange information with it. This model is highly asymmetrical: clients send requests and servers perform actions and return responses. It is illustrated in the figure below. 361/10000 · 361
4 ../../exercises/sockets.rst:96
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — Networked applications were usually implemented by using the :term:`socket` :term:`API`. This API was designed when TCP/IP was first implemented in the `Unix BSD`_ operating system [Sechrest]_ [LFJLMT]_, and has served as the model for many APIs between applications and the networking stack in an operating system. Although the socket API is very popular, other APIs have also been developed. For example, the STREAMS API has been added to several Unix System V variants [Rago1993]_. The socket API is supported by most programming languages and several textbooks have been devoted to it. Users of the C language can consult [DC2009]_, [Stevens1998]_, [SFR2004]_ or [Kerrisk2010]_. The Java implementation of the socket API is described in [CD2008]_ and in the `Java tutorial <http://java.sun.com/docs/books/tutorial/networking/sockets/index.html>`_. In this section, we will use the C socket API to illustrate the key concepts. 929/10000 · 929
4 ../../exercises/ex-sharing.rst:17
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — To understand the operation of Medium Access Control algorithms, it is often interesting to use a geometric representation of the transmission of frames on a shared medium. This representation is suitable if the communicating devices are attached to a single cable. Consider a simple scenario with a host connected at one end of a cable. For simplicity, let us consider a cable that has a length of one kilometer. Let us also consider that the propagation delay of the electrical signal is five microseconds per kilometer. The figure below shows the transmission of a 2000 bits frame at 100 Mbps by host A on the cable. 619/10000 · 619
4 ../../exercises/intro.rst:12
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — A good understanding of the topics covered by this e-book can only be obtained by solving the proposed exercises. Reading the e-book from the first to the last page is not sufficient to get a detailed knowledge of computer networking. 234/10000 · 234
4 ../../principles/network.rst:15
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — The main objective of the network layer is to allow hosts, connected to different networks, to exchange information through intermediate systems called :term:`router`. The unit of information in the network layer is called a :term:`packet`. 240/10000 · 240
4 ../../exercises/reliability.rst:15
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — Transmission links have sometimes different upstream and downstream bandwidths. A typical example are access networks that use ADSL (Asymmetric Digital Subscriber Lines). Consider two hosts connected via an ADSL link having an upstream bandwidth of 1 Mbps and a downstream bandwidth of 50 Mbps. The propagation delay between the two hosts is 10 milliseconds. What is the maximum throughput, expressed in frames/second, that the alternating bit protocol can obtain on this link if each data frame has a length of 125 bytes and acknowledgments are 25 bytes long. Same question if the protocol is modified to support 1500 bytes long data frames. 642/10000 · 642
4 ../../glossary.rst:180
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — AIMD 4/10000 · 4
4 ../../exercises/transport.rst:20
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — Open questions 14/10000 · 14
4 ../../principles/referencemodels.rst:19
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — The five layers reference model 31/10000 · 31
4 ../../principles/reliability.rst:31
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — To enable the two hosts to exchange information, they need to be linked together by some kind of physical media. Computer networks have used various types of physical media to exchange information, notably : 207/10000 · 207
4 ../../principles/security.rst:46
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — Alice and Bob 13/10000 · 13
4 ../../exercises/dns.rst:17
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — The Domain Name System (DNS) plays a key role in the Internet today as it allows applications to use fully qualified domain names (FQDN) instead of IPv4 or IPv6 addresses. When using the DNS, it is important to remember the role of the different types of DNS records. 267/10000 · 267
4 ../../exercises/email.rst:10
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — it is often easy to write a small prototype that implements a part of the protocol 82/10000 · 82
4 ../../exercises/http.rst:19
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — Instead of using `telnet` on port 80, it is also possible to use a command-line tool such as curl_. Use curl_ with the `--trace-ascii tracefile` option to store in `tracefile` all the information exchanged by curl_ when accessing the server. 241/10000 · 241
4 ../../exercises/ipv6.rst:15
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — An IPv6 forwarding table contains a list of IPv6 prefixes with their associated nexthop or outgoing interface. When an IPv6 router receives a packet, it forwards it according to its forwarding table. Note that IPv6 routers forward packets along the *longest match* between the destination address of the packet and the routes in the forwarding table. 350/10000 · 350
4 ../../exercises/lan.rst:26
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — Fig. 1. A small network composed of Ethernet switches 53/10000 · 53
4 ../../exercises/routing-policies.rst:56
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — In this network, what are the paths : 37/10000 · 37
4 ../../exercises/routing-protocols.rst:15
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — We first use IPMininet_ to explore the operation of OSPFv3, the version of OSPF that supports IPv6. We create a simple network with three routers and two hosts as shown in the figure below. 189/10000 · 189
4 ../../exercises/tcp-2.rst:16
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — Packet capture tools like tcpdump_ and Wireshark_ are very useful to observe the segments that transport protocols exchange. TCP is a complex protocol that has evolved a lot since its first specification :rfc:`793`. TCP includes a large number of heuristics that influence the reaction of a TCP implementation to various types of events. A TCP implementation interacts with the application through the ``socket`` API. 417/10000 · 417