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20 ../../exercises/sockets.rst:175
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — While the provided examples show the usage of a `char` array as the data buffer, implementers should **never** assume that it contains a string. C programs rely on the `char` type to refer to a 8-bit long value, and arbitrary binary values can be exchanged over the network (i.e., the ``\0`` value does not delimit the end of the data). 336/10000 · 336
20 ../../principles/sharing.rst:186
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — If `R1` has enough buffers, it will be able to absorb the load without having to discard packets. The packets sent by hosts `A` and `B` will reach their final destination `C`, but will experience a longer delay than when they are transmitting alone. The amount of buffering on the network node is the first parameter that a network operator can tune to control congestion inside his network. Given the decreasing cost of memory, one could be tempted to put as many buffers [#fbufferbloat]_ as possible on the network nodes. Let us consider this case in the network above and assume that `R1` has infinite buffers. Assume now that hosts `A` and `B` try to transmit a file that corresponds to one thousand packets each. Both are using a reliable protocol that relies on go-back-n to recover from transmission errors. The transmission starts and packets start to accumulate in `R1`'s buffers. The presence of these packets in the buffers increases the delay between the transmission of a packet by `A` and the return of the corresponding acknowledgment. Given the increasing delay, host `A` (and `B` as well) will consider that some of the packets that it sent have been lost. These packets will be retransmitted and will enter the buffers of `R1`. The occupancy of the buffers of `R1` will continue to increase and the delays as well. This will cause new retransmissions, ... In the end, only one file will be delivered (very slowly) to the destination, but the link `R1-R2` will transfer much more bytes than the size of the file due to the multiple copies of the same packets. This is known as the `congestion collapse` problem :rfc:`896`. Congestion collapse is the nightmare for network operators. When it happens, the network carries packets without delivering useful data to the end users. 1793/10000 · 1793
21 ../../exercises/ex-sharing.rst:491
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — In general, the links have a non-zero delay. This is illustrated in the figure below where a delay has been added on the link between `R` and `C`. 146/10000 · 146
21 ../../exercises/sockets.rst:177
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — Using this code, the program will read and print an arbitrary message received from an arbitrary peer who knows the program's socket address. If we want to know the address of the peer that sent us the message, we can use the ``recvfrom`` system call. This is what a modified version of ``bind_and_receive_from_peer`` is doing below. 333/10000 · 333
21 ../../principles/naming.rst:116
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — Each root nameserver maintains the list [#froot]_ of all the nameservers that are responsible for each of the top-level domain names and their addresses [#frootv6]_. All root nameservers cooperate and provide the same answers. By querying any of the root nameservers, a DNS client can obtain the nameserver that is responsible for any top-level-domain name. From this nameserver, it is possible to resolve any domain name. 422/10000 · 422
21 ../../principles/network.rst:103
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — the information to be sent 26/10000 · 26
21 ../../principles/referencemodels.rst:168
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — The upper layer of our architecture is the `Application layer`. This layer includes all the mechanisms and data structures that are necessary for the applications. We will use Application Data Unit (ADU) or the generic Service Data Unit (SDU) term to indicate the data exchanged between two entities of the Application layer. 325/10000 · 325
21 ../../principles/reliability.rst:59
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — To understand some of the principles behind the physical transmission of information, let us consider the simple case of an electrical wire that is used to transmit bits. Assume that the two communicating hosts want to transmit one thousand bits per second. To transmit these bits, the two hosts can agree on the following rules : 330/10000 · 330
21 ../../exercises/network.rst:185
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — The network operator uses would like to have the following paths in this network : 82/10000 · 82
21 ../../exercises/reliability.rst:41
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — Same question when using selective repeat instead of go-back-n. Note that the answer is not necessarily the same. 113/10000 · 113
21 ../../glossary.rst:155
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — The Border Gateway Protocol is the interdomain routing protocol used in the global Internet. 92/10000 · 92
21 ../../exercises/transport.rst:148
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — DATA segments contain a sequence number that is incremented for each DATA segment transmitted 93/10000 · 93
21 ../../principles/sharing.rst:188
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — Congestion collapse on the Internet 35/10000 · 35
21 ../../protocols/dns.rst:68
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — The DNS is mainly used to find the address that corresponds to a given name. However, it is sometimes useful to obtain the name that corresponds to an IP address. This done by using the `PTR` (`pointer`) `RR`. The `RData` part of a `PTR` `RR` contains the name while the `Name` part of the `RR` contains the IP address encoded in the `in-addr.arpa` domain. IPv4 addresses are encoded in the `in-addr.arpa` by reversing the four digits that compose the dotted decimal representation of the address. For example, consider IPv4 address `192.0.2.11`. The hostname associated to this address can be found by requesting the `PTR` `RR` that corresponds to `11.2.0.192.in-addr.arpa`. A similar solution is used to support IPv6 addresses :rfc:`3596`, but slightly more complex given the length of the IPv6 addresses. For example, consider IPv6 address `2001:1890:123a::1:1e`. To obtain the name that corresponds to this address, we need first to convert it in a reverse dotted decimal notation : `e.1.0.0.1.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.a.3.2.1.0.9.8.1.1.0.0.2`. In this notation, each character between dots corresponds to one nibble, i.e. four bits. The low-order byte (`e`) appears first and the high order (`2`) last. To obtain the name that corresponds to this address, one needs to append the `ip6.arpa` domain name and query for `e.1.0.0.1.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.a.3.2.1.0.9.8.1.1.0.0.2.ip6.arpa`. In practice, tools and libraries do the conversion automatically and the user does not need to worry about it. 1514/10000 · 1514
21 ../../principles/security.rst:279
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — This principle is important because it remains the basic assumption of all cryptographers. Any system that relies on the secrecy of its algorithm to be considered secure is doomed to fail and be broken one day. 210/10000 · 210
21 ../../exercises/dns.rst:57
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — Perform the same analysis for a popular website such as `www.google.com`. What is the lifetime associated to the corresponding IPv6 address ? If you perform the same request several times, do you always receive the same answer ? Can you explain why a lifetime is associated to the DNS replies ? 294/10000 · 294
21 ../../exercises/ipv6.rst:154
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — Routers ``R2`` and ``R3`` are buggy in this network. Besides the routes for their local interfaces (not shown in the figure), they only have a default route which is shown in the figure below. 192/10000 · 192
21 ../../exercises/lan.rst:211
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — By using :manpage:`brtcl(8)`, we can easily observe the state of the Spanning Tree protocol on the different switches. Let us start with ``s3``, i.e. the root of the Spanning Tree. 180/10000 · 180
21 ../../exercises/routing-policies.rst:191
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — Can you explain why at the eighth, ninth and tenth hopes several IPv6 addresses are reported in the ``traceroute6`` output ? 124/10000 · 124
21 ../../exercises/routing-protocols.rst:265
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — OSPFv3 also includes `LS Acknowledge` packets that acknowledge the correct reception of link state information. 111/10000 · 111