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146 ../../glossary.rst:232
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — root nameserver 15/10000 · 15
146 ../../principles/dv.rst:209
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — `A` sends its distance vector :math:`[A=0,B=\infty,C=\infty,D=1,E=\infty]`. `D` knows that it cannot reach `B` anymore via `A` 126/10000 · 126
146 ../../principles/reliability.rst:498
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — a `payload` that contains the user data 39/10000 · 39
146 ../../principles/transport.rst:852
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — Ambiguities caused by excessive delays 38/10000 · 38
146 ../../principles/sharing.rst:930
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — Let us first consider the simple problem of a set of :math:`i` hosts that share a single bottleneck link as shown in the example above. In this network, the congestion control scheme must achieve the following objectives [CJ1989]_ : 232/10000 · 232
146 ../../protocols/email.rst:258
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — N 1/10000 · 1
146 ../../protocols/ipv6.rst:492
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — The fragments of an IPv6 packet may arrive at the destination in any order, as each fragment is forwarded independently in the network and may follow different paths. Furthermore, some fragments may be lost and never reach the destination. 239/10000 · 239
146 ../../protocols/wifi.rst:23
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — 802.11a 7/10000 · 7
146 ../../protocols/tcp.rst:439
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — Example computation of the `rto` 32/10000 · 32
146 ../../bibliography.rst:356
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — Vyncke, E., Paggen, C., `LAN Switch Security: What Hackers Know About Your Switches <http://books.google.com/books?id=HkraAQAACAAJ>`_, Cisco Press, 2007 152/10000 · 152
147 ../../glossary.rst:234
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — A name server that is responsible for the root of the domain names hierarchy. There are currently a dozen root nameservers and each DNS resolver See http://www.root-servers.org/ for more information about the operation of these root servers. 241/10000 · 241
147 ../../principles/dv.rst:210
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — `D` sends its distance vector :math:`[D=0,B=\infty,A=1,C=2,E=1]` to `A` and `E`. `A` recovers routes towards `C` and `E` via `D`. 129/10000 · 129
147 ../../principles/reliability.rst:500
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — Some headers also include a `length` field, which indicates the total length of the frame or the length of the payload. 119/10000 · 119
147 ../../principles/transport.rst:856
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — To deal with this problem, transport protocols combine two solutions. First, they use 32 bits or more to encode the sequence number in the segment header. This increases the overhead, but also increases the delay between the transmission of two different segments having the same sequence number. Second, transport protocols require the network layer to enforce a `Maximum Segment Lifetime (MSL)`. The network layer must ensure that no packet remains in the network for more than MSL seconds. In the Internet the MSL is assumed [#fmsl]_ to be 2 minutes :rfc:`793`. Note that this limits the maximum bandwidth of a transport protocol. If it uses `n` bits to encode its sequence numbers, then it cannot send more than :math:`2^n` segments every MSL seconds. 755/10000 · 755
147 ../../principles/sharing.rst:932
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — The congestion control scheme must `avoid congestion`. In practice, this means that the bottleneck link cannot be overloaded. If :math:`r_i(t)` is the transmission rate allocated to host :math:`i` at time :math:`t` and :math:`R` the bandwidth of the bottleneck link, then the congestion control scheme should ensure that, on average, :math:`\forall{t} \sum{r_i(t)} \le R`. 372/10000 · 372
147 ../../protocols/email.rst:258
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — 30 2/10000 · 2
147 ../../protocols/ipv6.rst:494
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — The reassembly algorithm used by the destination host is roughly as follows. First, the destination can verify whether a received IPv6 packet is a fragment or not by checking whether it contains a fragment header. If so, all fragments with the some identification must be reassembled together. The reassembly algorithm relies on the `Identification` field of the received fragments to associate a fragment with the corresponding packet being reassembled. Furthermore, the `Fragment Offset` field indicates the position of the fragment payload in the original non-fragmented packet. Finally, the packet with the `M` flag reset allows the destination to determine the total length of the original non-fragmented packet. 717/10000 · 717
147 ../../protocols/wifi.rst:23
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — 5 GHz 5/10000 · 5
147 ../../protocols/tcp.rst:443
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — Advanced retransmission strategies 34/10000 · 34
147 ../../bibliography.rst:358
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — Wang XS, Balasubramanian A, Krishnamurthy A, Wetherall D. `How Speedy is SPDY ? <https://www.usenix.org/conference/nsdi14/technical-sessions/wang>`_ . In 11th {USENIX} Symposium on Networked Systems Design and Implementation ({NSDI} 14) 2014 (pp. 387-399). 256/10000 · 256