2. cnp3-ebook
3. English
4. Zen

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
148 ../../glossary.rst:123
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — round-trip-time 15/10000 · 15
148 ../../principles/dv.rst:211
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — `B` sends its distance vector :math:`[B=0,A=\infty,C=1,D=2,E=1]` to `E` and `C`. `C` learns that there is no route anymore to reach `A` via `B`. 144/10000 · 144
148 ../../principles/reliability.rst:502
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — The simplest error detection scheme is the checksum. A checksum is basically an arithmetic sum of all the bytes that a frame is composed of. There are different types of checksums. For example, an eight bit checksum can be computed as the arithmetic sum of all the bytes of (both the header and trailer of) the frame. The checksum is computed by the sender before sending the frame and the receiver verifies the checksum upon frame reception. The receiver discards frames received with an invalid checksum. Checksums can be easily implemented in software, but their error detection capabilities are limited. Cyclical Redundancy Checks (CRC) have better error detection capabilities [SGP98]_, but require more CPU when implemented in software. 742/10000 · 742
148 ../../principles/transport.rst:860
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — Connection release 18/10000 · 18
148 ../../principles/sharing.rst:933
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — The congestion control scheme must be `efficient`. The bottleneck link is usually both a shared and an expensive resource. Usually, bottleneck links are wide area links that are much more expensive to upgrade than the local area networks. The congestion control scheme should ensure that such links are efficiently used. Mathematically, the control scheme should ensure that :math:`\forall{t} \sum{r_i(t)} \approx R`. 417/10000 · 417
148 ../../protocols/email.rst:258
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — e 1/10000 · 1
148 ../../protocols/ipv6.rst:496
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — Note that the reassembly algorithm must deal with the unreliability of the IP network. This implies that a fragment may be duplicated or a fragment may never reach the destination. The destination can easily detect fragment duplication thanks to the `Fragment Offset`. To deal with fragment losses, the reassembly algorithm must bind the time during which the fragments of a packet are stored in its buffer while the packet is being reassembled. This can be implemented by starting a timer when the first fragment of a packet is received. If the packet has not been reassembled upon expiration of the timer, all fragments are discarded and the packet is considered to be lost. 676/10000 · 676