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170 ../../principles/reliability.rst:831
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — Sliding window example 22/10000 · 22
170 ../../principles/linkstate.rst:17
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — :math:`weight=\frac{C}{bandwidth}` where `C` is a constant larger than the highest link bandwidth in the network. If all link weights are configured this way, shortest path routing prefers higher bandwidth paths over lower bandwidth paths. 239/10000 · 239
170 ../../principles/sharing.rst:1020
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — Consider first a window of one segment. This segment takes 4 msec to reach host `D`. The destination replies with an acknowledgment and the next segment can be transmitted. With such a sending window, the transmission rate is roughly 250 segments per second or 250 Kbps. This is illustrated in the figure below where each square of the grid corresponds to one millisecond. 372/10000 · 372
170 ../../protocols/email.rst:270 • ../../protocols/email.rst:289 • ../../protocols/email.rst:303
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — Input data 10/10000 · 10
170 ../../protocols/ipv6.rst:548
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — The `Packet Too Big` ICMP messages enable the source host to discover the MTU size that it can safely use to reach a given destination. To understand its operation, consider the (academic) scenario shown in the figure below. In this figure, the labels on each link represent the maximum packet size supported by this link. 322/10000 · 322
170 ../../protocols/wifi.rst:102
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — The first part of the 802.11 header is the 16 bit `Frame Control` field. This field contains flags that indicate the type of frame (data frame, RTS/CTS, acknowledgment, management frames, etc), whether the frame is sent to or from a fixed LAN, etc [IEEE802.11]_. The `Duration` is a 16 bit field that is used to reserve the transmission channel. In data frames, the `Duration` field is usually set to the time required to transmit one acknowledgment frame after a SIFS delay. Note that the `Duration` field must be set to zero in multicast and broadcast frames. As these frames are not acknowledged, there is no need to reserve the transmission channel after their transmission. The `Sequence control` field contains a 12 bits sequence number that is incremented for each data frame and a 4 bits fragment number. 812/10000 · 812
170 ../../protocols/tcp.rst:557
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — a non-`SYN` segment was received for a non-existing TCP connection :rfc:`793` 77/10000 · 77
171 ../../glossary.rst:176
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — A relay operating in the datalink layer. 40/10000 · 40
171 ../../principles/reliability.rst:834
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — In practice, as the frame header includes an `n` bits field to encode the sequence number, only the sequence numbers between :math:`0` and :math:`2^{n}-1` can be used. This implies that, during a long transfer, the same sequence number will be used for different frames and the sliding window will wrap. This is illustrated in the figure below assuming that `2` bits are used to encode the sequence number in the frame header. Note that upon reception of `OK1`, the sender slides its window and can use sequence number `0` again. 529/10000 · 529
171 ../../principles/linkstate.rst:19
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — Usually, the same weight is associated to the two directed edges that correspond to a physical link (i.e. :math:`R1 \rightarrow R2` and :math:`R2 \rightarrow R1`). However, nothing in the link state protocols requires this. For example, if the weight is set in function of the link bandwidth, then an asymmetric ADSL link could have a different weight for the upstream and downstream directions. Other variants are possible. Some networks use optimization algorithms to find the best set of weights to minimize congestion inside the network for a given traffic demand [FRT2002]_. 579/10000 · 579
171 ../../principles/sharing.rst:1068
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — Consider now a window of two segments. Host `A` can send two segments within 2 msec on its 1 Mbps link. If the first segment is sent at time :math:`t_{0}`, it reaches host `D` at :math:`t_{0}+4`. Host `D` replies with an acknowledgment that opens the sending window on host `A` and enables it to transmit a new segment. In the meantime, the second segment was buffered by router `R1`. It reaches host `D` at :math:`t_{0}+6` and an acknowledgment is returned. With a window of two segments, host `A` transmits at roughly 500 Kbps, i.e. the transmission rate of the bottleneck link. 580/10000 · 580
171 ../../protocols/email.rst:270
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — `0x14fb9c03d97e` 16/10000 · 16
171 ../../protocols/ipv6.rst:570
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — If ``A`` sends a 1500 bytes packet, ``R1`` will return an ICMPv6 error message indicating a maximum packet length of 1400 bytes. ``A`` would then fragment the packet before retransmitting it. The small fragment would go through, but the large fragment will be refused by ``R2`` that would return an ICMPv6 error message. ``A`` can fragment again the packet and send it to the final destination as two fragments. 411/10000 · 411
171 ../../protocols/wifi.rst:105
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — The astute reader may have noticed that the 802.11 data frames contain three 48-bits address fields [#f4addresses]_ . This is surprising compared to other protocols in the network and datalink layers whose headers only contain a source and a destination address. The need for a third address in the 802.11 header comes from the `infrastructure` networks. In such a network, frames are usually exchanged between routers and servers attached to the LAN and WiFi devices attached to one of the access points. The role of the three address fields is specified by bit flags in the `Frame Control` field. 598/10000 · 598
171 ../../protocols/tcp.rst:558
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — by extension, some implementations respond with an `RST` segment to a segment that is received on an existing connection but with an invalid header :rfc:`3360`. This causes the corresponding connection to be closed and has caused security attacks :rfc:`4953` 258/10000 · 258
172 ../../glossary.rst:108
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — SYN cookie 10/10000 · 10
172 ../../principles/reliability.rst:841
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — Utilisation of the sliding window with modulo arithmetic 56/10000 · 56
172 ../../principles/linkstate.rst:24
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — When a link-state router boots, it first needs to discover to which routers it is directly connected. For this, each router sends a HELLO message every `N` seconds on all its interfaces. This message contains the router's address. Each router has a unique address. As its neighboring routers also send HELLO messages, the router automatically discovers to which neighbors it is connected. These HELLO messages are only sent to neighbors that are directly connected to a router, and a router never forwards the HELLO messages that it receives. HELLO messages are also used to detect link and router failures. A link is considered to have failed if no HELLO message has been received from a neighboring router for a period of :math:`k \times N` seconds. 751/10000 · 751
172 ../../principles/sharing.rst:1141
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — Our last example is a window of four segments. These segments are sent at :math:`t_{0}`, :math:`t_{0}+1`, :math:`t_{0}+2` and :math:`t_{0}+3`. The first segment reaches host `D` at :math:`t_{0}+4`. Host `D` replies to this segment by sending an acknowledgment that enables host `A` to transmit its fifth segment. This segment reaches router `R1` at :math:`t_{0}+5`. At that time, router `R1` is transmitting the third segment to router `R2` and the fourth segment is still in its buffers. At time :math:`t_{0}+6`, host `D` receives the second segment and returns the corresponding acknowledgment. This acknowledgment enables host `A` to send its sixth segment. This segment reaches router `R1` at roughly :math:`t_{0}+7`. At that time, the router starts to transmit the fourth segment to router `R2`. Since link `R1-R2` can only sustain 500 Kbps, packets will accumulate in the buffers of `R1`. On average, there will be two packets waiting in the buffers of `R1`. The presence of these two packets will induce an increase of the round-trip-time as measured by the transport protocol. While the first segment was acknowledged within 4 msec, the fifth segment (`data(4)`) that was transmitted at time :math:`t_{0}+4` is only acknowledged at time :math:`t_{0}+11`. On average, the sender transmits at 500 Kbps, but the utilization of a large window induces a longer delay through the network. 1390/10000 · 1390
172 ../../protocols/email.rst:272 • ../../protocols/email.rst:291 • ../../protocols/email.rst:305
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — 8-bit 5/10000 · 5