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138 ../../principles/dv.rst:121
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — `C` sends its distance vector `[C=0]` to `B` and `E` 52/10000 · 52
139 ../../principles/dv.rst:122
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — `E` sends its distance vector `[E=0,D=1,A=2,C=1]` to `D`, `B` and `C`. `B` can now reach `A`, `C`, `D` and `E` 110/10000 · 110
140 ../../principles/dv.rst:123
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — `B` sends its distance vector `[B=0,A=1,C=1,D=2,E=1]` to `A`, `C` and `E`. `A`, `B`, `C` and `E` can now reach all five routers of this network. 144/10000 · 144
141 ../../principles/dv.rst:124
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — `A` sends its distance vector `[A=0,B=1,C=2,D=1,E=2]` to `B` and `D`. 69/10000 · 69
142 ../../principles/dv.rst:126
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — At this point, all routers can reach all other routers in the network thanks to the routing tables shown in the figure below. 125/10000 · 125
143 ../../principles/dv.rst:203
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — To deal with link and router failures, routers use the timestamp stored in their routing table. As all routers send their distance vector every `N` seconds, the timestamp of each route should be regularly refreshed. Thus no route should have a timestamp older than `N` seconds, unless the route is not reachable anymore. In practice, to cope with the possible loss of a distance vector due to transmission errors, routers check the timestamp of the routes stored in their routing table every `N` seconds and remove the routes that are older than :math:`3 \times N` seconds. 573/10000 · 573
144 ../../principles/dv.rst:205
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — When a router notices that a route towards a destination has expired, it must first associate an :math:`\infty` cost to this route and send its distance vector to its neighbors to inform them. The route can then be removed from the routing table after some time (e.g. :math:`3 \times N` seconds), to ensure that the neighboring routers have received the bad news, even if some distance vectors do not reach them due to transmission errors. 439/10000 · 439
145 ../../principles/dv.rst:207
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — Consider the example above and assume that the link between routers `A` and `B` fails. Before the failure, `A` used `B` to reach destinations `B`, `C` and `E` while `B` only used the `A-B` link to reach `A`. The two routers detect the failure by the timeouts in the affected entries in their routing tables. Both routers `A` and `B` send their distance vector. 360/10000 · 360
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
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
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
149 ../../principles/dv.rst:212
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — `E` sends its distance vector :math:`[E=0,A=2,C=1,D=1,B=1]` to `D`, `B` and `C`. `D` learns a route towards `B`. `C` and `B` learn a route towards `A`. 151/10000 · 151
150 ../../principles/dv.rst:214
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — At this point, all routers have a routing table allowing them to reach all other routers, except router `A`, which cannot yet reach router `B`. `A` recovers the route towards `B` once router `D` sends its updated distance vector :math:`[A=1,B=2,C=2,D=0,E=1]`. This last step is illustrated in figure below, which shows the routing tables on all routers. 353/10000 · 353
151 ../../principles/dv.rst:295
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — Consider now that the link between `D` and `E` fails. The network is now partitioned into two disjoint parts: (`A` , `D`) and (`B`, `E`, `C`). The routes towards `B`, `C` and `E` expire first on router `D`. At this time, router `D` updates its routing table. 259/10000 · 259
152 ../../principles/dv.rst:297
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — If `D` sends :math:`[D=0, A=1, B=\infty, C=\infty, E=\infty]`, `A` learns that `B`, `C` and `E` are unreachable and updates its routing table. 142/10000 · 142
153 ../../principles/dv.rst:299
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — Unfortunately, if the distance vector sent to `A` is lost or if `A` sends its own distance vector ( :math:`[A=0,D=1,B=3,C=3,E=2]` ) at the same time as `D` sends its distance vector, `D` updates its routing table to use the shorter routes advertised by `A` towards `B`, `C` and `E`. After some time `D` sends a new distance vector : :math:`[D=0,A=1,E=3,C=4,B=4]`. `A` updates its routing table and after some time sends its own distance vector :math:`[A=0,D=1,B=5,C=5,E=4]`, etc. This problem is known as the `count to infinity problem` in the networking literature. 566/10000 · 566
154 ../../principles/dv.rst:301
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — Routers `A` and `D` exchange distance vectors with increasing costs until these costs reach :math:`\infty`. This problem may occur in other scenarios than the one depicted in the above figure. In fact, distance vector routing may suffer from count to infinity problems as soon as there is a cycle in the network. Unfortunately, cycles are widely used in networks since they provide the required redundancy to deal with link and router failures. To mitigate the impact of counting to infinity, some distance vector protocols consider that :math:`16=\infty`. Unfortunately, this limits the metrics that network operators can use and the diameter of the networks using distance vectors. 683/10000 · 683
155 ../../principles/dv.rst:306
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — This count to infinity problem occurs because router `A` advertises to router `D` a route that it has learned via router `D`. A possible solution to avoid this problem could be to change how a router creates its distance vector. Instead of computing one distance vector and sending it to all its neighbors, a router could create a distance vector that is specific to each neighbor and only contains the routes that have not been learned via this neighbor. This could be implemented by the following pseudocode. 510/10000 · 510
156 ../../principles/dv.rst:323
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — This technique is called `split-horizon`. With this technique, the count to infinity problem would not have happened in the above scenario, as router `A` would have advertised :math:`[A=0]` after the failure, since it learned all its other routes via router `D`. Another variant called `split-horizon with poison reverse` is also possible. Routers using this variant advertise a cost of :math:`\infty` for the destinations that they reach via the router to which they send the distance vector. This can be implemented by using the pseudo-code below. 550/10000 · 550
157 ../../principles/dv.rst:341
	English ↹ ↵ NBS … “ ” ‘ ’ ﹙-﹚ ‐ – — Unfortunately, split-horizon is not sufficient to avoid all count to infinity problems with distance vector routing. Consider the failure of link `A-B` in the four routers network shown below. 192/10000 · 192