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Distance Vector Routing Algorithm In Computer Networks | DV Routing Algorithm | Simplilearn — Transcript

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0:08dist inspector routing
0:10this is a protocol in data network that
0:13helps to determine the best route for
0:15data packets to travel in network
0:18channel
0:19hi guys and welcome to yet another
0:21interesting video i simply learn
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0:31update from us
0:32[Music]
0:34now without further ado let's take a
0:36look at the agenda for today's video
0:40in today's session we will look into the
0:44introduction for distance vector routing
0:48continuing with some key points for
0:50routing protocol
0:53moving on to distance vector routing
0:55algorithm
0:58in the end we will look into some
0:59examples to better understand the
1:01routing protocol
1:03now let's move on to the first heading
1:04for this session that is
1:07introduction to distance vector routing
1:11distance vector routing refers to the
1:13distance or the vector that is between
1:16the neighboring nodes
1:18in a network channel
1:20and the routing part refers to the
1:22established route through which data
1:25packets are transmitted in the network
1:28channel
1:29the other aspect of this routing
1:31protocol in a network channel is to
1:34determine the shortest path for a data
1:36packet in the network to reach its
1:39destination
1:41next we will look into some key points
1:43regarding the routing protocol
1:47for routing protocol three primary key
1:50points are to be remembered
1:54first one being
1:56each router in this protocol is designed
1:59to share the vector data of each node
2:03throughout the network
2:05this step is done to maintain proper
2:08circulation of data in the network
2:10channel
2:11the next key point is the routing
2:14pattern in the network protocol
2:16that is the data shared by the routers
2:19is transmitted only to those nodes that
2:22are directly linked to a particular
2:25router
2:28the last key point is
2:30that the routing protocol is designed in
2:33a manner that it shares the updated
2:36vector data periodically
2:38at the network channel
2:43now let's move on to the next heading
2:45that is
2:46the algorithm used in the protocol
2:52the algorithm used in this protocol
2:55is termed as
2:57bellman fold algorithm
3:00this algorithm defines the shortest path
3:04that the data packet can take to reach
3:06its destination node from an initial
3:09node
3:11where
3:12d of x y refers to the least distance
3:16from a node x to node y
3:19moving on we have
3:21c of x comma v
3:23where x is caused from each of its
3:26neighboring v node is taken into
3:29consideration
3:32then we have
3:33d of
3:34v y
3:35this is the distance of each neighbor
3:37from its initial node
3:40and lastly
3:42min of v
3:43this is an abbreviation used to
3:45determine the most shortest vector after
3:49the solution is obtained using the
3:51algorithm
3:53in the next heading that is network
3:55example we will understand all the
3:58points used in the algorithm
4:02now let's design a network model with
4:05five different nodes
4:07that are a b c and d
4:10with each node connected to each other
4:13using different network channel
4:18the numerical values in the network
4:21channel refers to the vector count that
4:23is the distance from node a to node b or
4:27to some other node in the network
4:29channel
4:36in a network model
4:37each node shares its vector data with
4:41its neighboring node at a regular
4:43interval
4:44as there are various updates in the
4:46network model
4:49this data is shared in a form of routing
4:52table where three different columns are
4:55used first being destination column
4:58second
4:59vector column that is the distance node
5:02and lastly the hop
5:06let's take a look at the neighboring
5:08nodes for this model
5:12for node a we have
5:14b and e as it is directly interlinked to
5:18the node a
5:20for node b we have
5:22node a and node c
5:25for node c we have b and d
5:28for d we have
5:30c and e
5:31and lastly for e we have node a and node
5:35d
5:36these nodes represents only the
5:38neighboring nodes
5:42now let's solve an actual example on
5:45routing protocol to clear all our doubts
5:49and understand the key points
5:54the example we will be taking
5:56to solve is
5:58this network model
6:01the initial step to start solving the
6:04network model using the distance vector
6:07routing protocol is to make routing
6:10table for each of the nodes
6:13let's make the routing table for node a
6:18this is the routing table for node a
6:20where we have
6:22destination as a
6:24that means destination is a from node a
6:28that means the distance vector should be
6:30zero
6:32then we have
6:33destination as b
6:36let's take a look
6:37in this given example there is no given
6:41connection between a and b
6:43that means
6:45this distance vector would be infinity
6:50then we have destination as c
6:54similarly as node b there is no direct
6:56connection between a and node c that
7:00means this is also infinity
7:03and same is the case with node d
7:07and in the end we have
7:09node e that is the only directly linked
7:12node
7:14to node a that means the distance given
7:17is 5 so the network
7:20vector is 5.
7:23let's make an another dropping table for
7:26node e
7:29this is the routing table for node
7:32let's take a look
7:33in this routing table the first is
7:36destination to node a
7:38according to the network model given
7:41distance is 5 so the vector value is 5
7:44with the hop to a
7:47then we have
7:49destination as node b
7:51according to the model the distance
7:53vector should be 4
7:56then we have destination as node c
7:59as given there is no direct connection
8:02between node e and node c
8:05that determines the value would be
8:07infinity
8:09then we have
8:10node d which represents
8:14direct connection with node e
8:16and the value is seven
8:19and lastly
8:21e destination from initial point e would
8:24be zero vector
8:27now let's make another routing table
8:29similarly for node c
8:33now using the same steps you can make
8:36the routing table for the remaining
8:38nodes that is node b and node d
8:45now let's move on to the next step in
8:47solving the routing issue
8:50this step is known as update step
8:53in this step we use the vector column
8:57from each of the node to update the
9:00vector column or the routing table for a
9:03particular node
9:05in this part we'll use
9:07node is vector column to update the
9:10routing table for node a
9:14and the steps are using the algorithm
9:17that is the bellman fold algorithm we
9:20used earlier
9:23as we already know
9:25moving from a to destination node a is
9:28always zero so we won't be doing this
9:30step let's move on to the next one that
9:33is moving from node a to node b
9:37that would be
9:41would be a comma e as we are using
9:43vector column of node e
9:47plus
9:48and the distance used to reach e from b
9:53a comma e represents
9:555
9:57and
9:57e to b represents 4 according to the
10:00network model so the value is
10:029. similarly let's solve
10:06for a to c
10:08to a to c we have
10:10a comma e plus destination required to
10:13move from e to c
10:15according to the network model we have
10:19node a is only directly linked with node
10:22e
10:23so the value would be 5
10:25but there is no connection between node
10:28e and node c so the value would be
10:31infinity
10:32moving on we have
10:34a to d
10:36in this case we have
10:38a comma e
10:40the value would be
10:415
10:42plus e to d that is a direct connection
10:45and the value would be 7 so the total
10:48distance is 12
10:50and the last one is
10:52a to e
10:53for this we have a comma e that is
10:56directly the neighboring node
10:58the value would be 5
11:01so the new routing table
11:04for node a is
11:060 9 infinity 12 and 5.
11:11similarly let's take another node for
11:14this
11:15for the next node we will take node c
11:20and we will update the routing table of
11:22node c
11:23using its neighboring nodes that are
11:27node b
11:28and node d according to the network
11:30model
11:32now
11:33how to do this step
11:35firstly it's
11:37c to destination a
11:40in case of node b we have
11:43c to a that would be
11:46c comma b that is a direct connection
11:49plus distance required from b to a
11:52according to network model we have c
11:54comma b as 6
11:56plus b to a does not have any connection
11:59that is infinity so the value would be
12:01null
12:03then we have node d
12:05in this case we have c comma d that is a
12:08direct connection 3 plus d to a that is
12:12again infinity
12:14as it is not linked directly to node a
12:19moving on we have
12:20destination as b
12:23for node b we have direct connection so
12:26c comma b value is 6
12:29whereas for node d we have
12:31c comma d that is distance 3 plus d to b
12:36that is infinity as it is not directly
12:39linked with the node b
12:42so again the value is null
12:46then we have
12:48destination as c
12:50so connection between c to c is zero
12:54moving on we have destination at node d
12:58that would be c to d
13:00as
13:01c to d
13:04as there is no direct connection between
13:06node b and node d
13:08so the value is infinity
13:10whereas for c to d is a direct
13:13connection for node d so the value would
13:15be 3
13:17now for the last node that is node e
13:21in case of node b we have
13:24c comma b
13:25plus
13:26b to e
13:28where the value for c to b node is 6 and
13:31b to a is 4 so the sum is 10
13:36and now in case of node d we have
13:393
13:39plus 7 that would be distance from c to
13:42d and then d to e
13:45and the value would be again 10. so we
13:47can use any of the side to reach the
13:50destination node e
13:53by completing all the steps we have the
13:56new table for node c
13:59similarly we can find the routing table
14:02for all the nodes present
14:07now to look into some key points we
14:09learned
14:11for the initial step use only the node
14:14distance of immediate neighbors
14:19nodes without distance that is directly
14:21linked to each other is always infinity
14:27for the update step use the vector table
14:30of the only
14:31neighboring node we have for the initial
14:34node
14:36and lastly to continue the update step
14:39roughly n minus 1 iterations are used
14:42where the value of n is equal to the
14:45number of nodes
14:46for example
14:48the model we took has
14:50five nodes that means we have to do a
14:52loop of four iterations
14:57with this we have reached the end of the
14:59session
15:00if you have any questions regarding the
15:02topic you can ask them in the comment
15:04section below thank you for watching
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