Consider the following network: Assume that at t=10 node 7 receives two LSA from
ID: 3812144 • Letter: C
Question
Consider the following network: Assume that at t=10 node 7 receives two LSA from node 4 regarding the link between nodes 1 and 4 (namely, LSA a. and LSA(b). LSA a has a sequence #242 and a cost=2, LSA b has a sequence #112 and cost=10. After processing, what will be the new cost of the link given that all nodes are using circular sequence numbers. Show the work, assume that an 8-bit field is used for sequence number. Node 7 does not receive any other LSA until t=15 where a new LSA c received from 4 with sequence #10, declaring the new cost for the link between nodes 1 and 4 being 3. No other LSA is received after that A. Find the route between nodes 7 and S1 at t=13. B. Find the route between nodes 7 and SI at t=16. C. Now assume that for some reason. Node 7 misses LSA b. What is the route between node 7 and SI at t=13 and t=16? Would this pose any issue? D. Assume that LSA a, LSA b. and LSA c are all received at the same time at t=10. Will router 7 face any issue finding its optimum route to S1 at t=13? Can the use of sequence numbers help find which of the three LSAs is most recent?Explanation / Answer
in circular sequence numbering
if a router receives LSA a and LSA b then the LSA is considered recent in these two condition
1)a>b and (a-b)<n/2
2)a<b and (b-a)>n/2
where n is the total number of sequence numbers that can generated.. for example if sequence number is 6 bits then n will be 2^6=64.
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in the given network the sequence number is 8 bits hence the value of n will be 256.
-->at time t=10 node 7 receives LSA a and LSA b with sequence numbers 242 and 112 and costs 2 and 10 respectively
so the recent LSA would be LSA b because according to 2nd condition.
so the node 7 at t=10 stores the cost of link between node 1 and 4 as 10.
and the cost remains same untill t=15.
at t=15 node 7 receives another LSA i.e LSA c with sequence number 10
now the LSA b and LSA c will be cosidered by node 7 and as LSA b with sequence number 112 is more recent than LSA c with sequence number 10 (from condition 1) 7 stores the cost as 10 at t=15.
a)at t=13
the cost of route from 1 and 4 is 10.
so the route from 7 and s1 would be 7-->2-->3-->1-->s1 because the total cost would be low(i.e 3+1+5)
b)
at time t=16 the cost of route from 1 and 4 will remain same i.e 10
so the path also would be same. 7-->2-->3-->1-->s1
c)
if node misses LSA b then at t=10 the cost of link between 1 and 4 would be 2.
so at time `t=13 the route from 7 to s1 will be
7-->4-->1-->s1(cost =2+2)
so at t=15 when LSA c was received with sequence number 10 it would be the recent LSA(basing on condition 2) , so the cost of link between 1 and 4 will be updated to 3 .
so now at t=16 the route from 7bto s1 will be
7-->4-->1-->s1(cost: 2+3);
d)
if all the three LSA are received at time t=10
then first compare any two LSA and find the recent one and again compare the recent one with third one to find the recent
all this can be done using sequence numbers.
--if we compare LSA a and LSA b first-->the recent would be LSA b. next compare LSA b with LSA c -->the recent woud be LSA b.
--now first compare LSA a and LSA c--> the recent would be LSA c.
next compare LSA c and LSA b -->recent would be LSA b.
--first compare LSA b and LSA c -->the recent would be LSA b.now compare LSA b with LSA a -->the recent would be LSA b.
see in all the three cases we get to same result.
hence the router will not face any issues finding optimum route to s1 at t=13.
The use of sequence numbers helped in finding the recent LRA.