distance vector routing query

Question

Consider a network system having 4 routers P,Q,R AND S as shownbelow.initially, the routersise Distance vector routing and use number of hops as distance metric.

Initially, link p-q was down and all other routers agree p is unreachable.thr link p-q comes back up. Let x be the number of exchanges needed by routers P to stabilize its table having distance to all the other routers.assume the routers all exchanges all the messages at the same instant.After reaching to stable state the link goes down again.let y be the number of exchanges need for all the routers to conclude that P is unreachable..what is the value of x and y??

I think x should be 2 or 3 and y should be 2 or 3.

When p-Q links comes back.first exchange Q and R sends information amd R updates path to P. Second exhange S and R sends information and S also updates path to P. 

And same thing will done when link P-Q goes down.
There will not be "count to infinity" problem as all routers are sending the information at the same instant.

How to solve this ques?

Comments

Table of p by using distance vector 

	Q	r	s
Initially/down	infinite	infinite	infinite
Up	1	infinite	infinite
2 up	1	2	infinite
3 up	1	2	3
4 up	1	2	3

As 4 time up or iteration is  done table become stable. . So the value of x is 4 .. 

 Similarly when P Q connection goes down again .. we apply the same thing .. that is completed in 7 iteration or 7 time down

 So the value of y is 7 . 

 => x =4 and y =7

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How seven iteration can you explain

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@Simmi Kindly provide a full solution please

Answer 1

Not 100% sure.please verify it.
For X it will take 2 exchanges.
And every router will know about P and will be in steady state.
And when PQ link goes down it will be the case like below( assume infinity as 16 as many books consider 16 as infinity for distance vector routing) Because all routers will send their path information at the same time. At the same time Q R and S will send their path information to their neighbours. And if and only if R gets the information sent by Q before sending its path table then there will not be any cpunt to infinity problem. And after 2 hops or 2 exchanges everyone will know about P's unreachability.

exchange number.
1. Q send info:" its path to p infinity (say 16)" to R and R will send to Q info :path to P 2 .Q update with 3 and R with 16.like this S with 3.

Exch.2 Q R S sends info each other as Q update with 16 R as 4 and S with 16.

Exch.3 Q R S sends info each other as Q update with 5 R as 16 and S with 5.

Exch.4 Q R S sends info each other as Q update with 16 R as 6 and S with 16.

Exch.5 Q R S sends info each other as Q update with 7 R as 16 and S with 7.

Exch.6 Q R S sends info each other as Q update with 16 R as 8 and S with 16.

Exch.7 Q R S sends info each other as Q update with 9 R as 16 and S with 9.

Exch.8 Q R S sends info each other as Q update with 16 R as 10 and S with 16.

Similarly like this upto 8+6
14 exchanges are required for 'y'
so x=2 and y=14

Answer 2

X=2  on P connecting

and Y=6 on P disconnecting