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For a finite size buffer of length N at a router the probability of b

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For a finite size buffer of length N at a router the
probability of buffer overflow is: p_n=(1-p)p^n/1-p^n+1 use
this to find n in terms of p and p_n. If p=0.3 and N is
100000 what is p_n? How does the ospf routing algorithm
work? Consult rfc 2328 for details-what general
information is in this rfc? Why are there error protection
techniques? and, how no they work in general?
Solution
3.a
value of N in terms of PN and P is as follow:
PN = ((1-P)PN)/1-PN+1
PN(1-PN+1) = (1-P)PN
PN/PN -PPN = 1-P
and solving this we got
N = PN/(1-P)(1-PN)logP
3.b
p=0.3 , n = 100000
0.3100000 will be infinity so you cannot calculate this
4.a
The router in this protocol uses a link-state data
After this a tree is created by the router and the root node
of the tree is router itself

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For a finite size buffer of length N at a router the probability of buffer overflow is: p_n=(1 -p)p^n/1-p^n+1 use this to find n in terms of p and p_n. If p=0.3 and N is 100000 what is p_n? How does the ospf routing algorithm work? Consult rfc 2328 for details-what general information is in this rfc? Why are there error protection techniques? and, how no they work in general? Solution 3.a value of N in terms of PN and P is as follow: PN = ((1-P)PN)/1-PN+1 PN(1-PN+1) = (1-P)PN PN/PN -PPN = 1-P and solving this we got N = PN/(1-P)(1-PN)logP 3.b p=0.3 , n = 100000 0.3100000 will be infinity so ...
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