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Consider a network consist of 4 interconnected nodes P,Q,R and S. Distance between them are as follows: PQ=254km, PR=243km,  PS=287km, QR=239km, QS=300km, RS=120km. This network uses CSMA/CD and signal travels at a speed of 2*10^5 km/sec, if sender sends at a rate 1Mbps, what could be the maximum size of packet?  Ans___?__ bits
asked in Computer Networks by (41 points)   | 72 views
is the answer 3000??
What is the source of ur question @LORD ofKINGS ??
This question is from textbook test series, @habib khan you are right,  this question is wrong.  Thanks for your valuable comment.

2 Answers

0 votes
D = 1443*10*3M

V = 2*10^8 m/s

Tp = d/v = 7215 microseconds

Tt = 2*tp = 14,430 microseconds

Tt = msg/B.W

msg = 14,430 bits
answered by Veteran (11.1k points)  

why are you summing up all the distances??....maximum distance is to be taken into account i.e 300Km.
Is it? i don't know? why will we consider only 300k?
If u consider the stations as nodes and links as the edges u will see that the above set of edges are constituting a complete graph of 4 vertices and hence a mesh topology as each of them is reachable to other directly..This is not true for bus topology which CSMA/CD follows..Hence the question at the first sight is wrong based on this criteria itself..
0 votes

Ans  3000 bits 

Max Distance =300Km 

RTT =300 Km/2*10^5 km/s  

RTT =3 ms


L>=3000 bits

answered ago by (429 points)  

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