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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 (59 points) | 117 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 (27.6k 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..
in this topology no collision will occur because it is mesh topology not bus topology, AND CSMA/CD is a bus topology protocol.

crct me if I am wrong?
if multipal distance is given for csma/cd then what we select(maximum distance or sum of all distance)
0 votes

Ans  3000 bits 

Max Distance =300Km 

RTT =300 Km/2*10^5 km/s  

RTT =3 ms


L>=3000 bits

answered by Junior (795 points)

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