CN -> CSMA/CD

Question

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

Comments

is the answer 3000??

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What is the source of ur question @LORD ofKINGS ??

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This question is from textbook test series, @habib khan you are right,  this question is wrong.  Thanks for your valuable comment.

Answer 1

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

Comments

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..

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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?

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if multipal distance is given for csma/cd then what we select(maximum distance or sum of all distance)

Answer 2

Ans  3000 bits 

Max Distance =300Km 

RTT =300 Km/2*10^5 km/s  

RTT =3 ms

L>=RTT *BW

L>=3000 bits