GATE CSE
First time here? Checkout the FAQ!
x
+1 vote
72 views
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)  
@manu00x

why are you summing up all the distances??....maximum distance is to be taken into account i.e 300Km.
@joshi_nitin
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>=RTT *BW

L>=3000 bits

answered ago by (429 points)  


Top Users Sep 2017
  1. Habibkhan

    6960 Points

  2. Warrior

    2416 Points

  3. Arjun

    2358 Points

  4. rishu_darkshadow

    2136 Points

  5. A_i_$_h

    2004 Points

  6. nikunj

    1980 Points

  7. manu00x

    1750 Points

  8. makhdoom ghaya

    1750 Points

  9. Bikram

    1744 Points

  10. SiddharthMahapatra

    1718 Points


26,059 questions
33,664 answers
79,738 comments
31,078 users