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+5 votes
Consider two nodes A and B on the same Ethernet, and suppose the propagation delay b/w the two nodes is 225-bit times. Suppose at the time both nodes A and B begin to transmit a frame.Both nodes transmit a 50-bit jam signal after detecting  collision.
For 10^7 bit/set find the time at which both nodes A and B sense idle channel?
in Computer Networks by Boss (11.7k points) | 512 views

2 Answers

+5 votes

Firstly, note that here time is given in bits not in sec. Now,

1: Both stn detect collision at t= 225.

2: after detecting collision, both will sent jam signal to inform other stn in that network about the occurrence of collision. here jam signal is 50 bits.

So both stn stop transmitting their jam signal at t= 225+50 = 275.

3: Last bit of jam signal will now propagate to other end. And Tp is given to be 225 bit time. So time at which both the jam signal will reach completely to opp end  and channel is sensed idle at t= 275+225 = 500 bits ( Final answer in bits)

Now "For 10^7 bit/sec. find the time at which both nodes A and B sense idle channel?" - Since Bandwidth is given, ques is asked in units of time (sec).

So now converting answer in bits to sec :-

time at which both nodes A and B sense idle channel = 500 bits  = 500 / BW = 500/ 10^7

     = 50 micro sec.

by Junior (505 points)
Hi @stanchlon, I have a query.

Let's say a collision occured at 225 bits then how much time will the sender takes to detect it?

@  Harish Kumar 2

If collision occured at 225bit, then answer would be different, ie. 2*225 + 50 + 225 bits = 725 bits (Final answer)

But here in the question your doubt may be, "Why we taking Tp and not 2*Tp for collision detection by both the station (step 1)?"

So the reason is, question given that- "Suppose at the time both nodes A and B begin to transmit a frame." Therefore assumed as it is Tp time only to detect collision.

Thanks @stanchlon, i got it now.

here is don't take 2*PT time for the collision because both are transmitting  simultaneously so that its take 225 bit times 

Why are we not considering the transmission delays?? Could anyone explain?
0 votes
Is it 27.5microsec.........??
by (81 points)
No, Answer is 50 micro sec.

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