+1 vote
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I am giving two questions along with answers

Q1 )

Suppose ‘A’ and ‘B’ are on same 10Mbps Ethernet segment and the propagation delay
between two nodes is 275 bit times. Suppose A and B are on two ends of the wire and tries to send a frame at time t=0 and frames collide. Then at what time (in bits) they finish transmitting a jam signal. Assume 48 bit jam signal.
A. 598 B. 323 C. 502 D. 227

Here in this question , time taken to send the jam signal is => Tp + (time taken to send 48 bit jam signal , i.e 48 bit times = (275 + 48 ) =323 bit times

this part is clear to me. Now the next question.

Q2)

Assume 10 Mbps Ethernet and two stations A and B on it’s same segment. The RTT between two nodes is 650 bit times. A and B start transmitting frame and collision occurs and both sends 30-bit jam signal. Find the time at which both nodes A and B sense an idle channel (in μsec).

Here , Tp = 0.5 * RTT = 325 bit times , so , add 30 bit jam signal = 325 + 30 = 355 bit times.

But they have also added Tp once again , 355 + 325 = 680 bit times.

Can you please justify the last Tp addition part ? Is it due to the fact that , in order to sense the channel idle , we need send one bit , so , one propagation delay is needed ?

edited | 379 views

1) Suppose ‘A’ and ‘B’ are on same 10Mbps Ethernet segment and the propagation delay
between two nodes is 275 bit times. Suppose A and B are on two ends of the wire and tries to send a frame at time t=0 and frames collide. Then at what time (in bits) they finish transmitting a jam signal. Assume 48 bit jam signal.

at t=0, A and B starts to transmit ===> at t=$\frac{Tp}{2}$ collision occurs, ===> at t=Tp collision Detects.

Now, Jam signal transmit by A and B,===> Tt of Jam signal = 48 bit times

then they finish transmitting a jam signal at t=Tp + Tt of jam signal = 275+48.

2) Assume 10 Mbps Ethernet and two stations A and B on it’s same segment. The RTT between two nodes is 650 bit times. A and B start transmitting frame and collision occurs and both sends 30-bit jam signal. Find the time at which both nodes A and B sense an idle channel (in μsec).

at t=0, A and B starts to transmit ===> at t=$\frac{Tp}{2}$ collision occurs, ===> at t=Tp collision Detects.

Now, Jam signal transmit by A and B,===> Tt of Jam signal = 30 bit times

then they finish transmitting a jam signal at t=Tp + Tt of jam signal = (0.5*650) + 30 .

But what time they sense an idle channel?

after the jam signal propagate the whole channel, another Tp requires to propagate the whole channel (Note that Tp is independent on the jam signal or Data, it is depend upon only distance and velocity. )

∴ at t=Tp ( collision detection ) + Tt of jam signal + Tp for jam signal = (0.5*650) + 30 + (0.5*650) = 680 bit times, channel is idle

answered by Veteran (61k points)
0

Shaik Masthan   can you tell me if  in this question they given half duplex instead of full duplex than what's the answer ?? https://gateoverflow.in/144462/cn-optimal-windows-size

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as per me,in your case the line changed

It is Full duplex  channel,  so total data send from x to y is = (2* Tp) * Band Width

should be as

It is Half duplex  channel,  so total data send from x to y is = (Tp) * Band Width

therefore answer changed as 1+(480/2) = 1+240 = 241

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then why it's not work in this case ..I can't figure it out  ??

https://gateoverflow.in/249949/testbook-computer-networks#c250058

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Why we need one extra propagation delay, this part is still unclear to me!

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channel sense idle, when jam signal also propagate one end to another end.

(Note that, when you send the packet(Tt), it takes Tp time to reach to another end, as same like that when you send jam signal(Tjam), it is also take Tp time to reach to another end. )

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@Shaik I am having one doubt now.. the question says that A and B "begin to transmit" frames at t=0 then why won't we consider their transmission time as well? Why are we considering that the first bit(which suffers the collision at first) is put into the channel at t=0? It needs time to transmit a bit onto the channel isn't it? Please clear me this..
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even before transmitting time completes collision occur. ===> $\frac{T_p}{2}$ is representating transmitting time only but not completely.
+1 vote
355 is the answer

I didn't know about  680
answered by (21 points)
+1 vote

Tp = 0.5 * RTT = 325 bit times , so , add 30 bit jam signal = 325 + 30 = 355 bit times.

Both nodes A and B need to sense an idle channel. This would need propogation of these Jamming signals. So, another Tp would be added to 355 bit times.

So, 680 bit times needed to sense an idle channel.

answered by (117 points)
0
Would the jamming signal not collide again as they are sent at the same instance and my understanding is that they would collide then in that case how do stations respond on detecting collision of jamming signal. I guess they must ignore it

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+1 vote