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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 ?

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    3 Answers

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    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

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    1 votes

    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.

     

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