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The round trip propagation delay for a 100Mbps Ethernet having 48-bit jamming signal is 64 ms. What is the minimum frame size (in bytes)?
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I am getting the answer as 6.49Mb. Please confirm is it correct or not.
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what the answer they given ? i am getting 0.8 M
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I don't have the answer of this.

I think you did like this,
Minimum frame size = RTT * Bandwidth
= 64 * 10-3 * 100 * 106 = 0.8 Mbps

My Query is that the 48 bit has to be included in Frame if we are considering TT = RTT and if this is the case we have to  delete that right? Please let me know your view.

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48 bit has to be included in Frame

why?

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As we are taking TT = 2PT that is we are talking about csma/cd and there the jamming bit has to be included in the frame right.
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according to me, frame means either data or jamming signal but not both at a time,

moreover, when collision identified by the station, then it transmit jam signal
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Might be i am misunderstanding the concept.

Please let me know one thing that what is the use of mentioning the Jam signal size. They can simply mention that the jamming signal without size. Isn't it?
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i hope there is no need of even talking JAM signal
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But if the question doesn't talks about the Jamming signal, how do we got to know that we have to consider TT = 2PT?
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how do we got to know that

for detect collision, we have L ≥ 2 * Tp

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Please share the way you solved it.

WHENVER THERE IS COLLISION THEN JAMMING SIGNAL IS USED UPTO 2PT.

FOR CALCULATING MIN FRAME SIZE WE USE TT = 2*PT

GIVEN RTT = 2*PT = 64 MILLISEC

LET FRAME SIZE = X BITS

-> X/100*10^6 = 64*10^-3

-> X = 64 * 10^5 BITS = 8*10^5 BYTE = 800000 BYTES
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It would be 8000006 bytes. The jamming-signal bits would be added, not because of the sender sending jamming signal in the frame but because of the increase in the Tt for sender.

Check out my solution on https://gateoverflow.in/270488/geeksforgeeks?show=277512#a277512

and Arjun Sir's response here- https://gateoverflow.in/1397/gate2005-74