Suppose the round trip propagation delay for a 10 Mbps Ethernet having 48-bit jamming signal is 46.4 μs. The minimum frame size is:

a) 94

b) 416

c) 464

d) 512

My approach:- I have read in the book that minimum frame transmission time is atleast twice the round trip time.So by that logic i can say that the answer will be 464.As soon as the first bit of jamming signal reaches the sender,it know about the collision. But I need to know whether time to transmit jamming signal over the link should also be included in this case?I mean whether the first bit of the jamming signal is sufficient to for the sender to detect the collision or the complete jamming signal needs to be received at sender side,which will make the answer as 512.

I dont have the answer as this question was in some exam in 2005 and there was no answers available to this.


3 Answers 3


This might not be the correct answer to your question(as I am not sure about jamming signal), but it might resolve the query.

As Ethernet uses CSMA/CD for error detection, the time taken to transmit the data must be greater than or equal to twice the time taken to propagate the data. Only then the station will come to know if there had been a collision.

Tt >= 2*Tp

The original size of ethernet IEEE 802.3 standard define the minimum ethernet frame size as 64 bytes .The minimum size of ethernet frame that carries on ICMP packet in 74 bytes.you can test this by installing a packet analyzer on your desktop and by running a ping packet with no options to distination inside your network.it will generate a 74 byte packet with a 60 byte IP header ,8 byte ICMP header,and ethernet frame size.

  • $\begingroup$ Your answer does not even mention the condition, "the round trip propagation delay for a 10 Mbps Ethernet having 48-bit jamming signal is 46.4 μs". Can you explain why? $\endgroup$
    – John L.
    Nov 14, 2018 at 17:07

2^n > 46.4*10^-6 * 10^6; n > log(464)/log(2) ~ 8.86...so set n = 9....2^9 = 512

  • 4
    $\begingroup$ We'd prefer that you explain your answer. Don't just give formulas -- share with us your reasoning and whe calculation came from. Thank you! $\endgroup$
    – D.W.
    Dec 5, 2019 at 6:42

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