Frame Size

Question

Suppose the round trip propagation delay for 10Mbps Ethernet has 24.2micro sec The network has 48bit jamming signal then what is minimum frame size in bits

Answer 1

RTT delay = 24.2 micro sec

 

So , we need to calculate time taken to send 48 bit jamming signal ::

 

10*(10^6) bits -----> 1 sec.

48 bit ------------->  48/(10*10^6) = 4.8 micro sec

So , total time = (24.2 + 4.8 ) microsec = 29 microsec.

Min frame size is the data to be sent in 29 microsec = 29*10^-6 * 10 * 10^6 = 290 bits

Comments

I have calculated this as
L>=2*propagation time* B
L=24.2 *10^-6 * 10*10^6 bits
L=242 bits

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I think we need to take transmission delay also into consideration.

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@arjun sir....

the jaming signals will be transmitted by the receiver and it is after the senders transmission....(after one propagation delay)

so maximmu transfer will be 242 bits