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| 99 views

+1 vote

For ethernet which follows CSMA / CD protocol , the following condition is necessary :

Transmission time  =  2 * Propogation time (or) round trip time

Given  round trip time  =   46.4 micro sec

Transmission time   =  Data size / Data rate

So we have to find data size basically

So equating T.T  =  Round trip time

x / 10 * 106   =   46.4 / 106

==>    x  = 46.4 * 10

==>     x  = 464

Hence C) is the correct option.

by Veteran (102k points)
0
okk..thats what I was getting also...

What they have given is (464-48)=416...(Removing bits for jamming signal).Do you think it is correct?
0
Following the definition of transmission time it should be C) option only..
0
Yes..Jamming is a seperate signal..i dont think there would be any relation between frame size and jamming signal
0
ethernet...means CSMA/CD has been used as access control....in this whenever there is a collision,the stations sends a jamming signal to stop all other transmissions....and here we are calculating minimum length of data packet for a successful transmission...means no collision should happen for 464 bits long data...so if no collision is happening i dont think jamming signal should come into picture...
+1 vote
Total Delay = RTT + Jam Signal Transmission Time

= 46.4 + 4.8 = 51.2 microsec

Min Frame Length = 51.2 * 10^-6 * 10 * 10^6 = 512 bits
by Active (3.1k points)

+1 vote