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2 Answers

1 votes
1 votes

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.

1 votes
1 votes
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

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My doubt -Whether # of subnet is 2bits for Subnet or 2bits for Subnet -2 ??