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Consider the Ethernet LAN shown below

The LAN has a data transfer rate of 100Mbps and signal propagates at the speed of the 2 x 108 m/s. If system A and the system E start transmitting the frames simultaneously how many bytes of data has node A transmitted before detecting collision?

Caption

A. 62.5

B. 62

C. 63

D. 64

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2 Answers

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option(A) is correct

if A and E started transmitting the frames simultaneously so collision will be in the middle,in that case TT>=PT =>L/B>=d/v =>L>=(d/v)*B

d=1000 meters ,propagation speed(v)=2*10^8 m/sec ,bandwidth(B) =100 Mbps =100*10^6 bps

so L=(1000/2*10^8)*100*10^6 =500 bits =>500/8 bytes =>62.5 bytes
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v = 2 * 10^8 mps

B = 100 Mbps

d = 1000 m (between A and E)

station A            1000m     station E

Tp = 1000/2 * 10 ^8 sec

at t = 0 Both start transmitting data

at t = Tp / 2 collide data packet (in middle collide data )

at t = Tp  Both senesce collision

so Station A transmit data  = Tp time   

transmitted data = Tp * B(bandwidth)

                          = (1000/2 * 10 ^8)*(100 * 10^6) bit

                         = 500 bits (by solving above question)

                       = 500 / 8 Bytes = 62.5 bytes

So , correct option (A)