7. (4 pointa) Cossider two nodes A and B consected vis an Etheret bo of 48Mbps S

Question

7. (4 pointa) Cossider two nodes A and B consected vis an Etheret bo of 48Mbps Suppone that the two nodes are the oaly sodes, and that the propsantion delay betwern temi 240 Suppose that both nodes each bas a frame to send, and st tve t After coll ision, node A chooses KA 0 and node B chocees Kp 1. Rl th, i the CSMA/CD 0 they bot algorithm, right after detecting a collislon, a node mrst seud a js signal of 4s bit me The CSMA/CD algorithm also requires tht a node, having Eranse to setd, sesee the bus ie or 96 bit times before it can transmit. The tame length ? 1200 bit. All answers mont be In microseconds. (a) (1 point) At what time does A detect the collision (in micreoda) time microseconds (b) (I point) At what time does the bus become idle (in microsecoode) time (c) (1 point) At what time does A begin retranmiscion (in microsecoods)? microseconds time (d) (1 point) At what time does A's frame get suocessfully delivered to B (in microseconds)? time

Explanation / Answer

formula to convert bits into second = bitTimes/Bandwidth (in Bps) seconds

a) A will detect collision at time = propagation delay/2 = 240/2 = 120 bit times =120/4.8*10^6= 25 microseconds (using formula)

b)

since passive medium i.e. ethernet bus cannot detect collision . Active devices i.e. sender and receiver can only detect it.

The collision will be detected after time = 2*propagation delay + transmission delay in jam signal
= 2*240 + 48 bit times = 528 bit times = 110 microseconds

after time t=110 microsecond bus will be idle for 96 bit times

C)

after sensing this collison, since A has choosen KA=0.

A will start retransmission after sensing bus for 96 bit times or 20 microsecond after collision detection

so at t = 110 + 20 = 130 microsecond A willl start retransmission

D)

frame transmission time by A = transmission delay + propagation delay

since the frame size = 1200 bits

so transmission delay = 1200/4.8*10^6 = 250 microseconds

time taken by A to sucessfully transmit the frame to B = 250 + 50= 300 microseconds

hence at t = 130 + 300 = 430 microseconds B will recieve all bits of frame successfully.

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