Question 6: Consider a 5 Km, 10Mbps, shared Ethernet segment running CSMA/CD protocol with Station A at one end, Station D at the other, and Stations B and C equally spaced between and D. Assume bits propagate through the link at speed 200,000Km/sec; and assume the frames sent on this segment have all 10,000 bytes. A B C D a) If A sends a frame to D, what is the transmission time of this frame? Show your detailed work b) What is the efficiency of the CSMA/CD protocol used in this LAN? Show your detailed work
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Dear Writer this is my calculation solution.
could you please double check it and let me know it if it soccrect or should i proceed with your solution: Please provide justification if correction is need and correct answer.
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a)
The transmission time of a frame can be calculated as follows:
Propagation delay (Tp) = distance/speed = 5km / 200000 km/s = 0.0025ms
Transmission time (Tt) = frame size/bandwidth =
10000 * 8bits / (1Mbps * 10 ^ 6) = 0.008sec = 8ms
Total time for transmission =
2 * Tp + Tt(since the frame has to travel from A to D and then D to A) =
2 * 0.0025ms + 8ms = 0.005ms + 8ms = 8.005ms
The transmission time of the frame from A to D is 8.005 ms.
- b)
The efficiency of the CSMA/CD protocol is given by the formula:
Tt is the transmission time of a frame
C is the number of collisions,
Tp is the propagation time of a signal from one end of the segment to the other.
Efficiency = Tt / (C * 2 * Tp + Tt + Tp)
Assuming that there are no other stations transmitting or attempting to transmit on the segment, the transmission from A to D will succeed without collisions. Therefore, C = 0. The propagation time can be calculated as
PropogationT(Tp) = distance/speed
Tp = 5km / 200000km/s = 0.0025ms
Tt = 8ms
Efficiency = Tt / (C * 2 * Tp + Tt + Tp) = 8ms / (8ms + 0.0025ms) = 0.9996875 99.96%
The efficiency of the CSMA/CD protocol is 99.96%
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