Please check my answer and correct the part which is need to be foxed. And provide justification for the answer. 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%

Operations Research : Applications and Algorithms

4th Edition

ISBN:9780534380588

Author:Wayne L. Winston

Publisher:Wayne L. Winston

Chapter20: Queuing Theory

Section20.10: Exponential Queues In Series And Open Queuing Networks

Problem 8P

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Please check my answer and correct the part which is need to be foxed.
And provide justification for the answer.

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.

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%

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 A
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
с
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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