17. A 1 KM 1long CSMA/CD with a propagation 10Mbps delay 5usec has bandwidth. Repeaters are not allowed in the system. Data bits of 224, header bits of 32 bits are in packet. One slot time is reserved for the sender to start its transmission and another slot time is reserved for the receiver to сapture the channel to send 32 bit acknowledgement frame. What is the effective data rate?

17. A 1 KM 1long CSMA/CD with a propagation 10Mbps delay 5usec has bandwidth. Repeaters are not allowed in the system. Data bits of 224, header bits of 32 bits are in packet. One slot time is reserved for the sender to start its transmission and another slot time is reserved for the receiver to сapture the channel to send 32 bit acknowledgement frame. What is the effective data rate?

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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Que.a) Consider a CSMA/CD network that transmits data at a rate of 400 x 106 bits per second. The maximum signal propagation time from one node to another node is 20 microseconds. The minimum size of a frame in the network is(. bytes).
Dear Writer could oyu please kindly explin why i get this soltuion and it is different that your solution? The transmission time of a frame can be calculated as follows: Propagation delay (Tp) = distance/speed = 5km / 200000 km/s = 0.025ms 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 * 0.025ms + 8ms = 0.05ms + 8ms = 8.05ms The transmission time of the frame from A to D is 8.05 ms. b) To calculate the efficiency of the CSMA/CD protocol in this LAN, we can use the formula: Efficiency = (1 / (1 + 5tprop / tframe)) tprop is the propagation delay tframe is the time it takes to transmit a frame. To calculate tprop, we need to determine the time it takes for a signal to propagate through the 5 Km distance between Station A and Station D. The speed of propagation is 200,000 Km/sec The propagation delay is: tprop = distance /…
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