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 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
Please check the formula logic and the formula for my solution for answer a
Alo please check the formula and calculation for the part b.
I got confused about the conversionts and not sure if i did it all correct way.
a)
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 * 0.0025ms + 8ms = 0.005ms + 8ms = 8.005ms
The transmission time of the frame from A to D is 8.005 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 / speed = 5 Km / 200,000 Km/sec = 25 microseconds = 0.025 ms
To calculate tframe, we need to determine the time it takes to transmit a frame of 10,000 bytes at a data rate of 10Mbps.
The frame size is 10,000 bytes = 80,000 bits
The time to transmit the frame is:
tframe = frame size / data rate = 80,000 bits / 10Mbps = 0.008 s = 8 ms
Efficiency = (1 / (1 + 5tprop / tframe))
= (1 / (1 + 5*25 microseconds / 8 milliseconds))
= (1 / (1 + 0.15625))
= 0.865
The efficiency of the CSMA/CD protocol used in this LAN is approximately 86.5%.
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