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

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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.

 

  1. 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%.




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
Transcribed Image Text: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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