Efficiency in ethernet

Question

Consider a 10 Mbps Ethernet LAN that has stations attached to a 2.5 km long coaxial cable.Given that the transmission speed is 2.3*10^8 m/s, the packet size is 128 bytes out of which 30 bytes are overhead,find the effective transmission rate and maximum rate at which the network can send data.

Here I am facing the following doubt:

1. Difference between effective transmission rate and maximum rate at which the network can send data.

My take:: maximum rate at which the network can send data is the throughput.And during its calculation I should consider the size of the entire packet(including overhead)i.e 128 bytes.
And for the calculation of the effective transmission rate I should consider just the date bytes(not the overhead) i.e 128-30=98 bytes.

Answer 1

In effective Transmission you can consider the overhead , as it is explicitly given otherwise it will be given as negligible.

And in the Maximum Transmission Rate just discard that overhead an take the Ideal Situation.

Here is the calculation that I made for the data you've given..

Prop.Delay =10 μsec (Taking Speed=2.5*10⁸ m/s)

Trans.Delay=102.4 μsec

Effective Data Rate= (128-30)/(10+102.4)= 0.871886 bits/μsec

Maximum Data Rate= (128)/(10+102.4)= 1.1387900 bits/μsec

Suggestions are welcome.

Comments

Your trans. delay is 128*102.4 us!

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In Effective Transmission does we consider overhead????

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@jason,in total time,why r u nt taking 2*prop delay??why just prop delay?

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@Akriti, actually I assume here the full duplex channel for CSMA/CD transmission.

Your point is also valid !!!

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@jason maximum rate means throughput and throughtput= efficiency * bandwidth

and efficiency of ethernet is 1/(1+6.44a). Source : https://gateoverflow.in/27636/formula-for-calculating-efficiency-of-ethernet

and effective transmission rate = throughtput * 98(usefull bit)/128(total)

plz correct if i m wrong ..i m confused

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This is from Local Area Networks 2E by Keiser.

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