Ethernet

Question

In standard ethernet with transmission rate of 20 Mbps, the length of the cable is 2500 m and the size of a frame is 1024 bits. The propagation speed of a signal in a cable is $2 \times 10^8$ m/s. The percentage of the time medium is occupied but not used by a station is?

Comments

19.6%??

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given 61
 

but explain your approach

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efficiency is 38.8%
it means u are using only 38.8% of the bandwidth
you are using bandwidth nly 38.8% of the time.
remaining 61.2% of the time is going waste

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in other way, u can transmit 20Mb in 1 secod
but u are transmitting only 0.388*20Mb in 1 second .. which can actually be transmitted in 0.388 seconds
means time wastage = 1-0.388 = 0.612 seconds

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whats the formula for efficiency? I used 1/(1 + 5a)  Using this, E = 44.44%

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Ohh..u used 1/(1+ 6.44a).

When to use 5a ? Any idea? In kurose-Ross, its 5a

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@Anusha Motamarri

add your both comments as answer ..you are correct

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Plz give the formula to use when such questions asked. Are there different cases in which these different formulas used???

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@Anusha Is efficiency of ethernet is taken as 1(1+6.44a) always or 1/(1+5a)?

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1/(1+6.44a)

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@Anusha But read Bikram sir's answer

https://gateoverflow.in/27636/formula-for-calculating-efficiency-of-ethernet?show=27636#q27636

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read the answer below it ...

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@Lokesh. Its given Ethernet and hence, it should be 5a.

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ethernet uses CSMA/CD only..so in both cases efficiency will be that 6.44 one...

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I am confused !!  CSMA/CD is ethernet technology and not vice-versa. Kurose-Ross has 5a and I think Forouzan has 6.44a. Which to use?

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Even I have been using 6.44a in case of both Ethernet and csma/cd,untill I came across the above mentioned link which confused me :(

Answer 1

Efficiency = 1/(1+6.44a)
a=Tp/Tt and Tp=d/v and Tt=L/B
Tp=2500/2*10^8
Tt=1024/2*10^7
You'll get efficiency as 38% approx
The percentage of the time medium is occupied but not used by a station is 62% approx