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What formula to use for calculating efficiency of ethernet is it 
$\frac{1}{1+6.44a}$ or $\frac{1}{1+5.44a}$ or $\frac{1}{1+5a}$​
I saw three versions for calculating the same thing in different places. I want to know which one is correct.

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Best answer
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27 votes

In CSMA/CD, for success, only 1 station should transmit while others shouldn't.

P(success)=nC1*p * (1-p)n-1 Binomial thm.

For max P(success), differentiate and equate to zero(line in limits to get maxima and minima).

We get P(max)=1/e
Number of times we need to try before getting 1st success= $\frac{1}{p(max)}$ = $\frac{1}{\frac{1}{e}}$=e

Efficiency=$\frac{transmission time(Tt)}{number of collision(C)*2*propagation time(Tp)+Tt+Tp}$  

Here number of times we need to try=e, ie C=e.

Put a=$\frac{Tt}{Tp}$ and divide by Tt, we get: $\frac{1}{e*2a+1+a}$

Put e=2.72, efficiency = $\frac{1}{1+6.44a}$

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4 votes

 For CSMA/CD it is taken as 1/ (1+6.44a)

For Ethernet it is taken as 1/ (1+5 a) reference : http://robotics.eecs.berkeley.edu/.../12203/ethernet%201.pdf ,

slide number 33 and 34.

There are 2 formulas basically ( 1/1+6.44a  ),      (1/1+5.44a ) or  1/1+5a  .

we consider  (1/1+5.44a ) or  (1/1+5a )  as same formula . 

The  difference between 1/1+6.44a  and   1/1+5.44a  is Propagation delay .

Propagation delay is consider in  (1/1+6.44a ) .  

While 1/1+5.44a don't consider propagation delay .

In case of Ethernet  we should use 1/1+6.44a .

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1 votes
1 votes
hi

its 1st one in case csma/cd is used

 

regards

Piyush
1 votes
1 votes

This formula is standard since it is derived from TP/(TP+CP+Propagation delay) It involves idle time period also.: 1 1 + 6.44 a

Second formula is approximate derived from TP/(TP+CP) It does not involve time period. 1 1 + 5.44 a

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