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I am computing the energy efficiency of a wireless system, given by the ratio of how many bits are received over how many energy is consumed.

$$EE=\dfrac{bits}{joules}$$

For a given scenario, I have an algorithm that outputs the value of \$bits\$ and the value of \$joules\$. My issue is when I compute the average \$EE\$ obtained across different (independent) scenarios.

Assume I run the algorithm for \$1000\$ scenarios and I obtain:

  • \$bits = 1\ \mathrm{Mb}\$ and \$joules = 1\ \mathrm{J}\$ for \$999\$ scenarios; and
  • \$bits = 1\ \mathrm{Mb}\$ and \$joules = 10^{-5}\ \mathrm{J}\$ for \$1\$ scenario.

Now, in average, I have \$\bar{bits} = 1\ \mathrm{Mb}\$ and \$\bar{joules} ≈ 1 = 999\times10^{-3}\ \mathrm{J}\$. This gives an average energy efficiency of \$\bar{EE} ≈ 1\ \mathrm{Mb/J}\$.

If I compute the \$EE\$ for each scenario (which I think is the correct way to do this) we will get:

  • \$EE = 1\ \mathrm{Mb/J}\$ for \$999\$ scenarios; and
  • \$EE = 10^5\ \mathrm{Mb/J}\$ for \$1\$ scenario.

This gives an average of \$\bar{EE} = 100\ \mathrm{Mb/J}\$.

Should I compute the average \$EE\$ as the average of \$bits\$ divided by the average of \$joules\$ or should I compute the average \$EE\$ as the average of individual \$EE\$ computed for each scenario.?

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  • \$\begingroup\$ What is \$E\$? How does it relate to \$EE\$? Where does the \$5j\$ come from? \$\endgroup\$
    – Andy aka
    Commented Apr 21 at 14:59
  • \$\begingroup\$ 1) j/mB makes more sense than mB/j imo. 2) The 1E-5j scenario is an outlier and shouldn't be counted \$\endgroup\$
    – InBedded16
    Commented Apr 21 at 15:14
  • \$\begingroup\$ @Andyaka E is only for exponent so $1E-5$ is $10^{-5}$ and the $j$ is joule. \$\endgroup\$
    – zdm
    Commented Apr 21 at 16:22
  • \$\begingroup\$ @InBedded16 but how to know if a scenario is an outlier or not? If I have 10_000 scenarios and some of them give below than 1 joule and some give above 1 joule. \$\endgroup\$
    – zdm
    Commented Apr 21 at 16:24
  • \$\begingroup\$ Energy is either written as \$joules\$ or \$J\$. Lower-case j is the complex operator used in complex math. \$\endgroup\$
    – Andy aka
    Commented Apr 21 at 16:24

1 Answer 1

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It depends on which average you want -- basically, what do you want to do with the average.

Basically, in the 1st computation, you are calculating the average assuming all 1000 scenarios are used, and you then know (using all 1000 equally often), how much energy is required per bit (on 'average') -- basically, if you need to transmit X bits, you would (probably) need Y joules.

In the 2nd computation, you are measuring the efficiency of each scenario independently. If you had to select one of those 1000 and use it exclusively (i.e. not getting access the other 999), you could expect (that's one interpretation of average) an efficiency of 100Mb/J. Of course, some instances will have 1e5, and some 1, but the average is 100.

Imagine you are selling communications boxes. In the 2nd case, you can claim that your average efficiency is 100, but most customers (who only but 1 box) will in fact have an efficiency of 1; there will be one 'lucky' customer who gets 1e5.

In the 1st case, if you have a system which uses all 1000 'channels' simultaneously, you can expect an efficiency of 1.

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