[olsr-dev] Olsr quality extensions metric

[Dries Naudts]

I think , you intuitively stated that 1/LQ1 + 1 /LQ2 is equal to 1/(LQ1 
* LQ2). There's the catch. ETX is not equal to 1/(LQ1 * LQ2). Let's 
clear this out.

Say,
ETXa:  ETX for two hop path1
ETXa1: ETX of first link of two hop path1
ETXa2: ETX of second link of two hop path1

ETXb:  ETX for two hop path2
ETXb1: ETX of first link of two hop path1
ETXb2: ETX of second link of two hop path1
LQ1 = ]0,1]
LQ2 = ]0,1]
LQ3 = ]0,1]
LQ4 = ]0,1]

So,

ETXa > ETXb   <=>
ETXa1 + ETXa2 > ETXb1 + ETXb2 <=>
1/LQ1 + 1/LQ2 > 1/LQ3 + 1/LQ4 <=>
(LQ1 + LQ2)/ LQ1*LQ2 > (LQ3+ LQ4)/ LQ3*LQ4  <=>
LQ1*LQ2 < LQ3*LQ4 * ((LQ1 + LQ2)/(LQ3 + LQ4))

In case (LQ1 + LQ2)/(LQ3 + LQ4) = 1 e.g., it is equal to LQ1 * LQ2 < LQ3 
* LQ4.
In case (LQ1 + LQ2)/(LQ3 + LQ4) > 1 e.g,  then you have eg LQ1 * LQ2 < 
LQ3 * LQ4 => ETXa > ETXb
                                                               
            BUT it is not true to state ETXa > ETXb => LQ1 * LQ2 < LQ3 * 
LQ4
But when (LQ1 + LQ2)/(LQ3 + LQ4) < 1 then it's NOT always true to state 
LQ1 * LQ2 < LQ3 * LQ4 => ETXa > ETXb

One could plot this in a graph, to see for which values it is true, but 
I think its outside the scope of this mail.. :-)

Kind regards,
Dries Naudts


Thomas Lopatic wrote:

>Ha! I got it. At least the 1-hop path versus 2-hop path thing. Suppose
>that we have two hops, one between node A and node B, another between
>node B and node C. Then node A re-sends its packets until it gets an
>acknowledgement from B. That's why the ETX for the complete 2-hop path
>is not 1 / (total link quality of the 2-hop path), I guess.
>
>The rest is still beyond me, though. :-)
>
>-Thomas
>
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-- 
Dries NAUDTS

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