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Squitter-Interrogation Collision Probability


The DF-11 squitter has a 60 ms duration and happens once a second. Since the UF-0 interrogation has 30 ms, the chance of any part of an interrogation happening while squittering is:

1)         p = (60+2x30 ms)/106 = 0.00012

There is a 12 hour test that fails when a collision is detected.  For this test, 10 interrogations are made each second.  Events with such a low probability as Squitter-Interrogation collisions follow the Poison probability distribution .


2)         p(n/N) = (Np)n e-Np/n!


Where N is the number of interrogations in 12 hours (432000) and n is the number of collisions. P(n/N) is the probability of  having precisely n collisions in N tries (interrogations).Poison distribution


Equation 2 is the limit of the binomial distribution for N -> , while the product Np stays finite… or in practical terms Np << N or  p << 1 (rare events).  In our case this condition is very much true:


3)         Np = 432000*0.00012 = 51.84

So, our Poison distribution will be:

4)         p(n) = (51.84)n * 3.06x10-23 /n!


But not all interrogations are observed, of the 432000 interrogations, only 2356 will be sampled by the computer (1 sample every 5 seconds), the probability of not observing a single collision in any of those samples, if n of them will happen is:


5)         Qo(n) = (1-n/432000)2356

Then the probability of observing one or more would be:

6)         Po(n) = 1 – Qo(n)


The probability of observing a collision when precisely n of them happen, will be the probability of observing Po multiplied by the probability P(n) of having precisely n collisions.  The probability of observing a collision, disregarding how many actually happened, would be the sum of the probabilities for each n form 1 to (in principle) 432000:


6)         P = S1432000 Po(n) x p(n) = 0.2457


The chance of getting a collision failure is ~25% in the mentioned 12 hour test.

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