Given a communication channel that transfers frames which are $n_{bits}=160$ bits long with a $l_{CRC}=32$ CRC checksum at the end of each frame. The CRC has a hamming distance of $h=6$.
This communication protocol goes into a system that has the option of stopping safely if it detects that something is wrong. Thus, the protocol is programmed to stop safely after detecting $c=2$ consecutive erroneous frames. Recognizing $c-1$ frames as erroneous and not considering them is not a big deal because of the high refresh rate and stopping after detecting $c$ consecutive erroneous frames is not considered a failure either because the system can stop safely. A failure happens when a wrong frame is recognized as okay (with a CRC falsely checking the frame as good).
The frame rate is $N=720000$ frames per hour. The error bit rate is $p_e = 10^{-2}$.
What is the failure rate of such a protocol ?
The result doesn't have to be algebraic and can be simply computed. Here is my attempt.
Probability of a frame with at least 1 erroneous bit giving a wrong frame : $$P(\text{wrong frame}) = 1 - (1 - p_e)^{n_{bits}}$$ $$P(\text{wrong frame}) = 0.7997 $$
Probability of a CRC recognized as okay : $$ P(\text{CRC recognized as okay}) = 2^{-l_{CRC}}$$ $$ P(\text{CRC recognized as okay}) = 2.3283\times10^{-10}$$
Probability of frame with k erroneous bits $$ P(\text{frame with k erroneous bits}) = \binom{n_{bits}}{k} {p_e}^k(1-p_e)^{(n_{bits}-k)}$$
Probability of wrong frame with h or more erroneous bits $$ P(\text{wrong frame with h or more erroneous bits}) = 1 - \displaystyle\sum_{k=0}^{h} P(\text{frame with k erroneous bits}) $$ $$ P(\text{wrong frame with h or more erroneous bits}) = 0.0057 $$
Probability of unrecognized wrong frame $$ P(\text{unrecognized wrong frame}) = P(\text{CRC recognized as okay})\times P(\text{wrong frame with h or more erroneous bits})$$ $$ P(\text{unrecognized wrong frame}) = 1.3371\times10^{-12} $$
Up to there I am quite sure of the result. It is in the next phase of the computation where has to be taken into account the number of wrong frames tolerated that I have doubts.
Expected value of length of communication : $$ E(\text{length of communication}) = P(\text{wrong frame})^{-c}$$ $$ E(\text{length of communication}) = 1.5636$$
Probability of unrecognized false frame during the length of communication $$ P(\text{unrecognized false frame during the length of communication}) = 1 - (1-P(\text{unrecognized wrong frame}))^{E(\text{length of communication})}$$ $$ P(\text{unrecognized false frame during the length of communication}) = 2.0905\times10^{-12} $$ Probability of unrecognized false frame during 1 hour, i.e. failure rate $$ \Lambda= P(\text{unrecognized false frame during the length of communication})\times\left(\frac{N}{E(\text{length of communication})} \right)$$ $$ \Lambda= 9.6267\times10^{-7}$$
when computed this result isn't a function of the number of false frame tolerated $c$ wich seems odd. Am I on the right track or should I look into conditional probabilities ? It seems to me that the failure rate should be much lower knowing that there are no 2 consecutive wrong frames rather than with an arbitrary number of wrong frames. Thank you for any help.