The Math Behind Bit Stuffing
We are encoding a packet data as a series of bits, for example: 0101000101. We want to denote the end of the packet with a "flag" of size J in the form of 0 followed by J 1s followed by a 0. So a flag with a flag size of 3 would be: 01110. To make sure no false flags appear in the packet, we insert a 0 after every J-1 occurences of a 1. To decode the packet after every J-1 occurences of a 1, we remove the following 0.
Our task is to find the optimal flag length if the expected size of a packet is E[K]
We assume P(0) = 1/2 and P(1) = 1/2
The probability of any bit being preceded by a 0 and being the first 1 in a sequence of j-1 1s is P(0)P(1)^(j - 1)=(1/2)^j. This does not apply to the very first bit (because it cannot be preceded by anything) and the final j-1 bits (they cannot be followed by j-1 1s), but as j is relatively small compared to E[K], we'll consider the probabilistic differences caused by these bits to be negligable.
Due to linearity of expectation, the expected number of stuffed bits is E[Stuffed Bits] ~= E[K]P(0)P(1)^(j - 1), so
E[Stuffed Bits and Ending Flag] ~= E[K]P(0)P(1)^(j - 1) + j + 2
To find the smallest optimum flag length
E[Stuffed Bits and Ending Flag] ~= E[K](1/2)^j + j + 2
Now we compare a flag of length j to that of j+1. We want to find a flag size s.t. increasing the flag size increases the total bits.
E[K](1/2)^j + j + 2 < E[K](1/2)^(j + 1) + (j+1) + 2
E[K](1/2)^j < E[K](1/2)^(j + 1) + 1
E[K](1/2)^j - E[K](1/2)^(j + 1) < 1
2E[K](1/2)^(j+1) - E[K](1/2)^(j + 1) < 1
E[K](1/2)^(j + 1) < 1
(1/2)^(j + 1) < 1/E[K]
log((1/2)^(j + 1)) < log(1/E[K])
-(j+1) < log(1/E[K])
-(j+1) < -log(E[K])
j > log(E[K]) - 1
As increasing the flag size increases the total bits, we want to minimize the flag size j. Therefore log(E[k]) - 1 is the optimum flag size.
This is much harder to do with calculus. Consider this: we want to minimize the expected total number of bits. So we want to minimize the function E[K](1/2)^j + j + 2 with respect to j. The derivative of the function would be E[K]log(1/2)(1/2)^j + j + 2. Setting that equal to 0, we get a complicated expression. So, the first solution is easier and more intuitive than the calculus one.