Computation Time.

For each $b_j$ (the last sequence number of data byte in the $j$th packet of $B$), the number of iterations in (4), (6), and (7) is $f(k,j)-f(k-1,j)+1$ every time the $k$th packet of $B$ is processed. So the total number of iterations for each $b_j$ when all the packets in $B$ are processed is at most $\sum_{k=1}^m \left(f(k,j)-f(k-1,j)+1\right) = n+m$, where $m$ is the number of packets in a packet stream $B$. This holds for any packet in $S$.

Suppose $O(m)=O(n)$, which is true for larger $n$ in most cases. The computation time in computing deviations for every packet stream in $S$ from $A$ is $O(nN)$, where $n$ is the number of packets in $A$ and $N$ is the number of packets in $S$.