I am struggling to develop a proof for an equation to compute the total bandwidth required for fragmented IP packets.
When a router forwards an IPv4 packet of size $n$ over a link with an MTU $m$ then the packet will be fragmented if $n>m$. Each fragment imposes an overhead of $o$ bytes (typically, $o=20$).
It looks like you could define this process recursively, where the size of each fragment is $F_i=m$ and $F_{i+1}=n-m+o$.
Total number of packets, with our without fragmentation, is $1 + \lfloor \frac{n}{m} \rfloor$. The first packet does not impose any additional overhead, so the total overhead from subsequent fragments should be $o \lfloor \frac{n}{m} \rfloor$.
For example, a 4500-byte packet traversing a link with an MTU of 1500 bytes will be fragmented into packets of sizes $F_0=1500$, $F_1=1480+20$, $F_2=1480+20$, and $F_3=40+20$.
The total number of fragments is $\lfloor \frac{4500}{1500} \rfloor = 3$ and the resulting overhead is $3\cdot20=60$ bytes.
We should be able to compute the total data transferred as
$ \begin{align} \displaystyle F_0 + F_1 + ... + F_{\lfloor \frac{n}{m} \rfloor} & = \\\\ \sum_{p\in F} p & = o\lfloor \frac{n}{m} \rfloor + n \end{align} $
This looks like something I should be able to prove with induction, but we never dealt with inequalities like this in school. I think the base case should be $m \lt n_1 \le 2m-o$ (ugly, I know). This should result in one fragment with $o$ bytes of overhead.
$ \begin{align} F_0 + F_1 & = \sum_{p\in F} p \\\\ (m) + (n - m + o) & = o\lfloor \frac{n}{m} \rfloor + n \\\\ n + o & = o \cdot 1 + n \end{align} $
How do I go about constructing the inductive step? I'm thinking you assume $cm < k < (c+1)m - o$ for $c \in \mathbb{Z}^+$ then the next step should be $k+m$, which adds one additional fragment and $o$ bytes of overhead, right?
