Computer Network | Fragmentation

Question

Consider PDU size of transport layer packet is 2260 Bytes at source host. MTU for the network is 400 Bytes and IPv4 header size is 20 Bytes.
The number of fragments for given PDU is _____.

Answer 1

\[
\begin{aligned}
&\textbf{Given:}\\
&\quad\text{Transport-layer PDU (IP payload)} = 2260\ \text{bytes}\\
&\quad\text{MTU} = 400\ \text{bytes (includes IP header)}\\
&\quad\text{IPv4 header} = 20\ \text{bytes}
\end{aligned}
\]

\[
\textbf{Step 1: Maximum data (payload) per IP fragment}
\]
\[
\begin{aligned}
\text{max\_payload\_per\_fragment} &= \text{MTU} - \text{IPv4 header} = 400 - 20 = 380\ \text{bytes}\\
\text{But each fragment's data (except possibly the last) must be a multiple of }8\ \text{bytes.}\\
\text{So use } \left\lfloor\frac{380}{8}\right\rfloor\cdot 8 &= 47\cdot 8 = 376\ \text{bytes per fragment (usable).}
\end{aligned}
\]

\[
\textbf{Step 2: Number of fragments}
\]
\[
\text{Number of fragments} = \left\lceil \frac{\text{total payload}}{\text{payload per fragment}} \right\rceil
= \left\lceil \frac{2260}{376} \right\rceil = \left\lceil 6.0106\ldots \right\rceil = 7.
\]

\[
\boxed{\text{Answer: }7\ \text{fragments}}
\]

\[
\textbf{Why the alternatives are wrong:}
\]
\[
\begin{aligned}
&\text{(a) }2260 - 20 \text{ is wrong: }2260\ \text{is already the transport-layer payload handed to IP; the 20-byte}\\
&\qquad\text{IPv4 header is \emph{not} part of that 2260 and is added to each fragment separately.}\\[6pt]
&\text{(b) } \frac{2240}{400} \text{ is wrong: }2240\ (=2260-20) \text{ is a mistaken payload value, and dividing by MTU}\\
&\qquad\text{(400) counts whole-fragment sizes including headers; fragmenting must be done on the payload}\\
&\qquad\text{with the payload-per-fragment limited to }376\ \text{bytes (a multiple of 8).}
\end{aligned}
\]
 

Comments

6 fragments will be the answer. Because, the last 4 byte fragment can be merged to 376 as MTU data = 380 bytes

Answer 2

6

Answer 3

6 number of fragments will be observed at the destination node,

 

because the whole IP datagram is 2260bytes, but MTU= 400bytes; now we can send that amount of data 6times because 2260/400 = 5.65= 6 fragments.

 

Fragment ID---------- Fragment Offset-----------data length------------header length-------M----------D

Fragment1    -               000                                      380                           20                      1             1

Fragment2    -               050                                      380                           20                      1             1

Fragment3    -               100                                      380                           20                      1             1

Fragment4    -               150                                      380                           20                      1             1

Fragment5    -               200                                      380                           20                      1             1

Fragment6    -               250                                      240                           20                      0             1