1e5091f1
From: barmar@alum.mit.edu
In article
<7486c924-f034-4c55-b583-88d7f7054c56@g19g2000vbi.googlegroups.com>,
acadian wrote:
> On May 11, 9:53 pm, Barry Margolin wrote:
> > In article
> > ,
> >
> >
> >
> > acadian wrote:
> > > Greetings-
> > > I am looking at rfc 791, and just do not follow the algorithm
> > > they use on page 25 to calculate the TL when a packet is fragmented.
> > > I specially do not follow step 9. Would someone please explain?
> > > IF TL =< MTU THEN Submit this datagram to the next step
> > > in datagram processing ELSE IF DF = 1 THEN discard the
> > > datagram ELSE
> > > To produce the first fragment:
> > > (1) Copy the original internet header;
> > > (2) OIHL <- IHL; OTL <- TL; OFO <- FO; OMF <- MF;
> > > (3) NFB <- (MTU-IHL*4)/8;
> > > (4) Attach the first NFB*8 data octets;
> > > (5) Correct the header:
> > > MF <- 1; TL <- (IHL*4)+(NFB*8);
> > > Recompute Checksum;
> > > (6) Submit this fragment to the next step in
> > > datagram processing;
> > > To produce the second fragment:
> > > (7) Selectively copy the internet header (some options
> > > are not copied, see option definitions);
> > > (8) Append the remaining data;
> > > (9) Correct the header:
> > > IHL <- (((OIHL*4)-(length of options not copied))+3)/4;
> >
> > Some IP options are copied into all fragments, some are only put in the
> > first fragment. If you don't copy all the options, the header may be
> > shorter. However, IHL counts in 4-octet words, while (length of options
> > not copied) is in octets. Step 9 figures out the new number of words
> > based on the decrease in octets.
> >
> > --
> > Barry Margolin, bar...@alum.mit.edu
> > Arlington, MA
> > *** PLEASE don't copy me on replies, I'll read them in the group ***
>
> Thanks. So what happens in step 9 is they take the Old IHL value
> multiply it by 4 to get the number of bit, subtract for the options
> that are not copied, then add 3 for 'copied flag' and 'option class'
> field, and then divide by 4. Is that correct?
Not quite. They add 3 to ensure that the result is rounded up properly.
Suppose the original IHL was 5, meaning 20 bytes, but 1 byte of options
is not copied. That reduces the header size to 19 bytes. If you just
divided by 4 you would get 4, which is only 16 bytes, which isn't big
enough. So first you add 3 to get 22, divide by 4, and that results in
IHL = 5.
> On a different note, how can I look at some real work values? Would
> you be able to suggest a small model that I can look at the fragmented
> values? I only have a couple of workstations, a hub, and an ISP
> facing router. I guess I need an inexpensive switch so can change the
> mtu, and sniff the traffic on two different ports? rfc 791 has some
> examples at the end, but it does not say which options they copied.
> The other values match.
IP options are not used very often.
--
Barry Margolin, barmar@alum.mit.edu
Arlington, MA
*** PLEASE don't copy me on replies, I'll read them in the group ***
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