IPV4 header checksum verification - networking

When a packet is sent from source to destination,it has to pass by through several routers and each router decreases the value of TTL of packet by 1. So header checksum is to be recomputed at each router since one of the header field i.e. TTL surely changes. Then how does the destination verifies the presence of error by calculating the checksum ,though the checksum it got has changed than that of checksum of source side?

When the packet leaves the source, it has some initial TTL and (hopefully) a valid checksum.
When the packet arrives at a router, the router checks only the IPv4 header checksum. If it is incorrect, it drops the packet. If it is correct, it (1) decrements the TTL; (2) checks that the TTL is higher than zero (otherwise the packet is dropped) and (3) computes and fills in the new IP header checksum.
Interestingly, the new checksum can be computed directly from the old checksum and the old and new value of the TTL, with some clever math: https://www.rfc-editor.org/rfc/rfc1624 so reading the whole IPv4 header again is not necessary.
Note that IPv6 has a hop limit (which works like the TTL) but no header checksum.

Every router calculates and verifies the checksum before routing, if there is a mismatch the router drops the datagram.

Related

Transparent IP fragmentation

From what I have understood so far, unlike non-transparent IP fragmentation; where packets are fragmented at the source and reassembled only at the destination; intermediate network systems in transparent IP fragmentation, reassemble and fragment the IP packet while in transit. Given the below example where two end-systems (A) and (B) communicate with each other, each of which is part of subnet1 and subnet3 respectively. Would router1 reassemble IP packet fragments sent by A and then send the whole non-fragmented packet to router2 who would fragment it before sending it to B ? Is this how transparent IP fragmentation works ?
(A) subnet1 ----- router1 ------ subnet2 ------- router2 ------- subnet3 (B)
I think you should read this post which gives you an idea of how it works.
For me the packet is rassemble by router1 sent to router2 and then arrive to B.
In all cases B recieve the whole paquet and not fragments made by A.
In normal IPv4 fragmentation, a router will fragment a packet if the MTU of the next interface is smaller than the packet size. Fragmentation occurs on the router(s), but reassembly of the packet fragments is the responsibility of the end-system.
The use of fragmentation and reassembly between two intermediate systems in the path is not directly covered by the IPv4 RFC, so there is no real standard for how this happens. Basically, the end of one link will fragment a packet to fit the MTU of the link, and the other end of the link will reassemble the fragments to the original packet. This is not something that is common because it places a large burden on the routers. In normal fragmentation, the burden of reassembly is place on the end-system.
From RFC 791, Internet Protocol:
The basic internet service is datagram oriented and provides for the
fragmentation of datagrams at gateways, with reassembly taking place
at the destination internet protocol module in the destination host.
Of course, fragmentation and reassembly of datagrams within a network
or by private agreement between the gateways of a network is also
allowed since this is transparent to the internet protocols and the
higher-level protocols. This transparent type of fragmentation and
reassembly is termed "network-dependent" (or intranet) fragmentation
and is not discussed further here.
The full description of fragmentation in the RFC:
Fragmentation
Fragmentation of an internet datagram is necessary when it originates
in a local net that allows a large packet size and must traverse a
local net that limits packets to a smaller size to reach its
destination.
An internet datagram can be marked "don't fragment." Any internet
datagram so marked is not to be internet fragmented under any
circumstances. If internet datagram marked don't fragment cannot be
delivered to its destination without fragmenting it, it is to be
discarded instead.
Fragmentation, transmission and reassembly across a local network
which is invisible to the internet protocol module is called intranet
fragmentation and may be used [6].
The internet fragmentation and reassembly procedure needs to be able
to break a datagram into an almost arbitrary number of pieces that can
be later reassembled. The receiver of the fragments uses the
identification field to ensure that fragments of different datagrams
are not mixed. The fragment offset field tells the receiver the
position of a fragment in the original datagram. The fragment offset
and length determine the portion of the original datagram covered by
this fragment. The more-fragments flag indicates (by being reset) the
last fragment. These fields provide sufficient information to
reassemble datagrams.
The identification field is used to distinguish the fragments of one
datagram from those of another. The originating protocol module of an
internet datagram sets the identification field to a value that must
be unique for that source-destination pair and protocol for the time
the datagram will be active in the internet system. The originating
protocol module of a complete datagram sets the more-fragments flag to
zero and the fragment offset to zero.
To fragment a long internet datagram, an internet protocol module (for
example, in a gateway), creates two new internet datagrams and copies
the contents of the internet header fields from the long datagram into
both new internet headers. The data of the long datagram is divided
into two portions on a 8 octet (64 bit) boundary (the second portion
might not be an integral multiple of 8 octets, but the first must be).
Call the number of 8 octet blocks in the first portion NFB (for Number
of Fragment Blocks). The first portion of the data is placed in the
first new internet datagram, and the total length field is set to the
length of the first datagram. The more-fragments flag is set to one.
The second portion of the data is placed in the second new internet
datagram, and the total length field is set to the length of the
second datagram. The more-fragments flag carries the same value as
the long datagram. The fragment offset field of the second new
internet datagram is set to the value of that field in the long
datagram plus NFB.
This procedure can be generalized for an n-way split, rather than the
two-way split described.
To assemble the fragments of an internet datagram, an internet
protocol module (for example at a destination host) combines internet
datagrams that all have the same value for the four fields:
identification, source, destination, and protocol. The combination is
done by placing the data portion of each fragment in the relative
position indicated by the fragment offset in that fragment's internet
header. The first fragment will have the fragment offset zero, and
the last fragment will have the more-fragments flag reset to zero.

Error in the tcp segment area inside a IP packet?

Inside an IP network IPv4 protocol is being used. When a datagram is being transported through the network (TTL=2), an error occurs inside the field which contains TCP-segment with the address of a web page. What will be the reaction of the router when reading such datagram?
I'm battling with this question for two days now.
What I know:
the router only reads the header of the datagram and if the checksum fails the datagram is being discarded by the router,
TTL is always decremented by one by the router if the header checksum is OK, after the -1, the datagram is being forwarded to the next router in the network
Based on this two points I conclude that the datagram will go further and TTL will be 1 and new checksum will be calculated by the router.
However there are three other possible answers to the questions:
The router sends ICMP to source with the error
the router repairs the mistake based on the header checksum and then forward the datagram
the router sends the datagram out of the network (discards the datagram)
Any help with clearing this problem?
The exact algorithms that must be implemented by IPv4 routers is described in RFC 1812 - Requirements for IP Version 4 Routers. According to the IETF RFC Index it still applies ("Status: BEST CURRENT PRACTICE").
The forwarding algorithm can be found in Section 5.2.1. The constraints determining the dependencies between the steps are (quoting):
(1) A router MUST verify the IP header, as described in section
[5.2.2], before performing any actions based on the contents of
the header. This allows the router to detect and discard bad
packets before the expenditure of other resources.
(2) Processing of certain IP options requires that the router
insert
its IP address into the option. [...]
(3) The router cannot check and decrement the TTL before checking
whether the packet should be delivered to the router itself, for
reasons mentioned in Section [4.2.2.9].
(4) More generally, when a packet is delivered locally to the
router,
its IP header MUST NOT be modified in any way [...].
The actual steps performed when receiving a packet are (quoting):
(1) The router receives the IP packet (plus additional information
about it, as described in Section [3.1]) from the Link Layer.
(2) The router validates the IP header, as described in Section
[5.2.2]. Note that IP reassembly is not done, except on IP
fragments to be queued for local delivery in step (4).
(3) The router performs most of the processing of any IP options.
As
described in Section [5.2.4], some IP options require additional
processing after the routing decision has been made.
(4) The router examines the destination IP address of the IP
datagram, as described in Section [5.2.3], to determine how it
should continue to process the IP datagram. There are three
possibilities:
o The IP datagram is destined for the router, and should be
queued for local delivery, doing reassembly if needed.
o The IP datagram is not destined for the router, and should be
queued for forwarding.
o The IP datagram should be queued for forwarding, but (a copy)
must also be queued for local delivery.
So it is clear that checksum verification of the IPv4 header is performed first. The exact steps are described in Section 5.2.2 IP Header Validation, but they are not important here. What matters is that only the IP header is checked, not the content. Therefore the router cannot detect the error.
Based on this two points I conclude that the datagram will go further and TTL will be 1 and new checksum will be calculated by the router.
Correct.
As for the other options:
The router sends ICMP to source with the error
No, there is no Time Exceeded error. As for other ICMP errors, there are none that signal to the sender packet corruption. So even if the router could detect packet corruption (say if the corrupted bit was in the IP header), it would still not send an ICMP message.
The router repairs the mistake based on the header checksum and then forward the datagram
No, the checksumming performed in IPv4 and TCP only allows error detection, not correction.
The router sends the datagram out of the network (discards the datagram)
It does not discard it because it does not detect the error.
Regarding what I said in the comment about link layer error detection, there are usually two sources of errors in transit: (1) from the transmission medium (interference, damaged cable, improperly connected plug etc.) and (2) from forwarding devices themselves (defective memory chips, firmware bugs, cosmic rays hitting a chip etc.). The link layer can usually detect and may even correct errors from (1), but obviously not from (2). So the scenario described in the question is indeed possible if a device malfunction corrupted the packet contents.

IP fragments failure on network?

Exam question (with no additional info):
When an bunch of IP datagram fragments are being sent over the network and only one of them does not get to it's destination, what will happen then?
I'm not sure if ICMP is involved here or not. Does ICMP send an error report reporting to the source that it needs to resend that same fragment (only this one fragment)?
The problem is here that I don't know if the IP fragments use UDP or TCP therefore I don't know the answer to the question.
(I've posted on the networkengineering.stackexchange but my question was rejected)
Points to cover:
After a timer triggered by the receipt of the first fragment has expired, the reassembling host will discard all the of fragments.
The reassembling host may generate an ICMP Time Exceeded (Fragment reassembly time exceeded).
The first fragment will need to have been received for the ICMP to include the first 8 bytes of the triggering payload. IPv6 will not generate the ICMP Time Exceeded unless the first fragment was received.
With IPv6 if the reassembled datagram would be larger than 1500 bytes then it may be silently discarded.
If a higher level protocol with reliable delivery was used (e.g. TCP), then the originating host should retransmit datagrams for which no acknowledgement has been received.

TTL field and IP checksum

If I change the time-to-live (TTL) field in an IP packet, should I also recompute the IP checksum?
And what about the checksum at the transport layer ? Will that have to change too?
Yes. If you change any field in the IP header, the checksum must be updated.
As for the checksum at the transport layer, it theoretically depends what the transport protocol is... however in practice I do not believe there exist any transport protocols whose checksums cover the IP TTL field. Certainly the pseudo headers for both TCP and UDP don't cover it.
Yes, if you change the TTL field, then you have to change the checksum also because checksum is related to TTL field.
Also, TTL will be changed at every hop. Since TTL is changing at every hop, so checksum will also be changed.

What is the Significance of Pseudo Header used in UDP/TCP

Why is the Pseudo header prepended to the UDP datagram for the computation of the UDP checksum? What's the rational behind this?
The nearest you will get to an answer "straight from the horse's mouth", is from David P. Reed at the following link.
http://www.postel.org/pipermail/end2end-interest/2005-February/004616.html
The short version of the answer is, "the pseudo header exists for historical reasons".
Originally, TCP/IP was a single monolithic protocol (called just TCP). When they decided to split it up into TCP and IP (and others), they didn't separate the two all that cleanly: the IP addresses were still thought of as part of TCP, but they were just "inherited" from the IP layer rather than repeated in the TCP header. The reason why the TCP checksum operates over parts of the IP header (including the IP addresses) is because they intended to use cryptography to encrypt and authenticate the TCP payload, and they wanted the IP addresses and other TCP parameters in the pseudo header to be protected by the authentication code. That would make it infeasible for a man in the middle to tamper with the IP source and destination addresses: intermediate routers wouldn't notice the tampering, but the TCP end-point would when it attempted to verify the signature.
For various reasons, none of that grand cryptographic plan came to pass, but the TCP checksum which took its place still operates over the pseudo header as though it were a useful thing to do. Yes, it gives you a teensy bit of extra protection against random errors, but that's not why it exists. Frankly, we'd be better off without it: the coupling between TCP and IP means that you have to redefine TCP when you change IP. Thus, the definition of IPv6 includes a new definition for the TCP and UDP pseudo header (see RFC 2460, s8.1). Why the IPv6 designers chose to perpetuate this coupling rather than take the chance to abolish it is beyond me.
From the TCP or UDP point of view, the packet does not contain IP addresses. (IP being the layer beneath them.)
Thus, to do a proper checksum, a "pseudo header" is included. It's "pseudo", because it is not actaully part of the UDP datagram. It contains the most important parts of the IP header, that is, source and destination address, protocol number and data length.
This is to ensure that the UDP checksum takes into account these fields.
When these protocols were being designed, a serious concern of theirs was a host receiving a packet thinking it was theirs when it was not. If a few bits were flipped in the IP header during transit and a packet changed course (but the IP checksum was still correct), the TCP/UDP stack of the redirected receiver can still know to reject the packet.
Though the pseudo-header broke the separation of layers idiom, it was deemed acceptable for the increased reliability.
"The purpose of using a pseudo-header is to verify that the UDP
datagram has reached its correct destination. The key to
understanding the pseudo-header lies in realizing that the correct
destination consists of a specific machine and a specific protocol
port within that machine. The UDP header itself specifies only the
protocol port number. Thus, to verify the destination, UDP on the
sending machine computes a checksum that covers the destination IP
address as well as the UDP datagram. The pseudo-header is not
transmitted with the UDP datagram, nor is it included in the length."
E. Comer - Internetworking with TCP/IP 4th edition.
Pseudo IP header contains the source IP, destination IP, protocol and Total length fields. Now, by including these fields in TCP checksum, we are verifying the checksum for these fields both at Network layer and Transport layer, thus doing a double check to ensure that the data is delivered to the correct host.

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