Chap 08 ip

CChhaapptteerr 88
IInntteerrnneett PPrroottooccooll
Objectives
Upon completion you will be able to:
• Understand the format and fields of a datagram
• Understand the need for fragmentation and the fields involved
• Understand the options available in an IP datagram
• Be able to perform a checksum calculation
• Understand the components and interactions of an IP package
TCP/IP Protocol Suite 1

Figure 8.1 Position of IP in TCP/IP protocol suite

8.1 DATAGRAM
A packet in the IP layer is called a datagram, a variable-lleennggtthh ppaacckkeett
ccoonnssiissttiinngg ooff ttwwoo ppaarrttss:: hheeaaddeerr aanndd ddaattaa.. TThhee hheeaaddeerr iiss 2200 ttoo 6600 bbyytteess iinn
lleennggtthh aanndd ccoonnttaaiinnss iinnffoorrmmaattiioonn eesssseennttiiaall ttoo rroouuttiinngg aanndd ddeelliivveerryy..

Figure 8.2 IP datagram

Figure 8.3 Service type or differentiated services

NNoottee::
The precedence subfield was designed,
but never used in version 4.

TTaabbllee 88..11 TTyyppeess ooff sseerrvviiccee

TTaabbllee 88..22 DDeeffaauulltt ttyyppeess ooff sseerrvviiccee

TTaabbllee 88..33 VVaalluueess ffoorr ccooddeeppooiinnttss

NNoottee::
The total length field defines the total
length of the datagram including the
header.

Figure 8.4 Encapsulation of a small datagram in an Ethernet frame

Figure 8.5 Multiplexing

TTaabbllee 88..44 PPrroottooccoollss

ExamplE 1
An IP packet has arrived with the first 8 bits as shown:
01000010
The receiver discards the packet. Why?
Solution
There is an error in this packet. The 4 left-most bits (0100)
show the version, which is correct. The next 4 bits (0010) show
the header length; which means (2 × 4 = 8), which is wrong.
The minimum number of bytes in the header must be 20. The
packet has been corrupted in transmission.

ExamplE 2
In an IP packet, the value of HLEN is 1000 in binary. How
many bytes of options are being carried by this packet?
Solution
The HLEN value is 8, which means the total number of bytes
in the header is 8 × 4 or 32 bytes. The first 20 bytes are the
base header, the next 12 bytes are the options.

ExamplE 3
In an IP packet, the value of HLEN is 516
and the value of the total length field is 002816 . How
many bytes of data are being carried by this packet?
Solution
The HLEN value is 5, which means the total number of bytes
in the header is 5 × 4 or 20 bytes (no options). The total length
is 40 bytes, which means the packet is carrying 20 bytes of data
(40 − 20).

ExamplE 4
An IP packet has arrived with the first few hexadecimal digits
as shown below:
45000028000100000102 . . .
How many hops can this packet travel before being dropped?
The data belong to what upper layer protocol?
Solution
To find the time-to-live field, we skip 8 bytes (16 hexadecimal
digits). The time-to-live field is the ninth byte, which is 01. This
means the packet can travel only one hop. The protocol field is
the next byte (02), which means that the upper layer protocol is
IGMP (see Table 8.4).

8.2 FRAGMENTATION
The format and size of a frame depend on the pprroottooccooll uusseedd bbyy tthhee
pphhyyssiiccaall nneettwwoorrkk.. AA ddaattaaggrraamm mmaayy hhaavvee ttoo bbee ffrraaggmmeenntteedd ttoo ffiitt tthhee
pprroottooccooll rreegguullaattiioonnss..
TThhee ttooppiiccss ddiissccuusssseedd iinn tthhiiss sseeccttiioonn iinncclluuddee::
MMaaxxiimmuumm TTrraannssffeerr UUnniitt ((MMTTUU))
FFiieellddss RReellaatteedd ttoo FFrraaggmmeennttaattiioonn

Figure 8.6 MTU

TTaabbllee 88..55 MMTTUUss ffoorr ssoommee nneettwwoorrkkss

Figure 8.7 Flags field

Figure 8.8 Fragmentation example

Figure 8.9 Detailed fragmentation example

ExamplE 5
A packet has arrived with an M bit value of 0. Is this the first
fragment, the last fragment, or a middle fragment? Do we
know if the packet was fragmented?
Solution
If the M bit is 0, it means that there are no more fragments; the
fragment is the last one. However, we cannot say if the original
packet was fragmented or not. A nonfragmented packet is
considered the last fragment.

ExamplE 6
A packet has arrived with an M bit value of 1. Is this the first
fragment, the last fragment, or a middle fragment? Do we
know if the packet was fragmented?
Solution
If the M bit is 1, it means that there is at least one more
fragment. This fragment can be the first one or a middle one,
but not the last one. We don’t know if it is the first one or a
middle one; we need more information (the value of the
fragmentation offset). See also the next example.

ExamplE 7
A packet has arrived with an M bit value of 1 and a
fragmentation offset value of zero. Is this the first fragment,
the last fragment, or a middle fragment?.
Solution
Because the M bit is 1, it is either the first fragment or a middle
one. Because the offset value is 0, it is the first fragment.

ExamplE 8
A packet has arrived in which the offset value is 100. What is
the number of the first byte? Do we know the number of the
last byte?
Solution
To find the number of the first byte, we multiply the offset
value by 8. This means that the first byte number is 800. We
cannot determine the number of the last byte unless we know
the length of the data.

ExamplE 9
A packet has arrived in which the offset value is 100, the value
of HLEN is 5 and the value of the total length field is 100.
What is the number of the first byte and the last byte?
Solution
The first byte number is 100 × 8 = 800. The total length is 100
bytes and the header length is 20 bytes (5 × 4), which means
that there are 80 bytes in this datagram. If the first byte
number is 800, the last byte number must be 879.

8.3 OPTIONS
The header of the IP datagram is made of two parts: aa ffiixxeedd ppaarrtt aanndd aa
vvaarriiaabbllee ppaarrtt.. TThhee vvaarriiaabbllee ppaarrtt ccoommpprriisseess tthhee ooppttiioonnss tthhaatt ccaann bbee aa
mmaaxxiimmuumm ooff 4400 bbyytteess..
FFoorrmmaatt
OOppttiioonn TTyyppeess

Figure 8.10 Option format

Figure 8.11 Categories of options

Figure 8.12 No operation option

Figure 8.13 End of option option

Figure 8.14 Record route option

Figure 8.15 Record route concept

Figure 8.16 Strict source route option

Figure 8.17 Strict source route concept

Figure 8.18 Loose source route option

Figure 8.19 Timestamp option

Figure 8.20 Use of flag in timestamp

Figure 8.21 Timestamp concept

ExamplE
10
Which of the six options must be copied to each fragment?
Solution
We look at the first (left-most) bit of the code for each option.
a. No operation: Code is 00000001; not copied.
b. End of option: Code is 00000000; not copied.
c. Record route: Code is 00000111; not copied.
d. Strict source route: Code is 10001001; copied.
e. Loose source route: Code is 10000011; copied.
f. Timestamp: Code is 01000100; not copied.

ExamplE
11
Which of the six options are used for datagram control and
which are used for debugging and management?
Solution
We look at the second and third (left-most) bits of the code.
a. No operation: Code is 00000001; datagram control.
b. End of option: Code is 00000000; datagram control.
c. Record route: Code is 00000111; datagram control.
d. Strict source route: Code is 10001001; datagram control.
e. Loose source route: Code is 10000011; datagram control.
f. Time stamp: Code is 01000100; debugging and management
control.

ExamplE
12
One of the utilities available in UNIX to check the travelling of
the IP packets is ping. In the next chapter, we talk about the
ping program in more detail. In this example, we want to show
how to use the program to see if a host is available. We ping a
server at De Anza College named fhda.edu. The result shows
that the IP address of the host is 153.18.8.1.
$ ping fhda.edu
PING fhda.edu (153.18.8.1) 56(84) bytes of data.
64 bytes from tiptoe.fhda.edu (153.18.8.1): ....
The result shows the IP address of the host and the number of
bytes used.

ExamplE
13
We can also use the ping utility with the -R option to
implement the record route option.
$ ping -R fhda.edu
PING fhda.edu (153.18.8.1) 56(124) bytes of data.
64 bytes from tiptoe.fhda.edu (153.18.8.1): icmp_seq=0 ttl=62 time=2.70 ms
RR: voyager.deanza.fhda.edu (153.18.17.11)
Dcore_G0_3-69.fhda.edu (153.18.251.3)
Dbackup_V13.fhda.edu (153.18.191.249) tiptoe.fhda.edu (153.18.8.1)
Dbackup_V62.fhda.edu (153.18.251.34)
Dcore_G0_1-6.fhda.edu (153.18.31.254)
voyager.deanza.fhda.edu (153.18.17.11)
The result shows the interfaces and IP addresses.

ExamplE
14
The traceroute utility can also be used to keep track of the
route of a packet.
$ traceroute fhda.edu
traceroute to fhda.edu (153.18.8.1), 30 hops max, 38 byte packets
1 Dcore_G0_1-6.fhda.edu (153.18.31.254) 0.972 ms 0.902 ms 0.881 ms
2 Dbackup_V69.fhda.edu (153.18.251.4) 2.113 ms 1.996 ms 2.059 ms
3 tiptoe.fhda.edu (153.18.8.1) 1.791 ms 1.741 ms 1.751 ms
The result shows the three routers visited.

ExamplE
15
The traceroute program can be used to implement loose source
routing. The -g option allows us to define the routers to be
visited, from the source to destination. The following shows
how we can send a packet to the fhda.edu server with the
requirement that the packet visit the router 153.18.251.4.
$ traceroute -g 153.18.251.4 fhda.edu.
1 Dcore_G0_1-6.fhda.edu (153.18.31.254) 0.976 ms 0.906 ms 0.889 ms

ExamplE
16
The traceroute program can also be used to implement strict
source routing. The -G option forces the packet to visit the
routers defined in the command line. The following shows how
we can send a packet to the fhda.edu server and force the
packet to visit only the router 153.18.251.4, not any other one.
$ traceroute -G 153.18.251.4 fhda.edu.

8.4 CHECKSUM
The error detection method used by most TCP/IP protocols iiss ccaalllleedd tthhee
cchheecckkssuumm.. TThhee cchheecckkssuumm pprrootteeccttss aaggaaiinnsstt tthhee ccoorrrruuppttiioonn tthhaatt mmaayy ooccccuurr
dduurriinngg tthhee ttrraannssmmiissssiioonn ooff aa ppaacckkeett.. IItt iiss rreedduunnddaanntt iinnffoorrmmaattiioonn aaddddeedd
ttoo tthhee ppaacckkeett..
CChheecckkssuumm CCaallccuullaattiioonn aatt tthhee SSeennddeerr
CChheecckkssuumm CCaallccuullaattiioonn aatt tthhee RReecceeiivveerr
CChheecckkssuumm iinn tthhee IIPP PPaacckkeett

NNoottee::
To create the checksum the sender does the following:
❏ The packet is divided into k sections, each of n bits.
❏ All sections are added together using 1’s complement
arithmetic.
❏ The final result is complemented to make the
checksum.

Figure 8.22 Checksum concept

Figure 8.23 Checksum in one’s complement arithmetic

ExamplE
17
Figure 8.24 shows an example of a checksum calculation for
an IP header without options. The header is divided into 16-bit
sections. All the sections are added and the sum is
complemented. The result is inserted in the checksum field.
See Next Slide

Figure 8.24 Example of checksum calculation in binary

ExamplE
18
Let us do the same example in hexadecimal. Each row has four
hexadecimal digits. We calculate the sum first. Note that if an
addition results in more than one hexadecimal digit, the right-most
digit becomes the current-column digit and the rest are
carried to other columns. From the sum, we make the
checksum by complementing the sum. However, note that we
subtract each digit from 15 in hexadecimal arithmetic (just as
we subtract from 1 in binary arithmetic). This means the
complement of E (14) is 1 and the complement of 4 is B (11).
Figure 8.25 shows the calculation. Note that the result (8BB1)
is exactly the same as in Example 17.
See Next Slide

Figure 8.25 Example of checksum calculation in hexadecimal

NNoottee::
Check Appendix C for a detailed
description of checksum calculation
and the handling of carries.

8.5 IP PACKAGE
We give an example of a simplified IP software ppaacckkaaggee ttoo sshhooww iittss
ccoommppoonneennttss aanndd tthhee rreellaattiioonnsshhiippss bbeettwweeeenn tthhee ccoommppoonneennttss.. TThhiiss IIPP
ppaacckkaaggee iinnvvoollvveess eeiigghhtt mmoodduulleess..
HHeeaaddeerr--AAddddiinngg MMoodduullee
PPrroocceessssiinngg MMoodduullee
QQuueeuueess
RRoouuttiinngg TTaabbllee
FFoorrwwaarrddiinngg MMoodduullee
MMTTUU TTaabbllee
FFrraaggmmeennttaattiioonn MMoodduullee
RReeaasssseemmbbllyy TTaabbllee
RReeaasssseemmbbllyy MMoodduullee

Figure 8.26 IP components

Figure 8.27 MTU table

Figure 8.28 Reassembly table

Chap 08 ip

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Chap 08 ip