Transcript The OSI Model

Chapter 8
Internet Protocol (IP)
Mi-Jung Choi
Dept. of Computer Science and Engineering
[email protected]
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Contents
8.1
DATAGRAM
8.2
FRAGMENTATION
8.3
OPTIONS
8.4
CHECKSUM
8.5
IP PACKAGE
8.6
KEY TERMS
8.7
SUMMARY
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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
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Position of IP in TCP/IP protocol suite
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Internet Protocol (IP)
 Packet delivery mechanism in TCP/IP
 Unreliable connectionless datagram protocol
Each datagram handled independently
Each datagram can follow a different route to destination.
datagrams sent by same source and destination could arrive out of order
 Best effort delivery
 No error checking or tracking
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8.1 IP Datagram
A packet in the IP layer is called a datagram, a variable-length packet
consisting of two parts: header and data. The header is 20 to 60 bytes in
length and contains information essential to routing and delivery
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8.1 DATAGRAM
 Version(VER)
4BIT field
Currently, the version is 4
 Header Length(HLEN)
4BIT field
The total length of the datagram header in 4byte words
No options : 5(5*4=20), with maximum option size : 15(15*4=60)
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8.1 DATAGRAM
 Differentiated Services (formerly
Service Type): 8bits
1. Service type
The first 3bit : precedence bit
Precedence

1(000) to 7(111)

The priority of the datagram
(congestion)

Not used in version 4
000
001
010
011
100
101
110
111
routine
Priority
Immediate
Flash
Flash override
Critical
Internetwork control
Network control
TOS Bits
0000
0001
0010
0100
1000
Description
Normal (default)
Minimize cost
Maximize reliability
Maximize throughput
Minimize delay
 The precedence subfield was designed,
but never used in version 4.
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Table 8.2 Default types of service
Protocol
ICMP
BOOTP
NNTP
IGP
SNMP
TELNET
FTP (data)
FTP (control)
TFTP
SMTP (command)
SMTP (data)
DNS (UDP query)
DNS (TCP query)
DNS (zone)
TOS Bits
0000
0000
0001
0010
0010
1000
0100
1000
1000
1000
0100
1000
0000
0100
Description
Normal
Normal
Minimize cost
Maximize reliability
Maximize reliability
Minimize delay
Maximize throughput
Minimize delay
Minimize delay
Minimize delay
Maximize throughput
Minimize delay
Normal
Maximize throughput
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8.1 DATAGRAM
2. Differentiated Services
First 6bit : code point
Last 2bit : not used
The codepoint subfield can be used in two different ways
1. When the 3bit right-most bits are 0s, the 3bit left-most bits are
interpreted the same as the precedence bits in the Service Type
Interpretation
2. When the 3bit right-most bits are not all 0s, the 6bits define 64 services
based on the priority assignment
Category
Codepoint
Assignment Authority
1
XXXXX0
Internet
2
XXXX11
Local
3
XXXX01
Temporary or experimental
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8.1 DATAGRAM (cont.)
 Total length
The total length field defines the total length of the datagram including the
header.
Total length of the IP datagram in byte(header + data)

Length of data = total length – HLEN

Maximum size : 65535 bytes (216 – 1)

Encapsulation is needed to transfer datagram in some case
(some padding added)
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8.1 DATAGRAM (cont.)
 Identification
Used in fragmentation
 Flags.
Used in fragmentation
fragmentation
 Fragmentation offset
Used in fragmentation
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8.1 DATAGRAM (cont.)
 Time to live
Datagram should have a limited lifetime



Decremented by each visited router
Discarded when zero
All the machine must have synchronized clocks and how long it takes for a
datagram to go from one machine to another
Cases


Corrupted router
Intentionally limit the journey of the packet
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8.1 DATAGRAM(cont.)
 Protocol
Define the higher level protocol that uses the services of the IP layer.

Encapsulate data from several higher level protocol(TCP,UDP,ICMP,IGMP)

Specify the final destination protocol to which datagram should be delivered
Value
1
2
6
8
17
41
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Protocol
ICMP
IGMP
TCP
EGP
UDP
IPv6
OSPF
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8.1 DATAGRAM (cont.)
 Checksum
Later present
 Source IP address
Define the IP address of the source
 Destination IP address
Define the IP address of the destination
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8.1 DATAGRAM (cont.)
 Example 1
: An IP packet has arrived with the first 8 bits as shown:

01000010
The receiver discards the packet. Why?
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?
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 main header, the next 12 bytes are the options.
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8.1 DATAGRAM (cont.)
 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?
 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?
 To find the time-to-live field, we should 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.
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8.2 FRAGMENTATION
 The format and size of a frame depend on the protocol used by the physical
network. A datagram may have to be fragmented to fit the protocol regulations.
 The topics discussed in this section include:
Maximum Transfer Unit (MTU)
Fields Related to Fragmentation
Ethernet formatted
frame
Token ring
formatted frame
router
Ethernet
network
Token ring
network
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8.2 FRAGMENTATION
 Maximum Transfer Unit (MTU)
Maximum size of the data field

The total size of encapsulated datagram in a frame must be less than
maximum size

Differ from one physical network protocol to another
IP datagram
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8.2 FRAGMENTATION
Table 8.5 MTUs for some networks
- MTUs for different network 20
8.2 FRAGMENTATION
 Fragmentation
Divide the datagram to make it possible to pass some networks

Each fragment has its own header
–
With most of the fields repeated, but some changed

A datagram can be fragmented several times

Only final destination can reassembly of the datagram

Required parts of the header must be copied by all fragments

Three fields(flags, fragmentation offsets, total length) changed
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8.2 FRAGMENTATION
 Fields related to fragmentation
Identification(16bits)

Identify a datagram originating from the source host

Combination of the identification and source IP address must be unique

IP protocol uses positive number counter(guarantee uniqueness)
–

Kept in the Main Memory
All fragment of a datagram have the same identification number
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8.2 FRAGMENTATION
Flags (3bits)

First bit : reserved

Second bit : do not fragment

–
1 : must not fragment the datagram
–
If cannot pass the datagram through any physical network,
discards the datagram and return ICMP error message to source
host
–
0 : the datagram can be fragmented if necessary
Third bit : more fragment
–
1 : more fragments after this one
–
0 : last or only fragment
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8.2 FRAGMENTATION
 Fragmentation offset (13bits)
Relative position of fragment with respect to the whole datagram
Offset of the data in the original datagram measured in units of eight bytes
(The size of first fragment is divisible by eight)
Reassemble step in final destination host:

First fragment offset value is 0.

Divide the length of 1st fragment by 8. This value is the offset value of
2nd fragment.

Divide the length of 2st fragment by 8. This value is the offset value of
3rd fragment.

Continue the process. The last fragment has more bit value of 0.
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8.2 FRAGMENTATION (cont.)
Fragmentation example
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8.2 FRAGMENTATION (cont.)
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8.2 FRAGMENTATION (cont.)
 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?
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?
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).
However, we can definitely say the original packet has been fragmented because the M
bit value is 1.
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8.2 FRAGMENTATION (cont.)
 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?
 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?
 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?
 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 879.
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8.3 Options
 The header of the IP datagram is made of two parts: a fixed part and a
variable part. The variable part comprises the options that can be a
maximum of 40 bytes.
 The topics discussed in this section include:
Format
Option Types
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8.3 OPTIONS
 Security, Source routing, Record route, Timestamp
 Used for network testing and debugging
 Not required for every datagram
 TLV (Type, Length, Value) Format
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8.3 OPTIONS
 Code (8bits) : Type
Copy(1bit)

Control the presence of the option in fragmentation
– 0 : only copied to the first fragment
– 1 : coped to all fragments
Class(2bits)

General purpose of the option
– 00 : datagram control
– 10 : debugging and management(01& 11 not defined)
Number(5bits)

Type of the option(only six types are in use)
 Length
Total length of the option(including code and length fields)
Not present in all of the option types
 Data : Value
The data that specific options require
Not present in all of the option types
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8.3 OPTIONS
 Option types(6 types)
Two types


1 byte
Do not require the length or the the data fields
Four types


Multiple bytes
Require the length and the data fields
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8.3 OPTIONS (cont.)
 No Operation (00001)
1 byte option
Used as a filler between options
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8.3 OPTIONS (cont.)
 End of Option (00000)
1 byte option
Used for padding at the end of the option field
Only one end of option can be used
Search payload(data) after option
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8.3 OPTIONS (cont.)
 경로 기록 - Record Route (00111)
Used to record the internet routers that handle the datagram
Nine router IP can be contained(4byte×9 = 36 bytes ≤ 40bytes)
Uses a pointer field containing the byte number of the first empty entry

Initialized by 4 and increased by 4 until over the length value
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8.3 OPTIONS (cont.)
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8.3 OPTIONS (cont.)
 Strict Source Route (01001)
Used by the source to predetermine a rout for the datagram
All of the routers defined in the option must be visited by datagram
If the datagram visits a router that is not on the list, the datagram is
discarded and an error message is issued
If the datagram arrives at the destination and some of the entries were not
visited, it is also discarded and an error message issued
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8.3 OPTIONS (cont.)
Strict source route concept
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8.3 OPTIONS (cont.)
 Loose Source Route (00011)
Similar to the strict source route
Each router in the list must be visited, but the datagram can visit other
routers
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8.3 OPTIONS(cont.)
 Timestamp (00101)
Used to record the time of datagram processing by a router

Milliseconds from midnight, Universal Time
Overflow field

Records the number of routers that could not add their timestamp because of
no more fields available
Flags field

Specify the visited router responsibilities
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8.3 OPTIONS (cont.)
Use of flag in timestamp
-0 : add only the timestamp in the provided field
-1 : add each router’s outgoing IP address and the timestamp
-3 : each router must check the given IP address with its own incoming IP address
If matched, the router overwrites the IP address with its outgoing IP address and
adds the timestamp
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8.3 OPTIONS (cont.)
Timestamp concept
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8.3 OPTIONS (cont.)
 Example 10: Which of the six options must be copied to each fragment?
We look at the first (left-most) bit of the code for each option.
No operation: Code is 00000001; no copy.
End of option: Code is 00000000; no copy.
Record route: Code is 00000111; no copy.
Strict source route: Code is 10001001; copied.
Loose source route: Code is 10000011; copied.
Timestamp: Code is 01000100; no copy.
 Example 11 : Which of the six options are used for datagram control and which are
used for debugging and management?
We look at the second and third (left-most) bits of the code.
No operation: Code is 00000001; control.
End of option: Code is 00000000; control.
Record route: Code is 00000111; control.
Strict source route: Code is 10001001; control.
Loose source route: Code is 10000011; control.
Timestamp: Code is 01000100; debugging
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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.
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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.
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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.
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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.
traceroute to fhda.edu (153.18.8.1), 30 hops max, 46 byte packets
1 Dcore_G0_1-6.fhda.edu (153.18.31.254) 0.976 ms 0.906 ms 0.889 ms
2 Dbackup_V69.fhda.edu (153.18.251.4) 2.168 ms 2.148 ms 2.037 ms
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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.
traceroute to fhda.edu (153.18.8.1), 30 hops max, 46 byte packets
1 Dbackup_V69.fhda.edu (153.18.251.4) 2.168 ms 2.148 ms 2.037 ms
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8.4 CheckSum
 The error detection method used by most TCP/IP protocols is called the
checksum. The checksum protects against the corruption that may occur
during the transmission of a packet. It is redundant information added to
the packet.
 The topics discussed in this section include:
Checksum Calculation at the Sender
Checksum Calculation at the Receiver
Checksum in the IP Packet
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8.4 CheckSum in IP
 To create the checksum, the sender does the following:
1. The packet is divided into k sections,
each of n bits.
2. All sections are added together using
one’s complement arithmetic.
3. The final result is complemented
to make the checksum.
 Checksum in one’s complement arithmetic
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Checksum concept
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An example of IP header checksum calculation
in binary and hexadecimal
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Note:
Check Appendix C for a detailed
description of checksum calculation and
the handling of carries.
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8.5 IP PACKAGE
We give an example of a simplified IP software package to show its components
and the relationships between the components. This IP package involves eight
modules.
The topics discussed in this section include:
 Header-adding module
 Processing module
 Routing module
 Fragmentation module
 Reassembly module
 Routing table
 MTU table
 Reassembly table
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8.5 IP PACKAGE
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8.5 IP PACKAGE
Header-adding module
• Receive : data, destination address
1. Encapsulate the data in an IP datagram
2.Calculate the checksum and insert it in the checksum field
3. Send the data to the corresponding input queue
4. Return
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8.5 IP PACKAGE
Processing module
1. Remove one datagram from one of the input queues
2. If(destination address is 127.X.Y.Z or matches one of the local addresses)
1. Send the datagram to the reassembly module
2. Return
3. If(machine is a router)
1. Decrement TTL
4. If(TTL less than or equal to zero)
1. Discard the datagram
2. Send an ICMP error message
3. Return
5. Send the datagram to the routing module
6. Return
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8.5 IP PACKAGE
Fragmentation module
• Receive : an IP datagram from the routing module
1. Extract the size of the datagram
2. If(size > MTU of the corresponding network)
1. If(D(do not fragment) bit is set)
1. Discard the datagram
2. Send an ICMP error message
3. Return
2. Else
1. Calculate the maximum size
2. Divide the datagram into fragments
3. Add header to each fragment
4. Add required options to each fragment
5. Send the datagrams
6. Return
MTU table
3. Else
1. Send the datagram
4. Return
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8.5 IP PACKAGE
Reassembly table
• Used by the reassembly module
• Five field
- State(FREE or IN-USE)
- Source IP address(source IP of the datagram)
- Datagram ID(uniquely define a datagram)
- Time-out(predetermined amount of time, in which all fragments must
arrive)
- Fragments(a pointer to a linked list of fragments)
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8.5 IP PACKAGE (cont.)
Reassembly module
Receive : an IP datagram from the processing module
1. If(offset value is zero and the M bit is 0)
1. Send the datagram to the appropriate queue
2. Return.
2. Search the reassembly table for the corresponding entry
3. If(not found)
1. Create a new entry
4. Insert the fragment at the appropriate place in the link list
1. If(all fragments have arrived)
1. Reassemble the fragments
2. Deliver the datagram to the corresponding upper layer protocol
3.Return
2. Else
1. Check the time-out
2. If(time-out expired)
1. Discard all fragments
2. Send an ICMP error message
5. Return
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