ip training

7/29/2019 IP training

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2012 Sasken Communication Technologies

INTERNET PROTOCOL

(RFC 791)-- Preetam Narayan


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An IP address is a 4 byte (32-bit) address.

*An IP addresses are unique

The address space of IPv4 is 232

or

4,294,967,296.


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IP

Address

Notation

BinaryNotation

Hexadecimal

Notation

Dotted decimal

Notation

01000000 00001011 00000011 00011111

0x800B021F

IP ADDRESS NOTATION


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PROBLEM 1

Find the Error if any in the following IP Address: 111.56.045.78

LEARNING

There are no leading zeros in a dotted decimal Notation


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PROBLEM 1

Find the Error if any in the following IP Address: 111.301.045.78

LEARNING

Max permissible value is 255; 301 is out of range


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CLASSFUL ADDRESSING


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In classful addressing, the address space is

divided into five classes:A,B, C,D, andE.

ADDRESS SPACE


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Decimal Notation Binary Notation

IDENTIFYING CLASS OF AN ADDRESS


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PROBLEM 1

Find the class of the below address

00000001 00001011 00001011 11101111

LEARNING

1st Bit zero implies class A


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PROBLEM 1


11000001 10000011 00011011 11111111

LEARNING

1st 3 bits binary representation 110; signifies class C


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Generalization


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PROBLEM 1


233.14.56.22

LEARNING

Check for the range of the 1st byte; 224


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No

CLASS A

Yes

No

CLASS B

Yes

No

CLASS C

Yes

No

CLASS DCLASS E

Yes

Generalization


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NETWORK ID & HOST ID


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PROBLEMS WITH CLASSFUL ADDRESSING

CLASS A -- EXAMPLE

Millions of class A addresses are wasted.


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CLASS B -- EXAMPLE

Many of Class B Addresses are wasted


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PROBLEMS WITH CLASSFUL ADDRESSINGCLASS C -- EXAMPLE

Class C addresses are fit for smaller organization only


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Class D addresses are reserved for multicasting;

Class E addresses are reserved for special

purposes; most of the block is wasted.



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Unicasting communication is one-to-one;

Multicasting communication is one-to-many;

Broadcasting communication is one to all;

COMMUNICATION MODE


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A number of Blocks in each class which are

assigned for private use only. These addresses

cannot be used to connect to the Internet

PRIVATE ADDRESSING

IP CLASS PRIVATE ADDRESS

CLASS A 10.0.0.0 10.255.255.255

CLASS B 169.254.0.0 169.254.255.255

CLASS B 172.16.0.0 172.31.255.255CLASS C 192.168.0.0 192.168.255.255


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NETWORK ADDRESS

Network Address is the first Address (of the

block) that is being assigned to an Organization.

Network Address identifies a network in theInternet.

Given a Network Address, we can find its

Network class, range of permissible addresses for

that network


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NETWORK ADDRESS


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PROBLEM 4

Find the network address 17.0.0.0, find its class and its address

range

SOLUTION

CLASS A; Address range 17.0.0.0 17.255.255.255


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NETWORK MASK

Network Mask is a 32-bit number which gives thenetwork Address when bitwise ANDed with an IP

Address


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NETWORK MASK

Default Network Mask of a network is formed by

setting the netid bits to 1 and hostid bits to 0

Default Network Mask of a class shouldnt be

applied to an address belonging to another class


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SPECIAL ADDRESSESDIRECTED BROADCAST ADDRESS


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SPECIAL ADDRESSESLIMITED BROADCAST ADDRESS


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SPECIAL ADDRESSESTHIS HOST Me


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SPECIAL ADDRESSESSPECIFIC HOST ON THE NETWORK


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SPECIAL ADDRESSESLOOPBACK ADDRESS


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APPLY YOUR LEARNINGS


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Default IP addresses are designed with

two levels of hierarchy.

All Hosts fall under the same level.

NETWORK DEPICTING 2 LEVEL HIERARCHY


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NETWORK DEPICTING 3 LEVEL HIERARCHY


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Subnetting: Process of dividing a Network

into various kutti-kutti networks calledsubnets

Subnetting creates 3 level hierarchy:

Network id, Sub-network id and Host id


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Subnet-mask: formed by setting network id andsub-network id bits to 1 and host id bits to 0


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Subnet-Address: formed by ANDing IP

address with the subnet mask


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PROBLEM 1

Given the destination address is 19.30.80.5 and mask is

255.255.192.0

Determine the sub-network address

SUBNETTING STEPS

SUBNET ADDRESS IDENTIFICATION

STEP1: if the network mask byte is 255 then copy the byte to the

address

STEP2: if the network mask byte is 0 then replace byte in the

address with 0

STEP3: if the byte in the mask is neither 0 nor 255 then perform

the AND by representing it in binary notation


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STEPS TO SUBNETTING


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PROBLEM 2


255.255.240.0Determine the sub-network address

SOLUTION

200.45.32.0


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Subnetting is done by borrowing bits from

the Host id section of an IP address


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The number of subnets = 2 (number of bits used for subnet) - 2

The number of Hosts per subnet =2 (number of bits used for Host) - 2

Broadcast address of a particular subnet =network address of following subnet - 1


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PROBLEM 2

A company is granted the site address 201.70.64.0 (Class C). The

company needs 6 subnets. Design the subnets

APPROACH

STEP1: Identify the number of subnets needed

STEP2: Identify the number of host bits to be borrowed

STEP3: ignore the subnet bits with all 0s and all 1s

STEP4: Find the subnet address space.Subnet address space = 2 (number of host id bits)

SOLVING A SUBNETTING PROBLEM



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SOLUTION

SUBNETTING A SUBNET


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SUBNETTING A SUBNET



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PROBLEM 2


255.255.240.0Determine the number of subnets and there address range


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Supernetting: combining several class C

blocks to create a large address space

Supernetting involves borrowing network bits

In supernetting we need the 1staddress of the

supernet and supernet mask to define therange of addresses

SUPERNETTING


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SUPERNETTING


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PROBLEM 2

A company needs 600 addresses. Which of the following set of

class C addresses can be used to for a supernet for this company ?

198.47.32.0 198.47.33.0 198.47.34.0

198.47.32.0 198.47.42.0 198.47.52.0 198.47.62.0

198.47.31.0 198.47.32.0 198.47.33.0 198.47.52.0

198.47.32.0 198.47.33.0 198.47.34.0 198.47.35.0


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RULES FOR SUPERNETTING

RULE1: The number of blocks must be power of 2 (1,2,4,8,26..)RULE2: The blocks must be contiguous in the address space.

RULE3: The 3rd byte of the 1st address in the superblock must be

evenly divisible by the number of blocks. In other words if the

number of blocks is N then the third byte must be divisible by N

SOLUTION

1: No, there are only three blocks.

2: No, the blocks are not contiguous.

3: No, 31 in the first block is not divisible by 4.

4: Yes, all three requirements are fulfilled.

COMPARISON BETWEEN SUBNET MASK


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COMPARISON BETWEEN SUBNET MASK,

DEFAULT MASK and SUPERNET MASK


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PROBLEM 6

We need to make a super-network out of 16 class C blocks. What is

the supernet mask?

SOLUTION

255.255.240.0


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PROBLEM 5

A supernet has a first address of205.16.32.0 and a supernet mask

of255.255.248.0. How many blocks are in this supernet and what

is the range of addresses?

SOLUTION

205.16.32.0 205.16.39.0


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PROBLEM 5

A supernet has a first address of 205.16.32.0 and a supernet mask

of 255.255.248.0. A router receives three packets with the

following destination addresses:

205.16.37.44

205.16.42.56205.17.33.76

Which packet belongs to the supernet?

SOLUTION

205.16.37.44


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CIDR NOTATION


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CIDR notation is also called slash notation

Classless Inter Domain Notation


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CHARACTERASTICS OF IP

SWITCHING METHODS


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SWITCHING METHODS

PACKET SWITCHING TYPES


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PACKET SWITCHING TYPES

CONNECTION TYPES


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Connection Types

Connectionless

service

Connection-oriented

service

CONNECTION TYPES

CONNECTIONLESS vs CONNECTION-ORIENTED


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http://courses.iddl.vt.edu/CS1604/media/connectOriented.swf

CONNECTIONLESS vs CONNECTION ORIENTED

ADDRESSING SCOPE
http://courses.iddl.vt.edu/CS1604/media/connectOriented.swfhttp://courses.iddl.vt.edu/CS1604/media/connectOriented.swf


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ADDRESSING SCOPE


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IP works in Packet switching technology

*Source and Destination IP addressdoesnt change during the journey of a

packet in an inter-network

IP is a Hop-to-Hop protocol


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IP packets are also referred as IP

Datagrams;

IP is a Best Effort delivery protocol;

POSITION OF IP IN NETWORK STACK


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IP HEADER

IP HEADER


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IP HEADER


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VER is the IP version, to which the below

packet belong.

Set to 4(0100) for IPV4


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Is requested

IP Version

Supported ?

IP PACKET

IP STACK

NO

YES

SILENTLYDISCARD THE PACKET

ETHERNET


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IP header size: Each bit is expressed in

terms of multiple of 4 octets

IP header size is variable 2060 bytes


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PROBLEM

An IP packet has arrived with the first byte as shown:

01000010

The receiver discards the packet. Why?


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PROBLEM

In an IP packet, the value of HLEN is 1000 in binary. How many

bytes of options are being carried by this packet?


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Total length of the datagram i.e. Headersize + Data size


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PROBLEM

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?


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IP ensures that the minimum data size ofunderlying technology is respected

ETHERNET HAS A MINIMUM DATA LENGTH OF 46 BYTES


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UPPER LAYER

DATA

IP HEADER

>= LAYER 3

NETWORK LAYER

DATA LINK LAYER

Total length < 46 bytes


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Protocol field: Denotes the upper layerprotocol being carried by the IP

IP multiplexes higher layer protocol


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Source IP Address and Dest IP Address

doesnt change in a packet journey in a inter-

network except in the case of source routing


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Checksum calculation is done as following:

1. Divide the packet into 16 bits chunks.

2. While calculation checksum, checksum field bits are

set to 0

3.Perform 1s complement sum ofeach chunk.

4.1s complement the result obtained in STEP2.

CHECKSUM CALCULATION


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Checksum is performed on the IP Header

only and not on IP data

Checksum is recalculated at each HOP

On detection of the checksum error, packet is

dropped thereby the HOP

TYPE OF SERVICE


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Service level parameter, important for

Routers. Prioritizes the packets based on this

parameter

TIME TO LIVE


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Maximum time a packet lives in an inter-

network, before it would gets destroyed

Set by the sender; decremented by 1 at each

HOP.

TIME TO LIVE


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MAXIMUM TRANSMITABLEUNIT


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MTU determines the maximum amount of

data that can be transmitted on an underlying

technology


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In case the Packet size is greater than the

MTU then packet need to be fragmented

Fragmentation is a process of dividing a

large packet (size > MTU) to packets (size


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Once the packet is reaches the destination or

a hop then it need to be reassembled

Reassembly: is a process is getting back the

original packet from the fragmented IP

packets


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IP Fragmentation

Intranet

Fragmentation

Internet

Fragmentation

Based on where the fragmentation-reassembly is done at Hop or final

destination, fragmentation are classified

under 2 types

INTRANET FRAGMENTATION


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INTRANET FRAGMENTATION


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fragmentation, transmission and reassembly

across a local network which is invisible to

the internet protocol module is called intranet

fragmentation

Efficient utilization of bandwidth

More overhead, processing delay andbuffering required

INTERNET FRAGMENTATION


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INTERNET FRAGMENTATION


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fragmentation, transmission and reassembly

across an inter-network all the way to the

final destination.

No buffering needed in intermediate nodes

Inefficient use of the network Bandwidth


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PROBLEM

Consider the journey of a packet of length 2048 octets and header

40 octets. It passes through the networks with maximum packet

sizes as shown:

Network Maximum Packet Length

A 2048 octets

B 512 octets

C 1024 octets

D 256 octets

E 1024 octets

Discuss the performance w.r.t internet and intranet fragmentation.*Performance can be measured in terms of no. of fragmentation performed and re-assembly and number of data units passed in eachnetwork

FRAGMENTATION FLAGS


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When SET (1) by the sender, then the packet wouldnt be

fragmented. If the MTU size is not respected then it would bedropped with an ICMP message back to sender

SET (0) implies there is permission to fragment

SET 0 for un-fragmented or last fragment packet

SET 1 for rest all fragments

FRAGMENTATION OFFSET


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Indicates the relative position of the fragment

w.r.t original IP datagram

Fragmentation offset are measured in units

of 8 octets

Fragment offset 0 for the 1stfragment


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EXAMPLE


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EXAMPLE


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PROBLEM

In an IP packet arrives with the fragment offset value as 100, the

value of the HLEN is 5 and value of the total length field as 100.

What is the number of the 1st and last byte in that fragment ?

SOLUTION

1st byte number is 100 * 8 = 800

Last byte = 879

OPTIONS


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Options are optional parameters that might

be sent by the sender; but not optional forimplementers. So all IP module need to

support all IP options

OPTIONS


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In case there is space enough for partial data

to be filled then the respective packet wouldbe dropped and an ICMP error message

would be generated

OPTIONS


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OPTIONS FORMAT


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END OF OPTIONS -- OPTION


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Optionally present at the end of all options if

option header doesnt coincide the 32 bit

boundary

NO OPERATION OPTION


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Optionally present in between the options; to

align the beginning of next option to 32 bit

boundary

RECORD ROUTE OPTIONS


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Route followed by the IP packet is recorded

in the IP header when this options is set;

Each router in the transit, adds its IP address

to the option

RECORD ROUTE OPTIONS FORMAT


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RECORD ROUTE OPTIONS


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On fragmentationonly the 1stfragment

would contain this information i.e. copied to

1stfragment only

It is the responsibility of the sender to define

the size of option as it is not changed in the

transit

WORKING PRINCIPLERECORD ROUTE


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STRICT SOURCERECORD ROUTE OPTIONS


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Means for the source to specify the route to

be followed for the IP datagram; This also

record the route in the transitStrict sourceeither follow the path or

discard the IP packet

Copied on each fragments

STRICT SOURCERECORD ROUTE OPTIONS

FORMAT


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FORMAT

WORKING PRINCIPLE

STRICT SOURCE RECORD ROUTE OPTIONS


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STRICT SOURCE-RECORD ROUTE OPTIONS

LOOSE SOURCERECORD ROUTE OPTIONS


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Identical to strict source route, except that the

gateway or host IP is allowed to use any

number of intermediate gateways to reach thenext address in the route

Copied on each fragments

LOOSE SOURCERECORD ROUTE OPTIONS

FORMAT


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FORMAT

TIMESTAMP OPTIONS


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Timestamp: Time at which the packet is

processed by the HOP

Measured in milli-secs

TIMESTAMP OPTIONS


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TIMESTAMP OPTIONS


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Over-Flow Flag: number of IP modules

which couldnt register there timestamp due

to lack of space

In case the overflow flag itself overflows then

an ICMP error message might be sent to thesender of the message

TIMESTAMP OPTIONS -- FLAGS


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WORKING PRINCIPLE -- TIMESTAMP


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