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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
Find the class of the below address
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
Find the class of the below address
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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PROBLEMS WITH CLASSFUL ADDRESSING
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.
PROBLEMS WITH CLASSFUL ADDRESSING
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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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APPLY YOUR LEARNINGS
PROBLEM 2
Given the destination address is 202.45.34.56 and mask is
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
SOLVING A SUBNETTING PROBLEM
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SOLVING A SUBNETTING PROBLEM
SOLUTION
SUBNETTING A SUBNET
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SUBNETTING A SUBNET
APPLY YOUR LEARNINGS
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APPLY YOUR LEARNINGS
PROBLEM 2
Given the destination address is 202.45.34.56 and mask is
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
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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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