network layer sc

8/3/2019 Network Layer Sc

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Network Layer:(Internet Protocol v.4,Addressing,subnetting)

Lecture No. 5.2

By: Engr. Mark Ryan S. ToECE and Cpe Department


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Responsible for host-to-host delivery

Main Network Protocol is the InternetProtocol (IP) Ex: IPv4, IPv6

How does a host get an IP address? DHCP: Dynamic Host Configuration Protocol

DHCP dynamically gets address from a server


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Internet as a Connectionless Network

The reason for this decision is that the

Internet is made of so manyheterogeneous networks that it is almostimpossible to create a connection fromthe source to the destination without

knowing the nature of the networks inadvance.


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The IPv4 addresses are 32 bits in length.

Newer versions like the IPv6 use 128 bitsaddresses.


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IPv4 Format


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A Computer is:

Version (VER)

A 4-bit field defines the version of the IPprotocol version. Currently the version is 4.

Header Length (HLEN)A 4-bit field defines the total length of the

datagram header in 4-byte words.

Services

Previously called service type but waslater changed by the Internet EngineeringTask Force (IETF) and now calleddifferentiated services.


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A Computer is:

Service Type

The first 3 bits are called precedence bits. The next4 bits are called type of service (TOS) bits, and thelast bit is not used.

Used by the host to specify a preference for howthe datagram would be handled as it makes itsway through the internet.

Precedence

A 3-bit field from 0 (000) to 7 (111) Used for priority

Datagrams with lowest precedence are discardedfirst


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A Computer is:

TOS Bits Similar to routing metric as

discussed in routingalgorithms.

One and only one of thebits can be set at a time.

Each router has a TOSvalue for each route in itsrouting table. TOS is used

to choose a destinationfor the packet. The routerchooses the route with thebest metric that matchesthe TOS.


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A Computer is: Differentiated Services The first 6 bits make up the codepoint subfield and

the last 2 bits are not used.

Used in two different ways

1.) When the 3 rightmost bits of the codepointsubfield are 0s, the 3 leftmost bits are interpretedthe same as the precedence bits

2.) When the 3 rightmost bits are not all 0s, the 6 bitsdefine 64 services based on the priority assignmentby the Internet or local authorities


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A Computer is:

Total Length 16-bit field that defines the total length of the datagram in bytes.

NOTE: HLEN together with the total length can be used to find thelength of the data. i.e. (Total lengthHLEN*4)

Identification, Flags, Fragmentation offset

A datagram can travel through different networks. Eachrouter decapsulates the datagram at the receiver,processes it, and then encapsulates it. The format and sizeof the received and transmitted frame depend on theprotocol used by the physical network through which theframe has just traveled and the frame is going to travel.

Identification is used helps in reassembly. Flags1st bit, reserved, 2nd bit, when set to 1, do not

fragment (0 fragment necessary), 3rd bit If its value is 1, itmeans the datagram is not the last fragment (0 = lastfragment)

Fragmentation offset shows the relative position of this

fragment with respect to the whole datagram


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A Computer is:

Time to Live

A datagram has a limited lifetime in its travelthrough the internet.

Formerly used a timestamp. Today, this field is usedmostly to control the maximum number of hops(routers) visited by the datagram.

The datagram is discarded when the valuebecomes zero.

Protocol An 8-bit field defines the higher-level protocol that

uses the services of the IPv4 layer.

Example: TCP (value =6), UDP (value =17),


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A Computer is:

IPv4 uses 32-bit addresses, which means thatthe address space is 2^32 or 4,294,967,296(more than 4 billion).

Addressing Notations

Binary notation

01110101 10010101 00011101 00000010

Dotted decimal 117.149.29.2


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A Computer is:

Example:

Change the following IPv4 addresses from binary

notation to dotted-decimal notation. a. 10000001 00001011 00001011 11101111

b. 11000001 10000011 00011011 11111111


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A Computer is:

Example:

Find the error, if any, in the following IPv4 addresses

a. 111.56.045.78 b. 221.34.7.8.20

c. 75.45.301.14

d. 11100010.23.14.67


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A Computer is:

Hierarchy

IP addresses, like telephone network addresseshave levels of hierarchy (area code, exchangecode, subscriber number)Others can have a

similar telephone number that you have, thedifference lies on the area code.

Two-Level Hierarchy (no subnetting)

Without subnetting, the address can be dividedinto two parts, the leftmost part defining thenetwork and the rightmost part the host (computeror router) to the network.


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A Computer is:

Classful Addressing

Uses the concept of classes

the address space is divided into five

classes: A, B, C, D, and E.


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A Computer is:

Example:

Find the class of each address

a. 00000001 00001011 00001011 11101111 b. 11000001 10000011 00011011 11111111

c. 14.23.120.8

d. 252.5.15.1l1


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A Computer is:

IP Address

Class Format

High-Order

Bit(s)

Address

Range

No. Bits

Network/

Host Max. Hosts

A N.H.H.H 0 1.0.0.0 to126.0.0.0

7/24 16777214

(224 - 2)

B N.N.H.H 1, 0 128.1.0.0 to191.254.0.0

14/16 65534 (216 -

2)

C N.N.N.H 1, 1, 0 192.0.1.0 to223.255.254.0

21/8 254 (28 - 2)

D N/A 1, 1, 1, 0 224.0.0.0 to239.255.255.255

N/A (not forcommercial

use)

N/A

E N/A 1, 1, 1, 1 240.0.0.0 to254.255.255.25

5

N/A N/A


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Exactly where the network part ends and thehost part begins is calculated by routers, using amask, a a 32-bit number in which the n leftmostbits are 1s and the 32- n rightmost bits are 0s.

In IPv4 addressing, a block of addresses can bedefined as w.x.y.z /n in which w.x.y.z defines oneof the addresses and the /n defines the mask.

Default masks for the following Classfuladdressing are:

A /8 B /16 C /24


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Problem with Classful addressing

Class A and B are rarely used because only afew requires as much as 65534 hosts. If a userneeds a little over 254 hosts, instead of beinggiven a Class C address, he would be given thenext class, class B which has 65,534 addresses.


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If an organization was granted a large block in classA or B, it cou1d divide the addresses into several

contiguous groups and assign each group to smaller

networks (called subnets)

Subnets are under local administration. As such, theoutside world sees an organization as a single

network and has no detailed knowledge of the

organization's internal structure.

Subnetting supports a three-level hierarchy.


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Subnet Mask A 32-bit subnet mask is used as a deciphering key

to determine how an IP address is to be divided

into extended network prefix (network+subnet)

and host part.

Internet routers use only the network number of the

destination address to route traffic to a subnetted

environment. It is the job of the local networkrouters to divide the communication out into

individual subnets and to the individual hosts on

these subnets.


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Example:

Suppose you are given a network number usingClassful addressing of 200.133.175.0

1.) Determine the class:

2.) Determine the subnet mask:


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Continuation

3.) Determine the addresses if you are to utilize thisnetwork to have 16 subnets with 14 hosts each.

In reality, 2 must be subtracted from all the numbers of hosts aboveto get the actual number of IP addresses available to use for hosts,because two addresses, namely the address which has all-0s in thehost bits (this network) and the address which has all 1s in the hostbits (broadcast), can not be assigned to hosts.


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Broadcast Address

The address where all the bits in the host

portion are set to 1. The broadcastaddress is used when you want to

communicate data to all the hosts on a

network.


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Classless AddressingCIDR: Classless Inter-Domain Routing

In this scheme, there are no classes, but theaddresses are still granted in blocks.

Restrictions:

1. The addresses in a block must be contiguous,one after another.

2. The number of addresses in a block must be apower of 2 (1, 2, 4, 8, ... ).

3. The first address must be evenly divisible bythe number of addresses.


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The addresses are contiguous. (32,33,34, 47)

The number of addresses is a power of 2 (16 = 2^4)

The first address is divisible by 16. The first address, when converted to a decimal number, is

3,440,387,360, which when divided by 16 results in215,024,210. 205.16.37.32


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Derivation for restriction 3 in previous problem

IPv4 addresses use base-256 to represent an address indotted decimal notation. The designers of the IPv4 addressdecided to use decimal numbers 0 to 255 as symbols and todistinguish between the symbols, a dot is used. The dot is used

to separate the symbols

205.16.37.32

205 * (256^3)= 3439329280

16* (256^2)= 104857637* (256^1)= 9472

32* (256^0)= 32

3,440,387,360


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The first address in the block can be found bysetting the 32- n rightmost bits in the binarynotation of the address to 0s.

Example:

A block of addresses is granted to a small organization. We

know that one of the addresses is 205.16.37.39/28. What isthe first address in the block?


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The last address in the block can be found by setting the 32 - nrightmost bits in the binary notation of the address to 1s.

Example:

A block of addresses is granted to a small organization. Weknow that one of the addresses is 205.16.37.39/28. What is thelast address in the block?

The number of addresses in the block is the difference

between the last and first address. It can easily be found usingthe formula2^(32-n)

Example: Find the # of addresses in the previous problem.


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The IP address 13.2.101.18/21 belongs to which network?

a.) 131.2.101.0

b.) 13.2.0.0

c.) 13.2.96.0d.) 13.2.128.0

e.) nota


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Alternative:

Another way to find the first address, the lastaddress, and the number of addresses is to representthe mask as a 32-bit binary (or 8-digit hexadecimal)

number.

11111111 11111111 11111111 11110000

Find:

a. The first address

b. The last address

c. The number of addresses.


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The first address can be found by AND-ing the givenaddresses with the mask.

The last address can be found by OR-ing the given

addresses with the complement of the mask.

The number of addresses can be found bycomplementing the mask, interpreting it as a decimal

number, and adding 1 to it.


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An ISP is assigned the network number193.1.1.0/24 and it needs to define at

least six subnets. The largest subnet is

required to support 25 hosts. Determine

the Broadcast Address for subnets 2 and

6.


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ABC Company is assigned the network

number 140.25.0.0/16 and it must create a

set of subnets that supports up to 60 hostswith more room for future expansion on

each subnet. Determine the broadcast

address for Subnet #3


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A corporation that has been assigned the

Class C network 165.214.32.0. The

corporation has to split this address range

into five separate networks each with thefollowing number of hosts:

Subnet 1: 50 hosts

Subnet 2: 50 hosts

Subnet 3: 50 hosts Subnet 4: 30 hosts

Subnet 5: 30 hosts


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