Review

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Settle down Review Qs - IP addressing _Basic Lecture/interactive discussion Revisit IP addressing _Basic

◦ Complete any unanswered questions Student presentation Task 1

◦ Problem definition, analysis and design(5 minutes/group)◦ Class discussion/ questions (10 minutes)

Group Feedback Homework : Review Qs - Subnetting

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Available on the module web pageTime: 10 minutes

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IPv4 address ◦Classful addressing◦Private and Public IP addresses

Subnet◦Need to subnet◦ Subnet Class C address◦ Task

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Network ID Host ID

8 16

Class A

32

Class B 10

Class C 110

Multicast AddressesClass D1110

Reserved for experimentsClass E 1111

248

IP v4 addresses are 32 bits long, given as a.b.c.d

IP addresses are divided into five classes, identified by the first group of numbers in the dotted decimal notation as

Class Range A 0-127 B 128-191 C 192-223 D 224-239 E 240-255

Addresses from classes A, B, C are assignable

0

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Generally, IP addresses have two parts◦ Network (Net id)◦ Host ID

Netid and Hostid in a given IP address are identified by Subnet mask

Default subnet masks are◦ Class A :

255.0.0.0◦ Class B :

255.255.0.0◦ Class C :

255.255.255.0

Network Host Host Host

Network Network Host Host

Network Network Network Host

1st octet 2nd octet 3rd octet 4th octet

Class A

Class B

Class C

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Loopback address◦ 127.0.0.0

Network address◦ IP address with all host bits set to 0

Example: 172.16.0.0 Broadcast address

◦ IP address with all host bits set to 1 Example: 172.16.255.255

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Public IP addresses◦Unique◦Used to connect to Internet. ◦ Use of an address class depends on

number of hosts / networks, required to be connected

Private IP addresses

◦Use to conserve public IP addresses◦ Three special ranges, one each in class A, B and

C.

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Assigned to hosts that do not connect directly to the Internet

Three blocks are available, one each from◦ Class A◦ Class B◦ Class C addresses

Addresses need to be ‘translated’ for connecting hosts to the Internet .

Class Range

A 10.0.0.0 – 10.255.255.255

B 172.16.0.0 – 172.31.255.255

C 192.168.0.0 – 192.168.255.255

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Inefficient Address Usage◦ In danger of running out of classes A and B◦ Why?

Class C too small for most domains Very few class A – very careful about giving them

out Class B poses greatest problem

◦ Class B sparsely populated But users refuse to give it back

◦ Need simple way to reduce the number of network numbers assigned

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Use Private Addresses Dynamic allocation of addresses

◦DHCP Subnet the given address Use Classless IP addressing schemes (CIDR) Use larger address space

◦IPv6 uses 128 bit address (32 bits for IPv4 addresses)

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Assign IP addresses to above network using appropriate subnet mask:

Class A

Class B

Class C

Device ?

Router

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Organizations have multiple networks which are independently managed

Subnetting allows us to break LANs into small sub-networks

Sub-networks created by borrowing bits from host-id. from the given IP address

What are the maximum number of bits that can be borrowed in a ◦ Class C address?◦ Class B address?

University NetworkUniversity Network

Business School

Library

EngineeringSchool

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When designing an address scheme, assign addresses to hosts, network devices and the router interface

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Steps◦ Know how many Different Networks are required ◦ Borrow bits from the host portion of the IP address◦ Find New Subnet Mask.◦ Calculate the number of sub-networks and the hosts

available corresponding to borrowed bits◦ Find the sub-network boundary

Network Address Find the broadcast address.

Let’s look at each of these steps in detail

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How many host bits CAN/DO I have to borrow to create required subnets◦ Depends on the class of your network address. ◦ How do you find the IP address class?

First octet of IP address◦ What are the host bits for the default subnet mask?

Class C: 8 host bits

Class B: 16 host bits

Class A: 24 host bits

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Class C Address: 210.93.45.0◦ Requirement: At least 5 subnets ◦ how many bits do we borrow (Bits Borrow (BB))?◦ How many bits in the host portion (HB) do we have for

default mask? Since it’s a Class C, we have 8 bits to work with.◦ 2 to what power will give us at least 5 subnets? 23 - 2 = 6 subnets◦ How many bits are left for hosts?

Bits left = Bits available – bits borrowed5 = 8-3

◦ Assignable host addresses 25 - 2 = 30 hosts

One network address, one broadcast address

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We determine the new subnet mask by adding up the decimal value of the bits we borrowed.

In the previous Class C example, we borrowed 3 bits. Below is the host octet showing the bits we borrowed and their decimal values.

128 64 32 16 8 4 2 1

1 1 1

We add up the decimal value of these bits and get 224 (128+64+32). NEW subnet mask is 255.255.255.224 (as against default subnet mask of 255.255.255.0)

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magic number : total_value_of_non-zero_octet – new_subnet_mask

In our Class C example, our subnet mask was 255.255.255.224.224 is our last non-zero octet.

Our magic number is 256 - 224 = 32◦Note: The last bit borrowed was the 32 bit.

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We now take our “magic number” and use it as a multiplier Our Class C address was 210.93.45.0. We borrowed bits in the fourth octet, so that’s where our

multiplier occurs.◦ 1st subnet: 210.93.45.32◦ 2nd subnet: 210.93.45.64◦ 3rd subnet: 210.93.45.96◦ 4th subnet: 210.93.45.128◦ 5th subnet: 210.93.45.160◦ 6th subnet: 210.93.45.192

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Now you can see why we subtract 2 when determining the number of host addresses.◦ Let’s look at our 1st subnet: 210.93.45.32◦ What is the total range of addresses up to our next subnet,

210.93.45.64? 210.93.45.32 to 210.93.45.63 or 32 addresses◦ .32 cannot be assigned to a host. Why? Because it is the subnet’s address.◦ .63 cannot be assigned to a host. Why? Because it is the subnet’s broadcast address.◦ So our host addresses are

.33 - .62 or 30 host addresses--just like we figured out earlier.

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Memorize this table. You should be able to:◦ Quickly calculate the last

non-zero octet when given the number of bits borrowed or...

◦ Determine the number of bits borrowed when given the last non-zero octet.

Bits Borrowed

Non-Zero Octet

1 1282 1923 2244 2405 2486 2527 2548 255

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Complete/correct answers Time: 10 minutes

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Task 1: Basic home networkObjectives covered 1. Problem analysis and definition2. Design

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