Review
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Settle down
Review Qs - IP addressing _Basic
Lecture/interactive discussion
Revisit IP addressing _Basic
◦ Complete any unanswered questions
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Student presentation Task 1
◦ Problem definition, analysis and design(5 minutes/group)
◦ Class discussion/ questions (10 minutes)
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Group Feedback
Homework : Review Qs - Subnetting
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Available on the module web page
Time: 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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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
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A
0-127
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B
128-191
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C
192-223
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D
224-239
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E
240-255
Addresses from classes A, B, C
are assignable
88
16
24
32
Host ID
Class A 0 Network ID
Class B 10
Class C 110
Class D 1110
Multicast Addresses
Class E 1111
Reserved for experiments
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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
1st octet
2nd octet
3rd octet
4th octet
Class A
Network
Host
Host
Host
Class B
Network
Network
Host
Host
Class C
Network
Network
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Host
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Loopback address
◦ 127.0.0.0
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Network address
◦ IP address with all host bits set to 0
 Example: 172.16.0.0
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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
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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
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
Addresses need to be
‘translated’ for connecting
hosts to the Internet .
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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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Device ?
Router
Assign IP addresses to above network using appropriate subnet
mask:
Class A
Class B
Class C
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University Network
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Engineering
School
Business
School
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Library
Organizations have multiple
networks which are
independently managed
Subnetting allows us to
break LANs into small subnetworks
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?
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When designing an address scheme, assign addresses to
hosts, network devices and the router interface
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Steps
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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?
3
2 - 2 = 6 subnets
◦ How many bits are left for hosts?
Bits left = Bits available – bits borrowed
5 = 8-3
◦ Assignable host addresses
5
2 - 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.
1
1
1
128
64
32
16
8
4
2
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
Non-Zero
Borrowed Octet
1
128
2
192
3
224
4
240
5
248
6
252
7
254
8
255
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Complete/correct answers
Time: 10 minutes
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Task 1: Basic home network
Objectives covered
1. Problem analysis and definition
2. Design
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