1. Introduction to Computer Networking
IP Addressing and subnetting
M Clements

2. Last Week . . . . Virtual week – multiple choice assignment
Reading of Network Media notes
Last time we met in person….
LAN, MAN, WAN, SAN each suit a particular group’s data requirement
Many devices exist for networking
Hubs, switches, routers, modems, cables etc.
All have a maximum bandwidth
BW is measured in bps Mbps Gbps etc.
Throughput is moderated by real world factors
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3. This Week. . . . IP addressing Reserved addresses Network identity
Host identity
Subnetworking
Subnet masks
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4. Internet Layer - IP Addresses
Hierarchical addressing scheme
32 bit address
Dotted decimal notation
Used for addressing and routing
Divided into 2 main sections
Network identity
Host identity
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5. IP version 4 4 bytes divided into 4 sections
Each section can range from 0 to 255
If any portion of the IP address contains a number greater than 255 it is ILEGAL !
Presented with dots in between 4 bytes
e.g
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6. For subnetting use BINARY
Always work in Binary
All solutions can be discovered in Binary
Eventually you will see shortcuts
Use Binary first
It will take time to understand
Practice makes perfect
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7. Address Classes (a history lesson)
Classes A, B, C, D, E were created
IP numbers that began with :
1 to 126 are class A addresses
128 to 191 are class B addresses
192 to 223 are class C addresses
224 to 239 are class D addresses
240 to 254 are class E addresses.
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8. Exercise 1 – ‘class’ identification
To which class do the following IP numbers legally belong ?
,98
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9. Class C addresses First 3 bytes is network portion
Last byte for hosts on network
How many hosts can class C have ? A
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10. Class B addresses First 2 bytes is network portion
Last 2 bytes for hosts on network
How many hosts can class B have ? A
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11. Class A addresses First byte is network portion
last 3 bytes for hosts on network
How many hosts can class A have ? A
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12. The end of Address class system
Classes wasted addresses through inflexibility
A better way of dividing network space was sought
This was known as Variable Length Subnet Masking or VLSM
Now we have to explicitly show the boundary of the host and network portion of an address
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13. Caveat to classless addressing
Many people and textbooks still refer to the address class system
No networks EVER use the class system in a production environment
Be aware of the translations between the classful and classless system
Remember ALWAYS to use the classless system despite the terminology of the question
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14. Subnet Mask identifies the boundary
Sometimes called the Extended Network Prefix
Identifies the two sections of the IP address, network and host fields
Used by routers to work out the network a particular host belongs to
Written either as dotted decimal or slash notation
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15. Subnet Mask – two notations
Dotted decimal is one of the notations
In binary this is
Slash is the second notation
Count the number of ‘1’s
There are 24 ‘1’s
So another way of expressing this subnet mask is /24
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16. Network (or wire) Address
This is the identity of the network and is used by routers to deliver packets across networks
No host is ever given this address
To discover the network address, convert IP address from decimal to binary
Replace the host section with 0s
Convert back to decimal
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17. Finding the Network Identity
/16
What address class was this ?
Which is the host field ?
Replace with 0s
Convert back to dotted decimal A
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18. Broadcast Address RFC 919 Each network needs a broadcast address
No host is ever given this address
A broadcast can message all hosts in a LAN
Useful when a host needs to find information without knowing exactly what other host can supply it e.g. ARP
When a host wants to provide information to a large set of hosts in a timely manner
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19. Finding the Broadcast Number
Substitute the host section of the binary version of the IP address with 1s
Convert back to dotted decimal
What is the broadcast address for the IP /16 A
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20. Reserved addressing numbers
Host bits set to 0 are Network address
Host bits set to 1 are Broadcast address
127 network numbers reserved for loopback
is ‘my NIC’
x.x & x.x & 10.x.x.x are non-routable private addresses – for NAT and private networks only
defined in RFC 1918
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21. Exercise 2 Write the network & broadcast number 12.123.14.235 /24
/24
/16
/24
/33 A
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22. Host numbers on networks
Original assignment of classes wasted many IP addresses ( 3 bears problem )
Class A for governments with 224 hosts
Class B can have hosts
more than most companies require
Class C can have 254 hosts
less than most companies require
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23. Subnetworking Solution is to subdivide networks Take a network number
Divide it into smaller networks
These are called subnetworks (subnets)
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24. Subnetting Take a /24 (class C) address 24 network bits & 8 host bits
Company may need different departments
We can borrow bits from host field to augment network field
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25. Subnetting /24 (Class C) address
e.g
Write network number
Borrow 4 bits from host field
4 bits remain for host identity
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26. Subnetted /24 (Class C)
How many hosts does this allow on each subnet ?
How many subnetworks can we use ?
First and last subnet numbers are reserved
First and last host numbers are reserved A
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27. More Subnetting Can borrow 2 host bits minimum
Maximum host bits that can be borrowed is all but the rightmost 2 bits of the IP address
Why is this so ?
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28. Calculating subnet mask
1. Express the subnetwork IP address in binary form
2. Replace the network and subnet portion of the address with all 1s
3. Replace the host portion of the address with all 0s
4. Now convert the binary expression back to dotted-decimal notation
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29. Example of subnet mask
Calculate the subnet mask for the subnetted class C address that has borrowed 3 bits from the host field
27 bits for network identity - substitute these 27 bits for 1s
Change remaining 5 host bits to 0s
Convert to dotted decimal
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30. Exercise 3
Calculate subnet masks for class B that has borrowed 4 bits from host field
Calculate subnet masks for class C that has borrowed 5 bits from host field A
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31. Use of Subnet Mask Mask is ANDed with IP address of host
This gives network identity
Can now be used to route the message
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32. Example
What is the network number for the IP address given that the subnet mask is
Convert to binary
AND both numbers together
Result is network identity A
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33. Exercise 3
IP network address You require at least 17 subnets. Calculate
1. The required subnet mask
2. The number of total subnets
3. The number of available host IP per subnet
4. The first four subnet addresses
5. The valid range of host addresses for the first four subnets
6. The Broadcast address for each subnet
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34. Conclusion 3 Classes available, A, B, C
Strictly not used any longer but well understood
Flexibility gained by subnetting
Subnets concealed from outside networks using subnet masks
Masks are ANDed with IP address of host to discover network identity
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35. Hosts on class A
There are a possible 224 = combinations available for the last 24 bits
Subtract the two combinations that cannot be used as host addresses
A class A address can have = different hosts
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36. Hosts on class B
There are a possible 216 = combinations available for the last 16 bits
Subtract the two combinations that cannot be used as host addresses
A class B address can have
= different hosts
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37. Hosts on class C
There are a possible 28 = 256 combinations available for the last 8 bits
Subtract the two combinations that cannot be used as host addresses
A class B address can have
= 254 different hosts
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38. Network Address
It means ‘ this network ’
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39. Broadcast number
Reaches all hosts on a network
Will not reach other hosts on different network
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40. Classes & numbers 12.123.14.0 12.123.14.255 (looks like Class A)
(Class B)
(Class C)
Illegal number
And illegal subnet mask
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41. Subnetted Class C 24 = 16 -2 = 14 hosts 24 = 16 subnetworks
Note that 14 hosts x 16 subnets is less than 256
Why is this ?
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42. Subnet masks
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43. Network mask discovery
Convert to dotted decimal
How many hosts on this subnet ?
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44. Exercise 3 Need 17 subnets. Calculate bits required for 17 subnets = 5
Leaves 3 bits for host identity on subnets
When calculating the first four subnet addresses, you will begin with:
= 0
= 8
= 16
= 24
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45. Exercise 3 contd
3 bits for the host leaves = 6 host addresses per subnet
Valid host range
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