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DCN286 INTRODUCTION TO DATA
COMMUNICATION TECHNOLOGY
IP Routing and subnet
TCP/IP model
Application
Application
Presentation
Session
Transport
Transport
Internet
Network
Network Interface
Data Link
Physical
TCP/IP protocol suite
TCP/IP Internet Layer
•
1.
2.
3.
4.
Defines how to deliver data from one host to
another on various physical networks:
Logical addressing
Routing and routing protocol
Main protocols: IP, ARP, RARP (Reverse Address
Resolution Protocol), ICMP (Internet Control
Message Protocol) and router protocols such as
RIP (Routing Information Protocol) and OSPF
(Open Shortest Path First).
PDU (Protocol Data Unit) is packet in layer 3
Switching
1. Switching is redirecting according to MAC
address.
2. Switching is functioning only in Ethernet.
(Please recall that Data Link header is also
called as Ethernet header.)
3. Switching is in layer 2 Data Link layer
4. Switching is faster than routing
5. Switch can separate collision domain.
Routing
• If all traffics are in the same LAN (subnet), no
routing is required. The computers talk to
each other over network cable directly.
• Only internetwork (between different
networks), the routing is required.
• Router can divide broadcast domain.
• Router can offer stronger security protection.
• Routing table could be updated by i) network
engineer manually, or ii) routing protocols
dynamically.
Routable and routing procotol
• Routable protocol: To redirect (forward)
traffic to other networks according to the
routing path defined by routing protocol.
Example: TCP/IP, IPX/SPX.
• Routing protocol: To dynamically define,
update and distribute the best (lowest cost)
routing path between networks. Example:
RIP, IGRP, EIGRP, OSPF and BGP, etc, etc.
• Routing table is to contain the final
information. It is possible to have multiple
entries with same routing costs.
Routing table
• is a small inmemory database
managed by the
router's built-in
hardware and
software
• Contains
destination
network
information
Example of routing table
Router
Router (pronounced /'rautər/ in the USA
and Canada, pronounced /'ru:tə/ in the UK
and Ireland, or either pronunciation in
Australia): a networking device whose
software and hardware are usually tailored
to the tasks of routing and forwarding
information.
To be able to route packets, a router must
know, at a minimum, the following:
• Destination address
• Neighbor routers from which it can learn
about remote networks
• Possible routes to all remote networks
• The best route to each remote network
• How to maintain and verify routing
information
Routing process (1)
1. Router receive frame
Ethernet Header
Preamble SFC Destination Source
7 bits
1 bit
6
6
IP Packet Trailer
Length
2
Data and
Pad
46-1500
2. Router de-encapsulate packet (TTL-1)
FCS
4
Routing process (2)
3. Router review routing table and find the right path
including the next interface to which the packet
needs to be sent to
4. Find out MAC address of destination interface by ARP
5. Re-encapsulate the IP packet in a new frame with the
new source and destination MAC address. In the new
frame, IP address information does not have change.
6. Send out the frame from the destination interface
Static routing
•
•
Routing table could be updated by i) network engineer
manually, or ii) routing protocols dynamically.
Static route is to use command manually create or update
routing entries in routing table.
Pros:
1. Securer (Only network engineer can modify)
2. Faster (No need for any further processing or calculation)
Cons
•
Hard to expand
•
Wrongly configure is hard to troubleshooting
Routing table fields
1.
2.
3.
4.
5.
Source
Subnet
Mask
Out Int (Output Interface)
Next-Hop (forward destination)
IP Route command (1) - Optional
To specify routes by output interfaces:
router>en
Router#conf t
Router(config)#ip route 10.1.10.0
255.255.255.0 s0/0
IP Route command (2) - Optional
Determine next hope
router>en
Router#conf t
Router(config)#ip route 10.1.10.0
255.255.255.0 10.1.11.1
IGP (Interior Gateway Protocols) /
EGP (Exterior Gateway Protocols)
Routing protocols could be divided in Interior
or Exterior routing protocols.
1. IGP (Interior Gateway Protocols) works in a
single Autonomous System (AS). Examples
are RIP, OSPF, IGRP, EIGRP, etc, etc.
2. EGP (Exterior Gateway Protocols) works
between Autonomous Systems. The only
viable example is BGP (Border Gateway
Protocol)
Route table calculation (optional)
• Each router contains two lists: Tentative
and Confirmed
• Each list contains a set of triples (Destination, Cost, NextHop)
• Note that "NextHop" is the first router on
the path from the source S to Destination
• After calculation, each router would build up
its own list.
Routing updates
• Periodically update
• Declare route unusable without updates
• Remove route entry after it was not usable
for sometimes.
VLSM (Variable Length Subnet
Mask)
• Different to traditional classful subnet mask
(class A – 255.0.0.0; class B – 255.255.0.0;
class C – 255.255.255.0), VLSM is to support
subnet and it is classless as its network mask
could be 255.255.224.0.
• Old routers only support classful routing. The
network mask is not included in routing update
information. New router supports VLSM and
mask is a part of update information.
Valid network mask
Binary value
00000000
10000000
11000000
11100000
11110000
11111000
11111100
11111110
11111111
Decimal value
0
128
192
224
240
248
252
254
255
IP address class
Traditionally, the IP address was classified in classes:
Class A: network 1 – 126 with subnet mask 255.0.0.0
(Initial bit starts as 0)
Class B: network 128 – 191 with subnet mask
255.255.0.0 (Initial bit starts as 10)
Class C: network 192 – 223 with subnet mask
255.255.255.0 (Initial bit starts as 110)
Class D: network 224 – 239 (Initial bit starts as 1110)
for multicast IP address
Class D: network 240 – 255 (Initial bit starts as 1111)
experimental use
The special network 0 and127 are not included in
those official classical IP address. 127.0.0.1 is the
loopback address which is used to test TCP/IP
stacks.
Subnet
Subnet is to logically divide your network into many
sub networks.
• In the same subnet, traffic is “local” and not gateway
(router) is required. Network hosts would use ARP
table for the MAC address of the destination machine
and send the packet to it accordingly. If it is not local,
packets would be forwarded to default gateway for
future routing (redirecting).
• In addition, the broadcast would be limited to the small
subnets leading to less “noise” in the network traffic.
• You can also use subnet to logically specify hosts for
different departments. (for security control, connection,
asset management, etc, etc)
Power calculation of 2
The powers of 2
Decimal value
Binary Value
2^0
1
00000001
2^1
2
00000010
2^2
4
00000100
2^3
8
00001000
2^4
16
00010000
2^5
32
00100000
2^6
64
01000000
2^7
128
10000000
2^8
256
100000000
2^9
512
1000000000
2^10
1024
10000000000
2^11
2048
100000000000
2^12
4096
1000000000000
2^13
8192
10000000000000
Subnet calculation
• Network address design would decide how
many subnets would be yield for the
additional subnet mask.
• Or, network design needs to make sure how
many hosts could be placed in each subnet.
• Based on subnet calculation, find out
network address and broadcast in each
subnet. (Those two IP address cannot be
used by any host.)
Several terms
• Subnet number: The numerically lowest number in a
subnet to present the IP Subnet.
• Subnet broadcast address: The numerically highest
number in a subnet. If a packet is sent to this address, it
would be forwarded to all hosts in the IP Subnet.
• Subnet zero (zero subnet): The numerically smallest
subnet number in any subnet scheme. It has all “0” in
subnet portion. With classful IP addressing, it is one of the
two reserved subnets and should not be used.
• Broadcast subnet: The numerically largest subnet number
in any subnet scheme. It has all “1” in subnet portion. With
classful IP addressing, it is one of the two reserved
subnets and should not be used.
Define subnet number
• Subnet bits (“1”) can tell how many subnet could be produced by the
mask. The formula is 2^(number of subnet “1” bits) - 2
Original classfull network mask:
• Class A network has the form N.H.H.H, the default subnet mask is 8 bits
long.
• Class B network has the form N.N.H.H, the default subnet mask is 16
bits long.
• Class C network has the form N.N.N.H, the default subnet mask is 24
bits long.
The additional subnet bits can generate subnet. For instance, You have an
IP of 156.233.0.0 with a subnet mask of 7 bits. How many hosts and
subnets are possible?
There is additional 7 bits to the default subnet mask. The total number of
bits in subnet are 16+7 = 23. This leaves us with 32-23 =9 bits for
assigning to hosts. 7 bits of subnet mask corresponds to (2^7-2)=128-2
= 126 subnets. 9 bits belonging to host addresses correspond to (2^92)=512-2 = 510 hosts.
Define host number in each subnet
• Subnet bits (“0”) can tell how many host could be contained in each
subnet. The formula is 2^(number of host “0” bits) - 2
Original classfull network mask:
• Class A network has the form N.H.H.H, the default host mask is 24 bits
long.
• Class B network has the form N.N.H.H, the default host mask is 16 bits
long.
• Class C network has the form N.N.N.H, the default host mask is 8 bits
long.
The additional subnet bits can generate subnet. For instance, You have an
IP of 156.233.0.0 with a subnet mask of 7 bits. How many hosts and
subnets are possible?
There is additional 7 bits to the default subnet mask. The total number of
bits in subnet are 16+7 = 23. This leaves us with 32-23 =9 bits for
assigning to hosts. 7 bits of subnet mask corresponds to (2^7-2)=128-2
= 126 subnets. 9 bits belonging to host addresses correspond to (2^92)=512-2 = 510 hosts.
Increment Number in Class C
• Increment number (magic #) is 2^(8number_of_subnet_bits). Or, 256 – subnet
mask.
For instance, You have an IP of 192.168.42.0 with
a subnet mask of 3 bits. What is the increment
number?
Beside the traditional class C network mask
255.255.255.0. The 3 bits means that the mask
is 255.255.255.224. Increment number is 256 –
224 = 32.
Or, 2^(8-3) = 32
Increment Number in Class B (1) optional
Class B – borrowing in the 3rd octet
156.233.0.0 Borrow 3 bits
Subnet Mask = 255.255.224.0
•
•
Number of networks created 2^3-2 = 6 (Microsoft way is 2^3=8)
Number of useable networks created 2^3 = 8 – 2 = 6
•
Number of host per network 2^5 X 256 = 8192
•
The increment for each network is 32 in the 3rd octet (the number of unmasked bits
in the 3rd octet is 5, this is the octet we borrowed from)
•
If you need to determine the network number of subnet 6, multiply 6 X 32 = 192.
The subnet 6 network address would be 156.233.192.0
•
If you are borrowing in the 3rd octet, just ignore the 4th octet to determine your
network numbers.
Increment Number in Class B (2) optional
Class B – borrowing all the 3rd octet
156.233.0.0
Borrow 8 bits
Subnet Mask = 255.255.255.0
•
•
Number of networks created 2^8 = 256
Number of useable networks created 2^8 - 2 = 256 – 2 = 254
•
Number of host per network 2^0 X 256 = 256
•
The increment for each network is 256 in the 3rd octet. (The number of unmasked
bits in the 3rd octet is 0, this is the octet we borrowed from. This means that the
value of the 3rd octet increases by one each time.)
•
If you need to determine the network number of subnet 26, simply insert that
number into the 3rd octet slot. The subnet 26 network address would be
156.233.26.0
•
If you need to determine the network number of subnet 100, simply insert that
number into the 3rd octet slot. The subnet 100 network address would be
156.233.100.0
Increment Number in Class B (3) optional
•
•
Class B – borrowing in the 4th octet
•
156.233.0.0
•
•
•
•
Number of networks created 2^11 -2 = 2046 (or 2^3 X 256 -2 = 2048 -2 = 2046)
Number of useable networks created 2^11 -2 = 2048 – 2 = 2046
Number of host per network 2^5 = 32
The increment for each network is 32 in the 4th octet. (The number of masked bits in the 4th octet is 3, the
number of unmasked bits in the 4th octet is 5, this is the last octet we borrowed from. This means that the
value of the 4th octet increases by the increment value.)
If you need to determine the network number of subnet 325, do the following:
Divide the desired subnet by 2 raised to the masked bits in the 4th octet
325 / 2^3 or 325 / 8 = 40 remainder 5
The 40 is the value of the 3rd octet -- 156.233.40.?
To find the value of the 4th octet multiply the remainder (5) times the increment (32)
32 X 5 = 160 This is the value of the 4th octet
The subnet 325 network address would be 156.233.40.160
-------------------------------------------------------------------------------If you need to determine the network number of subnet 40, do the following:
Divide the desired subnet by 2 raised to the masked bits in the 4th octet
40 / 2^3 or 40 / 8 = 5 remainder 0
The 5 is the value of the 3rd octet -- 156.233.5.?
To find the value of the 4th octet multiply the remainder (0) times the increment (32)
32 X 0 = 0 This is the value of the 4th octet
The subnet 40 network address would be 156.233.5.0
1.
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2.
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•
Borrow 11 bits
Subnet Mask = 255.255.255.224
Find the subnet address info
• Increment number (magic #) is 2^(8number_of_subnet_bits). Or, 256 – subnet mask.
• The zero subnet is as same as the original
classful network address
• Add increment number to the zero subnet will get
1st subnet number, adding increment number to
1st subnet will get 2nd subnet number, etc, etc,.
• Broadcast address of each subnet is the address
before next subnet number.
• After define the subnet number and broadcast
address in the subnet, the address range for the
host will be defined in the subnet.
Subnet calculation example
192.168.10.0/27 (255.255.255.224)
2^3 – 2 = 6 subnets
Increment number is 256-224=32
# of subnet
Subnet
Lowest HOST IP
address
Highest HOST IP
address
Broadcast
address
0
192.168.10.0
192.168.10.1
192.168.10.30
192.168.10.31
1
192.168.10.32
192.168.10.33
192.168.10.62
192.168.10.63
2
192.168.10.64
192.168.10.65
192.168.10.94
192.168.10.95
3
192.168.10.96
192.168.10.97
192.168.10.126
192.168.10.127
4
192.168.10.128
192.168.10.129
192.168.10.158
192.168.10.159
5
192.168.10.160
192.168.10.161
192.168.10.190
192.168.10.191
6
192.168.10.192
192.168.10.193
192.168.10.222
192.168.10.223
Broadcast
subnet
192.168.10.224
192.168.10.225
192.168.10.254
192.168.10.255
Boolean AND
When two bits (binary numbers) are in such
logical calculation:
If both bits are 1, final result is 1. Otherwise,
final result is 0.
For example, 1 + 1 = 1; 1 + 0 = 0; 0 + 1=0
0 + 0 =0
Find resident subnet
• Convert the questioned host IP address and subnet mask
into binary. Add them together by using Boolean AND
calculation. Convert binary result to decimal value and it
presents the resident subnet of the host.
Which subnet is 192.168.10.100 in when subnet mask is
255.255.255.224?
1st Octet
2nd Octet
3rd Octet
4th Octet
IP address
192.168.10.100
11000000
10101000
00001010
01100100
Mark
255.255.255.224
11111111
11111111
11111111
11100000
Calculation Result
11000000
10101000
00001010
01100000
Decimal of the
subnet number
192
168
10
96
Subnet calculation clarification
In real world, you may hear the argument of
subnet calculation:
• Microsoft: Number of subnets: 2^(numberof-subnet-bits) This is the number of
subnets that are created.
• Cisco: Number of subnets: 2^(number-ofsubnet-bits – 2) This is the number of
subnets you can use to connect devices.
Network address questions
Write the subnet, broadcast address, and valid
host range for each of the following:
• 1. 172.16.10.5 255.255.255.128
• 2. 172.16.10.33 255.255.255.224
• 3. 172.16.10.65 255.255.255.192
• 4. 172.16.10.17 255.255.255.252
• 5. 172.16.10.33 255.255.255.240
• 6. 192.168.100.25 255.255.255.252
• 7. 192.168.100.17, with 4 bits of subnetting
• 8. 192.168.100.66, with 3 bits of subnetting
• 9. 192.168.100.17 255.255.255.248
• 10. 10.10.10.5 255.255.255.252
Answer of q1
1. 172.16.10.5 255.255.255.128: Subnet is
172.16.10.0, broadcast is 172.16.10.127, and
valid host range is 172.16.10.1 through 126.
You need to ask yourself, “Is the subnet bit in
the fourth octet on or off?” If the host address
has a value of less than 128 in the fourth octet,
then the subnet bit must be off. If the value of
the fourth octet is higher than 128, then the
subnet bit must be on. In this case, the host
address is 10.5, and the bit in the fourth octet
must be off. The subnet must be 172.16.10.0.
Answer q 2 and 3
2. 172.16.10.33 255.255.255.224: Subnet is
172.16.10.32, broadcast is 172.16.10.63, and
valid host range is 172.16.10.33 through 10.62.
256 224 =32. 32 +32 =64—bingo. The subnet
is 10.32, and the next subnet is 10.64, so the
broadcast address must be 10.63.
3. 172.16.10.65 255.255.255.192: Subnet is
172.16.10.64, broadcast is 172.16.10.127, and
valid host range is 172.16.10.65 through
172.16.10.126. 256 192 =64. 64 +64 =128, so
the network address must be 172.16.10.64, with a
broadcast of 172.16.10.127.
Answer of Q4 -6
4. 172.16.10.17 255.255.255.252: Subnet is 172.16.10.16,
broadcast is 172.16.10.19, and valid hosts are
172.16.10.17 and 18. 256 252 =4. 4 +4 =8, plus 4 =
12, plus 4 =16, plus 4 =20—bingo. The subnet is
172.16.10.16, and the broadcast must be 10.19.
5. 172.16.10.33 255.255.255.240: Subnet is 172.16.10.32,
broadcast is 172.16.10.47, and valid host range is
172.16.10.33 through 46. 256 240 =16. 16 +16 =32,
plus 16 =48. Subnet is 172.16.10.32; broadcast is
172.16.10.47.
6. 192.168.100.25 255.255.255.252: Subnet is
192.168.100.24, broadcast is 192.168.100.27, and
valid hosts are 192.168.100.25 and 26. 256 252 =4.
4 +4 =8, plus 4 =12, plus 4 =16, plus 4 =20, plus 4
=24, plus 4 =28. Subnet is 100.24; broadcast is 100.27.
Answer of Q7-8
7. 192.168.100.17, with 4 bits of subnetting:
Subnet is 192.168.100.16, broadcast is
192.168.100.31, and valid host range is
192.168.100.17 through 30. 256 240 =16.
16 +16 =32. Subnet is, then, 100.16, with a
broadcast of 100.31 because 32 is the next
subnet.
8. 192.168.100.66, with 3 bits of subnetting:
Subnet is 192.168.100.64, broadcast is
192.168.100.95, and valid host range is
192.168.100.65 through 94. 256 224 =32.
32 +32 =64, plus 32 =96. Subnet is 100.64,
and broadcast is 100.95.
Question
Any question?
If you do not have question, please
search internet and collect more
information of router and its
manufacturers.
1. Please be comfortable to introduce
routing protocols.
2. Please be familiar with subnet
calculation.