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8/7/2019 Internet Working Addressing and Routing Updated
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Internetworking, Addressing and
Routing
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Topic & Structure of the lesson
� Internetworks
� Addressing
� Routing
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Learning Outcomes
� At the end of this session you should be
able to:-
± Understand the concept of interworking andthe need for the network layer in an
internetwork.
± Understand the type of switching used in the
Internet.
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Position of network layer
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Network layer duties
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InternetworksInternetworks
Need For Network Layer
Internet As A Packet-Switched Network
Internet As A Connectionless Network
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Fig1 Internetwork
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Fig2 Links in an internetwork
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Fig3 Network layer in an internetwork
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Fig4 Network layer at the source
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Fig5 Network layer at a router
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Fig6 Network layer at the destination
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Fig7 Switching
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Fig8 Datagram approach
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Switching at the network layer in the
Internet is done using the datagram
approach to packet switching.
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C ommunication at the network layer
in the Internet is connectionless.
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AddressingAddressing
Internet Address
Classful Addressing
Supernetting
Subnetting
Classless Addressing
Dynamic Address Configuration
Network Address Translation
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An IP address is a 32-bit address.
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T he IP addresses are unique
and universal.
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Fig9 Dotted-decimal notation
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T he binary, decimal, and hexadecimal
number systems are reviewed in
A ppendix B.
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E xample 1 E xample 1
Change the following IP addresses from binary notation to dotted-
decimal notation.
a. 10000001 00001011 00001011 11101111
b. 11111001 10011011 11111011 00001111
Solution Solution
We replace each group of 8 bits with its equivalent decimal
number (see Appendix B) and add dots for separation:
a. 129.11.11.239
b. 249.155.251.15
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E xample 2 E xample 2
Change the following IP addresses from dotted-decimal notation to
binary notation.
a. 111.56.45.78
b. 75.45.34.78
Solution Solution
We replace each decimal number with its binary equivalent
(see Appendix B):
a. 01101111 00111000 00101101 01001110
b. 01001011 00101101 00100010 01001110
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In classful addressing, the address
space is divided into five classes: A , B,C , D, and E .
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Fig10 Finding the class in binary notation
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Fig11 Finding the address class
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E xample 3 E xample 3
Find the class of each address:
a. 000000001 00001011 00001011 11101111 b. 111111110011 10011011 11111011 00001111
Solution Solution
See the procedure in Figure 19.11.
a. The first bit is 0; this is a class A address.
b. The first 4 bits are 1s; this is a class E address.
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Fig12 Finding the class in decimal notation
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E xample 4 E xample 4
Find the class of each address:
a. 227.12.14.87
b. 252.5.15.111
c. 134.11.78.56
Solution Solutiona. The first byte is 227 (between 224 and 239); the class is D.
b. The first byte is 252 (between 240 and 255); the class is E.
c. The first byte is 134 (between 128 and 191); the class is B.
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Fig13 Netid and hostid
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Fig14 Blocks in class A
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M illions of class A addresses are
wasted.
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Fig15 Blocks in class B
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M any class B addresses are wasted.
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T he number of addresses in class C is
smaller than the needs of most organizations.
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Fig16 Blocks in class C
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Fig17 Network address
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In classful addressing, the network
address is the one that is assigned tothe organization.
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E xample 5 E xample 5
Given the address 23.56.7.91, find the network address.
Solution Solution
The class is A. Only the first byte defines the netid.We can find the network
address by replacing the hostid bytes (56.7.91) with 0s. Therefore, the
network address is 23.0.0.0.
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E xample 6 E xample 6
Given the address 132.6.17.85, find the network address.
Solution Solution
The class is B. The first 2 bytes defines the netid.We can find the network
address by replacing the hostid bytes (17.85) with 0s. Therefore, the
network address is 132.6.0.0.
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E xample 7 E xample 7
Given the network address 17.0.0.0, find the class.
Solution SolutionThe class is A because the netid is only 1 byte.
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A network address is different from a
netid. A network address has both
netid and hostid,
with 0s for the hostid.
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Fig18 Sample internet
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IP addresses are designed with two
levels of hierarchy.
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Fig20 A network with three levels of hierarchy (subnetted)
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Fig21 Addresses in a network with and without
subnetting
Fi 22 Hi h t i t l h
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Fig22 Hierarchy concept in a telephone
number
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Table1 Default masks
Class In Binary In Dotted-
Decimal Using Slash
A 11111111 00000000 00000000 00000000 255.0.0.0 /8
B 11111111 11111111 00000000 00000000 255.255.0.0 /16
C 11111111 111111111 11111111 00000000 255.255.255.0 /24
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T he network address can be found
by applying the default mask to any
address in the block (including itself). It retains the netid of the block and
sets the hostid to 0s.
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E xample 8 E xample 8A router outside the organization receives a packet with destination
address 190.240.7.91. Show how it finds the network address to
route the packet.
Solution SolutionThe router follows three steps:
1. The router looks at the first byte of the address to find the
class. It is class B.
2. The default mask for class B is 255.255.0.0. The router ANDsthis mask with the address to get 190.240.0.0.
3. The router looks in its routing table to find out how to route the
packet to this destination. Later, we will see what happens if
this destination does not exist.
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Fig23 Subnet mask
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E xample 9 E xample 9
A router inside the organization receives the same packet with
destination address 190.240.33.91. Show how it finds the
subnetwork address to route the packet.
Solution Solution
The router follows three steps:
1. The router must know the mask.We assume it is /19, as shown in
Figure 19.23.
2. The router applies the mask to the address, 190.240.33.91. The subnet
address is 190.240.32.0.
3. The router looks in its routing table to find how to route the packet to
this destination. Later, we will see what happens if this destination does
not exist.
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� If a company wants to have a subnet 12, thethird byte of all machines on that subnet will be
00001100 (12).
� One byte dedicated to the subnet provides eight
bit positions.� Formula for calculating the maximum no. of
subnets and the maximum no. of hosts per
subnets.
� Max. no. of subnets= 2no.Of 1bits in subnet mask - 2
� Max. no. of hosts/subnet= 2no.Of 0bits in subnet mask - 2
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� NETWORK ADDRESS:
� 182 . 16 . 52 . 10/19
� SUBNETMASK:
� 11111111.11111111.11100000.00000000
� 255 . 255 . 224 . 0
� No. of subnets = 23 -2= 6
� No. of host/subnet= 213
-2= 8190 hosts per subnet.
� 1s in the third byte represent # of subnets.
� 0s represent # of hosts.
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� Q1: If you have a Class C Network with
a 6 bit Subnet mask, how many subnets
and how many hosts do you have?
� Q2: If you have a 19 bit subnet mask,how many subnets and hosts do have?
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� Ans1: Class C: Net . Net . Net . Node
� 11111111 .11111111 . 11111111.
11111100
� # of subnets= 26 - 2 =62� # of hosts = 22 - 2= 2
� Ans2:
� 11111111.11111111.11111111.11100000
� subnets = 219 - 2= 524288
� hosts = 25 - 2 = 30.
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� Q3: You have a network ID of 172.16.0.0 and you need to divide it
into multiple subnets. You need 600
host Ids for each subnet. Which subnet
mask should you assign that will allow
for growth?
� A. 255.255.224.0
� B. 255.255.240.0
� C. 255.255.248.0
� D. 255.255.252.0
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� It¶s a Class-B address:
� 2n -2 > 600
� n=10 210-2 > 600 therefore 10 bits
for hosts.� Remaining bits = 6 for subnets.
� 26 - 2 = 64 - 2 = 62
� 11111111.11111111.11111100.00000000
� Ans is D: 255.255.252.0
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Fig24 DHCP transition diagram
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Table2 Default masks
Range T otal
10.0.0.0 to 10.255.255.255 224
172.16.0.0 to 172.31.255.255 220
192.168.0.0 to 192.168.255.255 216
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Fig25 NAT
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Fig26 Address translation
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Fig27 Translation
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Table 3 Five-column translation table
Private
Address
Private
Port
External
Address
External
Port
Transport
Protocol
172.18.3.1 1400 25.8.3.2 80 TCP
172.18.3.2 1401 25.8.3.2 80 TCP
... ... ... ... ...
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RoutingRouting
R outing Techniques
Static Versus Dynamic R outing
R outing Table for Classful Addressing
R outing Table for Classless Addressing
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Fig28 Next-hop routing
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Fig29 Network-specific routing
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Fig30 Host-specific routing
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Fig31 Default routing
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Fig32 Classful addressing routing table
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E xample 10 E xample 10
Using the table in Figure 19.32, the router receives a packet for
destination 192.16.7.1. For each row, the mask is applied to the
destination address until a match with the destination address is
found. In this example, the router sends the packet through
interface m0 (host specific).
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E xample 12 E xample 12
Using the table in Figure 19.32, the router receives a packet for
destination 200.34.12.34. For each row, the mask is applied to the
destination address, but no match is found. In this example, the
router sends the packet through the default interface m0.
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Q & A
Question and Answer Session