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Transcript
```IP-adresses and
Figure 19.9
Dotted-decimal notation
• 32-bit address i today’s IP version 4 (IPv4).
• 128-bit in IPv6.
1100 0001 0000 1100 0101 1011 0001 11112 binary form.
• At punctured decimal form this is 193.12.91.31.
• Using hexadecimal form this is C10C5B1F.
1100 0001 0000 1100
C
1
193
0
.
0101 1011 0001 1111
C
12
5
.
D
91
F
1
.
31
Example 1
Change the following IP addresses from binary notation to dotteddecimal notation.
a.
10000001 00001011 00001011 11101111
b.
11111001 10011011 11111011 00001111
Example 1
Change the following IP addresses from binary notation to dotteddecimal notation.
a.
10000001 00001011 00001011 11101111
b.
11111001 10011011 11111011 00001111
Solution
We replace each group of 8 bits with its equivalent decimal
number and add dots for separation:
a.
129.11.11.239
b.
249.155.251.15
Example 2
Change the following IP addresses from dotted-decimal notation to
binary notation.
a.
111.56.45.78
b.
75.45.34.78
Example 2
Change the following IP addresses from dotted-decimal notation to
binary notation.
a.
111.56.45.78
b.
75.45.34.78
Solution
We replace each decimal number with its binary equivalent:
a.
b.
01101111 00111000 00101101 01001110
01001011 00101101 00100010 01001110
• Physical network address – A part used by the router
• Host address – Specific device, node or network interface
on the physical network
network node
3.1
2.1
A
1.1 1.2
1.3
1
1
2
3
2
1
3
1
Note:
space is divided into five classes: A, B,
C, D, and E.
Figure 9.5 IP address formats and
classes
Figure 19.10
Finding the class in binary notation
Figure 19.11 Finding the address class
Example 3
Find the class of each address:
a.
00000001 00001011 00001011 11101111
b.
11110011 10011011 11111011 00001111
Example 3
Find the class of each address:
a.
00000001 00001011 00001011 11101111
b.
11110011 10011011 11111011 00001111
Solution
a.
b.
The first bit is 0; this is a class A address.
The first 4 bits are 1s; this is a class E address.
Figure 19.12
Finding the class in decimal notation
Example 4
Find the class of each address:
a.
227.12.14.87
b.
252.5.15.111
c.
134.11.78.56
Example 4
Find the class of each address:
a.
227.12.14.87
b.
252.5.15.111
c.
134.11.78.56
Solution
a.
b.
c.
The first byte is 227 (between 224 and 239); the class is D.
The first byte is 252 (between 240 and 255); the class is E.
The first byte is 134 (between 128 and 191); the class is B.
Figure 19.13
Netid and hostid
Figure 19.14
Blocks in class A
Figure 19.15
Blocks in class B
Figure 19.16
Blocks in class C
Figure 19.17
Figure 19.13
Number of hosts and networks
N bit host ID allows 2N addresses,
or 2N-2 hosts in the network.
M free bit net ID allows 2M networks.
Example 5
Example 5
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
Example 6
Example 6
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 1s. Therefore, the
Example 7
Given the network address 17.0.0.0, find the class.
Example 7
Given the network address 17.0.0.0, find the class.
Solution
The class is A because the netid is only 1 byte.
• Then a network can be devided into several subnets, each
corresponding to a physical network. Arbitrary number of
host-bits can be used.
• A 0 in the subnet mask means that the corresponding bit in
the address belongs to the host-ID, and a 1 that it belongs
to the Net ID or subnet ID.
• Example: Subnet mask 255.255.0.0 = FFFF0000 (sixteen
ones and sixteen zeros) means that the first 16 bits in the i
IP-adressen are Net-ID or subnet ID, the rest are Host-ID.
Example:
192.16.4.0
192.16.4.3
172.16.4.255192.16.4.1 192.16.4.2
192.16.4.25
5
192.16.5.0
172.16.5.255
192.16.5.255
192.16.5.3
192.16.5.1
192.16.5.2
192.16.3.0
172.16.3.255
192.16.3.255 192.16.3.1 192.16.3.2 192.16.3.3 192.16.3.4 192.16.3.5
The mask is 255.255.255.0 for all hosts
Figure 19.23
Figure 19.21 Addresses in a network with and without subnetting
Class
In Binary
In Dotted-Decimal
Using Slash
notation
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
Note:
The 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.
Example 8
A router outside the organization receives a packet with destination
route the packet.
Example 8
A router outside the organization receives a packet with destination
route the packet.
Solution
The 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 ANDs
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.
Example 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
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
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.
sent to all the hosts on the same network
○ Convenient in broadcast networks (such as Ethernet)
replaced by 1s.
Unicast, Multicast and Reserved
• Unicast address is used for one-to-one communication
• Multicast address is used for one-to-many communication (group
communication) – D class
○
○
○
○
Network addresses – all host bits are 0
All network part 0 – host on this network
reserved for testing purposes. 127.0.0.1 = localhost.
○ 0.0.0.0 – default route
Figure 19.25
NAT
Figure 19.27 Translation
Table 19.3 Five-column translation table
Private
Private
Port
External
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
...
...
...
...
...
Range
Total
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
Useful Programs
• These programs use ICMP to probe the Internet
○ ping
• Sends packets that is echoed by remote computer
• Remote computer replies with echo packet
• Local computer reports receipt of reply
○ traceroute
•
•
•
•
Reports path to remote computer
Sends packets to the destination starting with TTL=1
Each successive packet identifies next router along path
Reports list of packets
○ ipconfig – shows network configuration info
• Displays all configuration information
Ping - Example
Four packets are sent. Each has different round-trip time
(RTT). Why?
Minimum, Maximum and average are also given.
Round-trip Time (RTT)
• Time for the packet to be sent and acknowledgement to
come back to the sender
• Why the packets have different RTT?
○ They might travel different paths
○ The load in some of the routers might be high. Therefore packet’s
waiting time at the routers can be different.
○ Ping also shows the percentage of lost packets.
Traceroute - Example
The source is sending three packets with TTL=1, then another three with TTL=2
and so on until TTL is by one bigger then the number of hops. A response is thus
obtained from each hop where the packets are dropped. RTT for each packet is
presented.
VisualRoute
• A program that displays visually (on a map) traceroute
• Trial version can be obtained free from
www.visualroute.com
Ipconfig /all - Example
Figure 20.15
Ipconfig
• Displays all the information about the IP configuration.
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