IP Internet Protocol Address_doc

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IP Internet Protocol Address

FT41224EN02GLA1 2011 Nokia Siemens Networks

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Contents1 IP Address 3 1.1 IP Address Format 3 1.2 Class Subdivision 4 1.3 IP Private Adresses 10 2 Subnetting 11 2.1 The Mask 11 2.2 Subnetting example on Class A 12 2.3 Subnetting example on Class C 17 3 IPCONFIG 18

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IP and Internetworking devices 19

4.1 IP and Bridge 19 4.2 IP and Router 21 5 Ipv4 and Ipv6 24 6 Exercise 25 7 Solution 29



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2011 Nokia Siemens Networks2


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1 IP Address

1.1 IP Address FormatThe IP address format is a decimal number composed by 32 bits (4 bytes)

Fig. 1 IP Address Format


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1.2 Class Subdivision

The most used addresses are subdivided in 3 main classes. Class A small number of networks and many hosts for each network.

Class B a medium number of networks and hosts inside each network.

Class C very high number of networks and a small number of hosts in eachnetwork.

Class D reserved for Multicast addresses

Class E future useWith the name "host" is possible to identify a PC or a general network element.

Other classes are in development.

Fig. 2 Main three IP Class subdivision


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1.2.1 Class A

Class indication depend on the first bit of the first byte The network number is identifiable by the first byte

On the first byte:Minimum binary digit 00000001 = 1 in decimal

Maximum binary digit 01111110 = 126 in decimal

all 0 and all 1 is not possible

Network ranges from 1 .0.0.0 to 126 .0.0.0

Network 127.0.0.0 ( all 1 in binary, cannot be used) Total number of network 126

Number of addressable hosts in 24 bits is 2 24 = 16777216

Fig. 3 Class A


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1.2.2 Class B

Class indication depend on the first two bits of the first byte The network number is identifiable by the first byte

Minimum binary digit 10 000000 = 128 in decimal

Maximum binary digit 10 111111 = 191 in decimal

All 0 and all 1 no possible


Total number of networks = 2 14 = 16384 -2 = 16382

Number of addressable hosts in 16 bits is 2 16 = 65536 -2 = 65534

Fig. 4 Class B


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1.2.3 Class C

Class indication depend on the first three bits of the first byte The network number is identifiable by the first byte

Minimum binary digit 110 00000 = 192 in decimal

Maximum binary digit 110 11111 = 223 in decimal

All 0 and all 1 no possible


Total number of networks = 2 21 = 2097152 -2 = 2097150

Number of addressable hosts in 8 bits is 2 8= 256 -2 = 254

Fig. 5 Class C


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1.2.4 Class Address Summary

The high order bits of the first octets, are used to determine how many bytes ofthe IP address identify the Network and how many identify the Host.

1110 224-239 D

E240-2551111

Fig. 6 First Byte and Address Class


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Fig. 7 Main Class address Summary


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2 SubnettingThe problem is not totally solved with the private network addresses. Due to thenature of the router and the protocols it is necessary a continuous change of network(normally one network for each router port), so in large organizations a lot of networkare necessary, and the private addressing fails to satisfy. In order to optimize betterthe network, the concept of subnetting (subnet-work) has to be introduced.

2.1 The MaskIn the three address classes (A, B, C) previously mentioned, there are only two field:Network and Host. How is it possible and where to introduce the Sub network field?For such purpose it is necessary to introduce the concept of Sub-net mask .

Class A default mask: one byte for network three bytes for hosts

Class B default mask: two bytes for network two bytes for hosts

Class C default mask: three bites for the network one byte for the host

Fig. 9 Default Masks


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2.2 Subnetting example on Class A

The modification of the default mask means the insertion of a new parameter called subnetwork .

Class A default mask: define one byte for network and three bytes for hosts

Adding a default mask class B on a class A the result is: one byte for networkidentification, one byte for subnetwork, two bytes for the host area.

Fig. 10 Subnetting example on a Class A with a Class B mask


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The result is the following:

Subnet bit map n = network s = subnetwork h = host

0nnnnnnn ssssssss hhhhhhhh hhhhhhhh

Mask bit 16

Subnet bits 8

Max number of subnetworks 254 (28 - 2)

Max hosts per subnetwork 65534 (216 - 2)

Subnet hosts address range from 10.10.0.1 to 10.10.255.254

Subnet ID10.10.0.0

In case more subnetworks are necessary:Adding a default mask class C on a class A one byte for network, two bytes forsubnetwork, one byte for host.

Fig. 11 Subnetting example on a Class A with a Class C Mask


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The result is the following:


10nnnnnn ssssssss ssssssss hhhhhhhh

Mask bit 24

Subnet bits 16

Max number of subnet 65534 (2 16 - 2)

Max hosts per subnetwork 254 (2 8 - 2)


Subnet ID10.10.10.0.


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The need of more subnetting can over cross the three bytes for the mask.

Fig. 12 Subnetting example on a Class A with exceeding the default format


10nnnnnn ssssssss ssssssss sssshhhh

Mask bit 28

Subnet bits 20

Max number of subnet 1048574 (2 20 - 2)

Max hosts per subnetwork 14 (2 4 - 2)


Subnet ID 10.10.10.0


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In the previous case how to distinguish the subnetwork from the host ?For example the following 3 IP addresses and their subnet mask:

10.134.76. 114 net.mask:255.255.255 .240 10.134.76. 116 net.mask:255.255.255. 240

10.134.76. 118 net.mask:255.255.255. 240

Which subnetwork are they part of?The basic rule is to take the last byte of the IP addresses and make an AND (logic) function with the last byte of their netmask.

114 ->01110010 AND with 240 ->11110000 = 01110000 -> 112

116->01110100 AND with 240 -> 11110000 = 01110000 -> 112

118->01110110 AND with 240 -> 11110000 = 01110000 -> 112The legacy among the three addresses and their mask is 112 so the subnetworkID is: 10.134.76.112

TIPremember that in AND Function the 0 wins over 1


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2.3 Subnetting example on Class C

The IP address is 192.168.1.32 /27.The notation /27 is an abbreviation of the mask 255.255.255.224 because in themask number there are 27 bits set at 1.Which is the subnetwork ID?

NE IP 11000000. 10101000. 00000001. 00100000

Mask 11111111. 11111111. 11111111. 11100000.

11000000. 10101000. 00000001 00100000 > ANDfunction

result

192 168 1 32 > decimalconversion

WARNING

In this particular example the AND function result shows that the subnetwork isthe same of the IP address . So it is not possible to use this IP as a hostidentification


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3 IPCONFIGA simple command on PC to obtain the full information according the IP address:

Fig. 13 IPCONFIG Command in a DOS environment


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4 IP and Internetworking devices

4.1 IP and BridgeHere an example of IP addressing in a LAN segment followed by a bridge extension.A collision domain (or Ethernet segment) is the set of LAN interfaces whose framescould collide with each other in the same collision domain. Only one host could senda frame at a time, so the hosts share the total LAN bandwidthAs the number of hosts increases, the probability of collisions increases.The bridge reduces the number of collisions. If you have only one Ethernet segmentor your complete network is bridged, all the devices are within a network area, withina subnet. The hosts don't know that there is a bridge. Transparent bridging.

Fig. 14 IP addressing example with Bridge


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4.1.1 Bridge Routing Table

The next example with the same LAN shows the routing table at MAC levelstored in the Bridge.This routing table is dynamically created during the traffic activity after that each hosthas sent at least one frame. In case one host is moving from one segment toanother, just sending a frame the bridge learns which part of the segment isconnected. This happens without change the IP address because the two segment ofthe bridge are part of the same network. In case of remote connection (two bridges)the routing table are stored in both bridges and contain exactly the same information.

Fig. 15 Bridge routing Table example


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4.2 IP and Router

In case of router connection, each router port is a new IP network or subnetwork.

Fig. 16 IP Addressing example with Router


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When a router receives a packet, it reads the destination IP address. Router looks forthe correspondent entry (comparing the destination IP address to network column) inthe routing table. After that, the router forwards the packet out, through the right

interface. Source and destination addresses inside the data frame are the basicinformation to build up the routing table.

Fig. 17 IP Routing Basic concept


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4.2.1 Routing Table in a Router

Simple network example with routing tables.

Fig. 18 Example of RoutingTables in Routers


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5 Ipv4 and Ipv6IPv4 is the Internet Protocol version now in use. The description is on ETF RFC 791published the first time in 1981. It is what explained in the previous chapters.

IPv6 is the Internet Protocol version designed to overcome the limitation of IPv4. It'smain characteristic is a wider area of addressing, up to 128 bits in 32 hexadecimalnumbers instead of the 32 decimal used on IPv4.

The official recommendations are RFC 2373 and RFC 2374


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6 Exercise

1. The first decimal numbers of a Class A Address are from 1 to 126, Class B from128 to191, Class C from 192 to 223. Why ?

2. According to the following subnet address and mask12.10.10.10/255.255.255224 Which number is associated to the Subnetwork?


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3. The following IP Address: 12.10.10.96/ 255.255.255.224 It is a host or asubnetwork ?

4. In most documents the IP address and its mask are written in the following way:12.10.10.75/21 What does it mean ?


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5. How many Networks (Subnetworks ) are present in the following networkexample?

bridge bridge

bridge bridgerouter

hub

Fig. 19


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7 Solution1. The first decimal numbers of a Class A Address are from 1 to 126, Class B from

128 to191, Class C from 192 to 223. Why ?

Is the first bit of the first byte in the binary count of the IP address.

With class A the first bit is always fix to 0, so the possibility decimal numbers arefrom 1 -> 0 0 0 0 0 0 0 1 to 127 -> 0 1 1 1 1 1 1 1 ( 127 is for special purpose andcannot be used so the last network in class a is 126

Wit class B the first two bits of the first byte are fixed to 1 0 so, the excursion isfrom 128 -> 1 0 0 0 0 0 0 0 to 191 1 0 1 1 1 1 1 1

with Class C the first three bits of the first byte are fix to 1 1 0, so the excursion isfrom 192 -> 1 1 0 0 0 0 0 0 to 223 -> 1 1 0 1 1 1 1 1

2. According to the following subnet address and mask12.10.10.10/255.255.255224 Which is the Subnetwork number ?

a) 12.10.10.0 /24

3. The following IP Address: 12.10.10.96/ 255.255.255.224 It is a host or asubnetwork ?

It is a subnetwork

4. In most documents the IP address and its mask are written in the following way:12.10.10.75/21 What does it mean ?

It is the sequential numbers of "1" in the subnetwork mask field starting from thefirst bit

5. How many Networks (Subnetworks ) are present in the following networkexample

6


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