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Unit 24 Network Design and Admin. IP Addressing (Part 2). Objectives for today…. Recap on last week’s work – an introduction to how computers share data (IP addressing and NetBIOS) the use of subnets and subnet masks What is the difference between a ‘Private’ and a ‘Real’ IP address?
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Unit 24 Network Design and Admin IP Addressing (Part 2)
Objectives for today… • Recap on last week’s work – an introduction to how computers share data (IP addressing and NetBIOS) the use of subnets and subnet masks • What is the difference between a ‘Private’ and a ‘Real’ IP address? • Private IP addresses used on LAN’s • The limitations of IP version 4 • New methods of IP addressing – CIDR (classless inter-domain routing)
Last week…Recap • How many bits in an IP address? • How do you identify a class A IP address? • Class B? • Class C? • What is a subnet? • What is a subnet mask? • What is the subnet mask for a class A network? 32 bits, 4 groups of 8 bits First octet 1-126 First octet 128-191 First octet 192-223 Network segment / small part of a larger network Network / host identifier 255.0.0.0
Private and Real IP Addresses Real IP addresses • Even though most networks route data using routing protocols (IP addressing), only a limited number of these networks use ‘real’ IP addresses – i.e. IP addresses that allow them to access the Internet. • It is only when a network needs access to the Internet that it must be allocated a ‘real’ IP address (and then this IP address must be assigned through IANA (Internet Assigned Numbers Authority), to avoid IP address duplication
Private IP Addresses • A network manager / administrator can allocate IP addresses as they see fit – there are no real rules when routing data around private networks, although all administrators must still allocate addresses according to the classes of IP address and appropriate subnet masks. • There are 3 main IP addresses that are always used for private networks, and are never routed to the Internet. • 10.0.0.0 (hosts from 10.0.0.1 to 10.255.255.254) • 172.16.0.0 (hosts from 172.16.0.1 to 172.16.255.254 • 192.168.1.0 (hosts from 192.168.1.1 to 192.168.1.254) • 169.254.0.0 (Automatic Private IP Addressing – APIPA)
Bury College • At Bury College there is one large network of over 1600 computers, broken down into manageable segments called subnets. • Each subnet may be for example Beacon, Woodbury, Millennium and Prospects. • Bury College’s IP address is 172.16.0.0 (private IP address). How might the subnets be addressed? • Beacon 172.16.1.0 • Millennium 172.16.2.0 • Woodbury 172.16.3.0 • Prospects 172.16.4.0 • How many hosts can be on each subnet? • 254 hosts on each subnet • Subnet Mask? • 255.255.0.0
Bury College • All computers in the college are Internet enabled, but they do not have direct access to the Internet – Why not? • All Internet traffic must pass through the Proxy Server (the only computer with direct access to the Internet) • Therefore, the Proxy Server is the only computer that requires a ‘real’ IP address • What class of IP address will the proxy server require? • A class C address (there is effectively one one host)
Limitations and Problems with Classful IP Addressing • Because the development of a routing protocol began over 30 years ago no-one had foreseen that IP addressing would dominate network routing the world over – but it has! • In its infancy many class A and class B IP addresses were sold without any regard for how many computers were on the networks on which they were being deployed. As a result we now have a severe shortage of classful IP addresses. • A company requiring an IP address to cover 2000 hosts, would be given a class B IP address, for it to effectively waste 63,534 possible IP addresses
Limitations and Problems with Classful IP Addressing Filling Routing Tables (with unnecessary entries) Because all of the initial batch of class B IP addresses were sold / allocated to unsuitable networks (size, number of hosts), there was a shortage of class B IP addresses. Therefore a company would have to use many class C IP addresses, and allocate each class C address to a particular subnet within the same overall network This meant that the routing tables of Internet routers were being filled unnecessarily with inappropriate entries
New Routing Protocols • But today the face of IP addressing has changed, as 2 new standards of IP addressing have emerged • CIDR (Classless Inter-domain Routing) • IP Addressing version 6 (128 bits)
What is CIDR? • CIDR is a replacement for the old process of Class A, B, and C addresses, with a generalised network ‘prefix’. • Instead of being limited to network identifiers of 8, 16 and 24 bits, CIDR currently uses prefixes anywhere between 13 and 27 • Therefore a CIDR address and prefix identifies that, within the standard 32 bit IP address, a specific number of bits are used to identify the network and the number hosts • An example of a CIDR IP address • 156.17.42.6 /20
CIDR • So what does this new IP address mean? Have a look again at the IP address156.17.42.6 /20 • The /20 indicates that within the structure of the 32 bit address we will only be using the first 20 bits to identify the network, whilst the remaining 12 bits will identify the hosts (computers?) on the network • In order to show this effectively we must convert the IP address to binary (either manually or using a Calculator)Look at the following table…
CIDR: Network ID, Host ID, Subnet Mask 00000110 00010001 00101010 10011100 10011100 00010001 00100000 00000000 12 bits 20 bits 0 17 32 156 00000000 11110000 11111111 11111111 0 255 255 240
Exercises… • Log onto the shared areaEmerson A\Unit 24\IP Addressing\Week 2\CIDR Exercisesonly attempt questions 1, 2 and 3
CIDR – How many hosts? • Once you have identified the network ID, the subnet mask and the number of bits allocated for hosts, how many hosts can be accommodated using the remaining bits? • Look at the Binary notation • 10011100.00010001.00100000.00000000 • From the binary bits, 20 are used to identify the network and 12 for the hosts. To calculate the number of hosts • (2^12) = 4096 hosts • But, you must subtract 2 hosts from this number (you can have any combination of 0’s and 1’s using the 12 bits but you cannot use all 0’s or all 1’s) • Therefore (2^12)-2 = 4094 Calculator
Exercises… • Log onto the shared areaEmerson A\Unit 24\IP AddressingWeek 2\CIDR Exercisescomplete question 4