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Chapter 10 IP Addressing and Subnetting

Chapter 10 IP Addressing and Subnetting. Paul Flynn. IP ADDRESSING. 32 Bit Address 2 32 = 4.2 billion possible addresses Decimal 192.5.5.0 Network.Network.Network.Host (N.N.N.H) (Class C Address) Binary 11000000.00000101.00000101.00000000 Hex C0

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Chapter 10 IP Addressing and Subnetting

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  1. Chapter 10IP Addressing and Subnetting Paul Flynn

  2. IP ADDRESSING 32 Bit Address 232 = 4.2 billion possible addresses • Decimal 192.5.5.0 Network.Network.Network.Host (N.N.N.H) (Class C Address) • Binary 11000000.00000101.00000101.00000000 • Hex C0 • Government broken addressing system down into classes A, B, C, D, E • A – very large organisations (1 - 127, 00000001 - 01111111) • Network.Host.Host.Host (N.H.H.H) • The first bit of the Network address is reserved for ease of recognition i.e. 0 by routers • 27 Networks (128) with 224 Nodes (16777214)

  3. IP ADDRESSING • Government broken addressing system down into classes • B – large organisations (128 - 191, 10000000 - 10111111) • Network.Network.Host.Host (N.N.H.H) • The two bits of the Network address are reserved for ease of recognition i.e. 10 by routers • 214 Networks (16384) with 216 Nodes (65534) • C – small organisations (192 - 223, 11000000 - 110111111) • Network.Network.Network.Host (N.N.N.H) • The three bits of the Network address are reserved for ease of recognition i.e. 110 by routers • 221 Networks (2097150) with 28 Nodes (256)

  4. IP ADDRESSING NOTE: • Cisco state that you cannot use the first and last of any of the network or node ranges as they are reserved for network or node addresses and broadcasts • Therefore: • A = 27 – 2 = 126 Networks 224 – 2 = 16777212 Nodes • B = 214 – 2 = 16382 Networks 216 – 2 = 65534 Nodes • C = 221 – 2 = 2097148 Networks 28 – 2 = 254 Nodes

  5. IP ADDRESSING Internet addresses • Assigned to us by the Network Information Centre upon application • A = 1 - 127* Binary Chart - all binary no.s beginning with 0 • B = 128 - 191Binary Chart - all binary no.s beginning with 10 • C = 192 - 223Binary Chart - all binary no.s beginning with 110 • Ref: Binary – Decimal Conversion Chart Appendix D P528 - 530

  6. IP ADDRESSING AND SUBNETTING Class C address N.N.N.H DecimalBinary • IP 192. 5. 5. 0 11000000.00000101.00000101.00000000 • Subnet Mask 255.255.255. 0 11111111.11111111.11111111.00000000 • Host range 192.5.5.0 – 192.5.5.255 • Cannot use 0 (0) Network address or 255 (11111111) Broadcast addresses • Usable Hosts 192.5.5.1 – 192.5.5.254

  7. IP ADDRESSING AND SUBNETTING Subnetting a Class C address N.N.N.H • Subnet 192.5.5.0 into 2 subnets • Borrow 2 bits from the Host part of the IP address (N.N.N.H 11000000.00000101.00000101.00000000) • This now gives you a network address that is now 26 bits long and leaves 6 bits to assign to node addresses. • You can borrow a minimum of 2 bits and a maximum of 6 bits from a class CIP address 192.5.5.0

  8. IP ADDRESSING AND SUBNETTING Borrow 2 bits (subnetting into 2 subnetworks) • (remember that you cannot use the first or last as they refer to the network address and the broadcast address) 22 – 2 = 2 • There are in fact 4 subnet address, but only 2 of them are usable • BINARY to DECIMAL • 00 cannot use • 01 64 • 10 128 • 11 cannot use • New Network address are 192.5.5.64 and 192.5.5.128 • The Subnet Mask also changes • To do this you must add the binary value of two bits borrowed (11000000 = 192) • The new Subnet Mask is 255.255.255.192 • IP Address Subnet Mask Range of Hosts No. of Hosts • 192.5.5.64 255.255.255.192 65 – 127 62 • 192.5.5.128 255.255.255.192 129 – 191 62

  9. IP ADDRESSING AND SUBNETTING Borrow 3 bits (subnetting into 4 subnetworks) • 23 – 2 = 6 • There are in fact 8 subnet address, but only 6 of them are usable • BINARY to DECIMAL • 000 Cannot use • 001 32 • 010 64 • 011 96 • 100 128 • 101 160 • 110 192 • 111 Cannot use • New address 192.5.5.32, 192.5.5.64, 192.5.5.96, 192.5.5.128, 192.5.5.160, 192.5.5192. • The Subnet Mask also changes • To do this you must add the binary value of two bits borrowed (11100000 = 224) • The new Subnet Mask is 255.255.255.224

  10. IP ADDRESSING AND SUBNETTING • IP Address Subnet Mask Range of Hosts No. of Hosts • 192.5.5.32 255.255.255.224 33 - 63 30 • 192.5.5.64 255.255.255. 224 65 – 95 30 • 192.5.5.96 255.255.255. 224 97 - 127 30 • 192.5.5.128 255.255.255. 224 129 – 159 30 • 192.5.5.160 255.255.255. 224 161 - 191 30 • 192.5.5.192 255.255.255. 224 193 – 223 30

  11. SUBNET MASKS Why do you need a subnet mask? Hosts and routers use the ANDing process to determine if a destination host is on the same network or not. The ANDing operation happens any time a host wants to send a packet to another host on an IP network. The result of the 1st AND is to identify the network where the source host resides. It will then compare the destination IP address to its own subnet mask (2nd AND) to determine the network address of the destination host.

  12. SUBNET MASKS Host x (Source) on network 200.1.1.0 (Class C Network) has an IP address of 200.1.1.5 and wants to send a packet to Host Z (Destination) on network 200.1.2.0 and has an IP address of 200.1.2.8. All hosts on each network are connected to hubs or switches and then to a router (Remember that with a class C network, ARIN assigns the first 3 octets (24 bits) as the network address, so these are two different networks. • Source Net 200.1.1.0 Dest. Net 200.1.2.0 • Subnet Mask 255.255.255.0 Subnet Mask 255.255.255.0 • Host IP 200.1.1.5 Host 200.1.2.8

  13. SUBNET MASKS • Host X compares it’s own IP address to its own subnet mask using the ANDing process • Host X IP 200.1.1.5 11001000.00000001.00000001.00000101 • Subnet Mask 255.255.255.0 11111111.11111111.11111111.00000000 • ANDing result 11001000.00000001.00000001.00000000 • The result of the 3rd step gives the network address 200.1.1.0 • Next Host X compares the IP address of the Host Z dest. to its own subnet mask using the ANDing process • Host Z IP 200.1.2.8 11001000.00000001.00000010.00001000 • Subnet Mask 255.255.255.0 11111111.11111111.11111111.00000000 • ANDing result 11001000.00000001.00000010.00000000 • The result of the 3rd step gives the network address 200.1.2.0

  14. SUBNET MASKS Host X now knows that host Z is not in its Local Area Network (LAN) and it must send the packet to its “Default Gateway” which is the IP address of the router interface of 200.1.1.1 on network 200.1.1.0. The router will then repeat the ANDing process to determine which router interface to send the packet out. Source Net 200.1.1.0 Dest Net 200.1.2.0 Subnet Mask 255.255.255.0 Subnet Mask 255.255.255.0 Host X Router Hub Host Z Hub Host IP 200.1.2.8 Host IP 200.1.1.5 Router Interface IP 200.1.1.1 Router Interface IP 200.1.2.1

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