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IP Addressing and Subnetting

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IP Addressing and Subnetting. Tony Madden. John Paul McCann. NEI. 2. IP ADDRESSING ... Government broken addressing system down into classes A, B, C, D, E ... – PowerPoint PPT presentation

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Title: IP Addressing and Subnetting


1
IP Addressing and Subnetting
  • Tony Madden
  • John Paul McCann
  • NEI

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 - 191 Binary Chart - all binary no.s
    beginning with 10
  • C 192 - 223 Binary 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
  • Decimal Binary
  • 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 11000000.00000101.00000101.000000
    00)
  • 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 C IP 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 its own IP address to its own
    subnet mask using the ANDing process
  • Host X IP 200.1.1.5 11001000.00000001.00000001.00
    000101
  • Subnet Mask 255.255.255.0 11111111.11111111.111111
    11.00000000
  • ANDing result 11001000.00000001.00000001.000000
    00
  • 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 11001000.00000001.00000010.00
    001000
  • Subnet Mask 255.255.255.0 11111111.11111111.111111
    11.00000000
  • ANDing result 11001000.00000001.00000010.000000
    00
  • 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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