csis 3723. we need to create some logic to the environment we want to keep like devices together ...
TRANSCRIPT
CSIS 3723
We need to create some logic to the environment
We want to keep like devices together We want to make money leasing the use of
the space Security
When designing the Internet it was decided that not all devices would need or want to be addressable from the Internet but the will still need to communicate using the network
Because of this private address space was created
These addresses are not accessible from the Internet without the network administrator doing something to give them a Internet address (NAT)
These addresses can be accessible in your intranet (corporate space)
RFC 1918 defines these◦ 10.0.0.0 - 10.255.255.255 (10.0.0.0/8)◦ 172.16.0.0 – 172.31.255.255 (172.16.0.0/12)◦ 192.168.0.0 – 192.168.255.255 (192.168.0.0/16)
These are the IP address spaces that can be used internally in an enterprise
RFC states a “link local” block◦ 169.254.0.0 – 169.254.255.255 (169.254.0.0/16)◦ To be used when a device can not get an IP address
through DHCP Also reserves lowest Class B
◦ 128.0.0.0 -128.0.255.255 (128.0.0.0/16)◦ Not able to be used under old class system but can be
assigned to someone Also defines loop back space (RFC 1700)
◦ 127.0.0.0 – 127.255.255.255 (127.0.0.0/8)◦ Used for a machine to communicate internally
Also defines multicast address space (RFC 5771)◦ 224.0.0.0 – 239.255.255.255 (224.0.0.0/4)
So you should never use these IP address spaces!
150.134.10.0/24
150.134.10.0/24
150.134.10.10
150.134.10.30 192.168.1.12
Internet
192.168.1.0/24
What are the IP addresses for the subnet 192.168.0.0/24?
192.168.0.0 through 192.168.0.255 Anything after the 24th most significant bit
can change and be in the same subnet
11000000 10101000 00000000 00000000
11000000 10101000 00000000 11111111
We use the CIDR as a binary number Every most significant bit is a one the rest
are zero So a /24 would be:
11111111 11111111 11111111 00000000
SubnetMask
255 255 255 00000000
What would subnet mask be for /20?
11111111 11111111 11110000 00000000
255 255 240 00000000
11111111128 64 32 16 8 4 2 1
128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255
192 168 0 0
11000000 10101000 00000000 10001010
/24
11111111 11111111 11111111 00000000
IP Address
SubnetMask
Logical AND
11000000 10101000 00000000 00000000
Logical AND
0 00 11 01 1
0001
Bitwise AND is used. Logical AND is done on each bit between the IP address and the subnet mask. If the result matches the network it is in the same subnet
192 168 0 138
Network
DestinationAddress
Network 11000000 10101000 00000000 00000000
192 168 0 0
10010110 10000110 00001010 00011100
/24
11111111 11111111 11111111 00000000
IP Address
SubnetMask
Logical AND
10010110 10000110 00001010 00000000
150 134 10 28
11000000 10101000 00000000 00000000
Network
DestinationAddress
Network
No match different subnet
If we look at just one octet we see a pattern
Mask Binary Ratio
0 0000 0000 1:256
128 1000 0000 2:128
192 1100 0000 4:64
224 1110 0000 8:32
240 1111 0000 16:16
248 1111 1000 32:8
252 1111 1100 64:4
254 1111 1110 128:2
255 1111 1111 256:1
192 168 10 0 /24If I start with:
This is the last octet:
128 1000 0000 2:128
0 0000 0000 1:256
If we change it to:
We would get two networks:
192 168 10 0 /25
192 168 10 128 /25
Lets look at what happens when the number change in the last octet
10000000 00000001
Mask Last Octet of IP address
00010010
01011010
10010100
As long as this bit does not become a one in the IP address it is in the first subnet
First Subnet
Second Subnet
10010100
10010100
What if an ISP owns a block of IP addresses like:
ISP's block 11001000 00010111 00010000 00000000 200.23.16.0/20
If I had 8 customers that want to buy subnets how could I change the subnet mask to get 8 subnets?
11001000 00010111 00010000 00000000 200.23.16.0/20
11001000 00010111 00010000 00000000
Each place I move I get a multiple of two
11001000 00010111 00010000 00000000
To get 8 in binary I would need 3 binary numbers
1114 2 1
4 + 2 + 1 = 7 ???
We start count from zero
ISP's block 11001000 00010111 00010000 00000000 200.23.16.0/20
Organization 0 11001000 00010111 00010000 00000000 200.23.16.0/23 Organization 1 11001000 00010111 00010010 00000000 200.23.18.0/23 Organization 2 11001000 00010111 00010100 00000000 200.23.20.0/23 ... ….. …. ….
Organization 7 11001000 00010111 00011110 00000000 200.23.30.0/23
Organization 0 11001000 00010111 00010000 00000000 200.23.16.0/24 Organization 1 11001000 00010111 00010001 00000000 200.23.17.0/24 Organization 2 11001000 00010111 00010010 00000000 200.23.18.0/24 ... ….. …. ….
Organization 16 11001000 00010111 00011111 00000000 200.23.31.0/24
What if I needed 11 subnets???
ISP's block 11001000 00010111 00010000 00000000 200.23.16.0/20
We need to create 16 subnets to get 11
What if I needed subnets that can have 56 hosts???
ISP's block 11001000 00010111 00010000 00000000 200.23.16.0/20
Where can we move from the right to get a number larger the 56 ???
11111111128 64 32 16 8 4 2 1
32 + 16 + 8 + 4 + 2 + 1 = 63 (plus one for zero) is 64
Organization 0 11001000 00010111 00010000 00000000 200.23.16.0/26 Organization 1 11001000 00010111 00010000 01000000 200.23.16.64/26 Organization 2 11001000 00010111 00010000 10000000 200.23.16.128/26 ... ….. …. ….
Organization 64 11001000 00010111 00011111 11000000 200.23.31.192/26
What if I needed subnets that can have 56 hosts???
ISP's block 11001000 00010111 00010000 00000000 200.23.16.0/20
We need to create 64 subnets each having 64 IP addresses
On the subnet 200.23.16.0/26 We only get 61 usable IP address One is used for the network
◦ 200.23.16.0 not used One must be used for the router interface
on the subnet◦ 200.23.16.1 is usual used as the router IP address
but does not have to (could be 200.23.16.62 or any other host IP address)
One must be used for the broadcast address◦ Is always the last IP address in the subnet
(200.23.16.63)