Cisco 1 Subnet Worksheet 02.08.07
Directions :
Make sure you show the default network mask, the subnet mask, and the range
of host IDs. Show the final work as xxx.xxx.xxx.xxx xxx / xx (prefix/length
notation).
Show all work show binary.
Use BLUE for SUBNET BITS and RED for HOST BITS
Restrictions : You cannot use all 1s for the host ID. (broadcast)
You cannot use all 0s for the host ID. (network id)
Prefix / Length Notation
Prefix / Length Notation indicates the subnet mask in a shorthand fashion.
For a given Class B network such as 172.16.0.0 the default subnet mask would
be 255.255.0.0. If we are going to use the 3rd octet for the subnet ID and use the
4th octet for the host ID then the subnet mask would now be 255.255.255.0.
Prefix / Length notation uses a single number to indicate the total number of bits that make up the default subnet mask and the additional bits for the subnet ID. In this case the number would be 24. IE: 172.16.0.0 / 24 which tells us that the subnet mask is 24 bits long (16 for the class B default + the additional 8 for the
subnet ID).
Problem Statement
Given the class B network of 172.16.0.0/16, subnet the network in order to create the network shown in the figure with the following host requirements.
Network ID Class B
172.16.0.0 = 10101100.00010000.00000000.00000000 /16
Net A : 19 hosts required Assign subnet ID = 45
1) How many bits are required to represent 19 hosts ?
24 = 16 so 4 wont work 25 = 32 so 5 bits will do it.
2) rewrite the IP with subnet and host bits identified :
10101100.00010000.00000000.00000000
(host bits= 5 and 16 5 = 11 bits used for subnet id)
3) assign minimum and maximum host id numbers
10101100.00010000.00000000.00000001 minimum host id
10101100.00010000.00000000.00011110 maximum host id
4) assign subnet id
10101100.00010000.00000101.10100001 subnet id = 45, host = 1
10101100.00010000.00000101.10111110 subnet id = 45, host = 30
5) rewrite ip address showing minimum and maximum host and subnet id
172.16.5.161 190 / 27 subnet mask = 255.255.255.224
Directions : Complete the problem using the following requirements:
Note : you do not have to write the steps out each time as long as all
the work from each step is shown as follows:
Net A : 19 hosts required Assign subnet ID = 45
Network : 172.16.0.0 / 16
10101100.00010000.00000000.00000000
(host bits= 5 and 16 5 = 11 bits used for subnet id)
10101100.00010000.00000000.00000001 minimum host id
10101100.00010000.00000000.00011110 maximum host id
10101100.00010000.00000101.10100001 subnet id = 45, host = 1
10101100.00010000.00000101.10111110 subnet id = 45, host = 30
172.16.5.161 190 / 27 subnet mask = 255.255.255.224
Net B : 43 hosts Assign subnet ID = 17
Network : 172.16.0.0 / 16
10101100.00010000.00000000.00000000
(host bits= 6 and 16 6 = 10 bits used for subnet id)
10101100.00010000.00000000.00000001 Minimum Host ID
10101100.00010000.00000000.00111110 Maximum Host ID
10101100.00010000.00000100.01000001 Subnet ID = 17, Host = 1
10101100.00010000.00000100.01111110 Subnet ID = 17, Host = 62
172.16.4.65 - 126 / 26 Subnet Mask = 255.255.255.192
Net C : 2 hosts Assign subnet ID = 139
Network : 172.16.0.0 / 16
10101100.00010000.00000000.00000000
(host bits= 2 and 16 2 = 14 bits used for subnet id)
10101100.00010000.00000000.00000001 Minimum Host ID
10101100.00010000.00000000.00000010 Maximum Host ID
10101100.00010000.00000010.00101101 Subnet ID = 139, Host = 1
10101100.00010000.00000010.00101110 Subnet ID = 139, Host = 2
172.16.2.45 - 46 / 30 Subnet Mask = 255.255.255.252
Net D : 151 hosts Assign subnet ID = 67
Network : 172.16.0.0 / 16
10101100.00010000.00000000.00000000
(host bits= 8 and 16 8 = 8 bits used for subnet id)
10101100.00010000.00000000.00000001 Minimum Host ID
10101100.00010000.00000000.11111110 Maximum Host ID
Net E : 445 hosts Assign subnet ID = 9
Net F : 325 hosts Assign subnet ID = 72
Reference