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 won’t 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