Subnetting Part-I

In network IP address (binary), we use '1' as network bit and '0' as host bit. Subnetting is breaking down the number of hosts to increase the number of networks.

Always remember the table:

| 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |

Sub-netting based on networks:

In the above network diagram we need 5 networks from a single network address : 216.21.5.0 / 24.
To subnet the network address into 5 different networks ( a. LAN1, b. LAN2, c. LAN3, d. WAN1, and e. WAN2);

i. We determine the number of networks and convert to binary.
    Here, we have to subnet into 5 networks, converting it into binary from the above table.
     we get. 5 = 00000101
    Here, the underlined 3 - bits are needed to make 5.

ii. Next, we reserve bits in subnet mask and find the incremental.
    In our case, our subnet mask is = 255.255.255.0
                                      In binary, = 11111111.1111111.11111111.00000000
    Now, we reserve the number of bits of subnet mask, equivalent to the bits to make of "5" coz 5       networks. We save the network bits.
                                    therefore, = 11111111.11111111.11111111.11100000
      Now, our new subnet mask will be = 255.255.255.224  (: 128 + 64 + 32)
   
   To find the increment ;  we take the lowest network bit from the customized subnet mask
   i.e. 11100000 (crossed one). So, the increment = 32 .

iii. Finally, we use increment to find the network range.
      Network range is found, starting from '0' subnet.
      so, first IP range will be = 216.21.5.0  - 216.21.5.31
                                                216.21.5.32 - 216.21.5.63
                                                216.21.5.64 - 216.21.5.95
                                              ------------------------------------
                                               --------------------------------- so on.

  
Sub-netting based on hosts:
In the above network diagram let's say, we need 30 hosts in each subnet. Now we have a network address : 216.21.5.0 / 24.
We have to divide the given network address into subnets with 30 hosts in each.
The Process is a like.

i. First we determine the number of hosts and convert it into binary.
   Here, we need 30 hosts per subnet. So, 30 = 00011110
   Here, the underlined 5 - bits are required to be 30.

ii. Next, we reserve bits in subnet mask and find the incremental.

    In our case, our subnet mask is = 255.255.255.0

                                      In binary, = 11111111.1111111.11111111.00000000

Now, we reserve the number of bits of subnet mask. We, here save the host. Because here, we focus on the number of hosts not network.

                                    therefore, = 11111111.11111111.11111111.11100000

      Now, our new subnet mask will be = 255.255.255.224  (: 128 + 64 + 32)  or in bit notation = /27

   To find the increment ;  we take the lowest network bit from the customized subnet mask
   i.e. 11100000 (crossed one). So, the increment = 32 .

iii. Finally, we use increment to find the network range.
      Network range is found, starting from '0' subnet.
      so, first IP range will be = 216.21.5.0  - 216.21.5.31
                                                216.21.5.32 - 216.21.5.63
                                                216.21.5.64 - 216.21.5.95
                                              ------------------------------------
                                               --------------------------------- so on.

I hope it is informative to you, and thank you !!

Default Route

It is configured for unknown destination from the local LAN network. If the network has to access to the internet, it is configured. It is configured in the edge router or border router from where the internet is accessed.

syntax:
HYD(config)# ip route 0.0.0.0 0.0.0.0 <destination ip address or own's exit interface> <enter>

Practice:
i. Initial Configuration.

ii. Do the following configuration on packet tracer.
iii. Test: Ping from PC1 or PC0 to Server1 or Server2 and vice Varese.
iv. You can download the completed file here.