How to construct VSLM subnet diagram

Q : Use variable-length subnet mask (VLSM) to support more efficient use of the assigned IP addresses and to reduce the amount of routing information at the top level.

You have been employed by America International Corporation as a Network Engineer to set up the corporate network with the following.

Network Address : 125.25.0.0/16 has been allocated.

B,C,D have a WAN connection to A.

a) B requires 7000 hosts.

b) A requires 4000 hosts.

c) C and D require 1800 hosts/each.



                          Router A
                    
                  S1        S2      S3
                 ↙          ↓        ↘
                S0          S0        S0

               Router B   Router C   Router D
                  │          │          │
                 PC1        PC2        PC3

You must design the IP addressing scheme using VLSM. You are allowed to use Subnet 0 for the design. You must be able to show your workings and allocate IP addresses to all the interfaces and PCs. First IP subset address should be given to default gateway (i.e. E0). You are allow to use calculator.

A) Show all workings (Tree diagram)

B) Allocate IP address to all the interfaces (S0,S1,S2,S3,E0) and PCs.


A) 125.25.0.0/16

   B = 7000 hosts

   2^13 - 2 = 8192-2
            = 8190

Since 11111111.11111111.11100000.00000000 where 13 '0' bits are for host

Subnet mask = 255.255.224.0

   A = 4000 hosts

   2^12 - 2 = 4096 - 2
            = 4094

Subnet mask = 255.255.240.0

  C and D = 1800 hosts

  2^11 -2 = 2048 -2
          = 2046

Subnet mask = 255.255.248.0


125.25.0.0 /16

Total bit for IP is 32 bits

32 - 13 ('0' bits for host) = /19 ('1' bits for network)

Magic Number = 256 - 224 (from subnet mask)
             = 32

Therefore 125.25.32.0/19

For b) 32 - 12 ('0' bits for host) = /20 ('1' bits for network)

Magic Number = 256 - 240
             = 16

Therefore 125.25.(32 + 16 =)48.0/20

For C) 32 - 11 ('0' bits for host) = /21 ('1' bits for network)

Magic Number = 256 - 248
             = 8

Therefore 125.25.(48 + 8 =)56.0/21

19/ - 16/ = 3

2^3 - 2 = 8 - 2
        = 6 branch for VSLM tree

20/ - 19/ = 1

2^1 - 2 = 2 - 2
        = 0 branch

21/ - 20/ = 1

2^1 - 2 = 0 branch


                      125.25.0.0/16

            ↙          ↙           ↓                ↘
B =
125.25.0.0/19  125.25.32.0/19 125.25.64.0/19  125.25.96.0/19
               ↙           ↓       ↓      
                             A to B = 125.25.64.0/30
             ↙             ↓ A to C = 125.25.64.5/30
                             A to D = 125.25.64.9/30
          ↙                ↓
     A = 125.25.32.0/20   
                           ↓
                    125.25.48.0/20   
                     ↙     ↓
     C = 125.25.48.0/21   D = 125.25.56.0/21     


B) Network Address :

A = 125.25.32.0/20

B = 125.25.0.0/19

C = 125.25.48.0/21

D = 125.25.56.0/21

A to B = 125.25.64.0/30

A to C = 125.25.64.5/30

A to D = 125.25.64.9/30


Subnet Address : 

A = 125.25.240.0/20

B = 125.25.224.0/19

C = 125.25.248.0/21

D = 125.25.248.0/21

A to B = 125.25.255.252/30

A to C = 125.25.255.252/30

A to D = 125.25.255.252/30


Broadcast Address :

A = 125.25.47.255

B = 125.25.31.255

C = 125.25.55.255

D = 125.25.63.255

A to B = 125.25.64.3

A to C = 125.25.64.7

A to D = 125.25.64.11


Usable IP Address :

A = 125.25.32.1

B = 125.25.0.1

C = 125.25.48.1

D = 125.25.56.1

A to B = 125.25.64.1

A to C = 125.25.64.5

A to D = 125.25.64.9


Default Gateway :

A = 125.25.32.1 to 125.25.47.254

B = 125.25.0.1 to 125.25.31.254

C = 125.25.48.1 to 125.25.55.254

A to B = 125.25.64.1 to 125.25.64.2

A to C = 125.25.64.5 to 125.25.64.6

A to D = 125.25.64.9 to 125.25.64.10

Read SICP.

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