Question

Hi, I need help with the following practice networking exam question

Acme is launching 18 new branches in cities where they currently have no existing branches, and consequently need a new subnetting plan to support these branches. The Acme CIO has asked that each subnet should host at least 550 devices. If a choice needs to be made between the number of subnets and the size of each subnet, then the number of subnets should be maximized as much as possible in order to allow growth in the total number of the branches.

Acme has obtained an IPv4 Network Adress of 123.45.0.0 /17

This address needs to be subnetted further.

The companies subnet plan should include address allocations for the following:

2 router interfaces in each subnet

18 managed devices in each subnet (e.g. servers or switches)

A DHCP pool for the remaining usable addresses.

Create an addressing plan for subnet 15 outlining the subnet ID, broadcast address and relevant addresses as outlined in the scenario.

Answer 1

As the priority is to keep the size of the subnet minimum and maximize the number of subnets, then first we need to find out the minimum number of IP addresses needed in the subnet.

The minimum number of IP address needed in the subnet = 550 (minimum number of devices) + 2 (subnet address and subnet broadcast address) + 2 (for the router interfaces) + 18 (for the managed devices) + 1 ( for the DHCP Pool) = 573 IP address in total.

So, 2⁹ = 512 < 573, hence the appropriate size for the subnet would be 2¹⁰ = 1024.

Hence, the possible number of subnets that we can have = 2⁵ = 32 subnets in total.

So, 123.45.0.0/17 can be written as :

01111011.00101101.00000000.00000000/17

In this, bits 1-17 are fixed network bits, next five bits from 18-22 will be our subnet bits, and remaining 10 bits from 23-32 are the host bits.

The subnet mask would be, 11111111.11111111.11111100.00000000 or 255.255.253.0

Taking IP address of the first subnet to be 01111011.00101101.00000000.00000000/22 or 123.45.0.0/22 and the IP address of the second subnet to be 01111011.00101101.00000100.00000000/22 or 123.45.4.0/22,

IP address of the 15th subnet (subnet ID) would be 01111011.00101101.00111100.00000000/22 or 123.45.60.0/22.

Broadcast Address of the subnet would be 01111011.00101101.00111111.11111111/22 or 123.45.63.255/22.