# 10.7.1 Skills Integration Challenge – Network Planning and Interface Configuration

30 Mar 2013
kenglong@gmail.com

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Binary reference charts

### The Requirements

This exercise requires us to create an efficient addressing plan using variable length subnetting given the following:

• `192.168.1.0 /24`
• One subnet with 60 hosts
• One subnet with 25 hosts
• One subnet with 10 hosts
• One subnet with 7 hosts
• Three subnets with 2 hosts each for router interconnections.

### First Subnet, 60 Hosts

The way to approach this is to start with the subnet that requires the most hosts, 60.
Since 60 is not an even power of 2, use 64 which is 26.
32 – 6 = 26 network + subnet bits, or /26.

`255.255.255.11000000`

Convert the partial binary number above to decimal. to give us a mask of `255.255.255.192`.
Notice the first "one" from the right. It's in the "64" placeholder making our magic number 64.
So our first two subnets will be:

```192.168.1.0 /26 192.168.1.64```     (We calculate this one only to make it easy to get the broadcast,
last host and first host as well as the network number to use for the next subnet calculation.)

### Second Subnet, 25 Hosts

Now let’s work on the next subnet which requires 25 hosts. We start with the network `192.168.1.64`.
Since 25 is not an even power of two, we’ll use 32 which is 25.
32 – 5 = 27 so our cidr is /27.

`255.255.255.11100000`

That gives us a mask of `255.255.255.224` with a magic number of 32. Our next two subnets will be:

```192.168.1.64  /27 192.168.1.96```     (We calculate this one only to make it easy to get the broadcast,
last host and first host as well as the network number to use for the next subnet calculation.)

### Third Subnet, 10 Hosts

The next one requires 10 hosts. Since 10 is not an even power of 2, we use 16 which is 24.
32 – 4 = 28 so our cidr is /28.

`255.255.255.11110000`

That gives us a mask of `255.255.255.240` with a magic number of 16. Our next two subnets will be:

```192.168.1.96  /28 192.168.1.112```   (We calculate this one only to make it easy to get the broadcast,
last host and first host as well as the network number to use for the next subnet calculation.)

### Fourth Subnet, 7 Hosts

Our next subnet requires 7 hosts. Since 7 is not an even power of 2, we will use 8 which is 23.
32 – 3 = 29 so our cidr is /29.

`255.255.255.11111000`

That gives us a mask of `255.255.255.248` with a magic number of 8. Our next two subnets will be:

```192.168.1.112  /29 192.168.1.120```   (We calculate this one only to make it easy to get the broadcast,
last host and first host as well as the network number to use for the next subnet calculation.)

### Last Three Subnets, 2 Hosts Each

Lastly, we need three subnets with two hosts each for our router interconnections. We start with the network `192.168.1.120`.
Two is an even power of two so we can use it directly.
32 – 2 = 30 so our cidr is /30.

`255.255.255.11111100`

That gives us a mask `255.255.255.252` with a magic number of 4. Our subnets will be:

```192.168.1.120  /30 192.168.1.124 192.168.1.128 192.168.1.132```   (We calculate this one only to make it easy to get the broadcast,
last host and first host.)

### Finished!

That completes our subnet calculations for Exercise 10.7.1. We can easily find the first host, last host and
broadcast addresses for all of our subnets using the above framework. As you can see, we have lots of
addressing space left over for future use.  The rest of the exercise is simply configuring the routers and hosts
then testing the network by pinging.