30 Mar 2013

kenglong@gmail.com

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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.

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 2^{6}.

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`

(We calculate this one only to make it easy to get the broadcast,

192.168.1.64

last host and first host as well as the network number to use for the next subnet calculation.)

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 2^{5}.

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`

(We calculate this one only to make it easy to get the broadcast,

192.168.1.96

last host and first host as well as the network number to use for the next subnet calculation.)

The next one requires 10 hosts. Since 10 is not an even power of 2, we use 16 which is 2^{4}.

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`

(We calculate this one only to make it easy to get the broadcast,

192.168.1.112

last host and first host as well as the network number to use for the next subnet calculation.)

Our next subnet requires 7 hosts. Since 7 is not an even power of 2, we will use 8 which is 2^{3}.

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`

(We calculate this one only to make it easy to get the broadcast,

192.168.1.120

last host and first host as well as the network number to use for the next subnet calculation.)

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`

(We calculate this one only to make it easy to get the broadcast,

192.168.1.124

192.168.1.128

192.168.1.132

last host and first host.)

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.