CCNA VLSM Subnet Question

Here’s a network layout for you. You are working for a Network consulting company. You’ve been told you when designing the network you need to:

• take the 192.168.10.0/24 network and use VLSM to break up the hosts into 3 different networks.
• make sure that when you break things up that you try and keep the networks as close to their needed sizes.
• The Default gateways for each network (A & B) will be the last valid IP Address. It will be the gig0/0 port on the router which is the port connected to the network.
• List the IP addresses and subnet masks of WAN Serial Interface ports se0/0
• ServA through ServC will Need to be assigned the first 3 valid addresses of the network.
• List the IP addresses, Subnet mask, Gateway of the first and last hosts in BOTH Network 1 AND Network 2. Network #1 Hosts start at the 10th valid address.

The first thing you need to do is look at how many IP addresses you will need in each network.

• Network 1 –> 3 server + 40 hosts = 43 IP addresses.
• Network 2 –> 10 Hosts = 10 IP addresses.
• Network 3 –> 2 Hosts = 2 IP addresses.

Now that you have the numbers you need it’s a good idea you write out these numbers:

We also know we already have 8+8+8 bits from the /24 subnet. We will just add these bits on to the /24.

Let’s look at Network 1

Find out “block size intervals”… Why is this important? It allows us to easily see where subnet starts and Ends! What is the closest number in the chart that is larger than 43. We can see it is 64! A crucial piece of knowledge is finding out the block size intervals starting at Zero. 0,64,128… etc.

Let’s first write out the subnet for the first network. Because we were told we need to use 192.168.10.0 we know we should start with this.

192.168.10.0 /26 (8+8+8+2=26)

Because the first network needs to use 64 hosts we know that the mask will be 255.255.255.192. (128+64). Since we wrote the block sizes out we can easily see the first network is from 0-63, the next (if we required it) is 64-127 etc.

We know the last IP address in the subnet is the Broadcast Address. This means 192.168.10.63 is the broadcast address.

We know the first address 192.168.10.0 is not usable by a host so we know that the usable range of IP addresses is 192.168.10.1 to 192.168.10.62.

Now that we’ve got everything we need let’s write it out in a nice condenses format:

• Subnet – 192.168.10.0/26
• Broadcast – 192.168.10.63
• Range – 192.168.10.1->62

Yay! The first network is done!

Because we’ve already carved out 0->63 we know the next VSLM network needs to start at 64. If we started at 63 or 62 the subnets would overlap and things just wouldn’t work.

Let’s do the same thing for Network 2!

What is the closest number that is larger that 10? We can see it is 16! Let’s find the block sizes starting at the first free network number: 64, 80, 96 …

Let’s first write out the subnet: 192.168.10.64/ 28 (8+8+8+4=28)

Because we of the /28 CIDR notation we know that the subnet mask will be 255.255.255.240 (We got the 240 from 128+64+32+16).

We wrote out the block earlier so we know 64-79 will be the block range for network #2.

The last address is the Broadcast address: 192.168.10.79

We know the usable range of the network will be 192.168.10.65->78

Similar to Network #1 let’s write out everything for Network #2

• Subnet – 192.168.10.64/ 28
• Broadcast – 192.168.10.79
• Range 192.168.10.65->78

Network Number 3

We only need two IP addresses for the WAN link. You should have this burned into your mind that point-to-point networks use a /30 mask.

Because of #2 we know that this net network starts at 80. Remember we don’t want to over lap with a prior network.

• So the Subnet will be 192.168.10.80/ 30.
• The subnet mask will be 255.255.255.252 (We got the 252 from –> 128+64+32+16+8+4)
• Let’s write out the block of 4. 80, 84, 88…81
• We know that the 3rd network will be 80-83.
• 83 is going to be the broadcast address.

From the previous lines we know that our valid IP addresses are 81->82. Look… It’s two host addresses just like we wanted!

Let’s write everything down nicely.

• Subnet –192.168.10.80 /30
• Broadcast – 192.168.10.83
• Range – 192.168.10.81->82

We know have everything we need to answer all the questions!

List the SE interfaces of the Routers:

• R1 – 192.168.10.81, 255.255.255.252
• R2 – 192.168.10.82, 255.255.255.252

ServA through ServC will Need to be assigned the first 3 valid addresses of the network.

• ServA – 192.168.10.1, 255.255.255.192, Default Gateway 192.168.10.62
• ServB – 192.168.10.2, 255.255.255.192, Default Gateway 192.168.10.62
• ServC – 192.168.10.3, 255.255.255.192, Default Gateway 192.168.10.62

What is the first and last IP address of HOSTS on the network. Remember Network 1 starts at the 10th valid IP address.

Network 1

• First Host – 192.168.10.10, 255.255.255.192, Default Gateway 192.168.10.62
• Last Host – 192.168.10.49, 255.255.255.192, Default Gateway 192.168.10.62

Network 2

• First Host – 192.168.10.65, 255.255.255.240, Default Gateway 192.168.10.62
• Last Host – 192.168.10.74, 255.255.255.240, Default Gateway 192.168.10.62