## Answers: Subnet Design Exercise 3

Today’s post shows the solution to subnet design exercise 3, specifically the IP subnets. This post isn’t all that meaningful without reading the other one first – after that, and after you take a crack at creating the design yourself, dive in! Here are some useful links:

First, let’s look at the design of the subnet mask.

The figure shows a need for 6 total subnets, and all subnets will come from class B network 172.24.0.0. To support 6 subnets, you need at least 3 subnet bits, because you need 2^S to be at least as big as the number of required subnets. So the minimum workable S (number of subnet bits), per the various requirements, is S=3.

The requirements also tell you to use a single mask throughout each classful network. In other words, avoid Variable Length Subnet Masks (VLSM). A lot of people miss this concept, so it’s worth a few words.  VLSM means: more than 1 mask is used inside one classful network. That is, more than 1 mask is used inside a single class A, B, or C network. In this case, the requirements tell us to use a single mask throughout each classful network, so VLSM is not used. From a design perspective, that means we need to pick one subnet mask to use in all subnets.

Next, look at the requirements for the number of subnets. Figure 1 repeats the requirements for the number of hosts per each subnet.

#### Figure 1: Network Diagram and Requirements

Because of the requirement to use one mask, and one mask only, you can ignore all the subnets except the subnet with the largest number of hosts. In this case, that subnet, R3’s LAN subnet, has a need for 90 hosts. Analyzing the various requirements:

• The largest subnet requires 90 hosts today
• The requirements said to plan for 20% growth in hosts
• We should plan for 90 + 18 = 108 hosts in the largest subnet
• Find the smallest value H for which 2^H – 2 is at least 108; in this case, that value, H=7, is the minimum number of host bits

To wrap up the mask design, note that the design uses one class B network. As a class B network, the network always has 16 network bits, by definition. So a mask that meets all requirements will have:

• Network bits: 16
• Subnet bits: at least 3
• Host bits: at least 7

Finally, the requirements also told us to use the fewest number of host bits. So, the mask has exactly 7 host bits.

How to write that mask? In prefix notation, it’s /25, calculated from 32 (the number of bits in the mask) minus H (the number of host bits). To find the dotted decimal notation (DDN) mask, just covert /25 to DDN format, or 255.255.255.128.

## Find All Subnets

Once you know the mask, the next step in the design process is to find all the subnets that exist. You can then compare the list to the requirements, which tell us to use the numerically low subnet IDs for the WAN subnets, and the numerically high subnets for the LAN links. I’ll leave the how as an exercise, but here’s the list, with low and high ends. Note that the pattern is to use a multiple of 128 in the 4th octet, and add by 1 in the 3rd octet.

172.24.0.0

172.24.0.128

172.24.1.0

172.24.1.128

.

.

.

172.24.254.0

172.24.254.128

172.24.255.0

172.24.255.128

## Assign Low Subnets to WAN, High to LAN

The exercise requirements state that the WAN subnets should use the numerically lowest subnet IDs from network 172.24.0.0, starting with the R1-R2 link, then the R2-R3 link, and then the R1-R3 link. Without any reason to avoid it, the zero subnet (172.24.0.0) should be used. So, just pick the first three subnets from the list of subnet IDs.

Similarly, the requirements state that the three numerically highest subnets should be the R1 LAN, R2 LAN, and R3 LAN, in that order. Just pick those from the list from the list.

The final choices for the subnets then are:

#### Figure 2: Chosen Subnet IDs

• R1-R2 WAN: 172.24.0.0/25
• R2-R3 WAN: 172.24.0.128/25
• R1-R3 WAN: 172.24.1.0/25
• R1 LAN:         172.24.255.128/25
• R2 LAN:         172.24.255.0/25
• R3 LAN:         172.24.254.128/25

The last big task is to create a configuration for each router. Per the requirements, each router should use the last IP address in each subnet. When two routers exist (the WAN subnets), the higher router number gets the higher router IP address. Examples 1, 2, and 3 show the resulting configuration on routers R1, R2, and R3.

#### Example 1: R1 Configuration

```interface F0/0
!
interface S0/0/0
description R2 uses the very highest address; R1 uses next lower
!
interface S0/0/1
description R3 uses the very highest address; R1 uses next lower```

#### Example 2: R2 Configuration

```interface F0/0
!
interface S0/0/0
description R3 uses the highest address; R2 uses the 2nd highest
!
interface S0/0/1
description R1 uses the 2nd highest address; R2 uses the very highest```

#### Example 3: R3 Configuration

```interface F0/0
!
interface s0/0/0
description R1 uses the 2nd highest address; R3 uses the very highest
!
interface S0/0/1
description R2 uses the 2nd highest address; R3 uses the very highest```
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wow that subnet design example has my head spinning, i will have to take some time out to analyze that, that is a trip, and if there are questions like that on the ICND1, i might be in trouble…that design seems a little off the wall.

Well, they’re not going to ask you anything that long on a single question. But you’ll need to know how to do what’s here so you can answer more specific questions. If it seems a little off the wall right now, I’d say make sure you’re good with the fundamentals before looking at this problem. IE, you should be able to choose/design the right mask based on a set of requirements; find all subnets of a network given one mask; find all addresses in a subnet given the subnet ID and mask; find the first/last IP address in a subnet, given the subnet ID and mask. This exercise just puts all those concepts together into a big picture, but if you’re still learning the pieces, the big picture can be a little scary.
Wendell

I think i see it now, see if i got this right the third octet counts up from 0 to 255
and the fourth octet alternates between 0 and 128, is that right?
that is a lot of subnets, what each with 126 hosts?mmmm
think outside the box man! wow!

Hi! Very good your example. Though I have a problem..I have a same net to design given IPs 146.124.0.0/16 and 146.124.10.0/24. My given requirements are : at least 60 more subnets and maximum 1000 hosts per subnet. I gave WAN1 : 146.124.10.0/30, WAN2 : 146.124.10.4/30 and WAN3 : 146.124.10.8/30 and sub1 : 146.124.0.0/22, sub2: 146.124.4.0 and sub3 : 146.124.8.0. When I tried to configure RIP protocol, it didn’t work (i use routerSim). Can you help me? What am I doing wrong?

It could be that you forgot to enable RIP version 2. IOS RIP sends RIP version 1 updates by default, so if the simulator you’re using does the same, it won’t support VLSM (because RIP-1 doesn’t). That may be all you need to do. Otherwise, I’d have to point you to the support channel for the product you’re using. Hope that helps..
Wendell

RIPv1 does not support VLSM. IF you want to use RIPv1 you would have to change your address assignment schemes.Another word you would have to use the same mask for all subnets. It would be easier if you change it to RIPv2.

I am confused. The subnet is listed as /25 to or 255.255.255.128.

However when coming up with the subnets, since the first three octects are 255. Those are network bits and do not change and only the 25th bit changes. so in my thinking I get
172.24.0.0 to 172.24.0.127
and
172.24.0.128 to 172.24.0.255

there is no changes to the 3rd octet since its set at 255. I am not sure where the 172.24.1’s are coming from. would the subnet mask for that be 255.255.254.0

Debbie

Hi Debbie,

First, on terms: you wrote “the subnet is listed as /25 or 255.255.255.128”. That’s actually the “subnet mask”, aka “mask”. The term “subnet” is typically short for “subnet number”, which is a different number. Sorry, not trying to be difficult, just making sure we’re on the same page.

When discovering the subnet numbers inside a network, in which all subnets use the same subnet mask (/25 in this case), the subnet part is the only part that differs when comparing each subnet number. So, the subnets of class B network 172.24.0.0 will always have the same first two octets (because it is a class B network), and differ in the next 9 bits (the subnet bits). The last 7 bits of every subnet number will be binary 0. Thinking in decimal, that means you could by the magic number in the 4th octet (128), and by 1’s in the other other that comprises the subnet part of the subnet numbers (the entire 3rd octet).

That either clicked, or it didn’t, which is generally true about most math, right? If it didn’t, feel free to post again.
Wendell

Thank you, Wendell for explaining. I had to read it a few times to get what you were saying. I am/was using subnet and subnet mask as the same thing.

one more question if I may? When the class B address is subnetted, these subnets are no longer considered classsful, correct?

Debbie

Hi Debbie,
Well, “classless” and “classful” is a big can of worms. There is no terminology police that require these words be used in a certain way, different people use them in different ways, and it’s frankly a mess. I’ve been trying to informally push for one view of them, just by being very explicit with them in my books. However, many people start CCENT/CCNA study with some prior opinions about those terms, and it’s just hard to make sense of them.

For real life, I’d say to not use them unless you have to. And you don’t have to use them for addressing and sub netting. You do for routing and for routing protocols.

As for your question “the subnets are not longer considered classful”… that’s a turn of phrase I would never use. (just an example of how to use these maddenly squishy terms.) The way I use the words, a subnet is neither classful nor classless. However, you can think about IP addressing, and IP sub netting, using classless or classful rules. It’s an issue of how you think.

All that to say: if you think about an IP address, or subnet ID, and see the network/subnet/host parts, you are thinking about class rules, and you are thinking class fully. If you think about that same address/subnet and see a prefix/host (or same structure, but call the left side “subnet”), basically ignoring the network part, you are thinking classlessly. But the address or subnet ID doesn’t change – only how you’re thinking about it changes.

Hope this helps! Do check out the section about those words classless/classful in the book (can’t check it from where I am right now).
Later,
Wendell

Thanks for this, this is a great help! I can finally wrap my head around this!

Ahh, i solved for mask/25 in the beginning of my answer. but then i went along for the rest of the question solved for /26 (Careless mistake), luckily those came out as they should be….for /26 but not 25. well it was good practice. Thanks for the question!

More great practice and I got it all right 1st time. Only thing which I need to brush up on is working out %s without using a calculator.

Hello Wendell,
I was able to workout the design and calculation of the Network. However, I found the wording for the exercise to be a bit misleading. For instance, the requirements state that we are to use one mask per “Classful Network”. The way in which it was worded, led me to believe that it was also requesting that I use separate mask for the subnets…as if I were being prompted to provide a mask for the LANS, and another for the WANS. Couldn’t there be a more straight-forward approach to better test our abilities for network-subnetting and calculation, as opposed to our ability to decipher exercise specfication criteria? In other words, wouldn’t it be easier to request that we “create one MASK for all networks in the exercise” in plain english?

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