Answer: Subnet Design Exercise 2

certskills
By certskills July 28, 2011 07:00

Today’s post shows the solution to subnet design exercise 2, 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!

The figure shows three WAN subnets, which should be allocated from network 172.30.0.0, and three LAN subnets, which should be allocated from network 172.20.0.0. You should think about these networks separately.

Focusing on the LAN subnets, one requirement stated that you should use a single mask throughout the network, so you can ignore the number of hosts per subnet for the smaller requirements, and just pay attention to the largest requirement: 128 hosts/subnet for R3’s LAN subnet.

To support 128 hosts, you need H host bits such that 2^H – 2 => 128. Some quick math tells us that H=7 is too few (2^7 – 2 = 126), but H=8 is enough, because 2^8 – 2 = 254.

The third requirement also told us to use the fewest host bits, so while H=9, H=10, and so on would work, H=8 is the lowest value that meets the requirements. The mask that has 8 host bits is /24, because /24 by definition means 24 prefix bits, leaving 8 more bits to round out the 32 bits in an IPv4 address.

Following the same logic for the mask in network 172.30.0.0 for the WAN subnets, all three WAN subnets require only 2 hosts. To support 2 hosts, you need H host bits such that 2^H – 2 => 2. Some quick math tells us that H=2 is exactly the right answer (2^2 – 2 = 2), with the prefix mask being /30.

The next step is to list all the subnet IDs, or at least enough to see the pattern. I leave the how/why as an exercise, but here’s a partial list for the LAN subnets in network 172.20.0.0:

172.20.0.0

172.20.1.0

172.20.2.0

172.20.3.0

.

.

.

172.20.253.0

172.20.254.0

172.20.255.0

Similarly, for network 172.30.0.0, for the /30 WAN subnets, you should at least list enough of the subnet IDs to see a pattern.  I leave the how/why as an exercise (you can always post a question here to ask), but here’s a partial list for the LAN subnets in network 172.30.0.0, mask /30:

172.30.0.0

172.30.0.4

172.30.0.8

172.30.0.12

172.30.0.16

.

.

.

172.30.255.248

172.30.255.252

The exercise requirements state that the WAN subnets should use the numerically lowest subnet IDs from network 172.30.0.0, 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 (172.30.0.0) should be used. So, just pick the first three subnets from the list of /30 subnets of network 172.30.0.0.

Similarly, the fourth requirement states 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:

R1-R2 WAN: 172.30.0.0/30

R2-R3 WAN: 172.30.0.4/30

R1-R3 WAN: 172.30.0.8/30

R1 LAN:         172.20.255.0/24

R2 LAN:         172.20.254.0/24

R3 LAN:         172.20.253.0/24

Subnet Design Exercise 2
New Books, but Not New Exams
certskills
By certskills July 28, 2011 07:00
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14 Comments

  1. activ July 28, 23:52

    Just want to point out that in the question you stated:

    The LAN subnets should come from class B network 172.16.20.0, and the WAN subnets should come from class B network 172.16.30.0.

    In the answer you have started using: 172.20.0.0 and 172.30.0.0

    Reply to this comment
    • Wendell Odom of Certskills July 29, 04:53

      I gotta slow down a bit! Thanks. I changed the problem statement to what I meant, rather than what I typed: two different class B networks, 172.20.0.0 and 172.30.0.0. Thanks for the heads up.
      Wendell

      Reply to this comment
  2. japalm July 29, 13:37

    Thanks for continually updating this site, I’m following along. My comment is actually site related: It would be great if each post had a tag (category) for the chapter in the book that it is relevant to. I often find myself going back to review various information contained in the book, and I have to dig around the site a bit to find what I’m looking for.

    Reply to this comment
    • Wendell Odom of Certskills August 2, 10:51

      Great idea. If I just did a tag, like ICND1-CHxx or somesuch, I think that’d do the trick. But I am trying to set these blogs up as helps when using the books, so it makes sense to make it more usable by chapter. Thanks for the suggestion, and the encouragement!
      Wendell

      Reply to this comment
  3. CyberGeekUS July 26, 20:55

    Wendell,
    The subnets IDs and subnet masks shown on the figure above are right but there is a typo under it for the R1-R2 WAN subnet ID: 172.16.0.0/30

    It should be 172.30.0.0/30 as it appears on the figure.
    CyberGeekUS

    Reply to this comment
  4. NetworkingChris March 27, 00:28

    Got it! I literally clapped aloud, haha. Tough, but it feels good.

    Reply to this comment
  5. ziad4unix May 16, 14:45

    Can you tell me why I got this wrong. I understand in your explanation the host numbering part. but I get shaky after that.(also just curious why even numbers are usually used?)

    This is what I got below:

    WANS: R1-R2: 172.30.0.0
    R2-R3: 173.30.2.0
    R3-R1: 172.30.4

    LANS: R1: 172.20.254.0
    R2: 172.20.252.0
    R3: 172.20.250.0

    IP for the Routers
    R1: 172.20.254.254
    R2: 172.20.252.254
    R3: 172.20.250.254

    for the IP addreses of the other outputs of the routers
    R1 S0/0/1 : 172.30.4.253
    R3 S0/0/0 : 172.30.4.254

    R1 S0/0/0: 172.30.0. 253
    R2: S0/0/1: 172.30.0. 254

    R2 S0/0/0: 172.30. 2. 253
    R3 S0/0/1: 172.30.2.254

    Thank you for your time!

    Reply to this comment
    • Bav October 4, 15:25

      For the LANs the mask is /24 so you are jumping on 1. In your answers you are jumping on 2. Remember, the magic number is 256 – the value in the interesting octet, this case 255 so the magic number is 1. The subnets should be:

      172.20.255.0 – R1
      172.20.254.0 – R2
      172.20.253.0 – R3

      Same story for the WAN links. It’s a /30 mask so you are jumping on 4, not 2.

      Hope this helps.

      Reply to this comment
  6. Mika July 21, 20:14

    Sorry to bother you Wendell with a question but I’m having trouble. First question: Would the WAN links broadcast be 170.30.255.254? I know it’s not part of the question but after reading the book I’m having trouble with the just-left part of learning not sure how far to go often. I’m not sure when to use the just left thing either.

    Like in your book there a 10.0.0.0 with a mask of /26 and it goes from 10.0.0.0 to 10.0.1.64 to 10.1.0. 192 to 10.1.1. 192 to 10.1.255.192 to 10.255.0.1. 192 to 10.255.255.192 but there like 9 different… slots I guess you could say, on fig 21-16 (basically I don’t understand the how that figure works.. I don’t know how I’ll remember which slots come first either.

    Btw using your CCENT/CCNA 2016 book.

    Reply to this comment
    • CCENTSkills July 25, 09:52

      Hi Mika,
      Sorry that it’s not so clear. To your questions…
      The subnet broadcast addresses for the WAN subnets are:
      172.30.0.3
      172.30.0.7
      172.30.0.11

      As to how you might have gotten the incorrect answer, if you were looking at figure 21-6 and in that neighborhood, that’s not where the book describes how to find the subnet broadcast address for one subnet. Rather than run through what you already have in the book, I’d suggest reviewing Chapter 16.

      On that figure 21-16, it’s more about how to find all subnets of a network, given a single subnet mask. That figure is my attempt to show something that won’t fit on a page just due to the sheer number of subnets (over 4 million subnets in that example). So, that figure is meant to show all subnets of network 10.0.0.0, if using mask /26, by showing the patterns, and leaving you to fill in the details. The figure just shows some of the main transition points.

      If that section is causing issues for you about a single subnet’s broadcast address, I noticed that your choice of subnet broadcast address was at the end of the entire network’s range of addresses. So as a way to keep your bearings, if you are thinking about ALL the subnets in a network when using one mask (as is represented by figure 21-16), remember that each and every subnet has a subnet broadcast address, and a subnet’s broadcast address is 1 less than the next subnet’s subnet ID.

      Hope this helps…
      Wendell

      Reply to this comment
      • Mika July 27, 14:15

        Oh! I wasn’t thinking straight on the broadcast addresses. I wasn’t thinking binary at that point I was seeing 255.255.252.0 vs the actual mask 255.255.255.252 for the WANs. And of course the confusion of the possiblity of multiple addresses due to what I just read in chapter 21.

        Another question: How do you know when you’re going to have multiple addresses? Is it when you have a question like subnets needed 300 and hosts 200? While reading I couldn’t grasp when they are needed, exactly. I keep finding the network bits and the subnets and host’s and when they’re over 8 bits (subnets and host bits) I think that they could be using more than one subnet mask. Correct or wrong?

        Oh okay. I think I’m getting that part a little more now about the broadcast address. I took a break and my mind’s a bit clearer now. I’ll keep that in mind.

        Thanks for your help!

        Reply to this comment
      • Mika July 28, 14:29

        Forget my reply question. I forgot that it had to do with the equation of N+S+H =32 for one subnet mask. And then the other equations for if it’s necessary to add more. Thanks though!

        Reply to this comment
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