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I've been trying to figure out how to divide a network IPv4 IP Address into 7 subnets, where each subnet requires a different number of hosts.

For example I have the question:

"You are an intern at a networking company and the company's Network Administrator has given you the task of subnetting the address 10.2.0.0/22 to maximize the IP addressing available for 7 different networks."

"The first network must be able to host 405 users."

"The second network must be able to host 200 users."

"The third network must be able to host 75 users."

"The fourth network must be able to host 16 users."

"Two networks must remain for future use and will remain unused for now and must end with a specific prefix in CIDR style."

"The last network must end with a specific prefix in CIDR style and be further sub-netted into three WAN networks needed to connect Site-1 to 3 other sites with 2 IP addresses in each WAN network."

From the material I have it says to determine the number of bits required to define 7 subnets but I have no idea how or where to go from there and how the bits needed are determined.

I am looking for any method or explanation on how to work out these types of questions, and a simplified answer if possible. I am having a really hard time understanding IPv4 subnetting in general and this question in particular.

Any help would be greatly appreciated. Thank you in advance.

Joseph Bay
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  • That does help, and I appreciate it, however I can't find anything in that post explaining how to calculate subnets with differing host numbers, nor with the high numbers in my original question. That is the part that is really tripping me up. – Joseph Bay Dec 02 '18 at 09:23
  • Basically, you can't only subnet /22 to four /24 but also to one /23 and two /24. Or one /23, one /24 and two /25, and so on. – Zac67 Dec 02 '18 at 09:38
  • We don’t do your homework. – Tommiie Dec 02 '18 at 10:38
  • "I can't find anything in that post explaining how to calculate subnets with differing host numbers," Apparently, you didn't see or read the entirety of the Subnet Sizing Based on Number of Hosts section of this answer that discusses a very similar problem. – Ron Maupin Dec 02 '18 at 14:38
  • I wasn't asking for anyone to do my homework, I was asking for help understanding the concept. Regardless, I stopped freaking out and re-read the post linked to this question with a calmer mindset and it made a lot more sense, so thank you Zac67 for the link, it helped me immensely. – Joseph Bay Dec 02 '18 at 23:45
  • @JosephBay Glad the materials helped. It's tricky until it clicks into place, then after it makes sense it becomes easy. – jonathanjo Dec 03 '18 at 00:02
  • I ended up subnetting the following way:

    405 required I did: Prefix; /23, IP; 10.4.0.0/23, which gave 512 IP's

    200 required I did: Prefix; /24, IP; 10.5.0.0/24, which gave 256 IP's

    75 required I did: Prefix; /25, IP; 10.5.0.128/25, which gave 128 IP's

    16 required I did: Prefix; /27, IP; 10.5.0.160/27, which gave 32 IP's

    1st unused I did: Prefix; /27, IP; 10.5.0.192/27, which gave 32 IP's

    2nd unused I did: Prefix /27, IP; 10.5.0.224/27, which gave 32 IP's

    Last I did: Prefix /27, IP; 10.5.1.0/27, which gave 32 IP's

    This makes sense to me, though corrections are good

     – Joseph Bay Dec 03 '18 at 00:08
  • Also jonathanjo, the link you provided was equally helpful, so thank you greatly for that. – Joseph Bay Dec 03 '18 at 00:09
  • @JosephBay you got all the sizes right, including the tricky 16-host one. But your addresses need checking. Your starting address is 10.2.0.0, and its /22 ends 10.2.3.255 so all of your answers need to be in that range. – jonathanjo Dec 03 '18 at 00:24
  • Am I calculating into the wrong octet then? – Joseph Bay Dec 03 '18 at 00:28
  • I'm looking at an online subnetting calculator and it looks like I am counting into the wrong octet. ] – Joseph Bay Dec 03 '18 at 00:30
  • Alright I re-calculated: 405 required: Prefix; /23, IP; 10.2.0.0/23. 200 required: Prefix; /24, IP; 10.2.2.0/24. 75 required: Prefix; /25, IP; 10.2.3.0/25. 16 required: Prefix; /27, IP; 10.2.3.128/27. 1st unused: Prefix; /27, IP; 10.2.3.160/27. 2nd unused: Prefix /27, IP; 10.2.3.192/27. WAN subnet: Prefix /27, IP; 10.2.3.224/27. The last usable IP is 10.2.3.254 with the broadcast being 10.2.3.255. The IP's are the beginning IP addresses for each network. How does this look? It looks like it makes more sense than what I had last. Any comments would be greatly appreciated. – Joseph Bay Dec 03 '18 at 01:43

1 Answers1

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(Question was marked as duplicate, so I moved this answer to the same place, specifically https://networkengineering.stackexchange.com/a/55146)

jonathanjo
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