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Haven't been able to solve this question this entire day. Which network is the best and why?!?

You are the network administrator for a new call center management company. The firm has 47 offices in various cities throughout the U.S. You have been tasked with producing a specification for the IP addressing on the network. The Internet Assigned Numbers Authority (IANA) has granted you a public Class B IP address of 165.45.0.0. You now need to create 47 networks (one for each location) and provide the maximum number of nodes per network. Which of the following subnet masks would you use?

Question 12 options:

255.255.255.0

255.255.240.0

255.255.248.0

255.255.252.0

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  • SuperUser is not the place where you get help with your homework. If you're stuck, you should ask your teachers.
    – LPChip
    Dec 13, 2015 at 14:41
  • @LPChip - I used to think this to, but not so - The site does allow helping with homework - see meta.superuser.com/questions/1505/…
    – davidgo
    Dec 13, 2015 at 18:27
  • Instructors take way too long to reply. I've been learning a lot and getting my hw done. I'm not on here trying to get answer abcd i'm trying to understand the problem so when the exam comes around I can pass. Dec 13, 2015 at 21:23
  • @davidgo You should read How do I ask and answer homework questions?: "Make a good faith attempt to solve the problem yourself first. If we can't see enough work on your part your question will likely be booed off the stage; it will be voted down and closed."
    – DavidPostill
    Dec 14, 2015 at 14:13

2 Answers 2

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I'll try provide a simple answer to this which does not focus too heavily on the math. For simplicity sake, I'm going to ignore broadcast and network addresses etc, as they are not really relevant and confuse the question.

What you need to do is break the network up into as close to 47 parts as practical. You also need to ensure that the number of parts is AT least 47.

If you took the solution 255.255.255.0 you would have 256 sets of 256 addresses. This would work, but is not optimal.

If you took the solution 255.255.240.0 you would have 16 sets of 256 * 16 addresses. This is not enough addresses. (256 - 240 = 16, which gives you 16 blocks of 16 * 256 addresses).

Similarly 255.255.248.0 would give you 32 blocks of 8 * 256 addresses (256 - 248 = 8 class C blocks, 256/8 = 32 ranges) which does not provide enough blocks.

Thus the answer must be 255.255.252.0, which gives you 64 * 4 class C blocks, which provides 256 * 4 = 1024 IP addresses per site, and allows you to have up to 64 sites.

Of-course, this answer is simplistic - In reality you are likely to have some bigger and some smaller offices, so you would not typically break the network up into equal blocks of this size, you would break it up to have a combination of larger blocks for larger offices - and the ability to further subnet those in the offices as well, and smaller blocks for smaller offices)

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  • How did you get the 64*4 class c blocks from 255.255.252.0.? Dec 13, 2015 at 20:46
  • I got 64 because 255.255.252.0 has 6 ones as a class b network so 2^6 but where do we get that 4 from? Dec 13, 2015 at 21:10
  • There are many ways of calculating this. The way I did it was to take the third octet, ie 252), and subtract it from 256 (2^8), which gives 4.
    – davidgo
    Dec 14, 2015 at 0:17
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I wrote this on Yahoo Answers years ago, and I've had countless people email/thank me for this explanation. Since you have an understanding of subnetting, read this and see if you can figure out your question. If you figure it out, leave an answer in a comment. If you get stumped, ask a question in a comment. It's good practice, and the only way to learn subnetting! So you're given your networks, and are told to break them up into smaller networks. I'll start with question A.

Question A says that we have the 172.18.0.0 /16 network, and we want to break it up into smaller networks. To do that, we need to borrow some host bits from our given network. Right now, our network bits are the 172.18 octets, and our host bits are in the 0.0 octets. In binary this would be a bunch of 1's and 0's but to help you visually see this, I'm going to use N for network, and H for host bits.

We would have nnnnnnnn.nnnnnnnn.hhhhhhhh.hhhhhhhh. To make our network smaller we need to take some of those "h" bits and make them "n" bits. The formula for this is 2^n. Then this will tell you how many networks you can make. So if we just borrow one "h" bit, we would take 2^1=2. So now out octets would look like nnnnnnnn.nnnnnnnn.nhhhhhhh.hhhhhhhh. Since the bit values of the octets go 128 64 32 16 8 4 2 1, our new network bit is using the .128. So our networks will be 172.18.0.0 /17 and 172.18.128.0 /17.

Since your questions asks to break it into 8 smaller networks we just need to find 2^n = 8. Since 2^3 = 8 we're going to borrow three host bits. So now our bits would look like nnnnnnnn.nnnnnnnn.nnnhhhhh.hhhhhhhh. Since our network bit is now using the 32 bit, our networks will go by 32. So your four networks would be 172.18.0.0 /19, 172.18.32.0 /19, 172.18.64.0 /19 and so on until you get to your last network of 172.18.224.0/19.

For question B, it wants us to subnet the 10.0.0.0 /8 into four subnets. Again, our bits would look like nnnnnnnn.hhhhhhhh.hhhhhhhh.hhhhhhhh. The now we need to find out 2^n = 4. Since 2^2 = 4, we'll borrow two host bits, and use them as network bits. So our octet would now look like nnnnnnnn.nnhhhhhh.hhhhhhhh.hhhhhhhh. And our networks would be the 10.0.0.0 /10 and the 10.64.0.0 /10.

Now that we know our network addresses, finding the broadcast address and number of nodes per subnet is easy. For example our first 172.18.0.0 network is the 172.18.0.0 /17 and the next network would be 172.18.32.0 /17. The broadcast address is always the last address of the subnet, so for the 172.18.0.0 network it would be 172.18.31.255. For the 172.18.32.0 network, it would be 172.18.63.255.

To find out the number of nodes, the long/hard way is to take 2^h. H being the number of hosts bits that you have. So for the 172.18.0.0 /17 network, we have 15 host bits, so we would take 2^15 = 32768.

I'm sorry that is extremely lengthy. I'm not sure on how comfortable you are with subnetting overall, but here are some shortcuts. If you're wanting to find out the number of subets, just take your bit value of either 128 64 32 16 8 4 2 1 and divide it by 256. So for example, question A, we needed eight subnets. All you would have to do is take 256/32 = 8. Then you would count your networks by 32, since you're using the 32 valued bit.

For finding the number of nodes per network, you just start doubling numbers. A /30 can have 4 hosts totals, 2 usable. a /29 has 8, /28 has 16 and so on. The last octet really goes by bit values so it would just go 128 64 32 16 8 4 2 1. That's the number of total hosts, if you minus 2, that would be the number of usable hosts.

Then once you get to the /23, you start doubling. A /24 is 256, /23 is 512, /22 1024. Then just double each time you increase one bit value

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