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I'm asked to divide up 10.64.0.0/15 into three subnets for

|Group  | Min # of Hosts|

|Group 1| 70,000        |

|Group 2| 30,000        |

|Group 3| 12,000        |

so I start by figuring out that 32-15 = 17 => 2^17 = 131072 are the total amount of addresses. Begin to split that number into powers of 2 in order to find the proper amount of addresses that can be pooled in order to get the proper network, broadcast netmask addresses.

I am then asked to to find the Network address, Broadcast Address, Netmask and total addresses.

I start by, for group 1, converting 10.64.0.0 to binary then adding 65k + 16k (got this by branching 65 -> 2x 32 -> 1x 32 -> 2x 16k). After adding them in binary I get 10.65.64.0 (not sure if that is right). Then to calculate the brodcast address i tried to |~ the network address I found previously with the netmask (15), which was wrong. I'm not sure how to find the netmask for that group as I assumed it was 15 for each group which again was wrong. And the total addresses I found above which was 65k+16k = 81,920 which again was wrong, haha.

Then from here I assumed for the next group I would just do the same adding 32k to the previous network address I found for group one and then do the same process as I did calculating group 1's data.

Any guidance on what I'm doing wrong or if there is an easier way to do this is welcome.

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  • Is this real world? Or cert prep / test stuff? Cisco and real world do not line up 100%, likewise another vendor may have another approach.
    – Austin T French
    May 4, 2013 at 16:34
  • First thought: Can't really work - your Bitmask leaves you with 2 x 255 x 255 possible combinations which if it has to be broken into 3 subnets means you have to move 2 bits to the network part leaving you with 127 x 255 (32385) combinations...how are you going to get a group with 70000 hosts?
    – M.Bennett
    May 4, 2013 at 16:54
  • @M.Bennett He can't do it with 3 equal sized /16 subnets but maybe there's another way? 'cos technically 17 bits is enough.
    – barlop
    May 5, 2013 at 0:51
  • 1
    Server Fault would close this question as a duplicate of How does IPv4 Subnetting Work?.
    – Michael Hampton
    May 5, 2013 at 1:22
  • 2
    @barlop Yes, they get closed fairly rapidly since a Q&A already exists which covers the topic in sufficient depth to answer the question.
    – Michael Hampton
    May 5, 2013 at 1:29

3 Answers 3

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Been a long time..

10.64.0.0/15

17 bits left. allows for 128K hosts (2^17) (memorize 2^7 is 128. 2^17 is 128K)

So it seems in theory that it might be possible.

If there was a way of creating network addresses to use all that space without waste.

But there might not be such a way!

Because as far as I know, you can only create a network size that is a power of two in size.

So your 70K group would have to be 128K in aize and that involves lots of waste and no room for any other group.

I was thinking maybe it's possible to create a network of 96K for the 70K one but I don't think it's possible. You can start a network at an address of 96K, by holding 64K and 32K as one. So 0 for 128. 1 for 64. 1 for 32. But the rest of it will just be 6 bits long (bits 0-5). So will just be ~(approx) 64 hosts in size. And anyhow that can't be done 'cos the network part has to be a sequential number of 1s.

If you did a network part of 192K, that sounds a bit more like you could do 12K too for example. But whatever the starting address of that network is, there are still only going to be 6 bits left in that octet so approx 64 variations in that octet. You can't get any network that is not a power of two in size. Even with any CIDR mask.

It seems so wasteful and strange that even with CIDR you are stuck with a group of 70,000 hosts in a network of 128K. But that's how it is.

And that also means it then can't be done. You only have 17 bits left. So 128K hosts If you made a group for the 70,000 you'd have no space for the other two groups. If you made a group for the other two groups you'd have no space for the 70,000 group.

CIDR as far as I know, is the best system around and the only system used nowadays for subnetting, or at least for IPv4. But it can't hack it, it leaves lots of waste that can't be used. So it looks like it can't be done. I hope i've explained why.

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You aren't doing anything wrong; the problem isn't solvable as given.

First, to accommodate 70,000 hosts you need at least a /15 (131072, 131070 usable) as a /16 (65536, 65534 usable) simply won't be large enough.

So you would have to give the entire 10.64.0.0/15 netblock to cover the 70,000 host group.

At this point you're out of netblocks and need to talk to the network guys to get more, for groups 2 and 3. For completeness, they will fit into a /17 and /18 respectively. But it appears you already have the math to figure that out, so I won't repeat it.

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  • can't they invent a way to do it 'cos in theory, if you could use part of a network, then you could. 17 bits is 128K addresses and he only needs to cover 112K hosts. I can see the issue that if each network has to be a power of two in size, and can't overlap then sure there is'n the space. but in theory it's kind of possible 'cos with 17 bits there are the hosts available. 12K addresses could be 8K+4K for example, but if all he can do is create a 16K network and not use those addresses in another group then it's wasteful
    – barlop
    May 5, 2013 at 1:28
0

Advanced IP Address Calculator 1.1 (Free Network Tool).

Advanced IP Address Calculator

May be get 8 subnet, 10.64.0.0/18 ?

Pools at 16,382 addresses is on the one hand more than 10,000 nearly Fibonacci number). On the other hand gives you the flexibility combined networks. And their number is foreseeable, only 8. Ask yourself what you need to do if you will have to extend one of the groups?

Sub ID                     Range                Broadcast
1   10.64.0.0   10.64.0.1    -  10.64.63.254    10.64.63.255
2   10.64.64.0  10.64.64.1   -  10.64.127.254   10.64.127.255
3   10.64.128.0 10.64.128.1  -  10.64.191.254   10.64.191.255
4   10.64.192.0 10.64.192.1  -  10.64.255.254   10.64.255.255
5   10.65.0.0   10.65.0.1    -  10.65.63.254    10.65.63.255
6   10.65.64.0  10.65.64.1   -  10.65.127.254   10.65.127.255
7   10.65.128.0 10.65.128.1  -  10.65.191.254   10.65.191.255
8   10.65.192.0 10.65.192.1  -  10.65.255.254   10.65.255.255
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