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In my problem I have a company to which was given the block of addresses 172.14.200.192/26 and the whole network is connected like this:

Network

The objective is to figure out the L1, L2 and L3 subnet addresses.

Since it's a 26 bit mask I assume I should be able to create 4 subnets with 62 hosts each.

What's throwing me off here is the 192, because 172.14.200.192 -> 172.14.200.|11|000000|, makes it so that the bits borrowed to create subnets are filled already.

I did another exercise where the address block was 44.44.44.0/26, so 3 subnets could be 44.44.44.64, 44.44.44.128 and 44.44.44.192. Did I do that one right? What am I getting wrong here?

I mean, can I even create subnets? Or do I have only one, the .192 with 62 hosts and broadcast at .255?

EDIT: I just thought that the point might be to borrow the first 2 from the last 6 bits to be able to create 4 networks. But those would only be able to have 14 hosts each, so even though the provided link below teaches to create subnetworks, I still can't answer the question.

EDITthe2nd: As sugested below, the answer could be (and most likely is) to borrow one bit and create two /27, each with 30 hosts and borrow another bit in one of those to create two /28, each with 14 hosts. The /27 would go to L2, with 29 computers and the /28's would go to L1 and L3.

L2 -> 172.14.200.|11|0|00000 / 27 -> 172.14.200.192/27

L1 -> 172.14.200.|11|10|0000 / 28 -> 172.14.200.224/28

L3 -> 172.14.200.|11|11|0000 / 28 -> 172.14.200.240/28

EDITthe3rd: The above was a nice step, but I failed to take into account the 172.14.200.212/30 segment, which takes away the .212 - .215 block. As such, the .192/27 block is the one to be divided:

L2: 172.14.200.224/27 -> gives me 30 hosts

L3: 172.14.200.192/28 -> gives me 14 hosts

The last block is divided again:

L1: 172.14.200.216/29 -> gives me 6 hosts

The remaining block, 172.14.200.208/29 will overlap with the .212 - .215 block, but I only need 3 subnets.

marked as duplicate by Ron Trunk, Craig Constantine Dec 29 '15 at 21:08

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    Homework questions are off-topic here, but you can look at the link above for some assistance. – Ron Trunk Dec 29 '15 at 15:59
  • This isn't a homework exercise, it's exam preparation. And I already looked at that link. Even if I divide the .192 block into 4 smaller blocks I wouldn't have enough hosts per block. It would work for L1 and L3, but not L2. – user2993349 Dec 29 '15 at 16:04
  • Yes u can subnet the given block into to smaller subnets. – user3476261 Dec 29 '15 at 19:01
  • All education questions are off-topic on this site. Check the help center for what you can ask here. – Ron Maupin Dec 29 '15 at 20:19
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You can divide a /26 into four /28 or one /27 and two /28, right? VLSM.

  • So, I would borrow one bit to create two /27 and then in one of those borrow one bit to create two /28? – user2993349 Dec 29 '15 at 17:05
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26 bit mask address only allow u to have maximum of 62 hots

  • A 26 bit mask would leave 6 host bits, which would be 64 - 2 hosts. – user2993349 Dec 29 '15 at 18:57
  • Yes my mistake i meant 64-2 – user3476261 Dec 29 '15 at 18:58

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