Networking Scenario

I've received a scenario based question that I'm trying to understand but I'm running into a wall. The question is as follows:

Scenario: Your environment is made up of 8 locations (subnets). 192.168.1.x - 192.168.8.X. You have a class C subnet (192.168.3.X) and 100 people are given three devices each. What is the problem and how do you fix it?


I believe this scenario revolves around VLSM but the 8 location subnets and the Class C "subnet" is throwing me off. I'm not sure if the listed class C subnet as written (192.168.3.X) is important or just informational. Any thoughts on how to potentially resolve this issue would be appreciated.
darinjwAsked:
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Dr. KlahnPrincipal Software EngineerCommented:
Subnets are no longer referred to as "Class A", "class B" or "class C".  This is a holdover from days past when the only subnet sizes assigned were /8, /16 and /24 blocks.  CIDR makes better use of the limited address space.

If you think of this as a /24 CIDR block the problem is evident.  A /24 block has only 256 addresses (254, depending on how you assign).  But either way you cannot cram 300 devices into a block that only holds 256.

You must (a) assign at least a /23 block to that location or (b) restrict the number of devices to less than 256 (or 254) or (c) use DHCP on that subnet and tell the users "It's first come, first served, and there will be times when you can't get on the network."
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darinjwAuthor Commented:
That makes sense if I was just looking at the single class C address listed (192.168.3.X), however the problem states that there are 8 subnets:

192.168.1.X
192.168.2.X
192.168.3.X
192.168.4.X
192.168.5.X
192.168.6.X
192.168.7.X
192.168.8.X

Assuming there isn't a use of zero-networks, that leaves 254 hosts per subnet (2,032 hosts total). More than enough for 300 devices, leading me to think about this as a scenario raising the problem of wasted addresses. But the address 192.168.3.X identified as a subnet in the question, I believe is just a Network, the network that is being subnetted across locations to account for the hosts requirement.

Am I over thinking this?
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darinjwAuthor Commented:
Actually, now that I'm reading this, I think the first part of the question was designed to throw people off. Your suggestion makes perfect sense now.

In effect, I don't care about any other of the subnets across the 8 locations except the 192.168.3.X subnet.

That actually makes this problem, exponentially easier.

I was WAY OVER THINKING THIS ONE!

Thanks, Dr. Klahn.
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AlanConsultantCommented:
Hi,

The simplest solution to not having enough IPs for 300 devices is to increase the subnet range to 512 IPs and make it a /23 instead of /24.

For example, make the network, say:

192.168.3.0 / 23

which gives:

192.168.2.0 through 192.168.3.255 (less the start and end)

so plenty of IPs to go around.

Strictly, I would write that as:

192.168.2.0 / 23

but it is the same either way.

I have had many clients with up to 400 devices on a single subnet behind a single router, and it has worked fine.  You would have to watch for capacity through the router, but that would depend on what the users were doing.


Alan.
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JustInCaseCommented:
192.168.3.0/23 can't be used since on some other location networks must not overlap. It is possible in the case below (3):

Depending on what are possible options:
1. add new subnet to location 192.168.9.0/24
2. break some of subnets on other locations (if possible and assign to this location) - for example break 192.168.4.0/24 into two subnets 192.168.4.0/25 and 192.168.4.128/25. So on one location you will have 192.168.3.0/24 and 192.168.4.128.0/25 on other location you will have 192.168.4.0.0/25
3. assign 192.168.3.0/23, but on current 192.168.4.0/24 perform break 192.168.5.0/24 into two subnets 192.168.5.0/25 and 192.168.5.128/25 and assign /25 to current 192.168.4.0/24 location and current 192.168.5.0/24 assign other part of subnet.
4. any variation of this above with different size of subnets since 300 hosts can fit into /24 + /26 (254 + 62 = 316 hosts)

Class C is address space starting with binary 110 on the first octet and by default 3 first bytes belong to network part, 4th byte belong to host portion of address. So, marking address space should not be bigger than /24 (which would eliminate solution 4).
Side note: on some lower end network devices you can't even configure /23 subnet mask.
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AlanConsultantCommented:
Hi Predrag,

Not sure what you mean by this:

192.168.3.0/23 can't be used since on some other location networks must not overlap

Your statement looks incorrect - there is no overlap unless you deliberately create it, and why would you suggest that?  It is just making life more complicated?


Alan.
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JustInCaseCommented:
Hi Alan,

There are consecutive non overlapping networks. Changing network mask on one of the existing /24 networks to /23 (except for 192.168.1.0/24 to 192.168.0.0/23, or 192.168.8.0/23)  will lead to overlapping network with network range that is already present somewhere else.
Changing 192.168.2.0/24 to 192.168.2.0/23 will overlap with network that is already present on other location where subnet is 192.168.3.0/24.

At least that's how I understood the issue - 8 locations with 8 subnets. Subnets can't overlap since routing will not work properly.
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AlanConsultantCommented:
Hi Predrag,

I don't think that is correct at all.

There is no overlap if you use 192.168.3.0 / 23 - that just means that you are taking up all of the 192.168.2.0 / 24 subnet that the question explicitly said is available to use (along with more, but you don't need those at this point).


Alan.
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JustInCaseCommented:
@Alan
I understood question that all 8 subnets are already assigned and in use (assigned to different locations).
You are obviously considering that other subnets are not in use and that can be assigned to location where 192.168.3.0/24 is configured.
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