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How To Subnet Class C 192.168.1.0/24


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#1 jude

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Posted 12 February 2005 - 04:22 AM

how to subnet 192.168.1.0/24 to get :

- 3 subnets for wan connections (2 hosts each)
- 2 subnets (70 hosts each)
- 1 subnet (40 hosts)
- 1 subnet (30 hosts)

i need just the sunets addresses without any details.

can anyone help?
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#2 skukade

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Posted 12 February 2005 - 04:53 AM

192.168.1.0/24 is itself is one subnet with 254 hosts...

I think, no more subnets are possible.

skukade
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#3 Ultraman

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Posted 12 February 2005 - 05:06 AM

jude, please confirm whether I've understood correctly: you want to be able to subnet the network 192.168.1.0/24 into all the subnets you mentioned, i.e. all of them co-exist within the same network address? Can't see how's that gonna be possible when you need two subnets with 70 hosts each... that alone will prevent you from creating further subnets.

Perhaps I've overlooked something...? :unsure:
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#4 hbbw063

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Posted 12 February 2005 - 04:45 PM

Jude,

Yes it is possible to subnet 192.168.1.0/24 for your needs bbut with only one subnet of 70 hosts not two because a 70 hosts subnet will have a at least 25 bits SM whcih gives you 126 hosts and it is the half of your whole subnet 192.168.1.0/24

1 subnet (30 hosts) = 192.168.1.0/27
Subnet Mask 255.255.255.224 (First Host .1 - last host .30)
1 subnet (40 hosts) = 192.168.1.32/26
Subnet Mask 255.255.255.192 (First Host .33 - last host .94)
1 subnet (70 hosts each) = 192.168.1.96/25
Subnet Mask 255.255.255.128 (First Host .97 - last host .222)
3 subnet for WAN connection = 192.168.1.224/30 - 192.168.1.228/30 - 192.168.1.232/30 Subnet mask 255.255.255.252.
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#5 Ultraman

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Posted 12 February 2005 - 08:19 PM

What you said is correct, hbbw063, but the subnets you derived didn't seem to make sense... Perhaps it's 'cos you did it in a rush? For example, take your subnet for the 70 hosts 192.168.1.96/25, while the subnet mask is correct, you're only using the first bit of the fourth octet, which mean 96 wouldn't be part of the network address. Therefore you can't have the host starting at 97 either. In fact, the subnets you derived appear to overlap... :unsure:

Anyway, assuming we only trying to derive one subnet of 70 hosts and not two, and assuming we can make use of the first and last subnets in each range, here's my solution:


3 subnets, 2 hosts each (max. 2)
- 192.168.1.0/30 (.1 - .2)
- 192.168.1.4/30 (.5 - .6)
- 192.168.1.8/30 (.9 - .10)

1 subnet, 30 hosts (max. 30)
- 192.168.1.32/27 (.33 - .62)

1 subnet, 40 hosts (max. 62)
- 192.168.1.64/26 (.65 - .126)

1 subnet, 70 hosts (max. 126)
- 192.168.1.128/25 (.129 - .254)
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#6 jude

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Posted 13 February 2005 - 12:09 AM

thank you all,

skukade, but we can borrow some bits form host bits to make subnets.

ultraman, yes i want them all to be within the same network address. when we create a subnet for 70 hosts, we do not use the full range. so we can use addresses within the range for another subnet.

hbbw063, you almost answered what i need, but you have a mistake when you made the subnet for 40 hosts. can't you see that with /26 we cannot get more than 32 addresses for hosts. so you need to recalculate again.

my last answer : it seems to be unpossible to get all the subnets i mentioned earlier, but we can do the following:

192.168.1.0 /27 ( 1 to 30 for hosts ) ; for subnet with 30 hosts

192.168.1.32 /30 ( 33, 34 for wan conn.)
192.168.1.36 /30 ( 37, 38 for wan conn.)
192.168.1.40 /30 ( 41, 42 for wan conn.)

192.168.1.44 /25 ( 45 to 126 for hosts) ; for subnet with 70 hosts

192.168.1.128 /26 ( 129 to 190 for hosts) ; for subnet with 40 hosts

192.168.1.192 /25 ( 193 to 254 for hosts) ; but for subnet with 62 hosts not with 70 hosts.

there is a question, is there any way we can know immediately whether it is possible or not before we waste time doing calculatiin?

so we can say : it is unpossible until we find someone saying the opposite.

jude...
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#7 jude

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Posted 13 February 2005 - 12:12 AM

there are other ways to subnet 192.168.1.0/24, but what i did - i think - is the most effective way that waste addresses as less as possible.

jude.
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#8 long_double

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Posted 13 February 2005 - 06:42 AM

jude, please confirm whether I've understood correctly: you want to be able to subnet the network 192.168.1.0/24 into all the subnets you mentioned, i.e. all of them co-exist within the same network address? Can't see how's that gonna be possible when you need two subnets with 70 hosts each... that alone will prevent you from creating further subnets.

Perhaps I've overlooked something...? :unsure:

Agree. May be the mask is not 24, but less, like 23 or 22?
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