I have class b address of 172.38.0.0/255.255.0.0 and one of the techicians came up with the following:

1st floor = 172.38.1.0 - 172.38.1.255/255.255.255.0 (max host/subnet 254). Additional subnets requiring a mask of 255.255.255.0 = 172.38.2.0 - 172.38.31.0

2nd floor = 172.38.32.0 - 172.38.32.127/255.255.255.128 (max host/subnet 126). Additional  subnets requiring a mask of 255.255.255.128 = 172.38.33.0 - 172.38.63.0

I know that he subnetted the orginal class B address using VSLM. However, what i don't understand is this. The 1st floor subnet mask should give me:

11111111.11111111.11111111.X and the third octet should allow me the following: 00000001 - 11111110 in subnets. So   the available subnets should be from 172.32.1.0 to 172.32.254.0 for the first floor. Why is it to .31 only?

As for second floor, the subnet mask produces the following 11111111.11111111.11111111.10000000. How does he end up with a subnet begin .32?

Can someone explain this to me? I am confused

Thanks

###### Who is Participating?

Commented:
Ok, let met start from the easiest first.

160.149.115.8/22 (mask 255.255.252.0) has 10 bits in the host portion and 22 bits in the network portion.  It's "naturally" a "class B" network (with 16 network bits) so, it's subnetted with 6 bits.  In other words, it's like this:
NNNNNNNN.NNNNNNNN.SSSSSSHH.HHHHHHHH where N=network, S=Subnet, and H=Host.

Now we're working on the 3rd octet, as you said.  We're borrowing 2 bits from the host portion.  A quick and easy cheat is that 2^2 bits borrowed = your "network incrementer".  So, starting with 0, the networks are going to go like this:
.0, .4, .8, .16, .20, ... .108, .112, .116.  Notice that last range.

This means your subnet goes from 160.149.112.1 through 160.149.115.254 (.0 is network, and .255 is broadcast).  So, this means that 160.149.115.8 and 160.149.114.66 both /22 are on the same subnet, which is 1022 addresses in size.

I'll seperate the other part into a different post.
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Commented:
it only goes to 31 b/c the address space 172.16.0.0 thru 172.31.255.255 is reserved for private IP addressing.  You could of course use 172.32.x.x but those addresses aren't reserved for private use (meaning they are "real" public IPs).  Bascially your technician messed up by using the 172.32.x.x range.

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Commented:
FYI, the standard for this is defined by RFC 1918.

http://www.faqs.org/rfcs/rfc1918.html
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Commented:
oops, i think i misunderstood your question after i re-read it.  I thought you were talking about private IPs but after i re-read it im not sure anymore.   But let me ask a few questions:

1.  you start off by saying you have class B address range of 172.38.0.0/255.255.0.0.  do you mean this was the range assigned to you by your ISP?  If so thats alot of addresses (over a million)

2.  later you start talking about the 172.32.1.0 address range which is a different range of addresses all together. are we dealing with 172.38.x.x pr 172.32.x.x????

3.  you later say "So the available subnets should be from 172.32.1.0 to 172.32.254.0 for the first floor. Why is it to .31 only?"... where are you getting the 31 from?  i think he stopped at 31 b/c of the RFC 1918 i mentioned earlier, but that RFC only applies if you are using the private class B range of 172.16.0.0 thru 172.31.255.255
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Author Commented:
My fault i wrote that wrong,

I meant on the first floor, the subnet mask is
11111111.11111111.11111111.X so the third octet should allow me the following: 00000001 - 11111110 in subnets. So the available subnets should be from 172.38.1.0 to  172.38.254.0. But instead the technican has this written down: Additional subnets requiring a mask of 255.255.255.0 = 172.38.2.0 - 172.38.31.0. Is there any reason why he stoped at .31?

Secondly, the second floor has a subnet mask of 255.255.255.128 on the same class b address 172.38.0.X The technican ended up with starting subnet of 172.38.32.0-172.38.32.127

And on the third floor he has the same subnet mask of 255.255.255.128 but with the addressing of 172.38.32.128 - 172.38.32.255

but the subnet mask is 255.255.255.128 -> 11111111.11111111.11111111.10000000, (borrowed 9 bits). How did he end up with a subnet starting at 172.38.32.X?

So far It looks like the techican took one of the 255.255.255.0 subnets and subnetted it again into two 126 host subnets. But i am not sure.

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Commented:
bascially it doesn't seem like he knew what he was doing. There is no reason to stop at 31 unless he was just confused about the RFC i mentioned earlier but that doesn't even apply here.

with a mask of 255.255.255.128 the starting point of the subnet isn't 172.38.32.128.  with a mask of  255.255.255.128 you are subnetting the 4th octet, not the third anyway.  why he didn't go to 172.38.2.x i dont know. it would make much more sense than jumping to 32 from 1 wouldn't it?   it would make much more sense to have 172.38.1.x for the first floor and 172.38.2.x for the second floor and so on.  that way the third octet number would match the floor number which would be easier to keep track of.

him stopping at 172.38.31.0 is totally incorrect. with a mask of 255.255.255.0 on 172.38.2.0 the range is 172.38.2.1 thru 172.38.2.254.  Again, he didn't what what he was doing.

you might want to use a subnet caclulator to do the math for you.  here is an easy web-based one that i use:

http://www.subnetonline.com/subcalc/subnet8.html

also, let me correct myself before someone else does...172.38.0.0/255.255.0.0 would give you 65534, not over a million like i said earlier,,, i was thinking about the 172.16.0.0 thru 172.31.255.255 block which has a /12 mask, not /16.
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Author Commented:
I rummaged around and I finally found some of his notes. (He's no longer here) according to his notes. He started with
the original class b address of 172.38.0.0 and subnetted it with 255.255.255.0 which of course gives him available subnets from 172.38.1.X - 172.38.254.X.

For whatever reason he decided to reserve the first 32 subnets for the first floor (can't read the reasoning behind it, smudged) which is .1-.31

He took the 172.38.32 subnet. This is where he got his 172.38.32.0 - 172.38.32.127 and 172.38.32.128 - 172.38.32.255 ip from.

Looks like he treated the 172.38.32.X subnet as if it was a class C address and resubnetting it by borrowing 1 bit from the 4th octet.

Regardless if he was insane or not in his approach to dishing out ips was his subnetting technically correct?

One more thing:

I was told that both of these are on the same subnet.

160.149.115.8  /255.255.252.0
160.149.114.66 /255.255.252.0

How is this possible? If you borrowed 6 bits from the third octet shouldn't the last remaining 2 bits be added to the host instead of the subnet?. Why is it added to the subnet instead?

X.011100 + 11.  = .115
X.011100 + 10.  = .114

If this is possible then I assume the very first subnet availabe on 255.255.252.0 would be 160.149.4.X (X.000001 + 00.X) - 160.149.7.X (X.000001 + 11.X) right?

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Commented:
Forget about it, it doesn't make sense to me.  It looks like he was trying to design for summarization, which is admirable, but there's chunks missing.

On the first floor, What happened to the 172.38.0.0/24 subnet?  Can't use it - no "subnet zero"?  That's fine.  Why did he leave 29 subnets with 254 hosts each left over?

On the second floor, subnetting more, he puts in 172.38.32.0/25.  Why the sudden drop in the number of hosts?  And then he forgets about the 172.38.32.128/25 subnet he created.

I'd like to know how many floors you have, how many hosts you have and what kind of growth room you need.
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Commented:
yes 160.149.115.8 and 160.149.114.66 are both on the same subnet.  if you use the subnet calculator that i posted a link to above you will see that.  If you want to do it the hard way by breaking everything down into binary you can, but there isn't a need for it unless you just really want to get into the nuts and bolts of it for your own understanding.

160.149.115.8  with a mask of 255.255.252.0 has a begining address of 160.149.112.1 and an ending address of 160.149.115.254 so yes they are both on the same subnet.  thats just the way it is. 2+2=4, thats just the way it is.  I think you are misunderstadning how to manually calculate subnets and that is why you are getting confused.

what you are doing by adding things to the IP addresses isn't correct at all (or either you aren't doing that method correctly).   I've never heard of any method where you add anything.  you have to remember, its called a MASK for a reason. it means that you mask out that part of the address, meaining it is irrellevant.  I think you are confusing yourself when you are thinking in terms of borrowing IPs when you are calculating subnets.

for example:
160.149.115.8 in binary is 10100000.10010101.01110011.00001000
255.255.252.0 in binary is 11111111.11111111.11111100.00000000
so the last 10 digits are irrlevant since they are masked out as far as subnetting goes
to figure out the network address you convert the host portion to 0s to end up with:
11111111.11111111.11111100.00000000

to figure out the broadcast address you convert the host portion to 1s to end up with:
11111111.11111111.11111100.00000000

i put the masks right below the net and broadcast addreeses so you could see whats going on.

all the numbers between the network address and broadcast address are useable IPs.  that is the nuts and bolts method and works for all masks regardless of which octed. subnetting the last octet is of course easy
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Author Commented:
I see where you are coming from, but i guess what i'm trying to say is this

we all now that each bit represent a value. there are 8 bits to one octet, each bit representing one value

1      1    1   1     1  1  1  1  = octet
128  64   32  16   8  4  2  1  = value for each bit

So something like this 10000000 would mean 128
and something like this 10000010 would mean 130 because it's

1      0    0   0  0   0  1   0
128   0    0   0   0   0  2  0  ---->  adding 128 and 2 gives you a number of 130

Now

IP: 160.149.115.8
10100000.10010101.01110011 = the first 6 bits together of the third octet (the subnet) has a numeric value of 112 not 115.  Beacuse it's 0 + 64 + 32 + 16 + 0 = 112.

In order to be 115, it needs the remaining two bits which if you notice would have belonged to the host. By adding 112 + 3 (the last 2 remaining bits in the 3rd octet) You get 115 for the value of the third octet.

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Commented:
iamuser,

yes, x.x.115.8 and x.x.114.66

x.x.115.8    = x.x.011100 11.00001000
x.x.114.66  = x.x.011100 10.01000010
^
y.y.252.0    = y.y.111111 00.00000000
^
now count 011100 & 00 = 01110000 = 112 is a decimal number  of subnet
A        B
here we don't care what is B (host part). To get next subnet we have to increase our subnet number (A) just on one

011101 & 00 = 01110100 = 116
A+1      ...
So your 114 and 115 are on the same subnet:

valuable hosts:      160.149.112.1 - 160.149.115.254

Hope this is clear. To answer your main question about subnetting your LAN - as mentioned by pseudocyber it's needed more information about number of addresses per floor you need (not only users, don't forget about servers, printers, firewalls and etc.), number of floors, any plans for the future growth.

Regards.
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Commented:
iamuser.

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