Subnetting Question

computer g33k

I need to create 14 subnets and the largest subnet requires 500 hosts. I'm starting out with 172.16.0.0/16. I'm not sure how many bits I need to borrow to satisfy 500 hosts. Would I need to borrow 9 bits to make a /25? (2^9-2=510)

172.16.0.0/25

255.255.255.128

This is that correct?

dustinmurphy

To have a total of over 500 hosts, you'd borrow 7 bits, making it a /23. That will give you 128 subnets(I believe) with 510 hosts each.

so...

172.16.0.0/23
255.255.254.0

Your blocks would be 2....so...
172.16.0.0/23
172.16.2.0/23
172.16.4.0/23

I'm pretty new to subnet math, but I'm pretty sure that's right...

a /25 would give you 2 (non-classful) or 512(classful) subnets with 126 hosts in each subnet.

etc....

If I'm not mistaken, this is the math for the correct answer:
172.16.0.0 is a class b subnet, so 255.255.0.0
2^7 = 128 (how many subnets)
256(8 bits)/128(total subnets) = 2 (subnet block size)
(2 (block size)*256 (hosts in each block)) -2 (for network/broadcast) = 510 (how many hosts per subnet)


Here's the math on your answer:
You wouldn't borrow 9 bits, you'd borrow 1 from a /24 or 255.255.255.0 (to make the math a bit easier)
2^1 = 2 (how many subnets)
256/2 = 128 (subnet block size)
(128 (block size)*1(since it's using the last octet))-2 (network/broadcast) = 126 hosts per subnet
Classful would give you 512 (2^9)subnets of 126 hosts per subnet.

therefore a 172.16.0.0/25 would give you 2 subnets with 126 hosts each.

computer g33k

Thanks dustin, I think I understand now.

drkat

^^

172.16.0.0/16

We need 500 hosts so... we know that 2^9 = 512 so since this is a class B network address we use the 3rd octet for working this out.

if we know we need 9 bits to cover the host requirement then we simply "turn on" 7 1's int he 3rd octet

172.16.0.0
===========
172.16.11111110.00000000 -> what it looks like in binary for the 2 last octets

128 64 32 16 8 4 2 1
1 1 1 1 1 1 1 0

172.16.0.0
255.255.254.0

2^9-2 = 510 hosts per subnet
2^7 = 128

so we have 128 subnetworks and 510 available hosts per network

Forsaken_GA

You're actually approaching it the wrong way. You should be considering how many bits you need to borrow to get the requisite number of subnets.

Borrow one bit, you have 2 subnets. Borrow two, you have 4. 3 you have 8, 4 you have 16. Alright, so if you borrow 4 bits, you can make the right number of subnets. That leaves you with 12 bits for host addressing. 2^12 = 4096. So each of your subnets can support 4094 hosts, leaving you with plenty of room for growth in the subnets, and they're big enough that you can chop a single one of those up into much smaller bits if you require further subnets later on (I have no idea what the other requirements are, but I suspect that if 500 hosts is what the largest of the subnets requires, I could chop a single /20 up into enough bits to satisfy the needs of the entire question, and have all the other ones for growth)

computer g33k

Ok, thanks guys. So for example, if I needed 20 subnets and largest subnet requires 1000 hosts. I would borrow 6 bits 2^6=32 since 20 falls in between 16 and 32. I would then be left with 10 bits for hosts 2^10-2=1022 host per subnet. Am I correct?