Help with subnetting

I'm practicing for a test, my teacher gave us an ip 172.16.0.0/16 he doesn't give us the number of subnets, but from the labs we've done, he uses two ,one private and one for a dmz, how do i go about sub-netting this so i can configure them in my dhcp server. because every time i put in the scopes ofr the private lane it works but when i input the ip addresses for the second scope(for the DMZ), i get conflict error between my ip and subnet masks, saying there's a scope like them already set up.

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Joedoc164

Goon, first off what ip address did you assign your private network and your dmz? The network that your teacher gave you is 172.16.0.0 to 172.16.255.255, with a network mask of 255.255.0.0. Keep in mind that the first address is your network address and the last address is your broadcast address, so you can't use either of those. I assume that your teacher wanted you to set the private lane and your DMZ in different networks, so one network can be the 172.16.x.x network and you would have to make your next network something like 172.17.x.x or 172.10.x.x. Hopefully this makes sense and answers your question.

TheGoon

Joedoc164 wrote: »

Goon, first off what ip address did you assign your private network and your dmz? The network that your teacher gave you is 172.16.0.0 to 172.16.255.255, with a network mask of 255.255.0.0. Keep in mind that the first address is your network address and the last address is your broadcast address, so you can't use either of those. I assume that your teacher wanted you to set the private lane and your DMZ in different networks, so one network can be the 172.16.x.x network and you would have to make your next network something like 172.17.x.x or 172.10.x.x. Hopefully this makes sense and answers your question.

thanks for responding i appreciate it.

ok i figured some stuff out. for my first subnet my N:172.16.0.0 F:172.16.1.1 L:172.16.62.254 B:172.16.63.255

i know that i can assign my clients in the private subnet ips from F and L, but what happens with the fourth octet? do they remain the same or what? for example, 172.16.1.1,172.16.2.1,172.16.3.1, until the third octect hits 62, then make the fourth octet 254. is that correct?

also, can i use 172.16.64.0 network address for my second subnet? he wants them on the same network.

Priston

@Goon when subnetting 172.16.0.0/16 is there any requirements for how many host or how many networks you need?

@Joedoc how do you get 172.10.x.x and 172.17.x.x? those are NOT in the 172.16.0.0/16 range.

Kinet1c

I used to be lost on this too. Please watch the CCNA videos on CBT nuggets by Jeremy, this nailed it for me and I've never forgotten it since.

TheGoon

Priston wrote: »

@Goon when subnetting 172.16.0.0/16 is there any requirements for how many host or how many networks you need?

@Joedoc how do you get 172.10.x.x and 172.17.x.x? those are NOT in the 172.16.0.0/16 range.

he didnt give any requirements. we have a nat a dc-dhcp-dns server on the private lane and an edns in the dmz.

Joedoc164

@Priston, I was using those as examples of other networks. I didn't mean that those would be in the 172.16.0.0 /16 range. Hopefully my response didn't cause any confusion. Subnetting can be tricky enough to pick up.

TheGoon

Joedoc164 wrote: »

@Priston, I was using those as examples of other networks. I didn't mean that those would be in the 172.16.0.0 /16 range. Hopefully my response didn't cause any confusion. Subnetting can be tricky enough to pick up.

hey jordoc mind helping me out with this question,

i know that i can assign my clients in the private subnet ips from F and L, but what happens with the fourth octet? do they remain the same or what? for example, 172.16.1.1,172.16.2.1,172.16.3.1, until the third octect hits 62, then make the fourth octet 254. is that correct?

[Deleted User]

You could just use a subnet calculator. Not recommending you depend on them if you plan on taking CCENT but still just so you can solve it and find out out to do it. Remeber, you find out the number of subnets by the number of bits on. Number of bits off minus 2 is your number of available hosts (2 taken away for broadcast and network address) Then subtract the number of bits off from 255 and you have your block size. Hope that helps you out.

thomas_

TheGoon wrote: »

hey jordoc mind helping me out with this question,

i know that i can assign my clients in the private subnet ips from F and L, but what happens with the fourth octet? do they remain the same or what? for example, 172.16.1.1,172.16.2.1,172.16.3.1, until the third octect hits 62, then make the fourth octet 254. is that correct?

In the example you gave:

172.16.0.0 is the subnet id
172.16.0.1 is the first usable ip address
172.16.63.254 is the last usable ip address(you put .62.254)
172.16.63.255 is the broadcast address

The next subnet would be 172.16.64.0.0 with:

172.16.64.1 being the first usable ip address
172.16.127.254 being the last usable ip addres
172.16.127.255 being the broadcast ip address

All of the above assumes that you borrowed 2 bits from the host portion which would be /18 or a 255.255.192.0 subnet mask. Each subnet would be a multiple of 64 in the thiird octet.