Subnetting Question

I'm doing a project and was given the address space of 200.0.5.2/23 and was told I could use VLSM where necessary. I'm aiming for a /25 mask for my first 3 VLANs since they all have 75 users. My question is, since a 200. address is in class C range, and the default mask is /24, if I adjust to a 200.0.5.2/25 address, am I borrowing 1 or 2 network bits? I'm honestly not sure...

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Comments

MrBrian

If you've got a class C address (/24) default, and you move it to a /25, then you're borrowing 1 bit from the last octet. That gives you 128-2= 126 hosts per subnet. You're right, a /25 would be good for vlans with 75 users because a /26 wouldn't give you enough, being 62 hosts per subnet.

What IP space were you given though... 200.0.0.5.2 ?? Is that an IP version 5 address? Lol

sentimetal

MrBrian wrote: »

If you've got a class C address (/24) default, and you move it to a /25, then you're borrowing 1 bit from the last octet. That gives you 128-2= 126 hosts per subnet. You're right, a /25 would be good for vlans with 75 users because a /26 wouldn't give you enough, being 62 hosts per subnet.

What IP space were you given though... 200.0.0.5.2 ?? Is that an IP version 5 address? Lol

It's a class C range (192-223) address, just not a private address if I remember correctly. I'm not sure why the instructor is giving us (what I'm assuming) is a class C address with a /23 mask to work with in the first place.

pham0329

I think what MrBrian is referring to is that you have one octet too many

sentimetal

pham0329 wrote: »

I think what MrBrian is referring to is that you have one octet too many

Ah, woops. I didn't even catch that. I typed it out and copy/pasted the second time. Thanks.

Assuming I'm borrowing I'm correct, does that mean my network addresses will simply be (with a /25 mask):

200.0.5.0
200.0.5.128

Or is it divided into 4 subnetworks?

5.0, 5.128, 6.0, 6.128? <- (which makes no sense to me)