Who knows how to subnet? Solve this puzzle >.<

kerbydogg

This question came up on a practice exam....

Which of the following network addresses enable you to support a network consisting of 60 subnets with up to 300 hosts on each subnet? (choose all that apply)

A. 10.48.0.0/14

B. 192.168.98.0/24

C. 172.30.0.0/25

D. 10.236.128.0/21

dynamik

Well you'll need 6 bits for the subnets (64 subnets) and 9 bits for the hosts (510 hosts).

Therefore, you'll 15 bits total. There's only one that satisfies that.

kerbydogg

Exactly what I was thinking!

I gave only one answer ...

The test gave 2 answers


Either I been studying too much lately and my brain is getting fried or something is wrong.

1MeanAdmin

I haven't subnetted for a while, but isn't (A) and (D) the correct answer?
Aren't (B) and (C) wrong because they only allow 256(-2) and 128(-2) hosts per subnet?

(A) and (D) have enough (6 and 13) bits for subnets and enough (18 and 11) bits for hosts?

Talic

(A) and (D) are right, download TE's subnet calc: http://www.techexams.net/ip-subnet-calculator/

dynamik

I see what you guys did, but they're not using the standard class masks. If you're given 10.236.128.0/21 to use, you only have 11 bits to work with.

1MeanAdmin

dynamik wrote:

I see what you guys did, but they're not using the standard class masks.

It depends on the interpretation of what's given (this question is very vague and I may not be correct). Your assumption could be correct if the question was something like:
"Your department is given a 10.236.128.0/21 network segment [meaning you only have this segment and, therefore, have to FURTHER subnet it into subnets/hosts]."

The actual question asks us which subnets could be a PART OF an original network address that was, at some point, subnetted into answer(A) or answer(D).
What I mean is if my math teacher gives me 2*X=Y+2 and wants me to find X, I should find X, not Y.

dynamik wrote:

If you're given 10.236.128.0/21 to use, you only have 11 bits to work with.

11 bits to work with hosts.
Since the range for private 10-dot network is 10.x.x.x, you can use any amount of bits from all octets except first 8 and last 11 bits. So, 21 - 8 = 13 bits for subnets.
10.0.0.0 /21 First subnet
10.0.8.0 /21
10.0.16.0 /21

---

10.236.128.0 /21 Given subnet

---

10.255.248.0 .21 Last subnet

dynamik

MeanAdmin wrote:

It depends on the interpretation of what's given (this question is very vague and I may not be correct). Your assumption could be correct if the question was something like:
"Your department is given a 10.236.128.0/21 network segment [meaning you only have this segment and, therefore, have to FURTHER subnet it into subnets/hosts]."

The actual question asks us which subnets could be a PART OF the original network address that was, at some point, subnetted into answer(A) or answer(D).

The former is how I'm interpreting it. It doesn't say "part of" anywhere in there; that's you reading into it. If someone told me to use those network addresses with their respective subnet masks to fulfill the requirements, I would say that A is the only valid choice. I'm not saying what you guys did is wrong. Upon knowing the answer, that seems to be what the question was after. I just don't think the question's phrased properly if that solution is what it's looking for.

Regardless kerby, I think you know what you're doing. I wouldn't get hung up over a question like this.