studying for MCSE, what is a /23 ??

C

cyr0n_k0r

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Here is the jist of the question.

You have 2 subnets
A is 10.1.1.0/24
B is 10.1.2.0 /24

each subnet has 200 workstations. Question setup is that your router and other equipment cannot handle another subnet, but you need to add an additional 200 workstations to subnet A. You ISP has given you another /24 of 10.1.3.0.
Choose the best setup to be able to fit 400 workstations on subnet A. Here are the options.

1) use 10.1.1.0/24 (obviously wrong)
2) use 10.1.1.0/23
3) use 10.1.2.0/23
4) use 10.1.2.0/24 (obviously wrong)

I keep choosing #2 as the answer but it keeps telling me I'm wrong. But it doesn't explain WHY I'm wrong.

Can anyone help me out on what I'm missing?
 
Maybe the question is wrong? I can see supernetting 10.1.2.x (Subnet B) and 10.1.3.x (C) using /23. Making the answer C.
 
I'd say #2...

If it needs to go on network A, it's got to be 10.1.1.0....
And it's got to use a different subnet, hence the 23.
However, that puts it using three subnets (Which negates the first part of your equipment not supporting more than 2).
A is 10.1.1.0/24
B is 10.1.2.0 /24
C is 10.1.1.0/23


Off the hip answer for me would be #2 as well.

But it's been awhile since I've done subnetting, so I may not know the best ;)
 
#2 is wrong because x.x.1.0/23 is not a network address. the network address for this subnet would be x.x.0.0/23, which you do not own, so you can't combine it.

the correct answer is 3, giving you 2 subnets -> 10.1.1.0/24 and 10.1.2.0/23
 
By my reasoning:

You have 2 subnets
A is 10.1.1.0/24
B is 10.1.2.0 /24

Each subnet can have a max of 254 hosts with a host range of 10.1.1.1 - 10.1.1.254 on subnet A.

This is sufficient for 200 hosts, but not for the 400 needed.

1) use 10.1.1.0/24 (obviously wrong) - Yes, 10.1.1.0/24 can only support 254 hosts.

2) use 10.1.1.0/23 - This will give you the correct number of hosts (510), but the subnet ID becomes 10.1.0.0, a subnet which you don't own.

A class A address has 8 bits for the network(N), and /23 means 23 bits are reserved for subnetting(S). Its confusing because this number includes the network part as well. With a 32 bit IP address that leaves 32 - 23 = 9 bits as host bits (H). This is the structure of the address in question 2.

NNNNNNNN . SSSSSSSS . SSSSSSSH . HHHHHHHH

So 10.1.1.0/23 is interpreted in binary as:

00001010 = 10 = The N's

00000001 = 1 = The second octet (first octet of S's)

00000001 = 1 = The third octet. Notice here that the S's are all 0. The only 1 is an H which is a host bit.

00000000 = 0 = The fourth octet (all H's).

A subnet address ignores all of the host bits (makes them 0's) and simply is the network bits and the subnet bits. That means the subnet address is 10.1.0.0 which was not allocated to you in the question.

3) use 10.1.2.0/23 - This is correct. You now get 510 hosts/subnet with the addresses being 10.1.2.1 - 10.1.3.254. Since you have been allocated 10.1.2.0 and 10.1.3.0 this works out fine. You would then use the 10.1.1.0/24 for the other 200 hosts.

You do not run into the same problem as answer #2 because the third octet is still:

SSSSSSSH

but it is now .2 or in binary:

00000010

Which makes the subnet address 10.1.2.0, which IS an address that was allocated to you.



For these kinds of problems thinking in terms of binary helps a lot.
 
i think answer should be C (it can't be B cuz it overlaps with with subnet B)

they expect you to move what was in net B to net A, and then have the 200 from A plus the 200 new ones into 10.1.2.0/23

either that or they DO expect you to move subnet B computers into 10.1.3.0/24 and expand A by using 10.1.1.0/23, but you say that is wrong (plus the guy above me seems to have found a technicality that makes the subnet ID bogus, he's right its gonna be an even number)

that is the way i look at it... but yea it is a retarded question... they should have said "add 200 more addresses to subnet B" but they probably didn't want to make it so obvious
 
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