Stumped on subnet question

bnturnbo

In subnetting, when a question asks to configure the router with the last IP in the 4th usable subnet, what exactly are they looking for? For example, in a Class B network 130.4.0.0 / 255.255.252.0 (in the question, subnet 130.4.4.0 has already been assigned and is unavailable), would the answer be 130.4.12.2 or 130.4.16.2? I hope this makes sense. [/b]

jezg76

130.4.0.0 - unusable (no ip subnet-zero)
130.4.4.0 - 1st usable subnet
130.4.8.0 - 2nd usable subnet
130.4.12.0 - 3rd usable subnet
130.4.16.0 - 4th usable subnet (host range 130.4.16.1 - 130.4.19.254, broadcast 130.4.19.255)
130.4.20.0 - 5th usable subnet
.
.
.

130.4.19.254 would be my answer for the last IP in the 4th usable subnet.

nel

so if the 130.4.4.0 subnet is used that will mean the range 130.4.4.0 - 130.4.7.255 is used making 130.4.8.0 your first usable subnet.

130.4.8.0 - 130.4.11.255 - 1st range
.12.0 - 15.255 - 2nd range
.16.0 - 19.255 - 3rd range
.20.0 - 23.255 - 4th range

So your last usable host would be 130.4.23.254 assuming the 130.4.8.0 is your first usable range. Obviously it cant be 23.255 as that is the broadcast. but if you can start at the 130.4.4.0 range then 130.4.19.254 would be the last usable host

I think that correct anyway - my subnetting is a little rusty!

miller811

based on the information provided

130.4.4.0 subnet cannot be used
130.4.8.0 1st avialable subnet
130.4.12.0 2nd available subnet
13.4.1.16.0 3rd available subnet
134.4.1.20.0 4th available subnet

based on that the first available address of the subnet is 134.4.1.20.1
the last useable address in the subnet would be 134.4.1.23.254
the broadcast address would be 134.4.23.255

the fifith subnet would be one address higher 134.4.24.0

as you can see the subnets are going up in blocks of 4
the mask is 252 and the max available in each octet is 256
256-252=4 which tells the increment for the subnet.

bnturnbo

This is great! Thanks! I did type in the wrong answers and meant to ask if it should be 130.4.19.254 or 130.4.15.254. So, would the answer be the same if 130.4.8.0 was unavailable instead of 130.4.4.0? Thanks again all!

nel

Yes as your just doing a straight swap for another subnet before the last usable subnet and the last host is still in the fourth usable subnet

jezg76

I guess it's too late to change my answer? :P

Kaminsky

VERY QUICK WAY... About 5 seconds ....

Step 1: Work out the subnet boudaries. (look straight at the subnet mask bearing in mind class a, b, c) Do not even look at the ip address execpt for the first octect to give you a clue about the class.

255.255.252.0 (with a 130 first octect of ip address) implies the class B so ....

... 256
... 252 -
...

---

... 4

as it is 252.0, that makes the subnet 4.0

so the boundaries are 4.0, 8.0, 12.0, 16.0, 20.0, etc...

4th subnet starts at 16.0 but they are asking for the last usuable address. Quickest way to do this is go for the start of the 5th subnet (20.0) and subtract 1 (19.255) for the broadcast of the previous subnet and then one more for the last usuable host address. 130.4.19.254

Do it this way and it is soo quick to work out you will save yourself major time in the exam.

jezg76

Kam, that was my thinking until I realized; why are they telling you 130.4.4.0 is already in use? I would assume 130.4.0.0 and 130.4.4.0 are, in fact, not usable, per the question. Hence my being the first to respond and the first to be wrong.

Damn you ambiguity!

Kaminsky

Red herrings are a biatch aint they !

i) question asks to configure the router with the last IP in the 4th usable subnet
ii) in the question, subnet 130.4.4.0 has already been assigned and is unavailable

Someone already did the first subnet..... You are being asked to do the 4th !

Above posters read this to mean unavailable so +1 on the subnet boundaries incorrectly.

bnturnbo

Crystal clear now...thanks.

adam-b

Since SUBNET ZERO IS ASSUMED unless otherwise specified i would argue that the correct answer to this question is 130.4.15.254 since this is the last usable ip in the fourth usable subnet of 130.4.12.0 /22

ConstantlyLearning

adam-b wrote:

Since SUBNET ZERO IS ASSUMED unless otherwise specified i would argue that the correct answer to this question is 130.4.15.254 since this is the last usable ip in the fourth usable subnet of 130.4.12.0 /24

I agree.