Sub netting question

Could someone explain this one

Which subnet does host 192.168.129.206/30 belong to?

Answer: 192.168.129.204
The subnet mask I calculated is 255.255.255.252?
So number of networks here are 2^8=256?
And the number of hosts =2^2-2=2?
I don't understand the answer here.

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Comments

TechGuru80

The subnet mask you have is correct. So the first three octets 192.168.129.x remain the same because 255.255.255.x.

Networks = 2^6 (you borrowed 6 bits 30 - 24 = 6) == 64 subnets

Range of networks = 2^8-6 == 2^2 = 4 per network

Hosts = range - 2 .... == 4-2 == 2 hosts per network.

I start at 192..
192-195
196-199
200-203
204-207 *********** 206 is in this network, 204 is the network address, and 207 is the broadcast address.
208-211

Each network will start on the previous plus 4.

sarahanand

Thanks TechGuru80,

I did get the entire thing right. But its after I found out the range that I stopped. What I do after getting the range is starting to count from "0". But could u explain how and y did u start to count from 192 range?

TechGuru80

128 and 64 are evenly divisible so especially on a 4, 8, 16, 32, 64 network setup you will have those points as a network start. When all else fails you can do the math but counting from 0 in increments of 4 probably took you what 1-2 minutes? I did that problem in about 10 seconds because of tricks like that. It takes a good amount of practice but when you have the fundamentals of the networks, ranges, bits borrowed you should be able to fly through the questions.

Checkout this website for tons of subnetting practice....Subnetting Practice Questions

smcclenaghan

Not sure how he does it, but I've memorized some "stopping points" in subnet math.

/24 = 256 hosts = 255.255.255.0
/25 = 128 hosts = 255.255.255.128
/26 = 64 hosts = 255.255.255.192
/27 = 32 hosts = 255.255.255.224
/28 = 16 hosts = 255.255.255.240
(of course you're subtracting two hosts: one for network address and one for broadcast address, but I still think in terms of 16, 32, 64, 128, etc...)

etc...

so if you're starting point is /30 (so, by 4s) well you know 4 is in 128 and 64, so 128+64=192 is a good (and close) place to start.

128 + 64 + 32 = 224 which is too much (and I hate counting backwards).

I'm not sure that's what TechGuru meant, but that's how I think about it.

sarahanand

Right, Thank u, but may i ask one more please

Whats is the last valid host on the subnetwork?
192.168.201 255.255.255.192

The Answer is 192.168.201.254.
But here's what I've calculated
CIDR = /26
So the range is 0-63 and 64-127.
Its wrong I guess...could u help?

smcclenaghan

sarahanand wrote: »

Right, Thank u, but may i ask one more please

Whats is the last valid host on the subnetwork?
192.168.201 255.255.255.192

The Answer is 192.168.201.254.
But here's what I've calculated
CIDR = /26
So the range is 0-63 and 64-127.
Its wrong I guess...could u help?

We haven't got enough info because the network is 192.168.201 and the mask is 255.255.255.192. We're not sure if you mean 192.168.201.0, .64, .128, or .192.

In either network, the answer is .63, .127, .191, or .255... except that you have to subtract one (since the last address is a broadcast address and isn't usable as a host address) so .62, .126, .190 or .254 is the last address usable as a host.

sarahanand

Oh my mistake...its 192.168.201.192

smcclenaghan

No worries.

So for 192.168.201.192, it's the same thing. You already figured out the range (192-255), now you just have to remember that you're not allowed to use the very last address as a host because the very last address (192.168.201.255) is a broadcast address for all of 192.168.201.192/26.

Does that help?

D-star

Great response. To add to this you know your incrementation is going to be 4(because of the /30) so you just need multiples of 4 and this is a default class C so you are only worried about the last octet. They closest you will get with very little thought is 200. So you can binary out x.x.x.200 and see that the subnet ends at x.x.x.203, note broadcast address will always be odd. Now you know where to increment in the subnets 204-207, 208-211, ect.

This method is not the fast but once you under stand the concepts you can breeze through.

sarahanand

Oh I think I hav overloaded myself with subnetting for today. I just did not see how silly I was being here..Need to rest I guess.

smcclenaghan

I think you're sweating it too much if you're worried about being silly.

Click through my post history to see real examples of silly

Good luck. And yeah, enjoy a much needed break.

D-star

sarahanand wrote: »

Oh I think I hav overloaded myself with subnetting for today. I just did not see how silly I was being here..Need to rest I guess.

Well don't feel silly as this is the place to get stuff wrong and learn, we are not going to grade you.

sarahanand

Right I am going to go watch Breaking Bad...a Well deserved badly needed break!

smcclenaghan

sarahanand wrote: »

Right I am going to go watch Breaking Bad...a Well deserved badly needed break!

AMAZING show. If you've avoided spoilers so far, log off the internet immediately. And stay off facebook for the next two weeks while they air the last two. Pretty much impossible to avoid some clown's comments like ten minutes before you finally sit down to watch it.

sarahanand

Its climax time for both Breaking Bad and Dexter....Dexter however has morphed itself into senseless stuff now.

smcclenaghan

sarahanand wrote: »

Its climax time for both Breaking Bad and Dexter....Dexter however has morphed itself into senseless stuff now.

I'm hearing that. I haven't watched the last two episodes though. I'll catch up this weekend. It's the last season and I stuck it out this far...