Subnetting questions

bon_chan

Hello happy people,

First of all I have a general question regarding subnetting during the exam.
Do you guys use paper? I mean do you write down your numbers to count,...
Or do you guys do it all quickly in your mind?

Second of all, I am confused regarding this example:

What is the first valid host on the subnetwork that the node 172.18.92.105/23 belongs to?

No worries on that, I found the right answer, which is: 172.18.92.1
But then I asked myself, the number of hosts that this subnetwork had.

I think it had 510 hosts, as there are 9 hosts bits, so (2^9)-2 = 510

I then wrote down the host range, which I think is: 172.18.92.1 to 172.18.93.254
But what confuses me is that, this range contains 2 unvalid addresses: 172.18.92.255 and 172.18.93.0

So I don't get it, since we can't have a host having a ".0" address.
Could somebody shed some light on this please.

fsanyee

No papers.
All the addresses from 172.18.92.1 to 172.18.93.254 are useable and valid!
You can have host with .0 or .255 address, it depends on the subnet mask.

Roguetadhg

The test place I had gave you a dry-erase marker (the tip Resembled writing with grade-school markers), nothing to wipe it clean (besides spit and my finger) and something to write on the size of half a sheet of paper. I wrote down the 128,64,32,16... and the hosts possible for each bit, just in case. I only needed it once, but it was nice to have down.

92.105 = 0101110|0.01101001
Net# is 92.0 = 0101110|0.00000000
Broadcast# 93.255 = 0101110|1.11111111

It's 510 hosts.

I'm not sure what you're asking otherwise. Everything between is usable.

bon_chan

Ok thx

I was asking whether 92.255 and 93.0 were valid hosts, as I always thought addresses ending with a .0 or .255 couldn't be.

Aman_saini

yes 92.255 and 93.0 are usable with respect to the given mask.
and for your question

First of all I have a general question regarding subnetting during the exam.
Do you guys use paper? I mean do you write down your numbers to count,...
Or do you guys do it all quickly in your mind?

i prefer to do it in mind, it take some time to get it but once you start it no number looks big. For example i look your example 172.18.92.105/23 as 9 host bits. That is 2 raise to 9, minus 2 are hosts 510 then 92.0 is the subnet. octet=92.0 to 93.255 with 92.1 to 93.254 usable hosts.
hope this helps

Roguetadhg

It's good to do it in your head, you'll need to be able to master subnetting. You'll be asked questions which being able to subnet quickly will eliminate or provide answers. I look at the subnet mask (using supplied address from http://faculty.valleycollege.net/rpowell/jscript/subnet2.htm):
1. 139.29.105.117 /16
2. 16 = 2 Octets
3. 29 is the network address, the rest are hosts.
4. 139.29.0.1 - 139.29.255.254 w/ .255 being broadcast.

For all intents and purposes I skip thinking about "What class is this?"

Another problem:
58.1.147.153/8
58.0.0.0 - 58.255.255.255

Somehow i'm getting lucky with these random questions. Toss in some 26s is where I have to sit down and do the math into binary to make sure I don't fudge it up.

Here we go, here's a good example: (4 more clicks later)
1. 50.114.170.27/12
2. 12 - 8 = 4 mask bits left over. The final mask being x.240.z.z
3. 240 = a mask layer of XXXX|ZZZZ. 16th bit. From constant practice I know that 240 and 248 are the center-two mask numbers. The thought immediately shoots through my head of "okay, the range is...", but I try to ignore it since I don't know the network number, making the range point moot.
4. 114 into binary = [I write down my 128,64,32,16,...]
5. [subtract 64 from 114 = 50]
6. 50 - 32 = I take 2 away from 30, making the subtraction easier (50 - 30) = 20. I subtract 2 more for 18 ([50 - 30]-2] = 20 - 2 = 18. The process is called Sutra. Like Karma Sutra
7. 18 - 16 = 2
8. [I skip the other numbers because I know there's a that'll fit]
9. 2 - 2 = 0. I jot down the 1s for each subtraction, forgetting the 0s (Sometime I forget all the zeros. I don't recommend this step)
10. binary: 01 11 00 10
11. I see it as 0111|0010.
12. Subtracting 2 from 114 = 112.
13. the network address is: 50.112.0.0
14. I look at the binary number and subtract one from 128 (1000|0000). Making the broadcast 50.127.255.255 Because the broadcast are all 1s [Im thinking 0111|1111],

That's my entire thought process, written for everyone to look at my colored binary mathmatics. It's written as I think about the question, shortcuts may make the math look weird, but it works for me.

Basicly, just keep doing subnetting. When you go to study, do 10. It takes a lot of practicing. I use that site which I linked above. I still do 10 to this day, to make sure my skills are sharp.


VLSM Practicing:
I also use this site http://www.random.org/sequences/?min=2&max=200&col=3&format=html&rnd=new at the same time to practice VLSM. The first number in each column equals hosts. I'm not sure if anyone else does this type of VLSM practice, i've struggled to find any way to think of numbers BESIDE the number 50. For whatever the reason there would be a 50-host or a 200-host lan each time I would try.


I always use a 3 router triangle topology. Think this picture:

Site for subnet: http://faculty.valleycollege.net/rpowell/jscript/subnet2.htm
Site for random numbers: http://www.random.org/sequences/?min=2&max=200&col=3&format=html&rnd=new

Subnet: 129.76.95.195 /18 **
Hosts per lans: 25**, 145**, 195**, 2, 2, 2.
VLSM for 6 networks (3 networks between the routers, 3 host networks). Find LEAST possible host.

** = Random numbers. Gotta love it.

fsanyee

This site is very good too:
subnettingquestions.com - Free Subnetting Questions and Answers Randomly Generated Online

bon_chan

Thank you very much guys