subnetting beyond one octet

I am fine with subnetting any class C,

but this one:

What is the first valid host on the subnetwork that the node 10.168.204.131/20 belongs to?

I have trouuble whit this kind, can anyone offer advice?

Find more posts tagged with

Save $250 on 2025 certification boot camps from Infosec!

Book now with code EOY2025

Button

Comments

ilcram19

10.168.204.131/20

10.168.204.131
255.255.240.0

10.168.192.0 network id
10.168.192.1
10.168.207.254 thats the ip range
10.168.207.255 broadcast

so it belongs to the 192.0 subnet

liven

hmmm

applying wildfires method from this post:

http://www.techexams.net/forums/viewtopic.php?t=15678

seems to be working....

ilcram19

sorry dats the first host 10.168.192.1
last host 10.168.207.254

r_durant

Here's how I would do it, subject to correction...

What is the first valid host on the subnetwork that the node 10.168.204.131/20 belongs to?

SO we have a 10 network, class A...which means it's using 12 subnet bits...

11111111.11111111.11110000.0000

SO we know from the 3rd octet, we're using a 240 mask, add the values of the 1's in the 3rd octet (128+64+32+16), which gives a 16-block of ip addresses (256 - 240 = 16), so your subnet addresses go...
The 1st and 2nd octets are fixed, since they are all 1's...3rd octet is being subnetted...

10.168.0.0
10.168.16.0
10.168.32.0
.
10.168.192.0
10.168.208.0
.
etc...

So your ip address falls in the 10.168.192.0 subnet and the first valid host would be 10.168.192.1 and your last valid is 10.168.207.254, with broadcast 10.168.207.255

wait2dominate

liven wrote:

I am fine with subnetting any class C,

but this one:

What is the first valid host on the subnetwork that the node 10.168.204.131/20 belongs to?

I have trouuble whit this kind, can anyone offer advice?

The way I see these is very similar to a class C, in sorts.

In a class C, you know to modify/subnet the last octet, the mask has to be a /24 to a /30.

To apply the VLSM in that order, for the 3rd octet to be modified/subnetted, the subnet mask needs to be a /16 to a /23. You get those numbers from the class b address, where. Same type of thing applys to a 'class A', where the second octet gets modified/subnetted if the mask is between a /8 and a /15.

Now....to properly subnet those like you have above.

You start with a /20 mask. You know because a class b has a /16 mask and a class c has a /24 mask by default that the subnet will be truely affecting the third octet. It has 4 more bits than the normal class b, so you know four bits will be modified.

4 bits make 240 for that octet, so you have a 255.255.240.0 subnet mask. You know from that subnet mask of 240 that each network will span '16' addresses. Starting from there
10.168.0.0
10.168.16.0
10.168.32.0
.......
10.168.192.0
10.168.208.0

You know it falls between the last two network addresses listed, so the network address is 10.168.192.0/20, the addresses usable are 10.168.192.1-10.168.207.254, and the broadcast address is 10.168.207.255.

Let me know if you need that clarified anymore, as I'm a bit under the weather right now and am sure I skipped a few steps.

rakem

i would do this question as follows:

What is the first valid host on the subnetwork that the node 10.168.204.131/20 belongs to?

1)
change subnet mask to binary

11111111.11111111.11110000.000000 (=255.255.255.240)

so now you know that the lowest bit value in the 3rd octec is 16. You can either count the bits or just do the 256 - 240 sum. Both will equal 16.

2)
So 16 is the value in which each network increments... (which also gives you 14 hosts per subnet and 4096 subnets)

so its
10.168.0.0 > 10.168.16.0
10.168.16.0 > 10.168.32.0
10.168.32.0 > 10.168.48.0
........
10.168.176.0 > 10.168.192.0
10.168.192.0 > 10.168.208.0 <-the address of 10.168.204.131 is in this range..

3)
First valid host is 10.168.192.1
Last valid host is 10.168.207.254
Network address is 10.168.192.0
Broadcast address is 10.168.207.255

easy!

r_durant

rakem...I think there are more than 14 hosts per subnet

rakem wrote:

which also gives you 14 hosts per subnet and 4096 subnets

You have 12 host bits...2^12 - 2 which gives 4094 hosts

NeonNoodle

Here's another one. You can find this in at least one CCNA book, I'm sure. To do this you need to practice converting from decimal to binary and back and you need to know what the 'and' function is.

With enough practice you will learn which octets, given a subnet mask, really need to be converted from decimal to binary and back. But until you really know what you're doing, I'd avoid any shortcuts because it will only come back to confuse you when you are taking the CCNA exam. Learn to do it right and learn to do it quick.

1. Convert the ip address to binary
2. Convert the subnet mask into binary
3. 'and' 1 and 2 to get the subnet address
4. Add 1 to the subnet address to find the first host
5. Invert the host bits to find the broadcast address
6. Subtract one to get the last host
7. Convert 3, 4, 5, 6 back to dotted decimal.

So, with IP address 10.168.204.131 and subnet mask 255.255.240.0

00001010.10101000.11001100.10000011 (Step 1)
11111111.11111111.11110000.00000000 (Step 2)
('and' the two to get)
00001010.10101000.11000000.00000000 (Step 3)
(add 1)
00001010.10101000.11000000.00000001 (Step 4)
(invert the host bits)
00001010.10101000.11001111.11111111 (Step 5)
(subtract 1)
00001010.10101000.11001111.11111110 (Step 6)

10.168.192.0 <---- subnet number (Step 7)
10.168.192.1 <---- first host
10.168.207.255 <---- broadcast address
10.168.207.254 <---- last host

rjbarlow

I don't understand why all these problems with non-standard masks... All You have to know is that the variable subnet number is into the ultimate subnetted octet and then You put down all remaining bits to 0 for the complete subnet number, surely You have to understand how to interpret ip address and subnet mask in bits and how they work in binary, however when You have learned the method of magic number is a joke.

NeonNoodle

rjbarlow wrote:

I don't understand why all these problems with non-standard masks... All You have to know is that the variable subnet number is into the ultimate subnetted octet and then You put down all remaining bits to 0 for the complete subnet number, surely You have to understand how to interpret ip address and subnet mask in bits and how they work in binary, however when You have learned the method of magic number is a joke.

Well, rather than talk about it show us. Maybe some of us aren't aware of your method.

rjbarlow

Well, rather than talk about it show us. Maybe some of us aren't aware of your method.

The method of the magic number is not mine is in the CCNA intro book of Odom, then is sufficient learn to him, however is pretty close to the method You described, only with a trick in decimal math, I have sense that much people want to learn subnetting without convert octets in binary, that is absolutely useless.

markzab

rjbarlow wrote:

Well, rather than talk about it show us. Maybe some of us aren't aware of your method.

The method of the magic number is not mine is in the CCNA intro book of Odom, then is sufficient learn to him, however is pretty close to the method You described, only with a trick in decimal math, I have sense that much people want to learn subnetting without convert octets in binary, that is absolutely useless.

It's useless if you don't know how to do it.