Subnet Question

Technology1

Question: What is the last valid host on the subnetwork 172.31.134.16 255.255.255.240?


This type of question is much easier in my opinion. The MASK OCTET is octet 4, which is 240. Taking 240 and subtracting it from 256 shows that the network increment is 16, in the last octet.

The Network Address is 172.31.134.16 (As you can see the network address is an even number at all times)

The Broadcast Address 172.31.134.31 (As you can see the broadcast address is an odd number at all times)

The last valid host on this network is 172.31.134.30

IRONMONKUS wrote: »

You're welcome! By explaining it, it helps me to better understand it myself as I am new at learning this.

Given: 172.18.0.0
using default mask: 255.255.0.0

Binary:
11111111.11111111.00000000.00000000

1s = 16, 2^16 = 65,536 subnets available
0s = 16, 2^16-2 = 65,534 hosts per subnet

The difference is that you are not subnetting a subnet, but are using the whole Class B subnet range.

IRONMONKUS

Technology1 wrote: »

65,000 subnets with 65,000 hosts? Are you sure about this?

Thank you for catching that.

I looked this up and I'm right about the hosts, but not the subnets. I forgot when doing it that you can't use the subnets that are special, reserved or off-limits, like 127.0.0.1, etc...


So it would be

Class A: (1.0.0.0 to 126.0.0.0)
126 subnets
16,777,214 hosts per subnet

Class B: (128.0.0.0 to 191.255.0.0)
16,384 subnets
65,534 hosts per subnet

Class C: (192.0.1.0 to 223.255.255.0)
2,097,152 subnets
254 hosts per subnet

IRONMONKUS

Technology1 wrote: »

How many subnets and hosts per subnet can you get from the network 10.0.0.0/20?

This seems to be a pretty good method in determining the subnets available. 20 bits - the default 8 bit mask would give us 12 bits.

2^12 is 4096 subnets.

That is correct!

Now for the hosts part using binary, count up the remaining zeros and you get 12 as well.
2^12 - 2 = 4094 hosts per subnet.

Easy way... /20-32 = 12, 2^12 - 2 = 4094 hosts per subnet.

Sometimes I use a subnet calculator to see if the answers I get are correct.

Technology1

Another one.

Question: What is the first valid host on the subnetwork that the node 172.28.138.132 255.255.240.0 belongs to?

The MASK OCTET is octet 3, which is 240. We take 240 and subtract it from 256, which is the network increment number of 16, in the 3rd octet.

172.28.128.0 is the network address that this node belongs to.

The first valid host on this network is 172.28.128.1

The last valid host on this network is 172.28.143.254

The broadcast address is 172.28.143.255

As we can observe, anything in between the network address and broadcast address is the IP Host Range.

IRONMONKUS wrote: »

Thank you for catching that.

I looked this up and I'm right about the hosts, but not the subnets. I forgot when doing it that you can't use the subnets that are special, reserved or off-limits, like 127.0.0.1, etc...


So it would be

Class A: (1.0.0.0 to 126.0.0.0)
126 subnets
16,777,214 hosts per subnet

Class B: (128.0.0.0 to 191.255.0.0)
16,384 subnets
65,534

Class C: (192.0.1.0 to 223.255.255.0)
2,097,152 subnets
254 hosts per subnet

veritas_libertas

This thread is getting very interesting to read. I'm horrible with math, so any simplification helps me tremendously...

Technology1

The beauty of it is that it is very possible to turn it completely around and get a good grasp of it. Sharing knowledge will help us all out, especially with a topic such as subnetting that many find confusing. Turning it from confusing to very doable is a very good feeling and I think we can all master this topic with practice and encouragement.

veritas_libertas wrote: »

This thread is getting very interesting to read. I'm horrible with math, so any simplification helps me tremendously...

IRONMONKUS

Technology1 wrote: »

Question: What is the last valid host on the subnetwork 172.31.134.16 255.255.255.240?


This type of question is much easier in my opinion. The MASK OCTET is octet 4, which is 240. Taking 240 and subtracting it from 256 shows that the network increment is 16, in the last octet.

The Network Address is 172.31.134.16 (As you can see the network address is an even number at all times)

The Broadcast Address 172.31.134.31 (As you can see the broadcast address is an odd number at all times)

The last valid host on this network is 172.31.134.30

That is correct, good job!

The way I do it is I make a decimal to binary comparison and find the last 1 for the subnet we are looking for. In this case, it's 240.

128 64 32 16 8 4 2 1
1 1 1 1 0 0 0 0

The last 1 is a 16, so it increments by 16.

Write down the starting network and increment by 16, first and last usable address in parenthesis:

172.31.134.0 (1) -(14) 172.31.134.15
172.31.134.16 (17) - (30) 172.31.134.31
172.31.134.32 (33) - (46) 172.31.134.47

I actually write this out on paper.

Your way is by far faster than mine, but I have a hard time working with it if I can't see it visually, so I take baby steps and right out the binary.

Thank you for sharing!

Technology1

Another one.

What valid host range is the IP address 172.31.216.6 255.255.252.0 a part of?

Again, the MASK OCTET is in octet 3, which is 252. Take 252 and subtract it from 256 to get the Network Increment number of 4, in the 3rd octet. Always focus on the MASK OCTET.

The network address (even number) of this IP address is 172.31.216.0

The first valid host is 172.31.216.1

The last valid host is 172.31.219.254

The broadcast address (odd number is always 1 less than the next subnet. The broadcast address of this network is 172.31.219.255

The next subnet would be 172.31.220.0

veritas_libertas wrote: »

This thread is getting very interesting to read. I'm horrible with math, so any simplification helps me tremendously...

Technology1

Thanks. We all have different methods of learning this interesting topic, but with all of these positive contributions, I believe all of us can fully understand this with enough practice and dedication.

Thanks, Iron.

IRONMONKUS wrote: »

That is correct, good job!

The way I do it is I make a decimal to binary comparison and find the last 1 for the subnet we are looking for. In this case, it's 240.

128 64 32 16 8 4 2 1
1 1 1 1 0 0 0 0

The last 1 is a 16, so it increments by 16.

Write down the starting network and increment by 16, first and last usable address in parenthesis:

172.31.134.0 (1) -(14) 172.31.134.15
172.31.134.16 (17) - (30) 172.31.134.31
172.31.134.32 (33) - (46) 172.31.134.47

I actually write this out on paper.

Your way is by far faster than mine, but I have a hard time working with it if I can't see it visually, so I take baby steps and right out the binary.

Thank you for sharing!

IRONMONKUS

veritas_libertas wrote: »

This thread is getting very interesting to read. I'm horrible with math, so any simplification helps me tremendously...

The fun part is to think that I understand it, with the hope that I am not doing it wrong.

veritas_libertas

IRONMONKUS wrote: »

The fun part is to think that I understand it, with the hope that I am not doing it wrong.

LOL, every time I think I have it down I get too confident and mess a question up. Usually it's just over confidence catching up.

The question that really is eating at me is how to speed the process up. I guess it just comes down to practice, practice, practice.

IRONMONKUS

Technology1 wrote: »

Another one.

What valid host range is the IP address 172.31.216.6 255.255.252.0 a part of?

Again, the MASK OCTET is in octet 3, which is 252. Take 252 and subtract it from 256 to get the Network Increment number of 4, in the 3rd octet. Always focus on the MASK OCTET.

The network address (even number) of this IP address is 172.31.216.0

The first valid host is 172.31.216.1

The last valid host is 172.31.219.254

The broadcast address (odd number is always 1 less than the next subnet. The broadcast address of this network is 172.31.219.255

The next subnet would be 172.31.220.0

Excellent! How did you know the network it was on was 216? I'm intrigued as to find a different way of doing it.

The way I've been doing it is with binary math.

Take the subnet 252 and the network of 216 and turn them into binary.

11111100 = 252
11011000 = 216

Now use ANDing (1+1 = 1, anything else is a 0)

11011000 = sits on the 216 network

Technology1

The MASK OCTET is in octet 3 which is the number 252. You take 252 and subtract it from 256, which gives us the network increment number of 4.

You are focusing on the mask octet, which is octet 3.

4 times 50 is 200. 4 times 53 is 216, which is where the network begins.

The network increment is 4, in the 3rd octet of the IP address since the MASK OCTET is also in octet 3. If the Mask Octet of the subnet mask is in octet 3, then you would focus on the same octet in the IP address.

IRONMONKUS wrote: »

Excellent! How did you know the network it was on was 216? I'm intrigued as to find a different way of doing it.

The way I've been doing it is with binary math.

Take the subnet 252 and the network of 216 and turn them into binary.

11111100 = 252
11011000 = 216

Now use ANDing (1+1 = 1, anything else is a 0)

11011000 = sits on the 216 network

Technology1

I can't stress enough to focus on the MASK OCTET of the subnet mask and take that number and subtract it from 256 to find out what the network increment is in the same octet of the IP address.

Remember, we are only dealing with 8 special numbers (128, 192, 224, 240, 248, 252, 254, 255) for the subnet mask, and once these 8 numbers are memorized these types of questions become much easier to understand and appreciate.

If the mask octet is octet 2 in the subnet mask, focus on octet 2 in the IP address.

IRONMONKUS wrote: »

Excellent! How did you know the network it was on was 216? I'm intrigued as to find a different way of doing it.

The way I've been doing it is with binary math.

Take the subnet 252 and the network of 216 and turn them into binary.

11111100 = 252
11011000 = 216

Now use ANDing (1+1 = 1, anything else is a 0)

11011000 = sits on the 216 network

IRONMONKUS

Technology1 wrote: »

The MASK OCTET is in octet 3 which is the number 252. You take 252 and subtract it from 256, which gives us the network increment number of 4.

You are focusing on the mask octet, which is octet 3.

4 times 50 is 200. 4 times 53 is 216, which is where the network begins.

The network increment is 4, in the 3rd octet of the IP address since the MASK OCTET is also in octet 3. If the Mask Octet of the subnet mask is in octet 3, then you would focus on the same octet in the IP address.

Excellent! Thank you for that explanation.

IRONMONKUS

Technology1 wrote: »

65,000 subnets with 65,000 hosts? Are you sure about this?

Going back to this, I'm glad that we will probably never end up actually subnetting a whole class range.

I am confused right now. The chart I found online said this:

Class B
16,384 subnets
65,534 hosts per subnet

---

I checked a subnet calculator and it said for the network 172.18.0.0 it would be:

1 subnet
65,534 hosts

This makes sense, because for the 255.255.0.0, all bits in the first two octets are turned on, meaning that you can't change the numbers 172.18, leaving only hosts left.

---

I then checked out a VLSM (variable length subnet mask) calculator and it is

65,536 subnets
65,534 hosts

---

Yeah, so my mind is now blown! I guess I'll try and figure out VLSMs when I get to that point in my studies.

Technology1

It was good we took time to practice. These are the types of learning experiences that will help us to continue to improve in the future with subnetting.

Thanks.

IRONMONKUS wrote: »

Going back to this, I'm glad that we will probably never end up actually subnetting a whole class range.

I am confused right now. The chart I found online said this:

Class B
16,384 subnets
65,534 hosts per subnet

---

I checked a subnet calculator and it said for the network 172.18.0.0 it would be:

1 subnet
65,534 hosts

This makes sense, because for the 255.255.0.0, all bits in the first two octets are turned on, meaning that you can't change the numbers 172.18, leaving only hosts left.

---

I then checked out a VLSM (variable length subnet mask) calculator and it is

65,536 subnets
65,534 hosts

---

Yeah, so my mind is now blown! I guess I'll try and figure out VLSMs when I get to that point in my studies.