Subnet Question

Can a Cisco Expert please explain the answers to these questions in detail.

Thanks.

Question 1: How many subnets and hosts per subnet can you get from the network 172.25.0.0/27?"

Question 2: How many subnets and hosts per subnet can you get from the network 10.0.0.0 255.255.240.0?

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Subnet Mask **** Sheet

Thanks for the info. Was hoping someone with expert knowledge could actually go into the details of explaining the answer to these questions in regard to finding how many subnets and hosts per subnet are available.

We all have our methods of what works for us when figuring subnet masks which is pretty cool. Finding out a particular subnet from a host address is not too much of a concern, but the other way around in finding how many subnets and hosts per subnet seems to get somewhat confusing at times. Could possibly be making it out harder than it really is. CIDR notation is not an issue. Finding a subnet from a given host address is not an issue. Finding a broadcast address is not an issue. IP host range is not an issue.

The 8 special numbers for the subnet mask are 128, 192, 224, 240, 248, 252, 254, 255. Taking any of these numbers and subtracting them from 256 would tell us what the network increments are.

Network Addresses are always even numbers and Broadcast Addresses are always odd numbers. Anything in between a network address and a broadcast address is the IP host range.

These particular questions seem somewhat confusing at times when it comes to finding out how many subnets and hosts per subnet are available.

Thanks to all the Cisco Experts who wish to go into detail in explaining the answers to these questions.

Question 1: How many subnets and hosts per subnet can you get from the network 172.25.0.0/27?"

Question 2: How many subnets and hosts per subnet can you get from the network 10.0.0.0 255.255.240.0?

saied45 wrote: »

Subnet Mask **** Sheet

ah okay. i just started studying for my ccna and havent gotten to subnetting( i hear its dreadful tho) but good luck

It's considered dreadful for many because simple tips are never discussed such as the 8 special numbers that keep showing up in any subnet mask available. I know most of it, but the questions just posted get somewhat confusing at times.

An absolute must to memorize the 8 special numbers of the subnet mask as they are the only numbers that will ever be used in an IPv4 subnet mask.

128, 192, 224, 240, 248, 252, 254, 255

saied45 wrote: »

ah okay. i just started studying for my ccna and havent gotten to subnetting( i hear its dreadful tho) but good luck

haha i have no idea what ur talking about but i guess i will in about 3 chapters

veritas_libertas

If you read this thoroughly you should be able to do those with no trouble:

https://www.techexams.net/forums/ccna-ccent/38772-subnetting-made-easy.html

veritas_libertas

Technology1 wrote: »

It's considered dreadful for many because simple tips are never discussed such as the 8 special numbers that keep showing up in any subnet mask available. I know most of it, but the questions just posted get somewhat confusing at times.

An absolute must to memorize the 8 special numbers of the subnet mask as they are the only numbers that will ever be used in an IPv4 subnet mask.

128, 192, 224, 240, 248, 252, 254, 255

Interesting, would you like to elaborate?

Elaborate about the 8 special numbers of the subnet mask?

Yes, they are 128, 192, 224, 240, 248, 252, 254, 255.

Special trick in finding out what the network increments are is by taking any of these 8 numbers and subtracting them from 256.

For instance, if the mask is 255.255.240.0 you would look at octet 3, which is considered the Mask Octet.

You take the 240 and subtract it from 256 which would give you the number of 16. You now know what the network increments are. The network increments are 16.

Take another subnet mask of 255.224.0.0

The MASK OCTET is octet 2, which is 224. You take the 224 from the mask octet and subtract it from 256, which gives you the network increment number of 32.

Look at any subnet mask and these 8 numbers will constantly show up as they are the only numbers that can ever be used in an IPv4 Subnet Mask.

veritas_libertas wrote: »

Interesting, would you like to elaborate?

And if you want to take it a step further, go into your wireless router and attempt to alter your subnet mask and enter a number outside of the 8 special numbers and you will be alerted by your router that only the 8 special numbers are available to use in the subnet mask.

veritas_libertas wrote: »

Interesting, would you like to elaborate?

I'm no expert, but here's how I would do it.

Given: 172.25.0.0/27

You know this is a class B address.
Default subnet mask: 255.255.0.0
Binary: 11111111.11111111.00000000.00000000

You are given the subnet of /27 or 255.255.255.224
Binary: 11111111.11111111.11111111.11100000

You borrowed 11 bits from the host portion to give you /27 from the default subnet of /16

We know that 1s equal the subnet (network) and the 0s equal the hosts.

Compare the default subnet to the new subnet.
11111111.11111111.00000000.00000000
11111111.11111111.11111111.11100000

Count the extra 1s and it equals 11. So, 2^11 = 2048 subnets available.

Count up the remaining 0s and it equals 5. So, 2^5-2 = 30 hosts per subnet.

You minus 2 when calculating hosts, because you cannot use the network or broadcast address for each network.

Check out this thread as well Subnetting made easy they have some really good information there.

Iron,

Thanks for your response. This is very interesting. Since the IP address is a Class B address and the mask octet is in octet 4, do you always have to compare the default subnet mask with the one that is given here, which is in octet 4?

Thanks.

IRONMONKUS wrote: »

I'm no expert, but here's how I would do it.

Given: 172.25.0.0/27

You know this is a class B address.
Default subnet mask: 255.255.0.0
Binary: 11111111.11111111.00000000.00000000

You are given the subnet of /27 or 255.255.255.224
Binary: 11111111.11111111.11111111.11100000

You borrowed 11 bits from the host portion to give you /27 from the default subnet of /16

We know that 1s equal the subnet (network) and the 0s equal the hosts.

Compare the default subnet to the new subnet.
11111111.11111111.00000000.00000000
11111111.11111111.11111111.11100000

Count the extra 1s and it equals 11. So, 2^11 = 2048 subnets available.

Count up the remaining 0s and it equals 5. So, 2^5-2 = 30 hosts per subnet.

You minus 2 when calculating hosts, because you cannot use the network or broadcast address for each network.

Check out this thread as well Subnetting made easy they have some really good information there.

Technology1 wrote: »

An absolute must to memorize the 8 special numbers of the subnet mask as they are the only numbers that will ever be used in an IPv4 subnet mask.

128, 192, 224, 240, 248, 252, 254, 255

With this in mind, I was reading Todd Lammle's book and he mentioned that when subnetting a class C address, you can only use 6 out of those 8 special numbers, because when creating a subnet, you have to have at least 2 bits available to have any hosts on that subnet.

For class C: 128, 192, 224, 248, 252

Another thing I was thinking is that we know right away that the default subnet mask is 16 bits, which is 255.255.0.0

If you take /27, which is 27 bits and subtract 16 from it, you immediately get the number 11.

Do you think that makes it any easier in determining the number 11 instead of going into binary to figure it out?

IRONMONKUS wrote: »

I'm no expert, but here's how I would do it.

Given: 172.25.0.0/27

You know this is a class B address.
Default subnet mask: 255.255.0.0
Binary: 11111111.11111111.00000000.00000000

You are given the subnet of /27 or 255.255.255.224
Binary: 11111111.11111111.11111111.11100000

You borrowed 11 bits from the host portion to give you /27 from the default subnet of /16

We know that 1s equal the subnet (network) and the 0s equal the hosts.

Compare the default subnet to the new subnet.
11111111.11111111.00000000.00000000
11111111.11111111.11111111.11100000

Count the extra 1s and it equals 11. So, 2^11 = 2048 subnets available.

Count up the remaining 0s and it equals 5. So, 2^5-2 = 30 hosts per subnet.

You minus 2 when calculating hosts, because you cannot use the network or broadcast address for each network.

Check out this thread as well Subnetting made easy they have some really good information there.

Good info. Thanks for the tip.

IRONMONKUS wrote: »

With this in mind, I was reading Todd Lammle's book and he mentioned that when subnetting a class C address, you can only use 6 out of those 8 special numbers, because when creating a subnet, you have to have at least 2 bits available to have any hosts on that subnet.

For class C: 128, 192, 224, 248, 252

wow thanks for the help. i mainly got it except for one thing how does he figure out which octate he borrowed it from. like he said here

We have borrowed from the last octet as the 29th bit is in the last octet. We start from zero and count up in our block size. Therefore it follows that the subnets are:

sorry if it looks like im stealing the thread OP, i just really wanted to know

Post as much helpful info as you want. We can never learn enough about this topic, and any helpful knowledge is absolutely embraced with appreciation.

saied45 wrote: »

wow thanks for the help. i mainly got it except for one thing how does he figure out which octate he borrowed it from. like he said here

We have borrowed from the last octet as the 29th bit is in the last octet. We start from zero and count up in our block size. Therefore it follows that the subnets are:

sorry if it looks like im stealing the thread OP, i just really wanted to know

Technology1 wrote: »

Iron,

Thanks for your response. This is very interesting. Since the IP address is a Class B address and the mask octet is in octet 4, do you always have to compare the default subnet mask with the one that is given here, which is in octet 4?

Thanks.

Yes, that is how I have learned it and it makes sense to me to see it visually using binary. You reference the default subnet masks, because you need to know how many bits you have or will borrow from the hosts.

In the subnetting made easy thread, they use binary math plus the default subnets 8, 16, 24, 32 to answer the questions. Check it out and see if that method works for you.

The method that works for me is a cross between CBT nuggets and Train signal. I only use the train signal method when trying to find what network a certain subnet is in. It's fast and easy, no counting up from zero. I use the CBT nuggets version to do everything else.

As I was saying earlier, please correct me if I'm wrong, but as we know the default was 16 bits and the /27 is 27 bits, would it not be easier to subtract 16 from 27 to get the number 11?

Thanks.

IRONMONKUS wrote: »

Yes, that is how I have learned it and it makes sense to me to see it visually using binary. You reference the default subnet masks, because you need to know how many bits you have or will borrow from the hosts.

In the subnetting made easy thread, they use binary math plus the default subnets 8, 16, 24, 32 to answer the questions. Check it out and see if that method works for you.

The method that works for me is a cross between CBT nuggets and Train signal. I only use the train signal method when trying to find what network a certain subnet is in. It's fast and easy, no counting up from zero. I use the CBT nuggets version to do everything else.

Technology1 wrote: »

Another thing I was thinking is that we know right away that the default subnet mask is 16 bits, which is 255.255.0.0

If you take /27, which is 27 bits and subtract 16 from it, you immediately get the number 11.

Do you think that makes it any easier in determining the number 11 instead of going into binary to figure it out?

Yes, that is a shortcut of getting it. I usually have to do and write things out visually for me to understand

/16 - /27 = 11 Take 2^11 = 2048 subnets available

Thanks for your help. There are aways multiple ways in figuring these things out, but any help with it is always appreciated.

IRONMONKUS wrote: »

Yes, that is a shortcut of getting it. I usually have to do and write things out visually for me to understand

/16 - /27 = 11 Take 2^11 = 2048 subnets available

Here's another one. If you don't mind going into this one, please do. Thanks.

How many subnets and hosts per subnet can you get from the network 172.18.0.0 255.255.255.224? 27 bits - 16 bits is 11.

IRONMONKUS wrote: »

Yes, that is a shortcut of getting it. I usually have to do and write things out visually for me to understand

/16 - /27 = 11 Take 2^11 = 2048 subnets available

One more.

How many subnets and hosts per subnet can you get from the network 172.29.0.0/26?

26 bits - 16 bits is 10.

IRONMONKUS wrote: »

Yes, that is a shortcut of getting it. I usually have to do and write things out visually for me to understand

/16 - /27 = 11 Take 2^11 = 2048 subnets available

saied45 wrote: »

wow thanks for the help. i mainly got it except for one thing how does he figure out which octate he borrowed it from. like he said here

We have borrowed from the last octet as the 29th bit is in the last octet. We start from zero and count up in our block size. Therefore it follows that the subnets are:

sorry if it looks like im stealing the thread OP, i just really wanted to know

An easy way of looking at it is change it to binary to look at it visually.

11111111.11111111.11111111.11111000
first octet.second octet.third octet.fourth octet

basically, count up the ones and they equal 29 bits or a /29 subnet.

You know you are in the last octet, because that's where you ended counting the 1s.

Since it says "we have borrowed from the last octet", you know that it had to have started with a class C address, because you cannot borrow from the subnets, you just borrow from hosts.

Default class C: 255.255.255.0, 0 being the hosts and the only octet you can borrow from.

Technology1 wrote: »

Here's another one. If you don't mind going into this one, please do. Thanks.

How many subnets and hosts per subnet can you get from the network 172.18.0.0 255.255.255.224? 27 bits - 16 bits is 11.

My way:

Default subnet mask: 255.255.0.0
Given subnet mask: 255.255.255.224

Binary:
11111111.11111111.00000000.00000000
11111111.11111111.11111111.11100000

1s taken from hosts = 11. 2 to the 11th (2^11) = 2048 subnets available
0s or hosts left = 5. 2^5-2 = 30 hosts per subnet

subnetting made easy way:

Find subnets:

default: /16
given: /27

/16 - /27 = 11, 2^11 = 2048 subnets available.

Find hosts:

given: /27
next octet total: 32 (Knowing 8,16,24,32)

/27 - 32 = 5, 2^5-2 = 30 hosts per subnet

Thank you. If we say that the mask is the default B mask of 255.255.0.0 with the IP address remaining the same at 172.18.0.0

How many subnets and hosts per subnet would there be?

Very much appreciate you taking the time to go into detail with this and explaining it.

IRONMONKUS wrote: »

My way:
Default subnet mask: 255.255.0.0
Given subnet mask: 255.255.255.224

Binary:
11111111.11111111.00000000.00000000
11111111.11111111.11111111.11100000

1s taken from hosts = 11. 2 to the 11th (2^11) = 2048 subnets available
0s or hosts left = 5. 2^5-2 = 30 hosts per subnet
subnetting made easy way:

Find subnets:

default: /16
given: /27

/16 - /27 = 11, 2^11 = 2048 subnets available.

Find hosts:

given: /27
next octet total: 32 (Knowing 8,16,24,32)

/27 - 32 = 5, 2^5-2 = 30 hosts per subnet

Technology1 wrote: »

One more.

How many subnets and hosts per subnet can you get from the network 172.29.0.0/26?

26 bits - 16 bits is 10.

My method:

Default: 255.255.0.0 /16
Given: 255.255.255.192 /26

binary:
11111111.11111111.00000000.00000000
11111111.11111111.11111111.11000000

1s borrowed from hosts = 10, 2^10 = 1024 subnets available.
0s left = 6, 2^6-2 = 62 hosts per subnet available.

subnetting made easy:

/16-/26 = 10, 2^10 = 1024 subnets available.
/26 - 32 = 6, 2^6-2 = 62 hosts per subnet

In these examples, I want to apologize if I messed up the subnetting made easy way of doing it. That is what I remember from reading over the thread. Please correct me if i'm wrong, so that I may better understand.

Technology1 wrote: »

Thank you. If we say that the mask is the default B mask of 255.255.0.0 with the IP address remaining the same at 172.18.0.0

How many subnets and hosts per subnet would there be?

Very much appreciate you taking the time to go into detail with this and explaining it.

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.

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

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.

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.

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.

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