Hi, How many subnets and hosts per subnet can you get from the network 172.20.0.0/28?

I tried using the 2^n and (2^n)-2 formula but I'm not getting
4096 subnet and 14 hosts answer.

Please help.

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172.20.0.0/28 or rather 172.20.0.0/255.255.255.240

Class B address.
Class B default mask = 255.255.0.0

172.20.0.0 is subnetted by borrowing 12 bits from the host portion of the address (28 - 16 = 12).

Number of subnets = 2^12 = 4096
Number of usable hosts = (2^4) - 2 = 16 - 2 = 14

closetgeek

16 subnets and 14 host -I'm pretty good at this so.... ask away....lol

closetgeek

SurferdudeHB wrote: »

I tried using the 2^n and (2^n)-2 formula but I'm not getting
4096 subnet and 14 hosts answer.

Please help.

The secret is in your subnet mask....once you find your mask the rest is easy... I did it in my head.Google subnetting secrets this is a very good way to learn subnetting.

SurferdudeHB

luke_bibby wrote: »

172.20.0.0/28 or rather 172.20.0.0/255.255.255.240

Class B address.
Class B default mask = 255.255.0.0

172.20.0.0 is subnetted by borrowing 12 bits from the host portion of the address (28 - 16 = 12).

Number of subnets = 2^12 = 4096
Number of usable hosts = (2^4) - 2 = 16 - 2 = 14

Got it thanks!

Morty3

closetgeek wrote: »

16 subnets and 14 host -I'm pretty good at this so.... ask away....lol

It is not classless and neither a class C address

blackninja

closetgeek wrote: »

16 subnets and 14 host -I'm pretty good at this so.... ask away....lol

Must try harder

sina2011

Firstly I know this an old thread but Ive been looking for a good explanation on this part of subnetting and you guys have made me 1 step closer to understanding subnetting completley.
I Just have one Question for:

@Luke Bibby

where did you get (Number of usable hosts = (2^4) from in the second part of your explanation I cant seem to figure it out.

Cheers.

KrisA

sina2011 wrote: »

Firstly I know this an old thread but Ive been looking for a good explanation on this part of subnetting and you guys have made me 1 step closer to understanding subnetting completley.
I Just have one Question for:

@Luke Bibby

where did you get (Number of usable hosts = (2^4) from in the second part of your explanation I cant seem to figure it out.

Cheers.

I am not Luke - However... that is the number of "Off bits" aka the 0's ...

/28 = 11111111.11111111.11111111.11110000 <- 4 Off bits..... Just count ( you will starting thinking of it binary after awhile) the 0's

pham0329

^ What he said. But rather than doing it in binary...if you have 32 bits, and 28 are used for the network address, than means 32 - 28 = 4 bits for hosts.

JockVSJock

I'm confused on how the answer for this was gotten too. I'm using Chris Bryant's method of subnetting and not getting the same answer:

IP Address: 172.20.0.0
Default Mask: 255.255.0.0
Subnet Mask: 255.255.255.224

I convert the default mask and the subnet mask to binary:

Default Mask = 11111111.11111111.00000000.00000000
Subnet Mask = 11111111.11111111.11111111.11100000

Using his method I take the bits turned on under the subnet mask (2^11) to find the subnets which gives me = 2048
And again, using his method to find the host, I use the following (2^11-2), which gives me = 2046

Not saying I'm correct, I'm just using the method that works for me the best and trying to generate discussion and understand what is going on.

thanks

Slowhand

JockVSJock wrote: »

I'm confused on how the answer for this was gotten too. I'm using Chris Bryant's method of subnetting and not getting the same answer:

IP Address: 172.20.0.0
Default Mask: 255.255.0.0
Subnet Mask: 255.255.255.224

I convert the default mask and the subnet mask to binary:

Default Mask = 11111111.11111111.00000000.00000000
Subnet Mask = 11111111.11111111.11111111.11100000

Using his method I take the bits turned on under the subnet mask (2^11) to find the subnets which gives me = 2048
And again, using his method to find the host, I use the following (2^11-2), which gives me = 2046

A /28 mask would give 12 network bits, not 11:
Subnet Mask = 11111111.11111111.11111111.11110000

That means the subnet mask is 255.255.255.240.

With 12 bits, you'd have 2^12 which equals 4096. For the hosts, you'd have (2^4)-2, which makes 14 hosts on each subnet.

JockVSJock

Slowhand wrote: »

A /28 mask would give 12 network bits, not 11:
Subnet Mask = 11111111.11111111.11111111.11110000

That means the subnet mask is 255.255.255.240.

With 12 bits, you'd have 2^12 which equals 4096. For the hosts, you'd have (2^4)-2, which makes 14 hosts on each subnet.

I don't get it.

The original subnet mask is 255.255.255.224, isn't that /27?

I don't understand where you are getting /28.

Slowhand

JockVSJock wrote: »

I don't get it.

The original subnet mask is 255.255.255.224, isn't that /27?

I don't understand where you are getting /28.

The OP asked for /28 in the title of the thread: Hi, How many subnets and hosts per subnet can you get from the network 172.20.0.0/28?

gramacorp

To find the number of subnets use 2^n for the on bits
172.20.0.0
255.255.255.224
11111111.11111111.11111111.11100000 (11 bits on) so 2^11=2048 subnets

To find the number of host per subnet use 2^n-2 for the off bits
172.20.0.0
255.255.255.224
11111111.11111111.11111111.11100000 (5 bits off) so 2^5-2=30 host per subnet

JockVSJock

Slowhand wrote: »

The OP asked for /28 in the title of the thread: Hi, How many subnets and hosts per subnet can you get from the network 172.20.0.0/28?

Ya, I really f****d that one up.

Sorry.

Slowhand

Heh, no worries. At least it was here on TE and not at work you got the wrong number. . . unlike me, who worked an entire graveyard shift, called Cisco SmartNet support twice, and probably tore out half my hair in frustration before realizing that it wasn't my VPN tunnel or my firewall settings that were wrong, it was the address and mask on a subinterface.

pham0329

Slowhand wrote: »

Heh, no worries. At least it was here on TE and not at work you got the wrong number. . . unlike me, who worked an entire graveyard shift, called Cisco SmartNet support twice, and probably tore out half my hair in frustration before realizing that it wasn't my VPN tunnel or my firewall settings that were wrong, it was the address and mask on a subinterface.

Eh, at least you weren't like closetgeek (3rd post) who claims he's really good at subnetting and then proceed to give the totally wrong answer

Steadfast Girl

i have a question. Answer me please. Question:How many subnet and hosts per subnet can you get from the network 174.20.0.0/23 Remember for host part at least 2 bits are required?

oli356

Steadfast Girl wrote: »

i have a question. Answer me please. Question:How many subnet and hosts per subnet can you get from the network 174.20.0.0/23 Remember for host part at least 2 bits are required?

Download this to check. IP Subnet Calculator

Subnet mask in binary:
11111111.11111111.11111110.00000000 (255.255.254.0)

Its /23 so there are 23 network bits.

This leaves us with 9 host bits (0s).. So 2^9=512 and minus the 2 so 510.
This leaves us with 7 host bits (1s) in the octet we're working with ..So 2^7=128.

128 subnets
510 hosts.

rphann

Steadfast Girl wrote: »

i have a question. Answer me please. Question:How many subnet and hosts per subnet can you get from the network 174.20.0.0/23 Remember for host part at least 2 bits are required?

The answer is:

(2^9)-2=510
510 hosts per subnet

(2^7)=128
128 subnet

Steadfast Girl

thanks for answering my question. GOD bless u allz.

fadhil

Steadfast Girl wrote: »

thanks for answering my question. GOD bless u allz.

most welcome

sizeon

luke_bibby wrote: »

172.20.0.0/28 or rather 172.20.0.0/255.255.255.240

Class B address.
Class B default mask = 255.255.0.0

172.20.0.0 is subnetted by borrowing 12 bits from the host portion of the address (28 - 16 = 12).

Number of subnets = 2^12 = 4096
Number of usable hosts = (2^4) - 2 = 16 - 2 = 14

This is correct

hardstylewon

rphann wrote: »

The answer is:

(2^9)-2=510
510 hosts per subnet

(2^7)=128
128 subnet

I couldn't have said it better.

gadav478

hardstylewon wrote: »

I couldn't have said it better.

How do you go about this method? Just curious.

gbdavidx

luke_bibby wrote: »

172.20.0.0/28 or rather 172.20.0.0/255.255.255.240

Class B address.
Class B default mask = 255.255.0.0

172.20.0.0 is subnetted by borrowing 12 bits from the host portion of the address (28 - 16 = 12).

Number of subnets = 2^12 = 4096
Number of usable hosts = (2^4) - 2 = 16 - 2 = 14

I understand that 2 can't be used because the first and last ip, but why it -2 a 2nd time?

ChickenNuggetz

gbdavidx wrote: »

I understand that 2 can't be used because the first and last ip, but why it -2 a 2nd time?

I think you've misunderstood what he was doing. He's basically writing out his thought process. The 16 he gets is from calculating 2^4 and then he is subtracting 2 which gives him the final answer of 14. It wasnt exactly clear, but I understand what he was trying to show.