Options
Hi, How many subnets and hosts per subnet can you get from the network 172.20.0.0/28?
SurferdudeHB
Member Posts: 199 ■■■□□□□□□□
in CCNA & CCENT
I tried using the 2^n and (2^n)-2 formula but I'm not getting
4096 subnet and 14 hosts answer.
Please help.
4096 subnet and 14 hosts answer.
Please help.
Comments
-
Optionsluke_bibby Member Posts: 162172.20.0.0/28 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 -
Optionsclosetgeek Member Posts: 15 ■□□□□□□□□□16 subnets and 14 host -I'm pretty good at this so.... ask away....lolLife is an enigma so lets figure it out....
-
Optionsclosetgeek Member Posts: 15 ■□□□□□□□□□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.Life is an enigma so lets figure it out.... -
OptionsSurferdudeHB Member Posts: 199 ■■■□□□□□□□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! -
OptionsMorty3 Member Posts: 139closetgeek 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 addressCCNA, CCNA:Sec, Net+, Sonicwall Admin (fwiw). Constantly getting into new stuff. -
Optionsblackninja Member Posts: 385closetgeek wrote: »16 subnets and 14 host -I'm pretty good at this so.... ask away....lol
Must try harderCurrently studying:
CCIE R&S - using INE workbooks & videos
Currently reading:
Everything. Twice -
Optionssina2011 Member Posts: 239 ■□□□□□□□□□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. -
OptionsKrisA Member Posts: 142Firstly 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'sWGU Progress BSIT:NA | Current Term:1 | Transfered To-Do In Progress Completed
EWB2 BAC1 BBC1 TSV1 WFV1 CLC1 LAE1 LUT1 LAT1 AXV1 TTV1 INC1 INT1 TPV1 SST1 SSC1 GAC1 HHT1 TNV1 QLT1 BOV1 LET1 ORC1 IWC1 IWT1 MGC1 ABV1 AHV1 AJV1 TWA1 CPW2 BRV1
Currently Reading
Darril Book -
Optionspham0329 Member Posts: 556^ 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.
-
OptionsJockVSJock Member Posts: 1,118I'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***Freedom of Speech, Just Watch What You Say*** Example, Beware of CompTIA Certs (Deleted From Google Cached)
"Its easier to deceive the masses then to convince the masses that they have been deceived."
-unknown -
OptionsSlowhand Mod Posts: 5,161 ModJockVSJock 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.
Free Microsoft Training: Microsoft Learn
Free PowerShell Resources: Top PowerShell Blogs
Free DevOps/Azure Resources: Visual Studio Dev Essentials
Let it never be said that I didn't do the very least I could do. -
OptionsJockVSJock Member Posts: 1,118A /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.***Freedom of Speech, Just Watch What You Say*** Example, Beware of CompTIA Certs (Deleted From Google Cached)
"Its easier to deceive the masses then to convince the masses that they have been deceived."
-unknown -
OptionsSlowhand Mod Posts: 5,161 ModJockVSJock 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.
Free Microsoft Training: Microsoft Learn
Free PowerShell Resources: Top PowerShell Blogs
Free DevOps/Azure Resources: Visual Studio Dev Essentials
Let it never be said that I didn't do the very least I could do. -
Optionsgramacorp Member Posts: 39 ■■■□□□□□□□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 -
OptionsJockVSJock Member Posts: 1,118The 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.***Freedom of Speech, Just Watch What You Say*** Example, Beware of CompTIA Certs (Deleted From Google Cached)
"Its easier to deceive the masses then to convince the masses that they have been deceived."
-unknown -
OptionsSlowhand Mod Posts: 5,161 ModHeh, 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.
Free Microsoft Training: Microsoft Learn
Free PowerShell Resources: Top PowerShell Blogs
Free DevOps/Azure Resources: Visual Studio Dev Essentials
Let it never be said that I didn't do the very least I could do. -
Optionspham0329 Member Posts: 556Heh, 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 -
OptionsSteadfast Girl Registered Users Posts: 4 ■□□□□□□□□□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?
-
Optionsoli356 Member Posts: 364Steadfast 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?
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.Lab:
Combination of GNS3 and Cisco equipment if required. -
Optionsrphann Member Posts: 76 ■■□□□□□□□□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 -
OptionsSteadfast Girl Registered Users Posts: 4 ■□□□□□□□□□thanks for answering my question. GOD bless u allz.
-
Optionsfadhil Member Posts: 200Steadfast Girl wrote: »thanks for answering my question. GOD bless u allz.
most welcome -
Optionssizeon Member Posts: 321luke_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 -
Optionshardstylewon Member Posts: 15 ■□□□□□□□□□The answer is:
(2^9)-2=510
510 hosts per subnet
(2^7)=128
128 subnet
I couldn't have said it better. -
Optionsgadav478 Member Posts: 374 ■■■□□□□□□□hardstylewon wrote: »I couldn't have said it better.
How do you go about this method? Just curious.Goals for 2015: CCNP -
Optionsgbdavidx Member Posts: 840luke_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? -
OptionsChickenNuggetz Member Posts: 284I 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.:study: Currently Reading: Red Hat Certified Systems Administrator and Engineer by Ashgar Ghori
Certifications: CCENT; CCNA: R&S; Security+
Next up: RHCSA