Subnet help please :(

[Bahumat]

Question
A company has 90 locations ranging in size from 5 to no more than 500 users. Their growth plans project adding another 75 similar locations over the next 5 years. They use a private Class B address 172.16.0.0. Which subnet mask best suits its needs?

Answer
The correct answer is A. A /23 or 255.255.254.0 subnet mask provides a total of 128 subnets with 512 hosts per subnet.
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Ok, I am confused on how to work this out. Could anyone tell me what I have to do in order to work this out please?

 

[old.Osy]

The question is very ambiguous - if that's intentional, it is also retarded.

That answer is correct only if the question would sound as:

Company A. has a private B class range 172.16.0.0. What subnet mask should they use for a location of the company which would have a max of 500 users, and a minimum of 5?

At which point you have to figure out which split gives you the needed pool of available addresses - A /23 makes sense, since it gives you 508 usable IPs.

However, if that location has Servers, network printers, routers, switches, wireless access points, a /23 does not suffice, since each of those active networking devices needs an IP address. I would recommend a /22, from the network engineer/architect perspective - It allows for flexibility and growth.

 

[Deebs]

We actually assign Class B's to each of our offices and then further assign Class C's to departments, servers, types of servers etc.

Makes it very easy to add ACL's between VLANs. Apart from the 2nd octet changing per site I know for a fact that Customer Services in France is 10.x.203.0/24 which is exactly the same as UK, 10.x.203.0/24 rinse and repeat. Nice to maintain a standard across the Group.

 

[Bahumat]

I'm doing my MCITP at the moment and this question came up. The problem I have is knowing the formula to use to figure out which is the correct answer.

If i am given the address, and I have a rough count on the amount of users (lets exclude other devices to make it easier). How do I work out which /number I need to use?

Sorry for my terrible explanation.

 

[Bahumat]

Just to add something else...

Mask---------------Notation---Subnet------Hosts
255.255.254.0--------/23--------128--------512


Ok I know on Class B that 255.255.255.0 is listed as /24, however I am unsure what maths/equation is in process to cause 255.255.254.0 to become /23

I 'heard' it's good to memorize the 'cheat sheet's', however it would be nice if I knew how to work out the conversion.

I hope my questions make sense, if not, please let me know and I will try to clarify.

 

[Krazeh]

I'm doing my MCITP at the moment and this question came up. The problem I have is knowing the formula to use to figure out which is the correct answer.

If i am given the address, and I have a rough count on the amount of users (lets exclude other devices to make it easier). How do I work out which /number I need to use?

Sorry for my terrible explanation.

The formula is quite simple. If you start with a /30 subnet you have 4 hosts, then each time you subtract one from the subnet (i.e. /29, /28, /27 etc) you double the number of hosts so a /29 has 8 address, /28 has 16, /27 has 32 and so on. You do however have to remember that you have to subtract 2 from the number of hosts available to find the actual number of usable hosts due to each subnet needing a network and broadcast adress.

In the case of your question you know that they need 500 usable addresses in each subnet so you just go through each subnet till you find one with more than 500 available hosts, which is a /23 with 512 (or 510 usable) hosts.

Just to add something else...

Mask---------------Notation---Subnet------Hosts
255.255.254.0--------/23--------128--------512


Ok I know on Class B that 255.255.255.0 is listed as /24, however I am unsure what maths/equation is in process to cause 255.255.254.0 to become /23

I 'heard' it's good to memorize the 'cheat sheet's', however it would be nice if I knew how to work out the conversion.

I hope my questions make sense, if not, please let me know and I will try to clarify.

The number after the / denotes how many bits of the 32bit address are 1 when viewed in binary (the rest are always 0). You always start from the lefthand side so a /24 mask in binary is 11111111.11111111.11111111.00000000 which in decimal is 255.255.255.0. Likewise a /23 mask in binary is 11111111.11111111.11111110.00000000 which in decimal is 255.255.254.0

 

[Bob007]

Attached a spreadsheet i used when i was learning to understand subnetting.

Well, Thats a lie, the one i used I lost. So i redid it. Hope you understand it. I can be a bit wierd in the way i do stuff

 

[Bahumat]

Thanks everyone, you've all helped alot. Also Krazeh, what you said about /24 equaling (11111111.11111111.11111111.00000000) makes perfect sense now.

I've one last thing eating away at me;
I understand that each time you reduce the /notation by 1, you double the hosts.

255.255.255.240 /28 16 16
255.255.255.248 /29 32 8
255.255.255.252 /30 64 4
/30 has 4, /29 has 8, /28 has 16 etc.

Lets say someone asks me what /notation value is for 255.255.255.240 - The answer is /28, but how could I get that via 255.255.255.240?

The only way I could find this out (to my knowledge), would be to memorise 255.255.255.252 which equals /30 and which has 4 hosts, then work my way down till I hit 255.255.255.240

This all works off me remembering 255.255.255.252 /30 4

 

[Bahumat]

Attached a spreadsheet i used when i was learning to understand subnetting.

Well, Thats a lie, the one i used I lost. So i redid it. Hope you understand it. I can be a bit wierd in the way i do stuff

Thanks for this. I will have a play around with it

 

[Krazeh]

Thanks everyone, you've all helped alot. Also Krazeh, what you said about /24 equaling (11111111.11111111.11111111.00000000) makes perfect sense now.

I've one last thing eating away at me;
I understand that each time you reduce the /notation by 1, you double the hosts.

255.255.255.240 /28 16 16
255.255.255.248 /29 32 8
255.255.255.252 /30 64 4
/30 has 4, /29 has 8, /28 has 16 etc.

Lets say someone asks me what /notation value is for 255.255.255.240 - The answer is /28, but how could I get that via 255.255.255.240?

The only way I could find this out (to my knowledge), would be to memorise 255.255.255.252 which equals /30 and which has 4 hosts, then work my way down till I hit 255.255.255.240

This all works off me remembering 255.255.255.252 /30 4

Only real answer is to learn binary so you can convert the mask you're given into binary. Then it's just a case of counting how many ones you have.

 

[Bob007]

best way i remember is that each octet = /8

so 255.255.255.240
gives me this to do.

8 + 8 + 8 + X =

now to find X. I would write down.

128, 64, 32, 16, 8, 4, 2, 1.

and make 240 from it.

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

128 + 64 + 32 + 16 = 240.

so thats 4 Bits (4 1's)

So X = 4

8 + 8 + 8 + 4 = 28

/28

Make sense ?

P.S. Missed some bits of first spreadsheet file. Update attached.

 

[Bob007]

Having not seen it posted. I think I better explain my 128, 64, 32, 16, 8, 4, 2, 1. Thing. Well this is an 8 bit octet used in IPV4 networking.

From this and some on/off (1 on, 0 off. Binary ) switches you can make any number from 0 to 255. Thats 256 combinations. 256 is a good number in networking. remember it.

An IP is made up of 4 8 bit octets.

128, 64, 32, 16, 8, 4, 2, 1, dot, 128, 64, 32, 16, 8, 4, 2, 1, dot, 128, 64, 32, 16, 8, 4, 2, 1, dot, 128, 64, 32, 16, 8, 4, 2, 1,

Sp to work out addresses from binary format to numbers you'd use a simalar system and 1's and 0's below each number. then add them up.

128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1
 1   0  0  0 0 0 0 0 .  0   0  0  0 0 0 0 0 .  0   0  0  0 0 0 0 0 .  0  0  0  0  0 0 0 0  /1
 1   1  0  0 0 0 0 0 .  0   0  0  0 0 0 0 0 .  0   0  0  0 0 0 0 0 .  0  0  0  0  0 0 0 0  /2
 1   1  1  0 0 0 0 0 .  0   0  0  0 0 0 0 0 .  0   0  0  0 0 0 0 0 .  0  0  0  0  0 0 0 0  /3
 1   1  1  1 0 0 0 0 .  0   0  0  0 0 0 0 0 .  0   0  0  0 0 0 0 0 .  0  0  0  0  0 0 0 0  /4
 1   1  1  1 1 0 0 0 .  0   0  0  0 0 0 0 0 .  0   0  0  0 0 0 0 0 .  0  0  0  0  0 0 0 0  /5
 1   1  1  1 1 1 0 0 .  0   0  0  0 0 0 0 0 .  0   0  0  0 0 0 0 0 .  0  0  0  0  0 0 0 0  /6
 1   1  1  1 1 1 1 0 .  0   0  0  0 0 0 0 0 .  0   0  0  0 0 0 0 0 .  0  0  0  0  0 0 0 0  /7

 1   1  1  1 1 1 1 1 .  0   0  0  0 0 0 0 0 .  0   0  0  0 0 0 0 0 .  0  0  0  0  0 0 0 0  /8 
 1   1  1  1 1 1 1 1 .  1   0  0  0 0 0 0 0 .  0   0  0  0 0 0 0 0 .  0  0  0  0  0 0 0 0  /9
 1   1  1  1 1 1 1 1 .  1   1  0  0 0 0 0 0 .  0   0  0  0 0 0 0 0 .  0  0  0  0  0 0 0 0  /10
 1   1  1  1 1 1 1 1 .  1   1  1  0 0 0 0 0 .  0   0  0  0 0 0 0 0 .  0  0  0  0  0 0 0 0  /11
 1   1  1  1 1 1 1 1 .  1   1  1  1 0 0 0 0 .  0   0  0  0 0 0 0 0 .  0  0  0  0  0 0 0 0  /12
 1   1  1  1 1 1 1 1 .  1   1  1  1 1 0 0 0 .  0   0  0  0 0 0 0 0 .  0  0  0  0  0 0 0 0  /13
 1   1  1  1 1 1 1 1 .  1   1  1  1 1 1 0 0 .  0   0  0  0 0 0 0 0 .  0  0  0  0  0 0 0 0  /14
 1   1  1  1 1 1 1 1 .  1   1  1  1 1 1 1 0 .  0   0  0  0 0 0 0 0 .  0  0  0  0  0 0 0 0  /15

 1   1  1  1 1 1 1 1 .  1   1  1  1 1 1 1 1 .  0   0  0  0 0 0 0 0 .  0  0  0  0  0 0 0 0  /16
 1   1  1  1 1 1 1 1 .  1   1  1  1 1 1 1 1 .  1   0  0  0 0 0 0 0 .  0  0  0  0  0 0 0 0  /17
 1   1  1  1 1 1 1 1 .  1   1  1  1 1 1 1 1 .  1   1  0  0 0 0 0 0 .  0  0  0  0  0 0 0 0  /18
 1   1  1  1 1 1 1 1 .  1   1  1  1 1 1 1 1 .  1   1  1  0 0 0 0 0 .  0  0  0  0  0 0 0 0  /19
 1   1  1  1 1 1 1 1 .  1   1  1  1 1 1 1 1 .  1   1  1  1 0 0 0 0 .  0  0  0  0  0 0 0 0  /20
 1   1  1  1 1 1 1 1 .  1   1  1  1 1 1 1 1 .  1   1  1  1 1 0 0 0 .  0  0  0  0  0 0 0 0  /21
 1   1  1  1 1 1 1 1 .  1   1  1  1 1 1 1 1 .  1   1  1  1 1 1 0 0 .  0  0  0  0  0 0 0 0  /22
 1   1  1  1 1 1 1 1 .  1   1  1  1 1 1 1 1 .  1   1  1  1 1 1 1 0 .  0  0  0  0  0 0 0 0  /23

 1   1  1  1 1 1 1 1 .  1   1  1  1 1 1 1 1 .  1   1  1  1 1 1 1 1 .  0  0  0  0  0 0 0 0  /24
 1   1  1  1 1 1 1 1 .  1   1  1  1 1 1 1 1 .  1   1  1  1 1 1 1 1 .  1  0  0  0  0 0 0 0  /25
 1   1  1  1 1 1 1 1 .  1   1  1  1 1 1 1 1 .  1   1  1  1 1 1 1 1 .  1  1  0  0  0 0 0 0  /26
 1   1  1  1 1 1 1 1 .  1   1  1  1 1 1 1 1 .  1   1  1  1 1 1 1 1 .  1  1  1  0  0 0 0 0  /27
 1   1  1  1 1 1 1 1 .  1   1  1  1 1 1 1 1 .  1   1  1  1 1 1 1 1 .  1  1  1  1  0 0 0 0  /28
 1   1  1  1 1 1 1 1 .  1   1  1  1 1 1 1 1 .  1   1  1  1 1 1 1 1 .  1  1  1  1  1 0 0 0  /29
 1   1  1  1 1 1 1 1 .  1   1  1  1 1 1 1 1 .  1   1  1  1 1 1 1 1 .  1  1  1  1  1 1 0 0  /30
 1   1  1  1 1 1 1 1 .  1   1  1  1 1 1 1 1 .  1   1  1  1 1 1 1 1 .  1  1  1  1  1 1 1 0  /31
 1   1  1  1 1 1 1 1 .  1   1  1  1 1 1 1 1 .  1   1  1  1 1 1 1 1 .  1  1  1  1  1 1 1 1  /32

thats a subnet mask run from 128.0.0.0 to 255.255.255.255 inclusive.

IP Addresses differ as they can be mixed ones and Zeros.

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

Thats an IP address. To find out what IP Address it is, we'd add up the numbers above the 1's

first octet
8 + 2 = 10

second Octet
2

third Octet
32 + 8 = 40

Forth Octet
8 + 1 = 9

So are Ip address would be.
10.2.40.9

We can extend on this and use are IP address and Subnet mask to find are network address.

Lets take an IP address of 172.19.195.57 with a subenet mask of 255.255.224.0 and find the network address for that subenet.

128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1
 1   0  1  0 1 1 0 0 .  0   0  0  1 0 0 1 1 .  1   1  0  0 0 0 1 1 .  0  0  1  1  1 0 0 1  IP Address in Binary
 1   1  1  1 1 1 1 1 .  1   1  1  1 1 1 1 1 .  1   1  1  0 0 0 0 0 .  0  0  0  0  0 0 0 0  Subnet Mask in Binary

Now we take this and And it down. Everywhere we see a 1 in the top and bottom row we put a 1 in the 3rd row.

128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1
 1   0  1  0 1 1 0 0 .  0   0  0  1 0 0 1 1 .  1   1  0  0 0 0 1 1 .  0  0  1  1  1 0 0 1  IP Address in Binary
 1   1  1  1 1 1 1 1 .  1   1  1  1 1 1 1 1 .  1   1  1  0 0 0 0 0 .  0  0  0  0  0 0 0 0  Subnet Mask in Binary
 1   0  1  0 1 1 0 0 .  0   0  0  1 0 0 1 1 .  1   1  0  0 0 0 0 0 .  0  0  0  0  0 0 0 0  Network adress in binary

Now we just add up the third row.

First Octet
128 + 32 + 8 + 4 = 172

Second Octet
16 + 2 + 1 = 19

Third Octet
128 + 64 = 192

Forth octet
all 0's so this = 0

172.19.192.0 is the network address for 172.19.195.57 with a subnet mask of 255.255.224.0

Whats the Broadcast Address ?

Hint 256 is a good number

 

[Krazeh]

172.19.223.255. Could've at least put something a little bit tricky

 

[Bahumat]

Haha this is awesome, thanks so much! Seeing as it's late though, I will give this a proper read tomorrow! Quite sleepy lol

 

[Bahumat]

Just been through this and it's a great help! Really appreciate all the help guys