I need help with this subnetting question...

CompUBug

You have been given the 172.16.0.0 network. Your boss has tasked you with creating 250 subnets that allow for up to 254 hosts each. What subnet mask would you use to achieve this goal?...

I don't want the answer... I want the method of solving this...

hermeszdata

CompUBug wrote: »

You have been given the 172.16.0.0 network. Your boss has tasked you with creating 250 subnets that allow for up to 254 hosts each. What subnet mask would you use to achieve this goal?...

I don't want the answer... I want the method of solving this...

The number of subnets within a given network is determined by the number of bits borrowed for the host portion of the address.

The default mask for 172.16.0.0 is 255.255.0.0 or CIDR /16. Therefore, the mask necessary to do what you need to do is going to be longer than 16 bits.

If using ip subnet zero the number of subnets equals 2^N where N+ borrowed bits. If no ip subnet zero, subtract 2.

miller811

the power of 2s is your friend

1 = 2
2 = 4
3 = 8
4 = 16
5 = 32
6 = 64
7 = 128
8 = 256
9 = 512
10 = 1024
11 = 2048
12 = 4096
etc

so with the address given (172.16.0.0 is a class b or a 16 bit subnet)
how many bits will you need to borrow to meet the objective?

8 = 256 = 250 subnets

172.16.0.0 /24

leaves 8 bis for host = 256 -2 = 254

IRONMONKUS

Given a class B address of 172.16.0.0, you know the default subnet mask is 255.255.0.0.
You need to find the subnet mask that will give you 250 subnets and 254 hosts per.

Solve for 250 subnets by finding it on the binary chart.
256 128 64 32 16 8 4 2 1
0 1
128 is the first number you can subtract from 250, so you count from there to the right. 128 = 1, 64 = 2, 32 = 3, 16 = 4, 8 = 5, 4 = 6, 2 = 7, 1 = 8. So, you have 8 bits that you borrow from the hosts.

Default subnet mask: 255.255.0.0
In binary: 11111111.11111111.00000000.00000000
1s = network, 0s = hosts
Borrow the 8 bits from the hosts.
New subnet mask: 11111111.11111111.11111111.00000000
Change to decimal: 255.255.255.0 or /24

CompUBug

I see, so in this question I'm not really looking for anything in the IP address or the current subnet mask that they provided me in the question, I'm just trying to see what subnet mask is necessary to host the amount of PC's they desired to host?

CompUBug

I have another question...

Question: What is the broadcast address of network 172.31.16.0/21?

/21 Network - Next Octet /24
24 - 21 = 3
172.31.16.0
172.31.16.3

I would have thought the broadcast address would be 172.31.16.2
... but instead its
172.31.23.255

How is this so? Howcome the third octet jumped up to 23? And I know 255 is a broadcast address but why this address instead of a subnet broadcast address???

mella060

Lets see...

172.31.16.0/21.../21 = 255.255.248.0

So work out what your increment or block size is...256-248 = 8

Or the method i use to find the increment is a lot quicker, works like this

Write out your subnet mask in binary

/21 = 11111111.11111111.11111000.00000000

Now the last bit position where the value is a 1 is your increment...

so in the 3rd octet...128 64 32 16 8 4 2 1

So basically your increment is 8 in the 3rd octet...

So your networks and their ranges will be as follows...

172.31.0.0 /21 = 172.31.0.1 - 172.31.7.255
172.31.8.0 /21 = 172.31.8.1 - 172.31.15.255
172.31.16.0 /21 = 172.31.16.1 - 172.31.23.255
172.31.24.0 /21 = 172.31.24.1 - 172.31.31.255
172.31.32.0 /21 = 172.31.32.1 - 172.31.39.255

and so on

Howcome the third octet jumped up to 23?

Because that is the interesting octet or the octet where the increment is

If the subnet mask was 255.255.255.248 or /29...then the increment would still be 8, but in the 4th or last octet

So for example, your networks would be...

172.31.0.0 /29
172.31.0.8 /29
172.31.0.16 /29
172.31.0.24 /29
172.31.0.32 /29 and so on

gosh1976

You may want to do some more studying. read this: http://www.techexams.net/forums/ccna-ccent/38772-subnetting-made-easy.html

and look at the resources in this post: http://www.techexams.net/forums/network/60791-subnetting.html#post477425

You could also watch professor messer's videos where he explains how to do ANDing http://www.professormesser.com/free-network-plus-training/professor-messers-free-comptia-network-certification-training-course


With 172.31.16.0/21 you would have a range of usable addresses of 172.31.16.1 - 172.31.23.254 with 172.31.23.255. The explanation above is good.

CompUBug

Thanks for all of your help... All of this was practice based only...wasn't used for any tests or homework... I am reading and reading everything I can get my hands on but I don't easily learn by reading, I have to see things and work through things before I can begin to understand the concepts. I am beginning to get some of this... I'll probably be back though >.>

CompUBug

Here's one thats confusing me now...

The network 10.0.0.0/8 that has been subnetted using the subnet mask of /16, which makes 256 subnets.
10.0.0.0/16
10.1.0.0/16
10.2.0.0/16
.
.
.
10.255.0.0/16

---

How so does that make 256 subnets when 2^16 is 65536???

I was guessing because we added 8 bits which borrowed an entire octet and an entire octet is 256?? I understand the concept... like this...

1 6 3 2 1 8 4 2 1
2 4 2 4 6
8

---

1 1 1 1 1 1 1 1 1

128+64+32+24+16+8+4+2+1 = 256...

But what I dont understand is unlike our previous problem, we're not using the power of 2... unless we're just talking 2^8 which IS 256... but 2^16 certainly isnt... so whats the logical formula for this??

hermeszdata

CompUBug wrote: »

Here's one thats confusing me now...

The network 10.0.0.0/8 that has been subnetted using the subnet mask of /16, which makes 256 subnets.
10.0.0.0/16
10.1.0.0/16
10.2.0.0/16
.
.
.
10.255.0.0/16

---

How so does that make 256 subnets when 2^16 is 65536???

I was guessing because we added 8 bits which borrowed an entire octet and an entire octet is 256?? I understand the concept... like this...

1 6 3 2 1 8 4 2 1
2 4 2 4 6
8

---

1 1 1 1 1 1 1 1 1

128+64+32+24+16+8+4+2+1 = 256...

But what I dont understand is unlike our previous problem, we're not using the power of 2... unless we're just talking 2^8 which IS 256... but 2^16 certainly isnt... so whats the logical formula for this??

you are only borrowing 8 bits from the host field.

CompUBug

And this one...

Question: What valid host range is the IP address 192.168.23.153/30 a part of?
11111111.11111111.11111111.11111100

1 6 3 1 8 4 2 1
2 4 2 6
8

---

1 1 1 1 1 1 0 0

128
64
32
16
8
4
---
252

Subnet Mask = 255.255.255.252

/30 Network - Previous Octet - 24
30 - 24 = 6

Network - First Host - Broadcast
192.168.0.0 192.168.0.1 192.168.5.252
192.168.6.0 192.168.6.1 192.168.11.252
192.168.12.0 192.168.12.1 192.168.17.252
192.168.18.0 192.168.18.1 192.168.23.252
192.168.24.0

Valid HOST range: 192.168.18.1 - 192.168.23.251

--- I know where I went wrong here... I considered a /30 network as a subnetted /24 network

The real host range is
Answer: 192.168.23.153-192.168.23.154

I was originally solving for this but I thought that this would be wrong because a /30 network doesn't represent a full octet between /24 and /30 so it had to be subnetted from the previous octet.

Now my question is... How do you know when and when not to subnet from the previous octet?

===

Also did I solve it correctly using this method?

What valid host range is the IP address 192.168.23.153/30 a part of?

/30 Network is a Class C network falling under 255.255.255.252
- Next Octet = 32

32 - 30 = 2

192.168.23.0 - 192.168.23.1
192.168.23.2 - 192.168.23.3
-- Use 2 time tables --
192.168.23.152 - 192.168.23.153
192.168.23.154 - 192.168.23.155

192.168.23.152 = NETWORK so it cannot be used
192.168.23.153 = HOST so it CAN be used
192.168.23.154 = NEXT NETWORK which concludes the range for the IP

Range - 192.168.23.153 - 192.168.23.154

2E151

CompUBug wrote: »

And this one...


192.168.23.152 = NETWORK so it cannot be used
192.168.23.153 = HOST so it CAN be used
192.168.23.154 = NEXT NETWORK which concludes the range for the IP

Range - 192.168.23.153 - 192.168.23.154

192.168.23.154 is the last usable host.
192.168.23.155 is the broadcast
192.168.23.156 is the next network

Also, I have a question of my own. I can figure the increment fairly quickly subtracting the subnet from 256 or converting it to binary and using the method mentioned in this thread.

However, and I know this is probably elementary but math has never been my strong point, whats the quickest way to get from .0 to .152? As it stands, I simply start listing out the networks: .0,.4,.8,ect,ect until I hit .152

I pretty much do this for every one. With an increment of 4 I can skip ahead some, but on the larger ones I simply list them all out.

Am I missing a short cut?

gosh1976

2E151 wrote: »

192.168.23.154 is the last usable host.
192.168.23.155 is the broadcast
192.168.23.156 is the next network

Also, I have a question of my own. I can figure the increment fairly quickly subtracting the subnet from 256 or converting it to binary and using the method mentioned in this thread.

However, and I know this is probably elementary but math has never been my strong point, whats the quickest way to get from .0 to .152? As it stands, I simply start listing out the networks: .0,.4,.8,ect,ect until I hit .152

I pretty much do this for every one. With an increment of 4 I can skip ahead some, but on the larger ones I simply list them all out.

Am I missing a short cut?

You could start counting from 128 instead of 0. No matter what the increment is you could start counting from 128 if the subnet you are looking for is a higher number. If the subnet you are trying to get to is higher than 192 then you could start from there or if it is higher than 224, 240, etc then you could start from there.

try watching this video to speed things: Creating Your Own Mental Subnet Calculator you don't have to use his whole method but you might get some ideas to make your own

miller811

2E151 wrote: »

192.168.23.154 is the last usable host.
192.168.23.155 is the broadcast
192.168.23.156 is the next network

Also, I have a question of my own. I can figure the increment fairly quickly subtracting the subnet from 256 or converting it to binary and using the method mentioned in this thread.

However, and I know this is probably elementary but math has never been my strong point, whats the quickest way to get from .0 to .152? As it stands, I simply start listing out the networks: .0,.4,.8,ect,ect until I hit .152

I pretty much do this for every one. With an increment of 4 I can skip ahead some, but on the larger ones I simply list them all out.

Am I missing a short cut?

you could simply take the range = 4 in this case
and multiply by 20, 30, 40 = 160, the go backwards....156, 152

mella060

CompUBug wrote: »

Here's one thats confusing me now...

The network 10.0.0.0/8 that has been subnetted using the subnet mask of /16, which makes 256 subnets.
10.0.0.0/16
10.1.0.0/16
10.2.0.0/16

First thing you need to look at is what class of address it is.

10.0.0.0/8 is a class A address which leaves you with 8 network bits and 24 host bits...

/8 = 255.0.0.0 = 11111111.00000000.00000000.00000000

A subnet mask of /16 means you are borrowing 8 bits from the host portion of the address and using them as subnet bits...

/16 = 255.255.0.0 = 11111111.11111111.00000000.00000000

So you still have a class A address but using a /16 subnet mask. The first 8 bits are still network bits...

11111111.11111111.00000000.00000000

the next 8 bits are subnet bits...

11111111.11111111.00000000.00000000

and the rest are your host bits...

11111111.11111111.00000000.00000000

To work out how many subnets you have, use the formula 2^n where n is the number of subnet bits...

2^8 = 256 subnets

To work out the number oh hosts you have, use the formula 2^n - 2 where n is the number of host bits...

2^16 - 2 = 65536 - 2 = 65534 hosts per subnet

CompUBug

I see... I tried to apply that method to this problem...

Question: You need to subnet the 192.168.4.0 network into 7 different subnets. What subnet mask would you use?


11111111.11111111.11111111.11110000

1 6 3 1 8 4 2 1
2 4 2 6
8 32

---

1 1 1 1 1 0 0 0
---
248

255.255.255.248 (/29)

-- but the real answer is
255.255.255.224 (/27)

Isnt this by default a /24 network???

How is that so when we're looking for 7 subnets, and a /27 network will give us 16 instead ??

phobophile

CompUBug wrote: »

I see... I tried to apply that method to this problem...

Question: You need to subnet the 192.168.4.0 network into 7 different subnets. What subnet mask would you use?


11111111.11111111.11111111.11110000

1 6 3 1 8 4 2 1
2 4 2 6
8 32

---

1 1 1 1 1 0 0 0
---
248

255.255.255.248 (/29)

-- but the real answer is
255.255.255.224 (/27)

Isnt this by default a /24 network???

How is that so when we're looking for 7 subnets, and a /27 network will give us 16 instead ??

/27 gives you 8 subnets.

/24 = 1 subnet
/25 = 2
/26 = 4
/27 = 8 subnets

Since you can not get 7 subnets, you have to round it up to 8.

2^? = 8

? = 3

You have to borrow 3 host bits from the last octet.

x.x.x.11100000 = /27

You end getting a total of 8 subnets with 30 (2^5 - 2) hosts.

CompUBug

Thank you SOOO much guys, I really do thank you for helping me through this when you could have just told me to go out and do harder reading / harder studying, re-reading, ect - (which I am doing too it's just this stuff is fustratingly hard) .. I'm still working on practice problems and I'm getting more and more right answers. Thanks!

2E151

I have acouple of questions of my own. I've gotten basic subnetting down fairly well (given this address and CIDR find which network it lays on, first usable, last usable, broadcast, next network, ect)

However, the harder problems are starting to stump me as I don't which process to solve the problem.

#1. Your building a network with 22 hosts with only one routable IP address provided by the ISP. Which IP address block can you use to address the network?

A. 10.11.12.16/28
B. 172.31.255.127/27
C. 192.168.1.0/28
D. 209.165.202.128/27

The correct answer is B; 172.31.255.128/27

I ruled out both /28s because I know that its increment is 16. But I can't tell how to differentiate between B & D...

#2. You have been assigned the address block 10.255.255.224/28 to create the network addresses for point-to-point WAN links. How many of these WANs can you support with this address block.

A.1
B.4
C.7
D.14

The correct answer is B; 4 networks at 224,228,232, and 236.

Evidently I'm missing a key step in here, but a /28 gives you an increment of 16...

miller811

2E151 wrote: »

I have acouple of questions of my own. I've gotten basic subnetting down fairly well (given this address and CIDR find which network it lays on, first usable, last usable, broadcast, next network, ect)

However, the harder problems are starting to stump me as I don't which process to solve the problem.

#1. Your building a network with 22 hosts with only one routable IP address provided by the ISP. Which IP address block can you use to address the network?

A. 10.11.12.16/28
B. 172.31.255.127/27
C. 192.168.1.0/28
D. 209.165.202.128/27

The correct answer is B; 172.31.255.128/27

I ruled out both /28s because I know that its increment is 16. But I can't tell how to differentiate between B & D...

#2. You have been assigned the address block 10.255.255.224/28 to create the network addresses for point-to-point WAN links. How many of these WANs can you support with this address block.

A.1
B.4
C.7
D.14

The correct answer is B; 4 networks at 224,228,232, and 236.

Evidently I'm missing a key step in here, but a /28 gives you an increment of 16...

The first question seems like a wrong answer. The 27 mask is right,

172.31.255.127/27 is not an address block, it is an address

On the second one you would take the original address and subdivide it.

so with the /28 it is 16 positions, 14 useable.
your usable space is .224 - 239

but if you changed it to a /30 it would become
so with the /30 you meet the criteria
10.255.255.224
10.255.255.228
10.255.255.232
10.255.255.236

Which is what you would prefer that would allow 4 distinct ip schemes for the remote networks.

gosh1976

That has got to be a typo on the first question - 172.31.255.127/27 is a broadcast address so it's not a subnet id/block or a usable address it's a broadcast address. you know it can't be the only other /27 address because that's not a private ip range and the question implies it should be. So without even looking at anything past the first octet and the subnet you know it has to be B and not D.

2E151

Another question. Given an address and a CIDR I can easily find the network its on, broadcast, 1st and last, ect, ect.

That's not typically a problem for me, however I'm confused when your given an address and a required host and have to subdivide it.

Ie: 10.20.100.50 and you need 512 hosts. I can divide it easily into the required subnets, however whats stumping me is what dictates if you start 100 and continue going up or when you start at 10.20.0.0 and start from there...

I hope I haven't worded this too confusingly, long day and I'm tired, but if I did I'll try rewording it tomorrow.

phobophile

2E151 wrote: »

Another question. Given an address and a CIDR I can easily find the network its on, broadcast, 1st and last, ect, ect.

That's not typically a problem for me, however I'm confused when your given an address and a required host and have to subdivide it.

Ie: 10.20.100.50 and you need 512 hosts. I can divide it easily into the required subnets, however whats stumping me is what dictates if you start 100 and continue going up or when you start at 10.20.0.0 and start from there...

I hope I haven't worded this too confusingly, long day and I'm tired, but if I did I'll try rewording it tomorrow.

I guess we need to know what the actual question is. Is it asking what subnet mask you should use? If so, I think your making it more complicated than what it is.

The only thing you would need to be concerned with is that 10.20.100.50 is a Class A address, with a default mask of 255.0.0.0. (/.

You have 24 host bits to play with. Since the question is asking that you need 512 hosts, how many host bits does it take to reach that number?

2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 16
2^5 = 32
2^6 = 64
2^7 = 128
2^8 = 256
2^9 = 512
2^10 = 1024

Keep in mind however, that you always have to subtract 2 from the available hosts, so this question is a bit tricky.

Saving 9 host bits would only give you 510 hosts, so you would have to save 10 host bits to satisfy this question.

Therefore, the subnet mask needed to have 512 usable hosts in this network is 255.255.252.0 (/22).

CompUBug

I have another question...

What is the broadcast address of the subnet that host 172.18.182.245/23 is a part of?

<MY Method of solving>
/23 - Next Octet = /24
24 - 23 = 1

172.18.0.0 - 172.18.0.1 - 172.18.0.255
172.18.1.0 - 172.18.1.1 - 172.18.1.255

Answer = 172.18.182.255
</MY Method of solving>

The real answer was 172.18.183.255

But how when the B Octet can only increase in increments of 1? 183 is on a different subnet than 182...

miller811

CompUBug wrote: »

I have another question...

What is the broadcast address of the subnet that host 172.18.182.245/23 is a part of?

<MY Method of solving>
/23 - Next Octet = /24
24 - 23 = 1

172.18.0.0 - 172.18.0.1 - 172.18.0.255
172.18.1.0 - 172.18.1.1 - 172.18.1.255

Answer = 172.18.182.255
</MY Method of solving>

The real answer was 172.18.183.255

But how when the B Octet can only increase in increments of 1? 183 is on a different subnet than 182...

you are missing the powers of 2....

one position = 2 numbers
2 = 4
3 = 8
4 = 16
5 = 32
6 = 64
7 = 128
8 = 256
etc

so if you took 172.18.0.0 with a /23
172.180.0.0 - 172.18.1.255
172.180.2.0 - 172.18.3.255
172.180.4.0 - 172.18.5.255
etc....
172.18.182.0 - 172.18.183.255

so the address 172.18.182.245/23
fits into the 172.18.182.0 - 172.18.183.255

starts on an even 172.18.0.... .2 .4 .6 .8 .182
ends on the odd 172.18..1... 3..5...7...9....183

hope that helps

mella060

CompUBug wrote: »

I have another question...

What is the broadcast address of the subnet that host 172.18.182.245/23 is a part of?

<MY Method of solving>
/23 - Next Octet = /24
24 - 23 = 1

172.18.0.0 - 172.18.0.1 - 172.18.0.255
172.18.1.0 - 172.18.1.1 - 172.18.1.255

Answer = 172.18.182.255
</MY Method of solving>

The real answer was 172.18.183.255

But how when the B Octet can only increase in increments of 1? 183 is on a different subnet than 182...

First of all you need to work out what the increment is for a /23 subnet mask...

/23 = 255.255.254.0 = 11111111.11111111.11111110.00000000

128 64 32 16 8 4 2 1...or 256-254 = 2

So your increment is 2 and in the 3rd octet...any octet with a 255 in it such as the second octet will always increment by 1...

256-255 = 1...or the method i use where you work out the last bit position where the value is a 1...

255 = 11111111...128 64 32 16 8 4 2 1

So your networks and their ranges and broadcast addresses would be...

172.18.0.0...network range = 172.18.0.1 - 172.18.1.254...broadcast address = 172.18.1.255
172.18.2.0...network range = 172.18.2.1 - 172.18.3.254...broadcast address = 172.18.3.255
172.18.4.0...network range = 172.18.4.1 - 172.18.5.254...broadcast address = 172.18.5.255
172.18.6.0...network range = 172.18.6.1 - 172.18.7.254...broadcast address = 172.18.7.255
172.18.8.0...network range = 172.18.8.1 - 172.18.9.254...broadcast address = 172.18.9.255
172.18.10.0 and so on right up to...

172.18.178.0...network range = 172.18.178.1 - 172.18.179.254
broadcast address = 172.18.179.255

172.18.180.0...network range = 172.18.180.1 - 172.18.181.254
broadcast address = 172.18.181.255

172.18.182.0...network range = 172.18.182.1 = 172.18.183.254
broadcast address = 172.18.183.255

next one would be

172.18.184.0...network range = 172.18.184.1 - 172.18.185.254
broadcast address = 172.18.185.255