Subnetting Question

I'm having a really difficult time grasping subnetting. I was told to go to learntosubnet.com for a clear explanation of things, so I'm going to do that tonight. In the meantime, I came across the following question and really don't know how to figure it out. Any help with an explanation would be greatly appreciated.

Your computer has an IP address of 172.18.56.17 with a subnet mask of 255.255.248.0. Which of the following IP addresses are on your local subnet? (Choose all that apply.)

a. 172.18.47.200
b. 172.18.60.100
c. 172.18.89.157
d. 172.18.54.3
e. 172.18.65.117
f. 172.18.57.42

Thanks in advance!

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EdTheLad

256-248 = 8
So now i know each subnet is using a block of 8 addresses.
(248 ) is in the 3rd octet so this is where i'm looking at the 8 blocks of addresses.
Multiples of 8 are 0,8,16.24,32,40,48,56,64......etc

Address given has 56 in the third octet, so this falls within block range 56->63
56 is the subnet address
57-62 are valid hosts
63 is the broadcast
The next block of 8 starts at 64 just for info sake.

Which ip addresses are in the local subnet,looking at the 3rd octet 57 and 60 are the only
addresses which fall within my block so the answer is B,F.

Johnny Johnson

Thanks, Ed! Here's another question.

Your computer has been assigned the IP address 192.168.148.72 with a subnet mask of 255.255.252.0. Which of the following IP addresses is on your local subnet?

a. 192.168.140.12
b. 192.168.150.55
c. 192.168.148.73
d. 192.168.151.250
e. 192.168.155.32

Using your example, I came up with each subnet using a block of 4 addresses. Address given has 148 in the third octet, so the range is 148->151.

148 subnet address
149 and 150 are valid hosts
151 is the broadcast

So the answer I came up with was b, c, and d. Did I do that right?

Big Jizay

Ed is somewhat right. That's a quick way to do it. Use the number 256 and minus the number that isn't 255 or 0 to get the correct number. Use that number to figure out how the subnets are set up. For this problem, 256-248=8. This is how I do it in my mind:

255.255.248.0 mask
172.16.56.17 given ip

The first two octets of the subnet mask are 255.255. Which ever octets have 255 in the subnet mask, those octets in the given ip will not change. Since in the first two octets of the subnet mask you have is 255.255, the first two octets in the given ip will not change, so the 172.16 part will still stay as 172.16. The third octet in the subnet mask however is not 255, it's 248, so the third octet in the given ip will not be 56. This is where you put the 8 that we got from the 256-248=8. So you end up with 172.16.8 for the first three octets. Since the last octet is 0 in the subnet mask, the last octet in the given ip will also be zero, 172.16.8.0.
If done this way, this ip will be the start of the second subnet, because the start of the first subnet always starts with 0.

172.16.0.0 - 172.16.7.255
172.16.8.0 - 172.16.15.255
172.16.16.0 - 172.16.23.255
172.16.24.0 - 172.16.31.255

and so forth. My list would eventually lead to the 172.16.56.0 - 172.16.63.255 subnet. Any ip's listed in the question that are a part of this subnet are correct answers. So B and F are correct.

Another good thing to know is that the first ip and the last ip of a subnet are not usable ip's. The first ip is the network id and the last ip is the broadcast. So in the first subnet, 172.16.0.0 is the network id and the broadcast is 172.16.7.255. So the ip range that is usable for that subnet is 172.16.0.1 - 172.16.7.254. Ed's explanation for valid hosts and broadcast addresses would work perfectly for class C subnets, but not for class B subnets.

EdTheLad

Big Jizay wrote: »

Ed's explanation for valid hosts and broadcast addresses would work perfectly for class C subnets, but not for class B subnets.

OPs! forgot i was working on the third octet.

rwwest7

The key to it is to find your increment. Work on finding the increment ( or "magic number") as fast as possible. Once you've gotten that mastered then everything will fall into place. There many ways, I prefer the CBT Nuggets way of using the last 1's position.

EdTheLad

Johnny Johnson wrote: »

Thanks, Ed! Here's another question.

Your computer has been assigned the IP address 192.168.148.72 with a subnet mask of 255.255.252.0. Which of the following IP addresses is on your local subnet?

a. 192.168.140.12
b. 192.168.150.55
c. 192.168.148.73
d. 192.168.151.250
e. 192.168.155.32

Using your example, I came up with each subnet using a block of 4 addresses. Address given has 148 in the third octet, so the range is 148->151.

148 subnet address
149 and 150 are valid hosts
151 is the broadcast

So the answer I came up with was b, c, and d. Did I do that right?

You got the block stuff down.
192.168.148.72 255.255.252.0
256-252 = 4
block starts 148->151
subnet is 192.168.148.0
host range is 192.168.148.1 -> 192.168.151.254
broadcast address 192.168.151.255

Same concept as i showed you just an additional step to remember if using class A or B.
You got the answer right though.

Big Jizay

EdTheLad wrote: »

You got the block stuff down.
192.168.148.72 255.255.252.0
256-252 = 4
block starts 148->151
subnet is 192.168.148.0
host range is 192.168.148.1 -> 192.168.151.254
broadcast address 192.168.151.255

Same concept as i showed you just an additional step to remember if using class A or B.
You got the answer right though.

Exactly like that!! Thanks Ed

Johnny Johnson

Thanks for all of your help! I have one more question about supernetting. Hopefully I will understand it more after visiting the learn to subnet website tonight, but I would like to get someone else's take on it as well. I'm not sure how to go about working on this problem:

How many bits are required to supernet seven Class C addresses?

a. 1
b. 2
c. 3
d. 4
e. 5

Thanks again!

Johnny Johnson

EdTheLad wrote: »

Same concept as i showed you just an additional step to remember if using class A or B.
You got the answer right though.

What's the additional step to remember?

EdTheLad

Johnny Johnson wrote: »

Thanks for all of your help! I have one more question about supernetting. Hopefully I will understand it more after visiting the learn to subnet website tonight, but I would like to get someone else's take on it as well. I'm not sure how to go about working on this problem:

How many bits are required to supernet seven Class C addresses?

a. 1
b. 2
c. 3
d. 4
e. 5

Thanks again!

You gotta know 2s complement

2^3 = 8
so 8 bits are needed.
Class C address is 255.255.255.0 i.e. /24
To supernet you need to move to the left of the natural mask, in this case you want 7 class C addresses which means 3 bits

24-3 = 21
255.255.248.0

EdTheLad

Johnny Johnson wrote: »

What's the additional step to remember?

My first example i went wrong specifying a subnet mask and broadcast address, i forgot about the last octet, should have been as follows:

256-248 = 8
So now i know each subnet is using a block of 8 addresses.
(248 ) is in the 3rd octet so this is where i'm looking at the 8 blocks of addresses.
Multiples of 8 are 0,8,16.24,32,40,48,56,64......etc

Address given has 56 in the third octet, so this falls within block range 56->63
56.0 is the subnet address
56.1-63.254 are valid hosts
63.255 is the broadcast
The next block of 8 starts at 64.0 just for info sake.