Given address identify 124th subnet

erfolg255

Hi i have a Task
Given the Class A network 10.0.0.0 create a subnet mask for 300 subnets
Identify 124th and VLSM for a further 75 subnets. Show all of your workin
Assume subnet 0 is used.

I created subnet mask which is 255.255.128.0
and the address 0 will be 10.0.0.1- 10.0.127.254)

but i have a problem how quickly on the exam identify 124 th subnet (

any suggestion will be appreciated

erfolg255

none? can help?

hiddenknight821

erfolg255 wrote: »

none? can help?

First of all, I want to say I'm surprised no one here was able to answer your question right away, even when we all should be familiar with subnetting.

I'm assuming that no one answers your post due to the question being slightly vague. I see no reason for doing VLSM for a further 75 subnets unless the question is asking us to have exactly 300 subnets out of 512 possible subnets, but again when I thought about it, I don't think it's possible, but I'm not good with calculating VLSM when I know there's a missing variable such as what's the minimum number of hosts needed for each subnet. If I know this then it would be very helpful.

Let me try brainstorming to see if I can help you get started. You have 2^15 (32,766) maximum hosts per subnet. Out of 512 subnets, you have 16,776,192 hosts. If you were to create 300 subnets, then you would get 9,829,800. So, 16,776,192 - 9,829,800 = 6,946,392 host addresses are being wasted along with 212 subnet IDs. From there, you can play around with the problem.

erfolg255

this task was on the exam in Open University in 2009 for CCNA

Finally after 5 hours, I managed to do it. below is a solution for it for some one if will have the same problem

so we have 10.0.0.0 and need 300 subnets.
10.0.0.0 is a class A Network (1-27) with a default subnet mask 255.0.0.0
we need 300 subnets it means that we need to borrow 9 bits - 2^9= 512

because we took 9 bits for a subnetting the mask for a new network will be 255.255.128.0 or /17 so give us 2^15 host for every subnet.

256- 128=128
the first subnet will be 10.0.0.1- 10.0.127.254

now we have to find 124th subnetwork , if count subnet 0 we actually need 124-1=123

123= 01111011 in binary

to find out the 124th subnet we need :

Convert the new number of the network to binary.
3 Place the binary address to the right-hand side of the subnet masked portion.
4 Fill the gaps (if any) on the left with 0s back to the network portion of the class of
address (as well as the right host portion, which must be 0).
5 Now calculate the result, which is the nth subnet.

11111111.11111111.10000000.00000000.
network. 00111101.1000000.00000000

so we have 10.61.128.0 -the 124th subnet (123)

I hope that someone will find this information useful. Do you know any other methods?

hiddenknight821

erfolg255 wrote: »

Identify 124th and VLSM for a further 75 subnets. Show all of your workin
Assume subnet 0 is used.

Oh I see you're asking for a shortcut but you didn't answer your own question wholly. Read again. It asked for VLSM. Did you compare your answer with the actual answer where you found the problem?

Hey, here's a rule of you must keep in mind. Let's say X represents the Class of the given IP address in the problem and Y represents the Class of the subnet mask you're using. Use the formula, Y - X where Y in the problem mentioned above is 3, because in the third octet, there are 1 and 0s. I called that the mixed octet or byte. The X is the number of octet where the default subnet mask is masked with 1s. 10.0.0.0 is Class A, so only 1 octet is masked by default. So you get 3 - 1 = 2. The result will always be the number of octet where you will see change or increment.

But you do not really need to memorize the formula since subnetting a Class A address with Class C mask will always use two octets. There are no other example such as subnetting Class A with Class D address, which is not possible.

So, because there are TWO octets you would have to look out for, it's always true that you would have to divide the given (subnet location you have to identify) by TWO. In your problem, 123/2, you get 61.5. Like you said, 124 - 1 = 123, which is the subnet ID of the 124th subnet.

61 represents the second octet, and if you multiply .5 by 256, then you get 128, which is the third octet. I hope that helps.

Forsaken_GA

Most folks probably aren't going to answer this because it seems too much like homework.

The answers really not that difficult, and you've already got it down. It's basically asking you to create 300 subnets, then find the 124th subnet, and within that 124th subnet, create another 75 subnets.

So you first chop it up into /17's, and on the 124th subnet, you'd take that and chop it up into 128 /24's.

Chopping larger blocks into smaller ones is pretty common. For example, I've had a few occasions where I've had to chop up a /24 into /30's for point to point links.

SubnettingGoddess

hiddenknight821 wrote: »

Oh I see you're asking for a shortcut but you didn't answer your own question wholly. Read again. It asked for VLSM. Did you compare your answer with the actual answer where you found the problem?

Hey, here's a rule of you must keep in mind. Let's say X represents the Class of the given IP address in the problem and Y represents the Class of the subnet mask you're using. Use the formula, Y - X where Y in the problem mentioned above is 3, because in the third octet, there are 1 and 0s. I called that the mixed octet or byte. The X is the number of octet where the default subnet mask is masked with 1s. 10.0.0.0 is Class A, so only 1 octet is masked by default. So you get 3 - 1 = 2. The result will always be the number of octet where you will see change or increment.

But you do not really need to memorize the formula since subnetting a Class A address with Class C mask will always use two octets. There are no other example such as subnetting Class A with Class D address, which is not possible.

So, because there are TWO octets you would have to look out for, it's always true that you would have to divide the given (subnet location you have to identify) by TWO. In your problem, 123/2, you get 61.5. Like you said, 124 - 1 = 123, which is the subnet ID of the 124th subnet.

61 represents the second octet, and if you multiply .5 by 256, then you get 128, which is the third octet. I hope that helps.

hiddenknight821

SubnettingGoddess wrote: »

I hope you find my clear, concise explanation entertaining. Thanks!

Forsaken_GA wrote: »

Most folks probably aren't going to answer this because it seems too much like homework.

The answers really not that difficult, and you've already got it down. It's basically asking you to create 300 subnets, then find the 124th subnet, and within that 124th subnet, create another 75 subnets.

So you first chop it up into /17's, and on the 124th subnet, you'd take that and chop it up into 128 /24's.

Chopping larger blocks into smaller ones is pretty common. For example, I've had a few occasions where I've had to chop up a /24 into /30's for point to point links.

First of all, I understand most folks are probably thinking this is homework question, but then what's the best way for him or someone to learn. I understand we should not answer a fill-in-the-blank question for anyone, but a math question like subnetting is not applicable at all since we all have to calculate and arrive to a solution at some points. He can't **** on the test if the teacher's gonna give him a different problem or else he's screwed. I am a peer tutor at school, and if he's my student, then I would probably give him an answer with an explanation on how I reach to a solution, but I would not give him answer again for the next similar problem. I will create my own problem similar to the given problem to guide him through the steps again.

Since you seem to be more familiar with his question, I wish the question was rephrased differently so I understand what it was asking in the first place. Your wording is much better. Thanks for the clarification, but just want to double-check with you to see if I and everyone else actually have the right idea. Is this what you mean?

128 subnets ( 2^7) will fit in the 124th subnet being further subnetting into 75 subnets with VLSM. I would need to borrow 7 more host bits. 17+7 = 24.

123rd subnet (122): 10.61.0.1 - 10.61.127.254 /17
124th subnet (123): 10.61.128.1 - 10.61.128.254 /24 = Answer to the original question
125th subnet (124): 10.61.129.1 - 10.61.129.254 /24
............
250th subnet (249): 10.61.255.1 - 10.61.254.254 /24
251th subnet (250): 10.61.255.1 - 10.61.255.254 /24
252nd subnet (251): 10.62.0.1 - 10.62.127.254 /17

So this means I would have 640 subnets altogether?

Am I getting the right idea? Not only I want to make sure I get the right idea, but so the others know what to do. Thanks.

Bl8ckr0uter

Man I suck at subnetting

notun1

what is subnetting??

erfolg255

Hi I wasn't going to **** at all. I am a new guy and not all rules are clear for me. As you've propably noticed allready English is not my mother tounge language,so sometimes I have a problem to ask a clear question. I am preaparing now for my firs certificate-Ccna that way I asked about how to quickly find right n subnet thx for your help