Subnetting Question on Quiz

[rcfant89]

"Given the network address of 120.120.0.0, subnet it for up to 100 usable subnets. What is the 50th usable subnetwork address (and the broadcast address for that subnet)?"

That is verbatim off the quiz. I'm not sure if the teacher wants to use the default subnet mask for a class A address (N.H.H.H.) or a non default subnet mask. The info above is verbatim so you just put down what you think the answer is. No other information is available.

 

[ex_bubblehead]

Everything you need to know is presented in the question. You need to determine the correct netmask necessary to satisfy the requirements in the first part of the question. Once you do that you can answer the second part of the question.

 

[rcfant89]

So what's the answer?

 

[ex_bubblehead]

So what's the answer?

That's for you to work out. I'm not going to do your homework for you.

 

[rcfant89]

Thanks for your insightful help...

 

[bill001g]

He is in a way correct. There is no way to know the subnet mask of the main network. It may be implied if you assume 0 in certain fields mean something or if you use the old assumptions based on classfull subnets. So the main network could range from a /14 to a /23 and still work for this question. So the question is invalid....but likely they want the answer based on the classful network mask.

 

[rcfant89]

He is in a way correct. There is no way to know the subnet mask of the main network. It may be implied if you assume 0 in certain fields mean something or if you use the old assumptions based on classfull subnets. So the main network could range from a /14 to a /23 and still work for this question. So the question is invalid....but likely they want the answer based on the classful network mask.

This is what I was talking about, thanks.

Class A networks use N.H.H.H by default (for subnet mask) but this one, the network is not using a default subnet mask since it is using N.N.H.H (as demonstrated by the 120.120.0.0 network address). If that was a host on the network, then I would assume subnet mask 255.0.0.0 or CIDR /8. I'm not sure if the teacher meant to write 128.120.0.0 or if he meant to use a non default subnet mask but if that's the case, he would have to say what it is since it could be either /13-16 or more I guess.

It would be:
01111000.01111000.00000000.00000000
11111111.11111000.00000000.00000000

I guess in binary but it could also be:
01111000.01111000.00000000.00000000
11111111.11111111.00000000.00000000

So that's the question.

 

[bill001g]

He is in a way correct. There is no way to know the subnet mask of the main network. It may be implied if you assume 0 in certain fields mean something or if you use the old assumptions based on classfull subnets. So the main network could range from a /14 to a /23 and still work for this question. So the question is invalid....but likely they want the answer based on the classful network mask.

This is what I was talking about, thanks.

Class A networks use N.H.H.H by default (for subnet mask) but this one, the network is not using a default subnet mask since it is using N.N.H.H (as demonstrated by the 120.120.0.0 network address). If that was a host on the network, then I would assume subnet mask 255.0.0.0 or CIDR /8. I'm not sure if the teacher meant to write 128.120.0.0 or if he meant to use a non default subnet mask but if that's the case, he would have to say what it is since it could be either /13-16 or more I guess.

It would be:
01111000.01111000.00000000.00000000
11111111.11111000.00000000.00000000

I guess in binary but it could also be:
01111000.01111000.00000000.00000000
11111111.11111111.00000000.00000000

So that's the question.

But it cannot be shorter than a /13 because they told you the network address must be 120.120.0.0 if you would go to a /8 the network address would be 120.0.0.0. But yes that is why there are multiple correct answers.

Lets say I use 128 (ie more than 100) subnets of size 4 ie /30. Then the networks would start at 120.120.0.0 and the second would be 120.120.0.4 and so on up to 120.120.1.254 or I could use 128 subnets of size 256 ie /17so the first would be 120.120.0.0 the second 120.120.1.0 and number 128 would be 120.120.128.0 There are many many other possibilities I am too lazy to calculate. But it just as easy go from 120.120.0.0 to 120.127.255.255