MS Press Question

Everlife

Hi guys,

I ran across a subnetting question that stumped me today when finishing up the subnetting chapter. I'm convinced the answer is very straightfoward, but I just can't figure it out and no explanation is offered.

Q: You have obtained the 192.168.0.0 /20 address block from the central network administrator. Within your network, you want the addresses 192.168.3.1 and 192.168.4.1 to belong to the same subnet, but you also want to maximize the number of subnets. Which subnet mask should you configure within your organization?

a. /20
b. /21
c. /22
d. /32

Answer: B

Can anyone run through the steps that would be involved to determine that?

Thanks!

Everlife

I read through this a few more times and I think I have it. If you're required to keep 192.168.3.0 - 192.168.4.255 within the same subnet you'd need to have a group value that meets that criteria.

If you decided to use 254, you would have increments of 2 (256-254) meaning...

192.168.0.0 - 192.168.1.255
192.168.2.0 - 192.168.3.255
192.168.4.0 - 192.168.5.255

This doesn't meet the requirements.

If you decided to use 252, you would have increments of 4 (256-252) meaning...

192.168.0.0 - 192.168.3.255
192.168.4.0 - 192.168.7.255
192.168.8.0 - 192.168.11.255

This doesn't meet the requirements.

To maintain the two ranges within the same subnet you'd want to use 255.255.248.0.

This would allow you to have increments of 8 (256-24 meaning you'd have ranges of...

192.168.0.0 - 192.168.7.255
192.168.8.0 - 192.168.15.255
192.168.16.0 - 192.168.31.255

This would meet keep both ranges within the same subnet as well as maintaining the maximum number of subnets available. Writing down this explanation definitely helps clear it up in my own mind. If my logic is wrong here, please let me know!

georgemc

OK, assume that 192.168.x.x is wrong and they meant 172.16.x.x.

D. /32 Throwaway answer

C. /22 subnets = 172.16.0.1 - 172.16.3.255
172.16.4.1 - 172.16.7.255
172.16.8.1 - 172.16.11.255
etc... through 172.16.252.1 - 172.16.255.255
172.16.3.1 and 172.16.4.1 are in different subnets, so this isn't your answer

B. /21 subnets = 172.16.0.1 - 172.16.7.255
172.16.8.1 - 172.16.15.255
etc.... through 172.16.248.1 - 172.16.255.255
172.16.3.1 and 172.16.4.1 are in the same subnet and this is the smallest subnet(32 subnets that are the equivalent of 8 class C's) that would be able to contain both, thus this is the correct answer.

A. /20 subnets = 172.16.0.1 - 172.16.15.255
172.16.16.1 - 172.16.31.255
etc... through 172.16.240.1 - 172.16.255.255
Each subnet(only 16 of them) is the equivalent of 16 class C's and and does not maximize the number of subnets.

Hope this helps...

BTW, I'm going through the same book and there are quite a few place where they give a not so obvious answer without any type of explanation.


You replied to yourself before I was able to complete this but here it is anyway. You are correct and seem to have explained it a lot more eloquently than I can.

Everlife

I know what you mean, you'd think they'd give a clear explanation instead of (B). Great thanks, that's extremely helpful Microsoft Press! /sarcasm off

Curse you Microsoft!

Thanks for the reply and explanation George. Hopefully I can return the favor.