I understand subnetting the long way, but that wont help on the exam. So I need to be able to get it quick for the exam. I understand the powers of 2 chart but I only know part of it and need some help and advice with the rest just to make sure I'm getting it right. Especially on the way they ask the questions or using cidr or not for th esubnet mask.

2^7 2^6 2^5 2^4 2^3 2^2 2^1 2^0

128 64 32 16 8 4 2 1

128

192

224

240

248

252

254

255

I understand (I think) this so far - How many subnets and hosts can this IP give you? 172.20.15.0 /23?

So a /23 uses 7 extra bits in the mask for the network, so looking at the chart 2^7=128, so they could have 128 subnets (we only minus 2 for hosts from what I learned). So, using 7 bits for the network we have 9 bits leftover for hosts, we look at the chart 2^9 is 512-2=510, then 510 is the number of hosts we can have.

Is that correct? How do I find the subnet mask based on the chart above? Is it if I used 7 bits for the network then I would count 7 on the column on the left and that is 254. So the subet mask for this /23 address is 255.255.254.0, correct?

Also, what subnet does this IP reside on 172.20.15.5 /16? This is where I need help since I dont know how to find the subnet increment.

Is this where I take the /23 as previously and see it uses 7 bits for the network and on the chart that would be increment of 128.

172.20.0.0

172.20.128.0

172.20.256.0

So it must live between the .0 and .128 networks and the range of usable hosts would be

172.20.0.1

172.20.127.254 (I might have messed that up?)

Also, if my logic is correct on this, is that I need to know for the exams?

How in the world can I use that chart for class A addresses with /20 bits used in the SM, seems impossible? How can it be done?

Thanks.

2^7 2^6 2^5 2^4 2^3 2^2 2^1 2^0

128 64 32 16 8 4 2 1

128

192

224

240

248

252

254

255

I understand (I think) this so far - How many subnets and hosts can this IP give you? 172.20.15.0 /23?

So a /23 uses 7 extra bits in the mask for the network, so looking at the chart 2^7=128, so they could have 128 subnets (we only minus 2 for hosts from what I learned). So, using 7 bits for the network we have 9 bits leftover for hosts, we look at the chart 2^9 is 512-2=510, then 510 is the number of hosts we can have.

Is that correct? How do I find the subnet mask based on the chart above? Is it if I used 7 bits for the network then I would count 7 on the column on the left and that is 254. So the subet mask for this /23 address is 255.255.254.0, correct?

Also, what subnet does this IP reside on 172.20.15.5 /16? This is where I need help since I dont know how to find the subnet increment.

Is this where I take the /23 as previously and see it uses 7 bits for the network and on the chart that would be increment of 128.

172.20.0.0

172.20.128.0

172.20.256.0

So it must live between the .0 and .128 networks and the range of usable hosts would be

172.20.0.1

172.20.127.254 (I might have messed that up?)

Also, if my logic is correct on this, is that I need to know for the exams?

How in the world can I use that chart for class A addresses with /20 bits used in the SM, seems impossible? How can it be done?

Thanks.

Network and collaborate with thousands of CTOs, CISOs, and IT Pros rooting for you and your success.

Andrew Hancock - VMware vExpert

See if this solution works for you by signing up for a 7 day free trial.

Unlock 1 Answer and 7 Comments.

Try for 7 days”The time we save is the biggest benefit of E-E to our team. What could take multiple guys 2 hours or more each to find is accessed in around 15 minutes on Experts Exchange.

Our community of experts have been thoroughly vetted for their expertise and industry experience.