Subnetting Question

paulaz

Hi all. Looking for some assistance with a subnetting question, looking for broadcast address and ranges.

206.82.73.146 /28

My answer:
Broadcast- 206.82.73.159-- All good here
Ranges- 206.82.73.147 -- 206.82.73.158--- All good here besides the minimum.

I run this through all the calculators and get a minimum address of 206.82.73.145. How in the world would the minimum address be 145, when the given address was 146? Everything I have learned has said the range would be the address between the subnet address and the broadcast address, that being 146-159 (with 147 being the first used and 158 being the last used). Very confused here folks.....please help me.

Welly_59

206.82.73.146 /28

A /28 is a 255.255.255.240, so count up in 16's until you get to a range which includes 206.82.73.146. 16 32 48 64 80 96 112 128 144 160

So start of the subnet is 206.82.73.144, broadcast is 206.82.73.159. Usable address space 206.82.73.145-206.82.73.158

Cisco Inferno

Youre doing it right. 32-28=4 , 2^4=16 block size.
0..16..32......144..160

The problem you are having is with the questioning. Most subnet mask questions like these give you a host.
Think of .146 as the host. Now what is the network and range that it is on?

You were thinking the .146 was the network ID. Why would they give you the network so easily?

Unless they explicitly said .146 was the network, I would think of it as a host.

paulaz

That is it, you are right and thank you! I got caught on wording!!!!! Really appreciate it.

paulaz

Perhaps ya'll can help again. I posted this lower in the forum, but I think that discussion may be dead:
Hey all. Hoping someone can help clear up what I sometimes get a little iffy on. With the following question What subnet does 10.34.67.234/12 sit on? The block size 16, so when I go to do my ranges, I just want to make sure the max is correct, that being the 3rd and 4th octet

10.0.0.0---10.15.254.255
10.16.0.0---10.31.254.255
10.32.0.0

---

10.47.254.255
10.48.0.0......etc

The answer is the 10.32.0.0 subnet. For that 3rd subnet, I put the max range of 10.47.254.255 but i don't exactly know why the third octet is 254. Is it correct that the subnet ID is 10.32.0.0 and the broadcast is 10.47.255.255? If so, I know that you cannot use the subnet ID and broadcast, but what is the "rule" that says the max third octet is 254? If I'm correct here, I'm only correct because I have seen it enough times but I don't why I'am right. Hopefully that makes sense.

huntert

https://www.udemy.com/ip-subnetting/

recommend that you take this course from udemy on subnetting. it is free and very good.

paulaz

I have watched countless hours of the videos. CBT, Chris Bryant, YouTube, Cisco Press, etc. Either I have missed a concept time and time again, or..I have no idea.

What I have gathered and what is confusing me can be explained in with these two addresses:
180.76.182.241 /29
59.247.181.211 /9

With 180.76.182.241 /29 the network is 180.76.182.240 and we are working in the 4th octet. The first usable is 180.76.182.241, and the last is .246. We can't use .240 as it is part of the network, so we go to .241. Cant use the first and last, got it.

With 59.247.181.211 /9 the network is 59.128.0.0 and we are working in the second octet. The first usable is 59.128.0.1 and the last is 59.255.255.254. Ok, but why is the first usable not 59.129 apposed to 128? Wouldn't 59.128 be the network, since we can't use the first or last, why is the first usable 59.128 when the network is 59.128?

damico

paulaz wrote: »

I have watched countless hours of the videos. CBT, Chris Bryant, YouTube, Cisco Press, etc. Either I have missed a concept time and time again, or..I have no idea.

What I have gathered and what is confusing me can be explained in with these two addresses:
180.76.182.241 /29
59.247.181.211 /9

With 180.76.182.241 /29 the network is 180.76.182.240 and we are working in the 4th octet. The first usable is 180.76.182.241, and the last is .246. We can't use .240 as it is part of the network, so we go to .241. Cant use the first and last, got it.

With 59.247.181.211 /9 the network is 59.128.0.0 and we are working in the second octet. The first usable is 59.128.0.1 and the last is 59.255.255.254. Ok, but why is the first usable not 59.129 apposed to 128? Wouldn't 59.128 be the network, since we can't use the first or last, why is the first usable 59.128 when the network is 59.128?

You need to think the first usable address. In this case it will be 59.128.0.1/9. You start to count on the last octet, since you already have the network address.

cholland89

paulaz wrote: »

Perhaps ya'll can help again. I posted this lower in the forum, but I think that discussion may be dead:
Hey all. Hoping someone can help clear up what I sometimes get a little iffy on. With the following question What subnet does 10.34.67.234/12 sit on? The block size 16, so when I go to do my ranges, I just want to make sure the max is correct, that being the 3rd and 4th octet

10.0.0.0---10.15.254.255
10.16.0.0---10.31.254.255
10.32.0.0

---

10.47.254.255
10.48.0.0......etc

The answer is the 10.32.0.0 subnet. For that 3rd subnet, I put the max range of 10.47.254.255 but i don't exactly know why the third octet is 254. Is it correct that the subnet ID is 10.32.0.0 and the broadcast is 10.47.255.255? If so, I know that you cannot use the subnet ID and broadcast, but what is the "rule" that says the max third octet is 254? If I'm correct here, I'm only correct because I have seen it enough times but I don't why I'am right. Hopefully that makes sense.

I'm interested in a response to this as well. I would have answered the range as 10.32.0.1-10.47.255.254. Not sure why the third octet wouldn't be 255.

stl4jsp

It's a class C address this is what I got and how I do subnetting (I'm taking a CCNA course) you would be hopping by 16 in till you get to 144 because your cider is 28

so the network would be 144 your first usable address is 145 and your last usable would be 158 and a broadcast of 159 so the range is 145-158 and the next network would be 160 hope this helps