Shortcut trick for finding broadcast address?

ehnde

Question: What is the broadcast address of the network 172.17.176.0 255.255.248.0?
Answer: 172.17.183.255


I can quickly determine that an interval of 8 is needed, so you count up like this:
172.17.0.0
172.17.8.0
.16.0
.32.0


Counting like this is ineffective going up to 176, but 176 divided by 8 is 22. An even number. Because of this, you know that 172.171.76.0 is the network address.



I'm taking the exam soon...should I be prepared for a dirty trick like being given a similar question, but instead I'm given 172.171.177.0 - what network does this address fall in and what is the broadcast address?


If this were the case, you could divide 177 by 8 and would be left with a remainder of 1, quickly showing you that 172.171.176.0 was the network you were dealing with.



Will this shortcut always work? Can I depend on it?

okplaya

I know subnetting quite well and I'm a bit confused at your method. Let me try to figure this out.

I tried a few problems and does seem to work, but I'm not telling you to depend on it. If you can divide well then go for it. If you are given a question like you posted, you really just need to find the next network. From there, you can get the broadcast in 1 second. No need to count up for that kind of question.

Jas21

I agree

I just remember networks are even, first address is odd, broadcasts are odd

ehnde

Jas21 wrote: »

I agree

I just remember networks are even, first address is odd, broadcasts are odd

If you are dealing with an interval of 4, 8, or 16 up to 200-something it's not quite as simple.

172.26.201.0 with a mask of 255.255.240.0. What network? We have to count in multiples of 16 up to that range to figure it out. Or, divide 201 by 16.
You get 12 with a remainder of 9.
201 - 9 = 192
Your network address is 172.26.192.0.
Counting in 4's, 8's, or 16's is a bit time consuming if you have 50 questions and only ....90 minutes, is it?

okplaya

If you're dealing with the interval 2,4,8, and 16 then multiplication is much faster and easier in my opinion. In my head it takes longer for me to figure out 201 divided by 16 versus 10 x 16 and just adding intervals of 16 until I get the answer. I guess each person is different in how they can perform calculations in their head.

For your question, I know the interval is 16, and the third octet is 201. No point in me going 16,32,48,etc. That takes forever. Here is where I know that 10 x 16 is 160. I also know that 2 x 16 is 32. I also know that 160 + 32 is 192. Again, that takes me all of 3-4 seconds tops in my head. The division would throw me off a little bit because I would feel the need to double check.

You have the fundamentals down packed, so I think you will be fine for the test. Go with the way that works best for you and good luck too. When are you taking it exactly?

ehnde

okplaya wrote: »

If you're dealing with the interval 2,4,8, and 16 then multiplication is much faster and easier in my opinion. In my head it takes longer for me to figure out 201 divided by 16 versus 10 x 16 and just adding intervals of 16 until I get the answer. I guess each person is different in how they can perform calculations in their head.

For your question, I know the interval is 16, and the third octet is 201. No point in me going 16,32,48,etc. That takes forever. Here is where I know that 10 x 16 is 160. I also know that 2 x 16 is 32. I also know that 160 + 32 is 192. Again, that takes me all of 3-4 seconds tops in my head. The division would throw me off a little bit because I would feel the need to double check.

You have the fundamentals down packed, so I think you will be fine for the test. Go with the way that works best for you and good luck too. When are you taking it exactly?

I hadn't thought of doing it that way never occurred to me. Now that you mention it 10x16+2x16 is easier than 201/16, taking the remainder and subtracting that from 201.

I'm taking the exam Friday (640-802). Hadn't did any serious subnetting since January and figured now is a good time to brush back up on it.

bermovick

In my "pre-exam" time, I quickly noted all the multiples of 16 on my "paper" to refer to. 16 seems to be the most difficult to quickly count up by in your head. Then if my interval was 32 I just needed to count every other 16. That way I got around without needing to have shortcuts to remember.

Forsaken_GA

ehnde wrote: »

Question: What is the broadcast address of the network 172.17.176.0 255.255.248.0?
Answer: 172.17.183.255


I can quickly determine that an interval of 8 is needed, so you count up like this:
172.17.0.0
172.17.8.0
.16.0
.32.0


Counting like this is ineffective going up to 176, but 176 divided by 8 is 22. An even number. Because of this, you know that 172.171.76.0 is the network address.

You over thought it.

They tell you flat out the network address and it's mask. From that you can determine the range. You found the range of 8, so 172.171.76.0 as the prefix, it ranges on up to 172.171.83.255. The broadcast address is always the last possible address in the subnet. The second you find the network address and the range, you can find the broadcast

Priston

@okplaya you don't even need to multiply


If I had a number like 177 and 248 mask I would think
128+64 = 192 (too big)
128+32 = 160 (ok, next)
160+16 = 176 (ding ding ding!)
so 176-183 must be the range

Nitro Mustang

ehnde wrote: »

Question: What is the broadcast address of the network 172.17.176.0 255.255.248.0?
Answer: 172.17.183.255


I can quickly determine that an interval of 8 is needed, so you count up like this:
172.17.0.0
172.17.8.0
.16.0
.32.0


Counting like this is ineffective going up to 176, but 176 divided by 8 is 22. An even number. Because of this, you know that 172.171.76.0 is the network address.



I'm taking the exam soon...should I be prepared for a dirty trick like being given a similar question, but instead I'm given 172.171.177.0 - what network does this address fall in and what is the broadcast address?


If this were the case, you could divide 177 by 8 and would be left with a remainder of 1, quickly showing you that 172.171.176.0 was the network you were dealing with.



Will this shortcut always work? Can I depend on it?

That confused the hell out of me. Maybe I'm oversimplifying it, but here's my take:

What is the broadcast address of the network 172.17.176.0 255.255.248.0?

The mask used indicates a network increment of 8
If 172.17.176.0 is the network address, add 8 to the 3rd octet to get 172.17.184.0. The IP "before" that is 172.17.183.255.

Done.

ehnde

Nitro Mustang wrote: »

That confused the hell out of me.

Sorry, my obfuscated math is confusing - it's definintely not you The whole key to the question was "network address". Talking through it helped me be a bit quicker with the reasoning you guys used. One of many great things about TE.

Welcome to the forums!