Summary address

What is the best summary route for the addresses 172.17.192.0/24 through 172.17.207.0/24?

In the third octet I get:

11000000
11000111

which gives a boundary between 5th and 6th bit.

So summary address i come up with is 172.17.192 /21 but this doesn't appear correct as it doesn't include all addresses between 192 and 207, but /20 does. Are there times when you might have to adjust the mask?

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Comments

Netstudent

11000111 = 199

11001111 = 207

thats what happened. I don't think you were using the right binary bit combination to derive the correct summary mask. Here you will see the boundry between the 4th and 5th bit, thus giving you a /20.

Mikdilly

Netstudent wrote:

11000111 = 199

11001111 = 207

thats what happened. I don't think you were using the right binary bit combination to derive the correct summary mask. Here you will see the boundry between the 4th and 5th bit, thus giving you a /20.

Thanks, guess I've been going at this for too long today.

Pash

Mikdilly wrote:

Netstudent wrote:

11000111 = 199

11001111 = 207

thats what happened. I don't think you were using the right binary bit combination to derive the correct summary mask. Here you will see the boundry between the 4th and 5th bit, thus giving you a /20.

Thanks, guess I've been going at this for too long today.

But you should be happy that you know this inside out now!

Good luck.

Tricon7

Btw, what's a quicker way of determining the summary address for a problem like this instead of writing out in binary all numbers between 192 and 207 to see what they all have in common? I won't have that kind of time for a cert. test.

Pash

Tricon7 wrote:

Btw, what's a quicker way of determining the summary address for a problem like this instead of writing out in binary all numbers between 192 and 207 to see what they all have in common? I won't have that kind of time for a cert. test.

Not really a quicker way, You find the interesting octet, ie where the binary's don't match and then find the boundary.

Netstudent

What i like to do is first, look at the first subnet number and last subnet number. Like in this case 192 through 207. That is the range you need to cover. So what is the difference between 207 and 192? It's 15. So is there a block that covers 15? NO, but there is a block that covers 16. So now you need to know if 192 is divisible by 16. IF it is then it makes your problem really easy. IN this case it is, so you know your summary address will be 172.17.192.0. NOw what mask gives you a block of 16? YOu should know that a 240 has to be in there somewhere because 256-240 = 16. And since we are talking about the 3rd octet here, your mask will be 255.255.240.0.

mobri09

Netstudent wrote:

What i like to do is first, look at the first subnet number and last subnet number. Like in this case 192 through 207. That is the range you need to cover. So what is the difference between 207 and 192? It's 15. So is there a block that covers 15? NO, but there is a block that covers 16. So now you need to know if 192 is divisible by 16. IF it is then it makes your problem really easy. IN this case it is, so you know your summary address will be 172.17.192.0. NOw what mask gives you a block of 16? YOu should know that a 240 has to be in there somewhere because 256-240 = 16. And since we are talking about the 3rd octet here, your mask will be 255.255.240.0.

Nicely said Netstudent! Thanks for the help!

dtlokee

Yeah really, a subnet address is nothing more than a summary address for a bunch of host addresses

If you think of it in those terms you may seem to have an easier time of it.

mobri09

I agree....I always seem to make things harder than they really are -