Wildcard mask

What would be the right wildcard mask for the address range of 192.168.5.0 to 192.168.17.0?

Will i be right in saying that this is
5.0 0.0.3.255
8.0 0.0.7.255
16.0 0.0.0.255
17.0 0.0.0.255

OR

192.168.5.0 0.0.15.255

I'm leaning towards the first solution, I need a bit of help please

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Comments

QUIX0TIC

0.0.15.255 is too many and would go beyond or before the range that you are requesting. I would stick to the first solution.

Yankee

In order to cover x.x.5.0 thru x.x.17.0 you need to cover what amounts to 32 /24 networks meaning x.x.0.0 thru x.x.31.0 so none of the answers appear right to me as I read your question. From this info you should be able to get the right answer.

Yankee

lordy

The range you specified covers 12 Class C networks.

As 12 is not a power of 2 there simply is no perfectly matching wildcard mask.

Did you make up this question yourself or where did you find this one ?

2lazybutsmart

If the range doesn't perfectly fit into a block size, then you'll always have to go for the next block size in the list. so since you have 12 networks, you'll need a block size of 16 to summarize those addresses.

Of course, a block size of 16 would range from 5.0 - 20.0, but then there's nothing you can do about that. You'll just have to include those three addresses in the summary. so 0.0.15.255 to the rescue!

2lbs.

Yankee

You must stay on the bit boundries. Yes he wants to cover 15 /24s but the borders for 16 are 0-15 and 16-31. Neither of which cover his range entirely so he needs to use the wild card mask of 0.0.31.255

Yankee

forbesl

Yankee wrote:

You must stay on the bit boundries. Yes he wants to cover 15 /24s but the borders for 16 are 0-15 and 16-31. Neither of which cover his range entirely so he needs to use the wild card mask of 0.0.31.255

Yankee

Yankee is absolutely correct...in fact, if you create any access-list with the following (I did this on a router to show you what happens):

router(config)#access-list 151 permit ip 192.168.5.0 0.0.15.255 any
router(config)#end
router#sh access-list 151
Extended IP access list 151
permit ip 192.168.0.0 0.0.15.255 any

Notice when you do a "sh access-list" that the router has automatically adjusted the bit boundries and assumed you want to permit 192.168.0.0 through 192.168.15.0....this does not cover through 17.0, so you have to use 0.0.31.255

Here's what is looks like with 0.0.31.255:

router(config)#access-list 151 permit ip 192.168.5.0 0.0.31.255 any
router(config)#end
router#sh access-list 151
Extended IP access list 151
permit ip 192.168.0.0 0.0.31.255 any

This would cover the range you want.

There is no single wildcard mask that will cover only 192.168.5.0 through 192.168.17.0...

2lazybutsmart

That's cool. I never thought about that actually.

thnx.
2lbs.

elcaminoguy

That would hold true for the fourth octet as well I assume. So if I had a class C network subnetted of say

192.168.1.64 mask 255.255.255.224

and I wanted 9 networks out of that I would have to use a block of 16 or would it still be 32. I guess my confusion is whether the subnet mask affects the block size you can use.

If so would it be like:

access-list 151 permit ip 192.168.1.64 0.0.0.31 any

because of the subnet mask bits or would it be:

access-list 151 permit ip 192.168.1.64 0.0.0.15 any

because of how big of a chunk of numbers I want. What defines the bit boundary? What if I want a range of networks that are higher numerically? like a network of 192.168.1.128 mask 255.255.255.224?

Just trying to get his straight in my head. I heard about inverting the subnet mask to get your wildcard mask but does that really fit.

255.255.255.224 would invert to 0.0.0.31 but that seems to totally discount the whol group size thing.

Yankee

In your example both masks are valid but the "tighter" one, the .15 would be more correct for your example of needing nine hosts (not networks).

Yankee

elcaminoguy

Oh,

I kept thinking the wildcard mask broke out networks instead of hosts. I guess I was making it harder than it had to be.

Thanks