Wildcard mask help

shadown7shadown7 Member Posts: 529
Hi,

I need to learn ACL's to complete CCNA 2 case study and I've read the reading material but I can't figure out wildcard masks. It is just not sinking in. I'm really confused by it to tell the truth. So far all I've learned in Cisco has been easy for expect for this. If anyone can help me that would be great!!

Thanks in advance!!

Comments

  • shadown7shadown7 Member Posts: 529
    No, but I'll read them now!!

    thank you very much!
  • WebmasterWebmaster Admin Posts: 10,292 Admin
    Happy reading then :) feel free to post some more details if you still have trouble understanding it after reading those.
  • shadown7shadown7 Member Posts: 529
    Ok, I'm still confused but let me see if I got this right.

    I'm doing the CCNA 2 case study and I'm not going to post the requirments for the ACL's because I want to try to get it myself. But, I will post the following information to see if I atleast got this right.

    The network ip is 199.16.7.0 and I have 8 available subnets.

    I figured out to have 8 subnets I would need a subnet mask of
    255.255.255.224

    Going by what I read about an invert mask I would take the broadcast of

    255.255.255.255
    -255.255.255.224 SNM
    ________________
    0.0.0.31 wildcard mask for my network and all the subnets on the network.


    Is what I have correct? Our instructor hasn't taught us ACL's yet but since I'm on spring break I wanted to try to get the case study done. Even after reading all those links I'm still very confused. I have no idea why this is so hard for me.


    Thanks again
  • WebmasterWebmaster Admin Posts: 10,292 Admin
    So what exactly got you confused? icon_wink.gif

    The inverse of the subnet address 255.255.255.224 is indeed 0.0.0.31

    I usually do this:

    255.255.255.224 = /27
    The first 27 bits are 1s, the last 5 zero (32 total)
    The inverse mask would have the first 27 set to 0, the last 5 to 1, thus:
    16+8+4+2+1 = 31

    In case of a subnet mask with the first three octets being 255, such as in a class C address in your example, 255-224 leads to the answer directly.

    Try making up some examples of your own, different classes different requirements, and keep practising as it will become easier once you've written it out a couple of times.
  • shadown7shadown7 Member Posts: 529
    What has got me confused is actually writting the ACL. I read the requirments but I just can't seem to figure out what I need to do. I'm going to try the simple ones you have in your technotes and see if I can get the hang of it.

    You do have some great technotes btw! :D

    Thanks for the help again,

    Keith
Sign In or Register to comment.