Subnetting - - please help

sarahanandsarahanand Member Posts: 52 ■■□□□□□□□□
Right, I'm lost here again.
What is the first valid host on the network that the host 172.24.7.218/22 is a part of?

So /22 the subnet mask is 255.255.252.0 ? right?

And the increment value would be 4? yeah?

The Answer though is 172.24.4.1 ? I dont get it?

Comments

  • bbarrickbbarrick Member Posts: 242 ■■■□□□□□□□
    172.24.7.218 /22

    /22 = 11111111.11111111.11111100.00000000 or 255.255.252.0

    So your subnets are 256 - 252 = 4

    So your valid networks would be 172.24.0, 172.24.4, 172.24.8 etc... You can pick any of those to use as the networks IP range.

    This IP happens to fall in that second set the 172.24.4 which has a subnet id of 172.24.4.0(can't be used as a host ip) and a broadcast address of 172.24.7.255(can't be used as a host ip either).

    So the first valid host would be as you said 172.24.4.1, these computers would be in the same subnet therefore they would be able to communicate via a switch. Now, if you added a third host and gave it the IP of 172.24.8.1 you would need a router to reach that subnetwork.
  • smcclenaghansmcclenaghan Member Posts: 139
    /22 is 255.255.252.0 like you said.

    We want to find out which /22 that 172.24.7.218 is a part of.

    I start at 172.24.0.0 because /22 is smaller than /16, and /16 is the class B (172.24.0.0/16). If your mask was something like /14, you would want to start counting at 172.0.0.0/8. (Does that make sense?)

    172.24.0.0 is the first and runs 172.24.0.0 through 172.24.3.255. Too low. We need to go higher.
    172.24.4.0 is the second and runs 172.24.4.0 through 172.24.7.255. This looks like it.

    172.24.4.0 through 172.24.7.255 is the full range (if that part isn't clear, let me know).

    If you agree that's the full range, then just remember we can't use the very first address (172.24.4.0) because it's the network address.
    And we can't use the very last address (172.24.7.255) because it's the broadcast address.

    That leaves the first USABLE host address as 172.24.4.1 and the last USABLE host address as 172.24.7.254.
  • sarahanandsarahanand Member Posts: 52 ■■□□□□□□□□
    Right , So just another day where I mastered being silly. I had scribbled the calculations in midst of other calculations, and kept incrementing by 2, thinking I was incrementing by 4....is incrementing not a word. Damn the spell-checker. Has a mind of its own now.
  • smcclenaghansmcclenaghan Member Posts: 139
    Lol, let's settle on incrementaciously as long as the spell check is going to yell at us anyway.

    Sounds like you've got the gist. The silly stuff isn't going to stop either. At least it hasn't for me. Caffeine does help tho. ;)
  • steveyeungsteveyeung Member Posts: 44 ■■□□□□□□□□
    =172.24.000001 | 11.218 (/22 means the network part is the first 22 bits, no need to care about the last octet)
    so the network it belongs to is 172.24.00000100, = 172.24.4.0
    and so the first valid host of this network is 172.24.4.1
  • TechGuru80TechGuru80 Member Posts: 1,539 ■■■■■■□□□□
    Increments always go in the octet that is not maxed out...255.
Sign In or Register to comment.