Subnetting help

Nettech_1Nettech_1 Registered Users Posts: 2 ■□□□□□□□□□
I am currently in a Cisco course and need input. I have an activity I'm doing and can't wait till next week for class. Basicaly, they give you the # of hosts needed and you input the subnet needed. I tried to attach as well as copy/paste, but they failed, so I will add it here:

Hosts needed 25

I said the subnet would be 255.255.255.224. They also want it in binary and /x. I input 11111111.11111111.11111111.11100000 with the /x at /27. They say this is wrong.

Their example was: 250 hosts. 255.255.255.0 11111111.11111111.11111111.00000000 and /24.

Am I missing something?

Comments

  • DeathmageDeathmage Banned Posts: 2,496
    looks good to me.

    Remember this (for a Class C):



    128
    192
    224
    240
    248
    252
    254
    255


    /25
    /26
    /27
    /28
    /29
    /30
    /31
    /32


    128
    64
    32
    16
    8
    4
    2
    1



    So you need 25 hosts, well 16 bits is too small and 64 bits would work but way to much so 32 is the correct amount and that would be a x.x.x.224.

    a /24 would be a range of 255 addresses and while your instructors addressing would work, it would be way more than you need which is /25 which would give you 32 (total) - 2 = 30 (usable) hosts. Remember always minus 2 from the host bits, one for the network ID and one for the broadcast ID.

    Incidentally, there example of 250 host's would be a 255.255.255.0 or a /24.

    just so you know your ranges would be this:

    I'll do a few of them and you just finish them. :)

    since there is no class C address, just presuming it's a Class C, I'll go with 192.168.104.15 as the address with a mask of 255.255.255.224. If they ask you want range does 192.168.104.67 reside in you just look in the table below and say the 3rd subnet.



    Network ID
    Range 1st Address
    Range Last Address
    Broadcast ID


    192.168.104.0
    192.168.104.1
    192.168.104.30
    192.168.104.31


    192.168.104.32
    192.168.104.33
    192.168.104.62
    192.168.104.63


    192.168.104.64
    192.168.104.65
    192.168.104.90
    192.168.104.91


    192.168.104.92
    192.168.104.93
    192.168.104.122
    192.168.104.123



    As you can see since it needs 32 hosts per range, you always need to minus 2 for the network ID and broadcast ID. I only did 4 ranges but you would need to go all the way up to 192.168.104.255 - I hope this helps!!!
  • Nettech_1Nettech_1 Registered Users Posts: 2 ■□□□□□□□□□
    Thank you. I thought maybe I was missing something. I feel a lot better now and can proceed with my studies. I think the netacad site may have an issue with this activity.
  • mikeybinecmikeybinec Member Posts: 484 ■■■□□□□□□□
    Which module? I'm a Netacad alumni
    Cisco NetAcad Cuyamaca College
    A.S. LAN Management 2010 Grossmont College
    B.S. I.T. Management 2013 National University
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