Ip Subnetting Question...

lAwBoYlAwBoY Member Posts: 30 ■■□□□□□□□□
which of the following address can be assigned to a host when using a subnet mask off 255.255.254.0

A : 175.33.3.255
B : 26.35.3.255
C : 113.10.4.0
D : 186.54.3.0
E : 152.135.7.0
F : 17.35.36.0

Anyone noe the answer?

Comments

  • EdTheLadEdTheLad Member Posts: 2,111 ■■■■□□□□□□
    None of them,with that mask your subnets are 0,2,4,6,8 ....254
    A,B are broadcast addresses in subnet 2, C,F are subnets
    Answer D,E


    A quick edit was called for!
    Networking, sometimes i love it, mostly i hate it.Its all about the $$$$
  • lAwBoYlAwBoY Member Posts: 30 ■■□□□□□□□□
    ed_the_lad wrote:
    None of them,with that mask your subnets are 0,2,4,6,8 ....254
    A,B are broadcast addresses in subnet 2,
    C,D,E,F have a 0 as the host identifier which isnt allowed.

    Ermm ... dont think so ...
  • UndertowUndertow Member Posts: 48 ■■□□□□□□□□
  • EdTheLadEdTheLad Member Posts: 2,111 ■■■■□□□□□□
    Actually now that i look again, D,E
    Networking, sometimes i love it, mostly i hate it.Its all about the $$$$
  • ziggyziggy Member Posts: 38 ■■□□□□□□□□
    YUP,

    Looks like D and E
  • evanderburgevanderburg Member Posts: 229 ■■■□□□□□□□
    255.255.254.0

    A : 175.33.3.255
    B : 26.35.3.255
    C : 113.10.4.0
    D : 186.54.3.0
    E : 152.135.7.0
    F : 17.35.36.0

    I agree. Just to help out those who are struggling and to prove it, here is why. We already agreed that A & B are broadcasts. C & F are network addresses. Here is my logic.

    NOTE: 255 octets are eliminated in my logic for simplicity because they are insignificant.

    C:
    Address: 00000100.00000000
    Subnet: 11111110.00000000
    Binary AND 00000100.00000000
    The address and the binary AND come to the same result so this address is the same as it's network address. C is invalid

    D:
    Address: 00000011.00000000
    Subnet: 11111110.00000000

    Binary AND 00000010.00000000
    The address and Binary AND differ so this address is ok.

    E:
    Address: 00000111.00000000
    Subnet: 11111110.00000000
    Binary AND 00000110.00000000
    The address and Binary AND differ so this address is ok.

    F:
    Address: 00100010.00000000
    Subnet: 11111110.00000000
    Binary AND 00100010.00000000
    The address and the binary AND come to the same result so this address is the same as it's network address. F is invalid
    "You can never know everything and part of what you know is always wrong. Perhaps even the most important part. A portion of wisdom lies in knowing that. A portion of courage lies in going on anyway. " - Lan, Winter's Heart by Robert Jordan
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