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Need help...Subnet...

Sbs1011Sbs1011 Member Posts: 12 ■□□□□□□□□□
New in CCNA, need somehow on this question:
Anyone can guide me?? Appreciate...

1. An organization is granted the block 16.0.0.0/8.
the administrator wants to create 500 fixed-length subnets.

a. Find the subnet mask.

b. Find the number of address in each subnet.

c. Find the first and last addresses in subnet 1.

d. Find the first and last addresses in subnet 500.

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    stevchastevcha Member Posts: 50 ■■□□□□□□□□
    Your mask is /8 = 255.0.0.0
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    Sbs1011Sbs1011 Member Posts: 12 ■□□□□□□□□□
    But, i think the question a:
    should be 255.255.128.0... am i right?
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    stevchastevcha Member Posts: 50 ■■□□□□□□□□
    Correct cause 2^9 = 512 so it would be 255.255.128.0. So if I'm not mistaken it would be 256-128 = 128. So your subnet ranges would be 0, 128, 256, 384, 512.
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    rebelratrebelrat Member Posts: 34 ■■□□□□□□□□
    stevcha wrote:
    Correct cause 2^9 = 512 so it would be 255.255.128.0. So if I'm not mistaken it would be 256-128 = 128. So your subnet ranges would be 0, 128, 256, 384, 512.

    This is incorrect - there is no 384 or 512 in the network addresses.

    The correct network addresses are (with subnet zero enabled)

    Network 16.1.0.0 Hosts 16.1.0.1 - 16.1.127.254 Broadcast 16.1.127.255
    Network 16.1.128.0 Hosts 16.1.128.1 - 16.1.255.254 Broadcast 16.1.255.255
    Network 16.2.0.0 Hosts 16.2.0.1 - 16.2.127.254 Broadcast 16.2.127.255
    Network 16.2.128.0 Hosts 16.2.128.1 - 16.2.255.254 Broadcast 16.2.255.255

    and so on. Hope that helps icon_wink.gif
    rebelrat :-)
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    EdTheLadEdTheLad Member Posts: 2,111 ■■■■□□□□□□
    16.0.0.0/8, need 500 subnets i.e. 9 additional bits to be part of the network portion.
    mask 255.255.128.0

    9 bits left for hosts 2^9-2 = 510 hosts

    subnet 1 is 16.0.0.0
    subnet 2 is 16.0.128.0

    Host Address range for subnet 1
    16.0.0.1 -16.0.127.254

    Host Address range for subnet 500
    16.255.128.1 -> 16.255.255.254
    Networking, sometimes i love it, mostly i hate it.Its all about the $$$$
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    NetstudentNetstudent Member Posts: 1,693 ■■■□□□□□□□
    EdTheLad wrote:
    16.0.0.0/8, need 500 subnets i.e. 9 additional bits to be part of the network portion.
    mask 255.255.128.0

    9 bits left for hosts 2^9-2 = 510 hosts

    subnet 1 is 16.0.0.0
    subnet 2 is 16.0.128.0

    Host Address range for subnet 1
    16.0.0.1 -16.0.127.254

    Host Address range for subnet 500
    16.255.128.1 -> 16.255.255.254

    Wouldn't 2^9 be for subnets? It looks like he has 15 host bits. Last 2 octects of the mask are 10000000.00000000

    so 32,766 hosts per subnet?
    There is no place like 127.0.0.1 BUT 209.62.5.3 is my 127.0.0.1 away from 127.0.0.1!
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    Sbs1011Sbs1011 Member Posts: 12 ■□□□□□□□□□
    Sbs1011 wrote:
    New in CCNA, need somehow on this question:
    Anyone can guide me?? Appreciate...

    1. An organization is granted the block 16.0.0.0/8.
    the administrator wants to create 500 fixed-length subnets.

    a. Find the subnet mask.

    b. Find the number of address in each subnet.

    c. Find the first and last addresses in subnet 1.

    d. Find the first and last addresses in subnet 500.

    Ok,guys...
    The final answer should be like this:
    a. 255.255.128.0, cause it needs additonal 9 bits to be part of the network portion.

    b. 32766. It left 15 bits at host portion. (2x15) - 2 = 32766.

    c. At subnet 1,
    Network address : 16.0.0.0
    First address : 16.0.0.1
    Last address : 16.0.127.254
    Broadcast add. : 16.0.127..255

    d. At subnet 500,
    Network address : 16.249.128.0
    First address : 16.249.128.1
    Last address : 16.249.255.254
    Broadcast add. : 16.249.255.255
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    uncamt26uncamt26 Member Posts: 5 ■□□□□□□□□□
    Actually subnet 1 should be
    Network address : 16.0.128.0
    First address : 16.0.128.1
    Last address : 16.0.255.254
    Broadcast add : 16.0.255.255

    Subnet 500 should be
    Network address : 16.250.0.0
    First address : 16.250.0.1
    Last address : 16.250.127.254
    Broadcast add : 16.250.127.255

    I believe that it was previously stated that subnet zero was enabled. If that's true that means you could use subnet zero when implementing your network addressing scheme, save for the network and broadcast addresses of course. Therefore subnet zero should look like this:
    Network address : 16.0.0.0
    First address : 16.0.0.1
    Last address : 16.0.127.254
    Broadcast add. : 16.0.127.255

    All your other calculations were correct. If you ever need to practice subnetting I recommend the following site. It help me immensely when I was learning subnetting. It provides many practice problems just like the one you presented. Hope this helps..

    http://www.nettechnet.org/POD/POD.html
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