Subnetting: Accommodate more than 225 hosts

anarchos78anarchos78 Posts: 3Registered Users ■□□□□□□□□□
Hi,

I have a subnetting question. First I’ll give an example:

An organization has the IP 200.1.1.0 and needs 4 subnets with the following characteristics:

Subnet A =>72 hosts

Subnet B => 35 hosts

Subnet C => 20 hosts

Subnet D => 18 hosts

Provide the IPs per subnet.


Solution

Subnet A =72 host

For 72 hosts we need 2^7 - 2 = 128 - 2 = 126 host. We need 7 bits for the hosts.

Network => 200.1.1.0/25

Mask => 255.255.255.128

First Host => 200.1.1.1/25

Last Host => 200.1.1.126/25

Broadcast => 200.1.1.127/25



Subnet B =35 host

For 72 hosts we need 2^6 - 2 = 64 - 2 = 62 host. We need 6 bits for the hosts.

Network => 200.1.1.128/25

Mask => 255.255.255.192

First Host => 200.1.1.129/26

Last Host => 200.1.1.190/26

Broadcast => 200.1.1.191/26



Subnet C =72 host

For 72 hosts we need 2^5 - 2 = 32 - 2 = 30 host. We need 5 bits for the hosts.

Network => 200.1.1.192/27

Mask => 255.255.255.224

First Host => 200.1.1.193/27

Last Host => 200.1.1.222/27

Broadcast => 200.1.1.223/27



Subnet D =72 host

For 72 hosts we need 2^5 - 2 = 32 - 2 = 30 host. We need 5 bits for the hosts.

Network => 200.1.1.224/27

Mask => 255.255.255.224

First Host => 200.1.1.225/27

Last Host => 200.1.1.254/27

Broadcast => 200.1.1.255/27



So, according the above solution, how do we create subnets for the IP 155.52.48.0/23 with the following characteristics (provide Network, Mask First Host, Last Host and Broadcast IPs foe each subnet)?

Subnet A =>200 hosts

Subnet B => 100 hosts

Subnet C => 40 hosts

Subnet D => 40 hosts



I’m really stuck with that.

Comments

  • bornwithbornwith Posts: 21Member ■■□□□□□□□□


    [TH]Subnet Name[/TH]
    [TH]Needed Size[/TH]
    [TH]Allocated Size[/TH]
    [TH]Address[/TH]
    [TH]Mask[/TH]
    [TH]Dec Mask[/TH]
    [TH]Assignable Range[/TH]
    [TH]Broadcast[/TH]


    A
    200
    254
    155.52.48.0
    /24
    255.255.255.0
    155.52.48.1 - 155.52.48.254
    155.52.48.255


    B
    100
    126
    155.52.49.0
    /25
    255.255.255.128
    155.52.49.1 - 155.52.49.126
    155.52.49.127


    C
    40
    62
    155.52.49.128
    /26
    255.255.255.192
    155.52.49.129 - 155.52.49.190
    155.52.49.191


    D
    40
    62
    155.52.49.192
    /26
    255.255.255.192
    155.52.49.193 - 155.52.49.254
    155.52.49.255



  • anarchos78anarchos78 Posts: 3Registered Users ■□□□□□□□□□
    Thank you very much about your reply.
    What I don't understand is the "Network Address" part.
    Is it valid from address 155.52.48.xxx to go to 155.52.49.xxx?
    Could you please explain me the "mechanics" of that approach?
  • rob42rob42 Posts: 423Member
    I don't know if it'll help you, but I'm working on tying to explain the "mechanics" of Subnets and have just put together this URL="http://rob42.net/IPv4_Subnets1.htm"]Rob42.net - IPv4 Subnets[/URL. I'd appreciate any feedback on my work.

    It's a work in progress...
    No longer an active member
  • volfkhatvolfkhat Posts: 944Member ■■■■■■■□□□
    anarchos78 wrote: »
    Thank you very much about your reply.
    What I don't understand is the "Network Address" part.
    Is it valid from address 155.52.48.xxx to go to 155.52.49.xxx?
    Could you please explain me the "mechanics" of that approach?

    To answer your question: YES.

    It's all about understanding how to identify the "octet of interest".
    Once you know the octet, everything else falls into place.

    155.52.48.0/23 = 155.52.48.0 - 155.52.49.255
    its not much to explain; you just have to practice subnetting /16 - /23.

    I learned subnetting from Danscourses:
    https://www.youtube.com/playlist?list=PL33E07ECCA73C0755
    Watch video# 47.
    (after you "get" it; give him a Thumbs-UP)


    As for the rest of your Q:
    200 hosts = a blocksize 256.
    aka: 2^8
    aka: /24
    This means it falls under the 4th octet.
    so your 200 hosts would fit on the subnet: 155.52.48.[0-255]

    Your next subnet starts at 155.52.49.0
    (because there is no more room inside 155.52.48.x ).

    B = 155.52.49.[0-127]
    C = 155.52.49.[128-191]
    D = 155.52.49.[192-255]

    just takes practice..
  • anarchos78anarchos78 Posts: 3Registered Users ■□□□□□□□□□
    Thank you very much!
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