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Classless Subnet Masking / Subnetting

Can someone please explain the following, if I am wrong in my thinking? I am meant to be learning this from an online resource, but I think the resource is wrong. Here is the hypothetical situation:

A company is assigned a Class A network address of 66.0.0.0/8. The default subnet mask is 255.0.0.0. The company has the following minimum requirements:

      1. Half the addresses must be reserved for future use.
      2. Twelve networks are required with 8,190 hosts per subnet.
      3. Ten networks are required with 2,046 hosts per subnet.
      4. Five networks are required with 254 hosts per subnet.

The following are simplified steps to achieve a goal:

1. It instructs me to divide the network into two to meet requirement #1. Fine. Ok.
2. It instructs me to divide 1/2 of the network into 19 equals sized portions, as 13 bits are used for the host address (2 ^ 13 = 8,192) leaving 19 bits for the network address. Using the first 12 segments (Total: 66.0.0.0/19 to 66.1.96.0/19) gives me requirement #2.

The unused subnets, beginning with 66.1.128.0/19, can be reserved for future use or to meet subsequent requirements.

This is where I am either confused, or material is wrong.

3. Next, you need to create 10 networks with 2,046 hosts per subnet. Each of the preceding subnets yields 8,190. When subnetted, each can yield three subnets with 2,046 host addresses. To meet this requirement, you would need to create 10 subnets, which would require taking four of the networks developed earlier and subnetting them.

In other words, you must start where you left off, since you cannot have overlapping network addresses. The starting network address in this case is 66.1.128.0/19. You need 11 host bits to create 2,046 addresses. This means that you can use 21 network bits (32 – 11) to create these subnets. Table [the following] defines several of the starting network addresses, as well as the ending network address. The bits that represent the network ID are shown in bold.

Class A Subnet Address          Binary Representation
66.1.128.0/21                         01000010.00000001.10000000.00000000
66.1.136.0/21                         01000010.00000001.10001000.00000000
to
66.127.248.0/21                     01000010.01111111.11111000.00000000


As before, you can select the first 10 subnets from this range to meet Requirement #3. The last subnet used in this configuration, then, is 66.2.32.0/21. The subnet mask is 255.255.248.0.

[Here is what I think: ] Surely using the first 10 subnets, starting at 66.1.128.0/21, in increments of 0.0.8.0 would result in the last subnet used for this configurating being 66.1.200.0/21.

Am I doing something wrong, or is the material wrong? I'm hoping the former, which would prove I need education and that I've paid for a good service that will teach me well. If the latter, it may be time to shout at some people.

Thanks in advance, sorry for the length of the question.
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Vampireofdarkness
Asked:
Vampireofdarkness
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1 Solution
 
dinkytoy101Commented:
Does the line;
3. Next, you need to create 10 networks with 2,046 hosts per subnet. Each of the preceding subnets yields 8,190. When subnetted, each can yield three subnets with 2,046 host addresses. To meet this requirement, you would need to create 10 subnets, which would require taking four of the networks developed earlier and subnetting them.

come from the resource of your own thoughts, as it contains a mistake four not three ;)? There are two ways imo of looking at this scenario, they either want you to subnet the existing subnets you previously created or use more of your original /8.

Frankly in scenarios and practise I really wouldn't worry about the design the idea is to practise the subnetting. My answer to the question would be akin to yours if it helps:

1. 66.128.0.0/9
2. 66.0.0.0.0/19
66.0.0.32.0/19
66.0.0.64.0/19
66.0.0.96.0/19
66.0.0.128.0/19
66.0.0.160.0/19
66.0.0.192.0/19
66.0.0.224.0/19
66.0.1.0/19
66.0.1.32/19
66.0.1.64/19
66.0.1.96/19

3. 66.0.1.128
66.0.1.136/21
66.0.1.144/21
66.0.1.152/21
66.0.1.160/21
66.0.1.168/21
66.0.1.176/21
66.0.1.184/21
66.0.1.192/21
66.0.1.200/21

4. 66.0.1.201
66.0.1.202/24
66.0.1.203/24
66.0.1.204/24
66.0.1.205/24

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VampireofdarknessAuthor Commented:
That part comes from the resource. My part is
[Here is what I think: ] Surely using the first 10 subnets, starting at 66.1.128.0/21, in increments of 0.0.8.0 would result in the last subnet used for this configurating being 66.1.200.0/21.

There is also the other thing you point out. You don't need to use four as three will suffice is, as you're not using the subnet with 8,190 (thus making it 8,192 / 4 = 2,048 - 2 = 2,046).
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dinkytoy101Commented:
lol just noticed my zero's are all over the place :). Corrected below for clarity's sake.

I agree with you tbh, the 66.2.32.0/21 being suggested would actually be the 21st /21 subnet. It ends with 66.1.200.0/21 as demonstrated in my list. I would query it if you have the facility it's either extremely badly worded or full of mistakes.

1. 66.128.0.0/9

2. 66.0.0.0/19
66.0.32.0/19
66.0.64.0/19
66.0.96.0/19
66.0.128.0/19
66.0.160.0/19
66.0.192.0/19
66.0.224.0/19
66.1.0.0/19
66.1.32.0/19
66.1.64.0/19
66.1.96.0/19

3. 66.1.128.0/21
66.1.136.0/21
66.1.144.0/21
66.1.152.0/21
66.1.160.0/21
66.1.168.0/21
66.1.176.0/21
66.1.184.0/21
66.1.192.0/21
66.1.200.0/21

4. 66.1.201.0/24
66.1.202.0/24
66.1.203.0/24
66.1.204.0/24
66.1.205.0/24
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VampireofdarknessAuthor Commented:
Thanks. I raised a query three days ago, and again today. No response as yet.
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