asked on

subnetting question

I have a practice lab where its asking me to take an IP address and find the ranges for it with the required number of hosts.

200.1.1.0 (Class C) and requires 40 hosts. I only know the long way to do this but my answer keeps coming out different than answer from the lab.

Lab Answer:
200.1.1.0, break into networks of 40 hosts each
40 hosts = 00101000 (6 bits) Subnet Mask = 255.255.255.192 Increment = 64
Ranges:
200.1.1.0 – 200.1.1.63
200.1.1.64 – 200.1.1.127
200.1.1.128 – 200.1.1.191
200.1.1.192 – 200.1.1.255

My answer (I hope you know the long way):

200.1.1.0, break into networks of 40 hosts each
40 hosts = 00101000 (6 bits)
255.255.255.0 = 11111111.11111111.11111111.00000000
255.255.255.252 = 11111111.11111111.11111111.11111100
(6 bits=252)
Since the last bit used in the last octet above converted from binary is 4, I add the 4th octet in increments of 4 for the results below. Where is my math wrong?
Ranges:
200.1.1.0 - 200.1.1.3
200.1.1.4 - 200.1.1.7
200.1.1.8 - 200.1.1.11

Thanks.

ASKER CERTIFIED SOLUTION

CompProbSolv

membership

This solution is only available to members.

To access this solution, you must be a member of Experts Exchange.

Start Free Trial

tolinrome

ASKER

Could it be that my results were mistakenly based on requiring 40 networks, instead of hosts?

SOLUTION

Infamus

membership

This solution is only available to members.

To access this solution, you must be a member of Experts Exchange.

Start Free Trial

SOLUTION

masnrock

membership

This solution is only available to members.

To access this solution, you must be a member of Experts Exchange.

Start Free Trial

tolinrome

ASKER

Sorry, in my question I wrote that I needed the required number of hosts, but I really needed the required number of networks as per the instructions.

As per the instructions in the lab, the question says:
C - 200.1.1.0 Break into 40 networks

Instructors answer to that question:
200.1.1.0, break into networks of 40 hosts each
40 hosts = 00101000 (6 bits) Subnet Mask = 255.255.255.192 Increment = 64
Ranges:
200.1.1.0 – 200.1.1.63
200.1.1.64 – 200.1.1.127
200.1.1.128 – 200.1.1.191
200.1.1.192 – 200.1.1.255

I still come up with

So with the IP of 200.1.1.0 (40 required networks), I still come up with
200.1.1.0 - 200.1.1.3
200.1.1.4 - 200.1.1.7
200.1.1.8 - 200.1.1.11

The video didn't even get into yet doing subnetting with the required number of hosts, that's the next video actually.

So, either the instructor mistakenly wrote hosts and might have meant networks. Or I'm still confused at this point?

Really all I care about is that I did the formula in the correct way depending on the requirement being the numbers of networks and not hosts.

masnrock

Something is very wrong....

The lab answer you're showing reflect the scenario of needing 40 hosts per subnet. Your answer reflects needing 40 subnets.

Which means either the instructor answered the wrong question or wrote the wrong problem. Or you're looking at the wrong part of the problem.

I normally do not like asking, but could you copy and paste the actual problem?

CompProbSolv

If it really is Class C (first three octets remain constant) and you need 40 networks, then your proposed example is correct.

The approach is similar to what was given above. You will need 6 consecutive bits (.11111100) to hold the varying part of the network address, which only leaves the last two bits for hosts. Mixing up decimal and binary for simplicity, it looks like this:

200.1.1.nnnnnnhh (where "n" is a bit used for the network address and "hh" is the two-bit host address)

The first network would have nnnnnn=000000 so it will range from:
200.1.1.0-200.1.1.3

The second network would have nnnnnn=000001:
200.1.1.4-200.1.1.7

And so on as you listed above.

Keep in mind that this will allow only two hosts because of the exclusions of all 0 and all 1 host addresses.

mansrock is correct about the instructor having made an error if it is as you described.