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Subnetting network into smaller networks.


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#1 AcesLight

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Posted 25 June 2015 - 12:44 AM

I was doing some practice question online and I'm not sure if I calculated the subnet correctly thus I have some questions:

 

Lets say example, I was given this IP address 192.168.1.0/24. My company requires me to do subnetting and provide enough hosts addresses for each department, with little address wastage.

 

There are 3 departments, and each department contain only certain number of hosts:

Human resource department have 2 hosts.

Sales department have 19 hosts.

Shipping department have 56 hosts.

 

1 ) If I do my so called "straight forward" calculations, I just check the department with the largest number of hosts and calculate the largest possible subnet that can accommodate it. So I use 6 bits from the host portion, leaving me 2 bits for subnet.

 

NNNNNNNN.NNNNNNNN.NNNNNNNN.HHHHHHHH (/24) default mask

NNNNNNNN.NNNNNNNN.NNNNNNNN.NNHHHHHH (/26) The new mask.

 

So I basically will have 4 (2^2) subnets, each capable of holding 62 hosts (2^6, minus away 2).

 

And if I assign the subnets to each department I will get:

 

Human resource department:

Network address: 192.168.1.0

Subnet mask: 255.255.255.192

Usable address range: 192.168.1.1 - 192.168.1.62

Broadcast address: 192.168.1.63

 

Sales department

Network address: 192.168.1.64

Subnet mask: 255.255.255.192

Usable address range: 192.168.1.65 - 192.168.1.126

Broadcast address: 192.168.1.127

 

Shipping department

Network address: 192.168.1.128

Subnet mask: 255.255.255.192

Usable address range: 192.168.1.129 - 192.168.1.190

Broadcast address: 192.168.1.191

 

Firstly, am I doing it correctly?

 

2 ) Second question, if I need to calculate the subnet WITH LITTLE WASTAGE, do I calculate the minimum number of hosts bits I need to use for each subnet and create subnet of different sizes for each department?

 

If so for:

HR department I need to borrow only 2 bits

Sales dept I need to borrow 5 bits.

Shipping dept I need to borrow 6 bits.

 

If I go according to this then my answer turns out to be these:

 

Human resource department:

Network address: 192.168.1.0

Subnet mask: 255.255.255.252

Usable address range: 192.168.1.1 - 192.168.1.2

Broadcast address: 192.168.1.3

 

Sales department

Network address: 192.168.1.4

Subnet mask: 255.255.255.224

Usable address range: 192.168.1.5 - 192.168.1.34

Broadcast address: 192.168.1.35

 

Shipping department

Network address: 192.168.1.36

Subnet mask: 255.255.255.192

Usable address range: 192.168.1.37 - 192.168.1.98

Broadcast address: 192.168.1.99

 

I get the new network address for each subnet by continuing from the broadcast address of the previous subnet (Sorry my english isn't the best but I'm trying very hard)

 

e.g In HR the broadcast address is 192.168.1.3 so the Sales department should start from 192.168.1.4 and from there I did my subnetting

 

Are my answers and understanding of calculation correct? 

 

I tried to post this in some other forums, some of the guys ain't very friendly.


Edited by AcesLight, 25 June 2015 - 12:47 AM.


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#2 LuminisIndia

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Posted 25 June 2015 - 03:02 AM

According to me this is not the correct way to calculate the subnetting.

Your requirements are: 

**************************************************************************************************************************

There are 3 departments, and each department contain only certain number of hosts:

Human resource department have 2 hosts.

Sales department have 19 hosts.

Shipping department have 56 hosts.

****************************************************************************************************************************

Step 1. Check the MAXIMUM no. of host required 56

so you need 2^6=64-2=62(Host)

The new network address will be: 192.168.1.0/26

Subnet mask:255.255.255.192

Usable IP:192.168.1.1-192.168.1.62

Broadcast address:192.168.1.63

 

Step 2Check the next  no. of host required 19

so you need 2^5=32-2=30(Host)

The new network address will be: 192.168.1.64/27

Subnet mask:255.255.255.224

Usable IP:192.168.65.-192.168.1.94

Broadcast address:192.168.1.95

 

Step 3Check the next  no. of host required 2

 

so you need 2^2=4-2=2(Host)

The new network address will be: 192.168.1.96/30

Subnet mask:255.255.255.252

Usable IP:192.168.97.-192.168.1.98

Broadcast address:192.168.1.99

 

With this way you can achieve " little address wastage".

I hope you will be able to understand it.

 

With thanks,

Team Luminis Consulting Services.


Edited by LuminisIndia, 25 June 2015 - 03:05 AM.


#3 AcesLight

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Posted 25 June 2015 - 03:11 AM

 

According to me this is not the correct way to calculate the subnetting.

Your requirements are: 

**************************************************************************************************************************

There are 3 departments, and each department contain only certain number of hosts:

Human resource department have 2 hosts.

Sales department have 19 hosts.

Shipping department have 56 hosts.

****************************************************************************************************************************

Step 1. Check the MAXIMUM no. of host required 56

so you need 2^6=64-2=62(Host)

The new network address will be: 192.168.1.0/26

Subnet mask:255.255.255.192

Usable IP:192.168.1.1-192.168.1.62

Broadcast address:192.168.1.63

 

Step 2Check the next  no. of host required 19

so you need 2^5=32-2=30(Host)

The new network address will be: 192.168.1.64/27

Subnet mask:255.255.255.224

Usable IP:192.168.65.-192.168.1.94

Broadcast address:192.168.1.95

 

Step 3Check the next  no. of host required 2

 

so you need 2^2=4-2=2(Host)

The new network address will be: 192.168.1.96/30

Subnet mask:255.255.255.252

Usable IP:192.168.97.-192.168.1.98

Broadcast address:192.168.1.99

 

With this way you can achieve " little address wastage".

I hope you will be able to understand it.

 

With thanks,

Team Luminis Consulting Services.

 

 

Thanks for the reply. Can I just say, so you always start from the subnet / department with the largest number of hosts / nodes first? And follow by the next largest, working your way down? Cause from your way of calculation I see it as the "inverted" version of mine. Is that correct?

 

Please enlightenment me if I'm wrong.



#4 LuminisIndia

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Posted 25 June 2015 - 03:21 AM

Yes start from the largest no. of hosts and then the next largest, so on. This method will help you to avoid the mistakes in subnetting. 



#5 Kilroy

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Posted 25 June 2015 - 05:54 AM

Subnetting is something I wish the certifications would just let die.  I learned it over a decade ago for my Network+ and haven't used it since.  In the real world you'll use a Subnet Calculator and IPv6 doesn't have subnetting.  Most networks in the real world will use an entire private IP range per need, be it department or device.  So, understand it well enough for your exam, but don't beat yourself up if you don't get it completely.

 

The main points to understand about subnetting is that you can divide a network into smaller portion and you lose the first and last IP address of each portion when you do.


Edited by Kilroy, 25 June 2015 - 06:02 AM.


#6 CaveDweller2

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Posted 25 June 2015 - 08:37 AM

The world of IT is vast. You can find jobs doing layer 1 all the way up to layer 7 and still be under the IT umbrella. And while I will agree the vast majority of IT jobs do not require you to have more than a rudimentary understanding of subnetting there are many jobs where you need to know it like the back of your hand. And to pass most certifications involving subnetting, you need to be able to dissect and reassemble IPv4 subnetting quickly.

 

When the questions reads: "Subnet these with as little wasted addresses as possible" or something like that, they are talking about Variable Length Subnet Mask(VLSM). The way LuminisIndia did it is correct. Biggest requirement to smallest.


Hope this helps thumbup.gif

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#7 AcesLight

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Posted 25 June 2015 - 09:16 AM

Thank you all of you. I'm taking an I.T. class now trying to sharpen my understanding on computer networking. I want to take the CompTIA Network+ exam in the near future. I tried to post this on another forum (shall not say the name), one guy says I'm trying to get them to do the homework for me. If I was trying to ask others to do the homework I wouldn't even do the calculation myself in the first place. Sigh..


Edited by AcesLight, 25 June 2015 - 09:16 AM.


#8 AcesLight

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Posted 29 June 2015 - 02:25 AM

One more question regarding subnetting. The term Classful vs Classless.

 

Does term Classful means all subnets after subnetting, so long as each of the subnets uses the SAME SUBNET MASK, they are classful networks? If I were to take 172.16.0.0 / 24, is this a classful or classless network? The reason I asked that is because by default the mask for 172. should be 255.255.0.0, and now the mask becomes 255.255.255.0.

 

So is it still a classful or classless network?



#9 Wand3r3r

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Posted 29 June 2015 - 09:57 AM

Your question is easily googled. Basically when you apply a subnet beyond the ip addresses class its classless.  For example if you use 10.0.0.0 and apply 255.255.255.0 to in instead of 255.0.0.0 its classless.



#10 CaveDweller2

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Posted 29 June 2015 - 01:35 PM

Each class of subnets has it's own mask by default. A = 255.0.0.0, B = 255.255.0.0, C = 255.255.255.0. so if you use those masks with that class you are using the classful subnet mask. Once you start subnetting you by design change to classless masks.

 

 

If I were to take 172.16.0.0 / 24, is this a classful or classless network?

That is a class B address with a class C subnet mask. So what do you think? classful or classless?


Hope this helps thumbup.gif

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#11 AcesLight

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Posted 29 June 2015 - 02:45 PM

Each class of subnets has it's own mask by default. A = 255.0.0.0, B = 255.255.0.0, C = 255.255.255.0. so if you use those masks with that class you are using the classful subnet mask. Once you start subnetting you by design change to classless masks

 

 

If I were to take 172.16.0.0 / 24, is this a classful or classless network?

That is a class B address with a class C subnet mask. So what do you think? classful or classless?

 

Classless?



#12 CaveDweller2

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Posted 29 June 2015 - 03:30 PM

Correct =)


Hope this helps thumbup.gif

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#13 LuminisIndia

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Posted 31 July 2015 - 12:24 PM

Classful- we are using the default subnet mask                                  Classless- we are not using the default subnet mask, we are                                                                                                                                                 doing subnetting.

 

Class          IP Range      Default Subnet mask/ Prefix                     Prefix (Classless)      

A                 1-126             255.0.0.0/8                                                  9-to-32 anything as per the requirement

B                 128-191         255.255.0.0/16                                            17-to-32 anything as per the requirement

C                 192-223         255.255.255.0/24                                        25-to-32 anything as per the requirement

 

With thanks,

Team Luminis Consulting Services.


Edited by LuminisIndia, 31 July 2015 - 12:31 PM.


#14 Wand3r3r

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Posted 31 July 2015 - 01:03 PM

Though this thread is a month old I wanted to mention that classless is not just about subnetting but also supernetting.






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