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IP Range Assigning

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Old Unix and Linux 05-31-2016
TryllZ TryllZ is offline
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IP Range Assigning

THIS IS A SAMPLE PRACTICAL EXAM QUESTION, COMPLETE FILE HAS BEEN ATTACHED AS WELL.

Hi All,
I'm a bit confused about assigning IP address from IP Ranges.
I am using this scenario below to understand.
Scenario
Adatum.com an international IT solutions company, is launching 12 new branches in a new country where they currently have no existing branches. The sWin CIO has asked that each subnet should host at least 1000 devices.
sWin has obtained a network address of 192.168.0.0/18. This address needs to be subnetted further. The internal subnet mask will be /22
The company’s subnet plan allows for:
- Router Pool: 2 router interfaces for each subnet – the first 2 IP addresses from each subnet.
- Managed Network Device Pool: The first 20 IP addresses that follow the Router Pool are reserved for switches and other network devices
- Server Pool: The first 20 IP addresses following the Managed Network Device Pool are reserved for servers.
- Workstation Static Pool: The first 10 IP addresses following the Server Pool are reserved for Workstations that require static IP addresses.
- Workstation DHCP Pool: The remaining IP addresses are reserved for automatic IP configuration
You have been asked to use subnet 9 for your device addressing.
Configure the first IP address from the Server Pool to Lon-DC1.
Configure the first IP address from the Workstation Static Pool to sWinPC1.
Configure the first IP address from the Router Pool to be the default gateway.

So I have the IP address range as such:
192.168.0.0 is the Network ID and 192.168.3.255 is the Broadcast address.
Now can some one explain me how the IP ranges will be assigned having found the IP ranges being that since the last borrowed bit is place holder of 4 the IP ranges will also be in 4, for example 192.168.0.0 - 192.168.0.3, which means that if I use the 9th subnet the IP Range is 192.68.0.32 - 192.168.0.35, however my teacher has explained me (which I did not understand for some reason) in a way that IP range 9th subnet will be 192.168.32.1 etc..
I'm confused as to when to use my method for IP ranges and when to use my teachers.

My basic question is how to know when does the subnet range changes and which is the subnet range the first one of the teacher method.
Please, Thank You

University Of Swinburne , Kuching, Malaysia, Bachelors in Information Communication Technology Networking Major, Second Year, Professor Mujahid, Bachelor of Information and Communication Technology | Swinburne University of Technology Sarawak Campus
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File Type: pdf NetworkAdminPracticeSkillsExam3-S12015.pdf (135.8 KB, 3 views)
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Old Unix and Linux 05-31-2016
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In 192.168.0.0 - 192.168.0.3, you have the last two digits of 192.168.0.3 reversed, likewise for the range that you mention after that, maybe that will make you see the logic..
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Hi,

Thanks for the reply, I did not understand what do you mean by reversed, if you could elaborate.

The thing is that my method I have got it from a CISCO practice book and that is where is the book explains this method but the professor says otherwise that is why I wanted to know where the mistake it.
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It should be 192.168.3.0 instead of 192.168.0.3
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That's all about how the 32 address bits are being distributed between network address and host address. /18 means the most significant 18 bits designate the network part, and the remaining 14 bits the host part. 192.168.0.0/18 will give you 4 subnets with 16384 (= 2^14) addresses each:
Code:
192.168.0.0
192.168.64.0
192.168.128.0
192.168.192.0

For minimum 1000 devices you'll need 10 bits to accommodate 1024 addresses, so the internal /22 netmask comes in handy (32 bit - 22 bit = 10 bit addresses in each subnet). The 16 (= 2^4; 4 = 14 - 10) sub-subnet ranges are


Code:
192.168.0.0
192.168.4.0
192.168.8.0
192.168.12.0
192.168.16.0
192.168.20.0
192.168.24.0
192.168.28.0
192.168.32.0
192.168.36.0
192.168.40.0
192.168.44.0
192.168.48.0
192.168.52.0
192.168.56.0
192.168.60.0

, the 9th of which - depending how you count them, starting from 0 or 1 - is the network address 192.168.32.0 with the broadcast address 192.168.35.255 (hopefully, as my binary calculus has become somewhat rusty).

Make sense? And, reread the CISCO book...

Last edited by RudiC; 05-31-2016 at 02:20 PM..
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Hi Guys,

Thanks for the detailed information, please let me know if I'm understanding correct.

So the Subnet changes on every 4, as in 192.168.0.0, then 192.168.4.0 as per Rudic explanation.

1 - Does that mean that the first subnet IP Range is 192.168.0.0 - 192.168.0.255 and the next range is 192.168.1.0 - 192.168.1.255 ?

2 - Does this also mean that the incremental octet will be based on the subnet mask octet, for example in this case the subnet mask octet is the 3rd octet which is .252 so the increment subnet is also the 3rd octet, meaning that if the octet was 255.252.0.0 then the IP subnet would be (ONLY FOR UNDERSTANDING) 192.0.0.0 then 192.1.0.0 ?

Please let me know if I have understood it correct.

Thank You once again..

NB: If this is true means the CISCO book has not been explaining properly because in all of their examples I have come across they have only been using the last octet and that's the reason I have been stuck in that mindset.
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Forget the "octet" (or byte) concept for a moment. Look at it as 32 binary digits, all 1s and 0s.

Code:
192.168.0.0    =    11000000.10101000.00000000.00000000
x.x.x.x/18     =    11111111.11111111.11000000.00000000

That last one illustrates the subnet mask, and you perform a binary AND operation of this mask and the IP address to obtain the network part of the latter. The mask can be (in principle) any value, be it 16 bits, or 24 bits, or 30 bits. The remainder is used to address the hosts on that special subnet.

To answer your questions:
1 - If you are talking of the /18 subnets, NO. The first subnet's available IP addresses range from 192.168.0.1 - 192.168.3.254, and the next range is 192.168.4.1 - 192.168.7.254.

2 - Not sure I understand your question. A subnet mask on its own is pointless. 255.252.0.0 is equivalent to /14, so the subnets 192.168.0.0/14 would be (given you want four with 262142 IP addresses in them)
Code:
192.168.0.0
192.172.0.0
192.176.0.0
192.180.0.0

Please be aware that you are leaving the non-routeable Class C private range 192.168.0.0/16 with these.

Last edited by RudiC; 05-31-2016 at 04:21 PM..
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