IP Range Assigning Post: 302974543

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Homework and Emergencies Homework & Coursework Questions IP Range Assigning Post 302974543 by RudiC on Tuesday 31st of May 2016 02:06:27 PM

05-31-2016

Registered User

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 03:20 PM..

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