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control wrote: » Your address is a Class A address... (default Mask 255.0.0.0) So with the given mask you have the 8 default bits, plus another 12 "borrowed" bits (255.240). 2^12 - gives you the 4096 This then leaves 12 bits left over for the host bits - again 2^12 (-2) gives you 4094
Reck_ wrote: » With this kind of question, I use this as a guide.2 (NSM-DSM) = # of Subnet2 (32-NSM) -2= # of Hostwhere: NSM= New Subnet MaskDSM= Default Subnet Mask32 = IPv4 in bitsHow many Subnets ?2 (NSM-DSM) = # of Subnet2 (20-= 2 (12) =4096 SubnetsHow many Host ?2 (32-NSM) - 2= # of Host2 (32-20) - 2= # of Host2 (12) - 2= 4096-2= 4094 HostSubnettingquestions.com is my breakfast as I prepared for myCCNA exam, so I think you are in the right track.The foundation of subnetting in my opinion, relies ondefault subnet masks, so make sure you know them by heart. Good luck.
mohamedshajid wrote: » Ohhh. But i'm reading Todd lammle's 7th Edition CCNA But there is no formula that i've encountered.
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