subnetting

james35

Help! I am a little lost. I have this ip address of 172.16.0.0/16 and I need to subnet it for network with 750 hosts and another network for 512 hosts then three seperate routers. Can someone help me or give me a step by step refresher subnetting for multi hosts

dave0212

Hi

You could probably go the simple route and just use either

/18 - 4 Subnets + 16382 Hosts
/19 - 8 Subnets + 8190 Hosts
/20 - 16 Subnets + 4094 Hosts
/21 - 32 Subnets + 2046 Hosts
/22 - 64 Subnets + 1022 Hosts

I think you get the idea

Or you could use VLSM if you want to get the best use of your available IPs although you will need /22 to meet both subnet requirements your list leaving you 62 subnets to play with

itdaddy

you learned in subnetting




2 to the power of 5 is equal to 32 then double it each time you add 1 to the power

2^6=64
2^7=128
2^8=256
2^9=512 9 bits for hosts
2^10=1024 10 bits for hosts

172.16.(6)subnet bits(2+ the next 8 bits) = 10 bits 750 (1024 covers it)
YES 2^n-2 = host

172.16.11111100.0000000

172.16.(7)subnet bits(1 + the next 8 bits) = 9 bits 512
YES 2^n-2 =hosts this can only be 510 so you miss the mark of 512
172.16.11111110.00000000

warraning: yes you need the minimum of /22 to get both hosts to fit in range.

2^n-2 = hosts n = host bits
2^n=subnets n = subnet bits



but you have to think in host bits and subnet bits....not to scare you but
IPV6 is similar but way way harder...got to www.subnettingquestions.com
and practice daily your subnetting. great daily practice for net engineers
Good luck.

but if you have 3 routers you might need 4 subnets. you have alot to work with
with the 172.16.0.0 /16 so go to BOSON and get the IP subnet calc for free to
test your manual work of the subentting.. you need to do it by hand/brain
and then check your work..serious the CCNA exams you have to be able to
jam on VLSM and Subnetting host and subnets I mean like 2nd nature.
I took the 1 exam 802 and trust me when your CCIE tellls you need to jam
you need to jam with it.