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Saturday, June 25, 2016
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Posted on 25-02-10 22:58
We're gong over subnetting in class and its really not clicking.
I understand we barrow bits from the host portion and put it on the network bits, but thats as far as I could understand.
Lets say i wanted two networks. One with 30 hosts and the other with 50.
Class C IP.
I believe i need to take two bits from the host side, making the new subnet 255.255.255.129?
and the first rang of IP's 192.168.1.0-192.168.1.64?
second range being 192.168.1.65-192.168.1.129?
Posted on 15-04-10 09:23
moshbat is right you should have your teacher explain.....here is the correct answers for your example if you were wondering
your two networks would be 192.168.1.0 - 192.168.1.63
and the second would be 192.168.1.64 - 192.168.1.127
subnet mask /26 or 255.255.255.192
so you have 192.168.1.xxxx xxxx x being the bits
the values of x is as follows 128, 64, 32, 16, 8, 4 , 2 , 1
and you need a max number of 50 hosts. So we will pick 64. 64 is able to hold 50 hosts plus a couple more....
your range of hosts is 64
and 0 counts so we get the networks
192.168.1.0 - 192.168.1.63
192.168.1.64 - 192.168.1.127
*+2 more networks if you want to work it out*
For the subnet mask.... since we used 64
we had to barrow 2 bits from the 4th octet
to find your subnet mask add those two bits you borrowed
so it would be 128+64 which = 192
so your subnet mask is 255.255.255.192
ehhh i hope this helped lol it may seem hard at first but once you understand it, it is fairly easy
Posted on 15-04-10 19:52
Thanks it does help, and it showed that I was right.
The teacher said it was an introduction, and that we will be going over it more in depth the third semester next year, along with VLAN.
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