CCent exam or CCNA exam

gbdavidx

Do you have to know how to find what subnet mask to use for a given network if you need X many subnets and X many hots per subnet?

EdTheLad

Do not even consider paying for a networking exam if you cant subnet, subnetting it the core skill you need to have. If you cant master subnetting, its time to choose a different career path. Just give it one or two weeks and it will click.

ande0255

Yes you absolutely do.

gbdavidx

i can understand basic subnetting but when trying to do it reverse by given a number of subnets is when I get stuck

jcarrillo26

the way i learned is that each octet has 8 bits an ipv4 address has 32 bits total. so starting from left to right in each octet you have 128,64,32,16,8,4,2,1

with a default subnet mask you have /8 /16 /24 in a, b, and c ip addresses

so with the default subnet mask you have only 1 network so that wont work you must segment the network

so by placing a 1 in the 128 field you have 2 networks

placing a 1 in both the 128 and 64 you have 4 network

placing a 1 in both the 128 and 64 and 32 you have 8 networks

placing a 1 in both the 128 and 64 and 32 and 16 you have 16 networks

when was learning subnetting if i wanted to find how many networks i needed i work from left to right if i wanted to find how many hosts i worked from right to left and still do

jcarrillo26

also by placing a 1 in the last bit shows what you increment in the network by. i hope this helps!

aftereffector

I make a quick subnetting table on my scratch paper as soon as I start the test. It looks something like the one I attached. Going in order by columns from the left, you have the CIDR or slash notation (/24, /25, and so on), the last octet of the subnet mask (.0, .128, .192, etc), and then the number of subnets, number of IPs, and number of hosts. To check your math, if you multiply the number of IPs and the number of subnets for each row, it will be 256.The last octet of the subnet mask goes in binary order. 00000000 is 0, of course; 10000000 is 128; 11000000 is 128+64=192, 11100000 is 128+64+32=224; and so on to 11111100, which is 128+64+32+16+8+4=252. I stop there because that gives us only 4 IPs; subtract the subnet ID and the broadcast address and you have two assignable host addresses, which we often use for the IPs on a point-to-point link such as a serial line. The next subnet would only have 2 IPs and no assignable host addresses, so it's not relevant. The number of subnets and number of IPs are easy to derive mathematically and even easier to get if you just go through the logical order. With a /24, 255.255.255.0 Class C mask, you have one subnet with 256 addresses (everything from .0 to .255). If you split that in half, you get a /25 with a 255.255.255.128 mask, and you will have two subnets, each splitting the number of addresses evenly in half, for 128 addresses per subnet. The /26, 255.255.255.192 mask, has four subnets and further divides the previous ranges in half: 64 per subnet. And so it goes: 1, 2, 4, 8, 16, 32, 64 subnets, and 256, 128, 64, 32, 16, 8, 4 IPs per subnet. And remember, the total number of IPs in your Class C range is still 256, whether you split that up into two subnets of 128 each or 32 subnets of 8 each, so you can check your math that way.