A Class C network uses 5 bits for a subnet id.
A. How many possible subnets are there?
B. How many bits will be in the host id?
C. How many possible hosts are there on each subnet?
D. What is the mask in binary notation?
Basically where do I start?
If anyone could help answer this for me, or guide me along, it would greatly appreciated as it will help me understand the other similar problems like this for my Review.
Thank you!
i am not sure at the moment(i am rusty), but perhaps the answer to D. is 11111111 11111111 11111111 11100000. let me know if that's helpful at all. i always forget and get confused with subnets.
i am not sure at the moment(i am rusty), but perhaps the answer to D. is 11111111 11111111 11111111 11100000. let me know if that's helpful at all. i always forget and get confused with subnets.
I came up with 11111111 11111111 11111111 11111000
But I will recheck my work and get back to you, for both my sake and yours :P.
I came up with this:
a. How many possible subnets are there? - 30 2^5-2 = 30
b. How many bits will be in the host id? 29 class c = /24 + 5 = /29
c. How many possible hosts are there on each subnet? - 6 32-29 = 3, 2^3-2 = 6 hosts
d. What is the mask in binary notation?-11111111 11111111 11111111 11111000
well.. as long it is just a midterm PRACTICE..go ahead and ask your instructor.. or perhaps we can get more ppl here on daniweb involved....i always get into massive arguments over math stuff because when you start taking calculus and whatnot...it gets harder and harder to explain the simpler stuff....
I came up with this:
a. How many possible subnets are there? - 30 2^5-2 = 30
b. How many bits will be in the host id? 29 class c = /24 + 5 = /29
c. How many possible hosts are there on each subnet? - 6 32-29 = 3, 2^3-2 = 6 hosts
d. What is the mask in binary notation?-11111111 11111111 11111111 11111000
made me think on this one, been a while for subnetting since the real world makes tools for us anymore unlike when i took my CCNA test years ago.
A is correct minus the broadcast and loopback as you did
B is not so much you took 5 bits for your subnet leaving you with only 3 host bits
C possilbe hosts 2^3 don't recall subtracting 2 here so i would relook at that one
D is correct taking bits left to right or turning on the bit from a zero to a one
hope that helps
i'm totally confused
i thought 255.255.255.0 is equal to 11111111 11111111 11111111 00000000
like this?
0 =1
1=2
10=3
11=4
100=5
101=6
110=7
111=8
1000=9, no 1000=8 , so zero is still zero, i guess.
i'm totally confused
i thought 255.255.255.0 is equal to 11111111 11111111 11111111 00000000like this?
0 =1
1=2
10=3
11=4
100=5
101=6
110=7
111=8
1000=9, no 1000=8 , so zero is still zero, i guess.
bits count like this for 8 bits in each octet which there is four 255.255.255.255 counting in 8 bits of 00000000 and they count down from left to right 128 64 32 16 8 4 2 1 and we borrow 5 so you add 128-8 to get your 255.255.255.248 subnet and all the rest that was typed above. if that makes sense
not really if you add the total of all 5 binary digits, you get 32+16+8+4+2+1= 63.
and 255 minus 63 is 192. so i'm thinking it's 255.255.255.192...
anyway, does anding fit in this thread somewhere?
You add left to right just as you read. Add the first 5 bits for the subnet and the last 3 left of the 8 give you the host.
Minimee120 has their confirmation from my earlier post. Subnetting is never fun
oh i know why i was confused, i thought they meant 5 bits for the subnets THEMSELVES, not for the subnet ID's as is the case. i simply misread the question...oops
-edit but then, wouldn't that be 29 bits for the ID's?
Would think that way but you actually take the 2^5 and you get 32 and you take away 2 bits for the broadcast address and loopback. You don't really add them as you would think. I always hated Subnetting but I wouldn't have built the networks I have over the years without it.
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