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Subnetting questions
Im taking some CCNA practice exams. The ONLY part thats giving me trouble are these two types of subnetting questions:
1. Enter the number of networking bits that would give a network with 2 host addresses.
I know the answer, but I dont know how I got it lol
its 255.255.255.252
or 30bits. I only know that from subnetting in real environments. How do you properly answer this question?
2. Enter the subnet mask dotted decimal value that would give a network with 2,046 host addresses
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ASKER
Leo, thanks. Where did you get the numbers "3=11" in questions 1?
3 will be 11 in binary
2 will be 10 in binary
255 wil be 11111111 in binary
You know the joke: there is 10 types of people in the world: who understand the binary and who is not :)
ASKER
oh 3 is 00000011
2 is 00000010
Gotcha. thanks man
2 is 00000010
Gotcha. thanks man
ASKER
so if you want 510 hosts, you would need 16 host bits?
255.255.0.0
1111111.11111111.00000000. 00000000
Is this correct?
255.255.0.0
1111111.11111111.00000000.
Is this correct?
Nope.
510 and add 2 (network and broadcast) = 512
512 = 2 to the 9th power = 9 bits for hosts
11111111.11111111.11111110 .00000000 - hosts are zeros
subnet mask is
255.255.254.0
Do the math for 30 hosts. I'll check you.
510 and add 2 (network and broadcast) = 512
512 = 2 to the 9th power = 9 bits for hosts
11111111.11111111.11111110
subnet mask is
255.255.254.0
Do the math for 30 hosts. I'll check you.
ASKER
Wait I think I see. When ever they ask:
"enter the number of networking bits that would give a network with 1022 hosts"
The first thing you do is add +2 to 1022. This is because one is reserved for network, the other reserved for broadcast. This gives us 1024.
Now, I just have to do the formula 2^x=y to find the 1024.
2*2*2*2*2*2*2*2*2*2*2 =1024
Thats eleven "2s", or eleven host bits
11111111.11111111.11111000 .00000000
255 255 248 0
eh, is this right?
"enter the number of networking bits that would give a network with 1022 hosts"
The first thing you do is add +2 to 1022. This is because one is reserved for network, the other reserved for broadcast. This gives us 1024.
Now, I just have to do the formula 2^x=y to find the 1024.
2*2*2*2*2*2*2*2*2*2*2 =1024
Thats eleven "2s", or eleven host bits
11111111.11111111.11111000
255 255 248 0
eh, is this right?
Close.
2 to the 10th power is 1024, not eleventh. So what would the mask be now?
2 to the 10th power is 1024, not eleventh. So what would the mask be now?
ASKER
11111111.11111111.11111100 .00000000
255.255.252.0
22 networking bits 10 host bits
Is this correct?
255.255.252.0
22 networking bits 10 host bits
Is this correct?
Absolutely.
ASKER
thanks! finally the light has come on. Let's hope it doesnt go out LOL
ASKER
q2 is actually trial and error? You play with 2^x-2=your answer
Thats what I figured... I guess you just try all types of combos with questions like this.