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
dissolvedAsked:
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minmeiCommented:
Q1

subnet mask defines what is network and what is host. If only last 2 bits are host, then 00 01 10 11 are the only possibilities. Since each subnet has a network address (00) and a broadcast address (11) , that only gives two addresses left (01 and 10)

if subnet mask is

11111111.11111111.11111000.00000000

or

255.255.248.0

then you have 2 to the 11th power hosts (11 zeros) which is 2048. Subtract the two for network and broadcast (00000000000 11111111111) and you have 2046.

Make more sense?
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dissolvedAuthor Commented:
q1: is there a formula for this? I noticed they said number of hosts and not host bits. Im only used to questions that ask something in bits etc.

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.
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-Leo-Commented:
Q1: I would not say its a formula, but I do like this:

1. Convert number of hosts to the binary
2. Check if this number consists of all "1", if yes - add one more bit
3. Substract number of resulting bits from 32 (total possible number of networking bits)
4. Convert this to the dotted binary or decimal notation, if you need (question was how many networking bits ...)

If you need 2 hosts:

1. 2 = 10
2. All "1" - ? No!, number of bits = 2
3. 32 - 2 = 30
4. 11111111.11111111.11111111.11111100 = 255.255.255.252

if you need 3 hosts:

1.  3 = 11
2.  All "1" - ? Yes!, number of bits = 3 (2 + 1)
3. 32 - 3 = 29
4. 11111111.11111111.11111111.11111000 = 255.255.255.248

And so on ...

Q2: Same calculation:

1. 2046 = 11111111110
2. All "1" = ? No!, number of bits = 11
3. 32 - 11 = 21
4. 11111111.11111111.11111000.00000000 = 255.255.248.0

Hope this will help you!
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dissolvedAuthor Commented:
Leo, thanks. Where did you get the numbers "3=11" in questions 1?
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-Leo-Commented:

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 :)
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dissolvedAuthor Commented:
oh 3 is 00000011
2 is 00000010

Gotcha. thanks man
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dissolvedAuthor Commented:
so if you want 510 hosts, you would need 16 host bits?

255.255.0.0


1111111.11111111.00000000.00000000
                                     
Is this correct?
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minmeiCommented:
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.
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dissolvedAuthor Commented:
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?
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minmeiCommented:
Close.

2 to the 10th power is 1024, not eleventh.  So what would the mask be now?
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dissolvedAuthor Commented:
11111111.11111111.11111100.00000000
255.255.252.0

22 networking bits  10 host bits

Is this correct?
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minmeiCommented:
Absolutely.
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dissolvedAuthor Commented:
thanks!  finally the light has come on. Let's hope it doesnt go out LOL
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