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Math Formula to determine Network/Host subnet

Math Formula to determine Network/Host subnet

Let's say I have a subnet 192.168.10.0/24 , if I need to have 400 Hosts in the network , I can just change the subnet to 192.16.10.0/23.. this can stem from the formula (2^9) -2=510 ,this will be the correct one that covers 400 Hosts , because if I use (2^8)-2=254 , it will be less than 400
***2^9 means I already have 8 bits for the Host in the /24 and will just borrow 1 bit from the Network portion and will be /23

Sometimes the request is much more complex, for instance when they ask for 1113 hosts, then I will have to use the calculator and start from 2^1 and add each time 1 to the power to get closer to the desired number , ex: 2^10,2^11,etc.....

I wonder if there is a formula, like (2^x)-2= "a number close to 1113", and then I will just determine the value of x.

Thank you

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Qlemo

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jskfan

ASKER

is it ld(n+2) or ld (n-2) ?

log(n+2) / log(2) can be much more involved... than counting 1,2,4,8,16,32,64,128,256,512,.....etc....

using 1,2,4,8,16,32,64,128,256,512 will be

1,2,4,8,16,32,64,128,256,512
0,1,2,3,  4,  5,   6,    7,    8,     9

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So 400 hosts will be (2^9)-2 =510
then from the right of the subnet (the host portion) we'll use 9 bits for the hosts , and we'll be left with 23 bits for the Network portion

***in the calculator , I do not see the ld function.. where can I find it ?

David Johnson, CD

in excel
=ROUNDUP(((LOG(A1+2))/LOG(2)),0)

as stated ld(n) = log(n+2)/log(2)
ee29113185.xlsx

Qlemo

You asked for maths, not counting ;-). As stated, ld(n) = log(n) / log(2).
n+2 is correct - n=510 hosts, you need to calculate for 512 addresses then.

J0rtIT

We used this:
2^n -2 > #Hosts
the minus 2 because you have to remove the Network Address and the Broadcast address

So for 400 Hosts

testing for #host =400
n=1 => 2 -2 > 400 (not apply)
n=2 ....
...
n=8 => 256 -2 > 400 (still not apply)
n=9 => 512 -2 > 400 (woks)

(for #host = 1113)
n=10 => 1024 -2 > 1113 (not apply)
n=11 => 2048 -2 > 1113 (apply)

So this applies for n=9 (for 400) and n=11(for 1113)

then count from right to left 9 bits. from the 32 of a network
_ _ _ _ _ _ _ _ . _ _ _ _ _ _ _ _ . _ _ _ _ _ _ _ x . x x x x x x x x

And you turn on all the "_" bits for the submask:
255.255.253.0 so /23 for the 9 bits

Or count n=11 (for 11113)
_ _ _ _ _ _ _ _ . _ _ _ _ _ _ _ _ . _ _ _ _ _ x x x . x x x x x x x x
255.255. 248.0

------------------------------------------------------------------------------------------> this is the network part.
Now you can have as many subnets as you want to apply the same formula.
2^n -2 > #Subnets

and add them to the actual count of bytes, for example (15 subnets)
2^n -2 > 15
n=3 => 8 -2 > 15 (not apply)
n=3 => 16 -2 > 15 (not apply)
n=4 => 32 -2 > 15 (apply)
The would be 4 bits of network

So this applies to n=9 (for 400) and n=11(for 1113)

then count from right to left 9 bits. from the 32 of a network
_ _ _ _ _ _ _ _ . _ _ _ _ _ _ _ _ . _ _ _ s s s s x . x x x x x x x x
Where you do the math for each subnet
0 0 0 0
0 0 0 1
0 0 1 0
0 0 1 1
0 1 0 0
...
1 1 1 1

And for count n=11 (for 11113) it's the same 4 bits
_ _ _ _ _ _ _ _ . _ _ _ _ _ _ _ _ . _ s s s s x x x . x x x x x x x x
And the same logic

Qlemo

Jose, that is what the OP does already, but wanted a direct method.

SOLUTION

masnrock

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jskfan

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Thank you

masnrock

Thanks, but in fairness, Qlemo deserved at a minimum the lion's share of the points to this. He did provide the solution, while I provided the additional detail of what was necessary to do it in Windows.