# Number of subnets = x^2 or x^2-2

Discussion in 'MCSE' started by Vernon, May 19, 2004.

1. ### VernonGuest

Which of the following network address spaces can be devided into at least
four subnets with atlest 100 hosts per subnets ?
a. 10.25.0.0/22
b. 10.25.0.0/23
c. 10.25.0.0/24
d. 10.25.0.0/25
Now for minimum 4 subnets do I need 2 bits or 3 bits for Microsoft ? (Never
mind the study guides)

Thanks

Vernon, May 19, 2004

2. ### RichGuest

Any of the options would give you the desired result.
option c for example would give you 254 available subnets
and 254 available host per subnet. Not sure what the
question is wanting past that.

Rich

Rich, May 19, 2004

3. ### NeilGuest

try this:

http://www.telusplanet.net/public/sparkman/netcalc.htm

oh, and Rich was incorrect about 1 peice in his post. with a /24 network
and this address you would have 64K networks not 254.

(and MS has historically done the 2^n-2 and Cisco the 2^n)

Neil, May 19, 2004
4. ### RichGuest

Neil
If this was starting out as a true Class A address then
you would be correct. I assumed that the class A address
(10.25.0.0) And that the question was to subnet it
networks by adding the subnet range to the 3rd octet. Of
course I am doing it the Cisco way and not MS's way.

I hope I am looking at this correctly.

Thanks for the input!

Rich

Rich, May 19, 2004
5. ### NeilGuest

I believe you are off base here. a class A is a class A is a class A. if
it was subnetted with /16 how would making the mask /24 change that. if
the address was a 10.x.x.x (something defined in RFC 1918) then changing
the /16 to /24 would mearly increase the number of netowrks from 253 to
64k and reduce the number of hosts from

you may want to look up RFC 1878 and RFC 1519 as some useful
references....

Neil, May 20, 2004
6. ### NeilGuest

pfffffttt! I had a nap. should have said "networks from 254" and "reduce
the number of hosts from 64k to 254"

Neil, May 20, 2004
7. ### VernonGuest

10.25.0.0/22 gives you 10 remaining bits to play with. Since we have min 100
hosts per subnet we need those 7 bits for host portion. That leaves us
32-22-7 = 3 bits. My question is how many subnets we can have with these 3
bits for Microsoft exam? 8 or 6? IF it is 8 then answer (10.25.0.0/23)
would have 4 subnets. Answer c and d are not correct since the remaining
bits are insuffecient to meet 4 subnets and 100 host demand.

Over to you Neil...I looked up the website the link you provided. It sort of
.........
So I guess it is calssless routing as far as Microsoft is concerned so it is
2^n and not 2^n-2

Vernon, May 20, 2004
8. ### NeilGuest

you are doing fine up till here. yes 7 bits will give you 126 host (even
if you do the 2^n, the 0 and 255 address are resevred for the network
this you have decided to subnet your subnet and this was not part of the
original question. I think you will find that if you dig out the original
question (I will assume that this is not live test question) you will
read that it say 4 networks and at least 100 hosts or at least 4 networks
and 100 hosts (one statement of at least not 2). class A addresses have a
defualt mask of 255.0.0.0 or /8 you have 24 additional bits to play with.
If you need 7 bits for hosts then you can have anything up to
255.255.255.128 or /25 for a mask. converting this to Cisco terminolgy
means that /22, /23, /24, or /25 is fine as they will all give you at
least 100 hosts. I think the one that closest matches the 100 hosts
recommendation is the /25

HTH

Neil, May 20, 2004
9. ### Andy FosterGuest

The original question was ...
"Which of the following network address spaces can be devided into at least
four subnets with atlest 100 hosts per subnets ?
a. 10.25.0.0/22
b. 10.25.0.0/23
c. 10.25.0.0/24
d. 10.25.0.0/25
Now for minimum 4 subnets do I need 2 bits or 3 bits for Microsoft ? (Never
mind the study guides)"

"Which of the following address spaces" implies that in a. 10.25.0.0/22 is
your entire address space - you do not have the entire Class A 10.0.0.0/8
range to yourself.
CIDR is the new black.

To the OP, M\$ used to say 2^n-2 subnets - they now say 2^n - but you should
be aware that both are possible answers. You should never be expected to
chose only one as a correct answer though.

HTH

Andy

Andy Foster, May 21, 2004