+2 Rehman . · June 9, 2015
I was doing a problem: ": An organization is granted the block The administrator wants to create 500 fixed-length subnets.
a.  Find the subnet mask.
b.  Find the number of addresses in each subnet.
c.  Find the first and last addresses in subnet 1.
d.  Find the first and last addresses in subnet 500. "

As by some data given, i found the subnet mask by 32-8=24 which means
and other thing how many bits we need to create 500 fixed-length subnets should be 2(powe)9 as its equal to 512
what i could understand by searching is we have:

0001000 00000000 00000000
nnnnnnn sssssssss shhhhhhh

the part 's' is of 9 bit as we figured it out can be changed from 000000000 to 111111111 & part of host as well from all 0 to all 1.
still what are the total number of address in each as if the 's' part is whole '1' how would we count it? because its exceeding 255?
So because of it part c & d are stuck for me 

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+1 Pere Garau Burguera · June 10, 2015
So, the subnet mask will take the first 8+9 bits = 17 bits

So 11111111.11111111.10000000.00000000 =
+1 Pere Garau Burguera · June 10, 2015
Sorry if I confused you with the mask at the beginning, it should be the one taking 17 bits
+1 Pere Garau Burguera · June 10, 2015

You are given 8 bits for the general net (The same as saying that you have to use 16 in your first octave)

Then you have to add to those already assigned bits the ones to create subnets (In this case you have to add 9)
0 Rehman . · June 10, 2015
well pere i didn't get the question right when they say they want 500 subnets its not just coming into my mind
The slash number they have given is from the class ful address and the 500 subnet they are asking for will become class less address.

As we have figured out there must be 2^9 bits for makind it 500 subnets. we may say that 32-8-9 =15 bits for the ip, 
and it may like 2^15 = 32786 total number of networks ?

i have been searching on it what i get is so simple because they haven't asked doing any number subneting
example: |28
as its a class less address, we can simply find the subnet mask which is
by converting it into binary we will get 11111111.1111111.1111111.11000000

as the number of '1' bit goes on the power of 2 which will become : 2^2 = 4 total number of networks 
and number of '0' bits goes on the power of 2 as well we'll get: 2^6 = 64 address en each network

But here the 500 shit is un understandable for me :/
0 Pere Garau Burguera · June 10, 2015
As I said it's been a while since I did this, so I'm not fresh in this area.

With 9 bits to determine the network you can make 512 networks.

Then, each of those 512 networks will have 32-8-9 = 15 bits for the host.
So each one of the networks will have 2^15 addresses = 32768 addresses
So number of addresses in total = 500*2^15 = 16384000

So by that you will create 12 networks you are not going to use

Then the first network will be:

First address = 00010000.00000000.00000000.00000000 =
Last address = 00010000.00000000.01111111.11111111 =

Network #500

First address = 00010000.11111001.10000000.00000000 =
Last address = 00010000.11111001.11111111.11111111 =

If we used 512 networks, the Network #512 would then be

First address = 00010000.11111111.10000000.00000000 =
Last address = 00010000.11111111.11111111.11111111 =

So all the bits from the 2nd block and the first one of the third block determine the number of the network you are
0 Rehman . · June 10, 2015
I think i have found the answer 2^9 = 512. We needed 9 bits here 8 are from the actual class A. 1 bit is extra so the remaining 2^7 bits are 128 addresses in each subnet.
c) as the first address is the last address would be
d)the first address of last subnet is and last address would be
0 Rehman . · June 10, 2015
@Pere its from class A shouldn't it be ?
And sorry that answer was before your answer as here net was not working so i couldn't reply :/
0 Pere Garau Burguera · June 10, 2015
I don't see your answer

If you say you can only use one extra bit aside from the 8 bits given to the net, you could only create 2 subnets (bits 0 and 1)

What you're doing there is creating 2 subnets only. 

We need to create 500 subnets, so we will create 512.

If we had to create 2 subnets, the addresses would be like this:

First subnet : -
Second subnet :

You forgot to add the 255's there
If you do this, it can be clearly seen that you only create 2 subnets
0 Rehman . · June 10, 2015
Sorry for that, i meant " ---
should we go over second octave because of class A?
And, 128+128 = 256 addresses"

The formula for subnet addressing says the number of "zeros" should be the power of 2.
0 Pere Garau Burguera · June 10, 2015
So you are designing 2 subnets?
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