Subnetting

+2 Rehman . · June 9, 2015
I was doing a problem: ": An organization is granted the block 16.0.0.0/8. 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
a) 255.0.0.0
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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0 Pere Garau Burguera · June 10, 2015
Because you use the 7 first bits of the 2nd octave to set the subnet
The other ones are from the host

Correct me if I got those addresses wrong
0 Pere Garau Burguera · June 10, 2015
Adresses wouldn't be those in the 100 subnets example

They would be

First one
16.0.0.0 - 16.1.255.255

100th one
16.198.0.0 - 16.199.255.255
0 Rehman . · June 10, 2015
i can say if question says you have an ip address 16.0.0.0 | 8 and you have to make 100 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 100

a) 255.0.0.0

b) 2^7 = 128 the nearest. 32-8-7 = 17 bits for the host.
2^17 = 131072 address 

c)subnet would be
11111111.11111110.00000000.00000000
255.254.0.0

Address would be 16.0.0.0 ---- 16.0.128.255

d) 16.100.128.255 --- 16.100.255.255
+1 Pere Garau Burguera · June 10, 2015
Yeah

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
okay okay, what i can say is when ever we are given any number (e.g. 500) to make subnets we will convert them into the power of 2 and then we will add those bits with the already given bits and proceed further ? :3 Am i right ?
(General formula)
+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
So, the subnet mask will take the first 8+9 bits = 17 bits

So 11111111.11111111.10000000.00000000 = 255.255.128.0
0 Pere Garau Burguera · June 10, 2015
Look at my first answer

The bits of the 2nd octave and the first one of the 3rd octave are the ones indicating the subnet
0 Pere Garau Burguera · June 10, 2015
If you are designing 2 subnets, the subnet would be indicated with the first bit of the second octave:

First subnet

00010000.0000000.00000000.00000000 - 00010000.01111111.11111111.11111111 (The last address has 1's on the host part, while the first one has 0's in it) That would be 16.0.0.0 - 16.127.255.255

Second subnet

00010000.10000000.00000000.00000000 - 00010000.11111111.11111111.11111111 That is 16.128.0.0 - 16.255.255.255

As I said, you're only designing 2 subnets, you have to design 512 subnets (only 500 will be used)

When you're granted 16.0.0.0/8 you can do whatever you want with your 2nd, 3rd, and 4th octaves
0 Pere Garau Burguera · June 10, 2015
So you are designing 2 subnets?
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