I was doing a problem: ": An organization is granted the block 16.0.0.0/8. The administrator wants to create 500 fixedlength 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 328=24 which means
a) 255.0.0.0
and other thing how many bits we need to create 500 fixedlength 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
Subnetting 
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Pere Garau Burguera
· June 9, 2015
I get that yeah the subnet mask is 255.0.0.0
When you do the question b) You said you need 9 bits for the subnet So are you saying that it will be like this? 00010000.00000000.00000000.00000000  00010000.00000000.01111111.11111111 00010000.00000000.10000000.00000000  00010000.00000000.11111111.11111111 00010000.00000001.00000000.00000000  00010000.00000001.01111111.11111111 00010000.00000001.10000000.00000000  00010000.00000001.11111111.11111111 ... 00010000.11111111.10000000.00000000  00010000.11111111.11111111.11111111 ? 
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 3289 =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: 192.168.1.55 28 as its a class less address, we can simply find the subnet mask which is 255.255.255.192 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 3289 = 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 = 16.0.0.0 Last address = 00010000.00000000.01111111.11111111 = 16.0.127.255 Network #500 First address = 00010000.11111001.10000000.00000000 = 16.249.128.0 Last address = 00010000.11111001.11111111.11111111 = 16.249.255.255 If we used 512 networks, the Network #512 would then be First address = 00010000.11111111.10000000.00000000 = 16.255.128.0 Last address = 00010000.11111111.11111111.11111111 = 16.255.255.255 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 16.0.0.0 the last address would be 16.127.0.0 d)the first address of last subnet is 16.128.0.0 and last address would be 16.255.0.0 
0 
Rehman .
· June 10, 2015
@Pere its from class A shouldn't it be 16.127.0.0 ?
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 : 16.0.0.0  16.127.255.255 Second subnet : 16.128.255.255 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 "16.0.0.0  16.127.0.0
16.127.0.0  16.255.0.0 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?

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
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 
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