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## An organization is granted the block 16.0.0.0/8.

An organization is granted the block 16.0.0.0/8. The administrator wants to create 500 fixed-length subnets.

255.255.128.0
b. Find the number of addresses in each subnet.

2^15-2= 32766

c. Find the first and last address in the first subnet.

d. Find the first and the last address in the last subnet (subnet 500).

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Cisco Employee

## Hi uuuugeshforu

I could solve it by excel, you can analize that, I think it is the solution, please check the attached file :)

network

 16 249 128 0

 16 249 128 1

 16 249 255 254

regards!

VIP Blue

## Well, it's not "an

Well, it's not "an organization", 16.0.0.0/8 is granted to Hewlett Packard.

255.255.128.0

As 255.255.128.0 will create 512 subnets I assume the "500" is not exact number of subnets and any mask forming 500 or more subnets will be considered correct answer.

So you missed following correct answers: 255.255.192.0, 255.255.224.0, 255.255.240.0, 255.255.248.0, 255.255.252.0, 255.255.254.0, 255.255.255.0, 255.255.255.128, 255.255.255.192, 255.255.255.224, 255.255.255.240, 255.255.255.248, 255.255.255.252

In short, any mask /M in the range  from /17 to /30 will create more than 500 fixed-length subnets.

b. Find the number of addresses in each subnet.

2^15-2=32766

Incorrect.

There's 2^(32-M) addresses in each subnet. For M=17 (subnet size of your choice) it is 32768 addresses.

Yes, the first and last address of each subnet can't be used as regular IP address, but you has not been asked to calculate number of usable addresses of each subnet but just addresses of subnet. In such case you can't forget first and last address - it still exist, despite it can't be used.

c:Same problem as in b.

While first usable address is 16.0.0.1 regardless the subnet size, first address (also known as subnet address) is 16.0.0.0.

d: At the first, subnet 500 is last subnet for no mask. For M=17 (e.g. mask of your choice) the last subnet is 512.

Subnets are subnet size each. Thus they starts on (16.0.0.0+i*subnet size) where i is subnet numberAlgorithm for last address is described in c. Even here you need to distinguish first/last address and first/last usable address.

For M=17 and i=499 (e.g. subnet 500) it's 16.0.0.0+499*2^(32-M) = 160.249.0.0. Last subnet is subnet 512 (i=511) thus first address of last subnet is 160.255.128.0

You know the size of subnet is 2^(32-M) thus last address of subnet 160.249.0.0/17 is 160.249.127.255 and  last address of subnet 160.255.128.0/17 is 160.255.255.255

Also note that "subnet zero" and "subnet all one" may not be considered usable on some systems thus first/last subnet may not be the same as first/last usable subnet

If all-ones subnet is not usable in your particular environment, then last subnet is subnet 511 e.g. from 160.255.127.0 to 160.255.127.255

While it may look so complex, it's easy. It's just necessary to see binary IP address as concatenation of  net:subnet:host numbers. First/last subnet/address question is most easy one, as particular subnet or host is either all-zero or all-one, thus it's simple to calculate with them. The rest of task it is about your ability to work with binary numbers (and convert them to/from decimals). It's just about a lot of training to gain "it's clearly visible" level of experience.

I used no calculator or even pen and paper to calculate all those addresses mentioned above (I hope I didn't missed).

And final note - it seems not to be real task So you are off topic here. Learning questions should be asked on appropriate community of Cisco Learning Network.

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