• # CCENT subnetting questions

Hey fellow CCENT/CCNA travelers,

Two subnetting problems:

1 - Given the following address: 10.100.48.27/19 what is the valid host range?

2 - Given the following address: 172.16.10.22 255.255.255.240 what is the valid host range?

Just trying to confirm my own approach and answers.

Just trying to confirm my own approach and answers.

1 - /19 indicates two things: interesting octet is the third, and increment size is 32.
So the network address should be 10.100.32.0 with usable host range of 10.100.32.1 - 10.100.63.254 and broadcast of 10.100.63.255

2 - Subnet mask of 255.255.255.240 is a /28 so interesting octes is 4th and increment size is 16.
Network address should be 172.16.10.16 with host range of 172.16.10.17 - 172.16.10.30 and broadcast of 172.16.10.31.

Is this correct, or am I missing something?

Thanks.

William

• Just to clarify,

These problems were NOT taken from the CCENT EXAM. They do pertain to CCENT material.

Thanks,

William

• Both answers look good, remember the key for preparing for the exam is to look get these answers in about a minute per question.

Cordially,
Ronnie Wong
Edutainer, ITProTV

*if the post above has answered the question, please mark the topic as solved.
**All "answers" and responses are offered "as is" and my opinion. There is no implied support or guarantee by the ITProTV team.

• Learning to do these in your head is the key and isn't hard with practice.

If you can multiply by 2, you can write down a quick table to show you each segment.
I know that each octet is 8 bits. So, I write down an 8 bit decimal equivalent:

128 64 32 16 8 4 2 1

Now, when I see a /19, I know that there are all 1's in the first 2 octets (16 places), and all 0's in the last octet (places 25 through 32). The third octet is where I need to find the network number, and I do that by deciding how many bits to use, starting from the left (19 minus 16). That leaves me 3 bits for the third octet, which is 128 + 64 + 32 = 224. So, my decimal mask is 255.255.224.0 if I need that to answer an exam question. If I'm given the decimal mask instead of the network mask bitcount, I just convert it to binary as usual.