# Subnetting help

cleanwithit
Posts:

**63**Member ■■□□□□□□□□
in CCNA & CCENT

Hello, all. I am trying to get this subnetting down. Here's my problem:

1. 12.0.0.0 15 subnets 128 64 32 16 8 4 2 1--I use this to help.

2. Class A

3. 2 to the power of 5, minus 2= 30

4. Custom Mask= 255.248.0.0

5. 256-248= 8 increment, correct?

6. 12.8.0.0, 12.16.0.0, 12.24.0.0, 12.32.0.0, etc, etc. 12.248.0.0= Broadcast, yea?

7. This is where I get confused. On this sheet I have it says 2^19-2=524286 How did they derive at that answer. Where did the 19 come from? Thanks in advance.

***I think I know where they get the 19 from. 255.248.0.0= 11111111.11111|000.00000000.00000000.

So, add up all of the 0's(hosts) and it equals 19, then do 2^19-2. Am I on the right track?

1. 12.0.0.0 15 subnets 128 64 32 16 8 4 2 1--I use this to help.

2. Class A

3. 2 to the power of 5, minus 2= 30

4. Custom Mask= 255.248.0.0

5. 256-248= 8 increment, correct?

6. 12.8.0.0, 12.16.0.0, 12.24.0.0, 12.32.0.0, etc, etc. 12.248.0.0= Broadcast, yea?

7. This is where I get confused. On this sheet I have it says 2^19-2=524286 How did they derive at that answer. Where did the 19 come from? Thanks in advance.

***I think I know where they get the 19 from. 255.248.0.0= 11111111.11111|000.00000000.00000000.

So, add up all of the 0's(hosts) and it equals 19, then do 2^19-2. Am I on the right track?

A+, Network +, Linux +, MCP, MCTS, CCENT

A.S Network Administration

A.S Network Administration

0

## Comments

12,314Banned63Member ■■□□□□□□□□Right. For this exercise, can't use the first and last subnets. Right. I get that -2 should be subtracted from the result of 2^19. I was just typing fast. I appreciate it. I will be posting another one here shortly so I can make sure I'm getting it right.

Thanks, Dynamik!

A.S Network Administration

933MemberYeah you catch the drift. That appears to be right. This would mean that no ip subnet-zero is in effect. As far as 2^19

It comes from the host bits in your binary 000.00000000.00000000 there are 19 bits

So we take 2^19 = 524,288 then subtract two "reserved subnet address and broadcast address" 524,288-2 = 524,286 and there you go!

Post your next I'd be glad to help