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Subnetting help
cleanwithit
Member Posts: 63 ■■□□□□□□□□
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
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dynamik
Banned Posts: 12,312 ■■■■■■■■■□
So are you not allowed to use the first and last subnets? Usually you only need to subtract 2 for the hosts section, so you'd only need four bits for the subnets. If you can't use those subnets, that is correct. They got 19 for subtracting 5 (your subnet bits) from the total available (24 bits), and that's your number of hosts. However, you shouldn't subtract 2 from 19, but rather the result of 2^19. -
Optionscleanwithit
Member Posts: 63 ■■□□□□□□□□
So are you not allowed to use the first and last subnets? Usually you only need to subtract 2 for the hosts section, so you'd only need four bits for the subnets. If you can't use those subnets, that is correct. They got 19 for subtracting 5 (your subnet bits) from the total available (24 bits), and that's your number of hosts. However, you shouldn't subtract 2 from 19, but rather the result of 2^19.
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+, Network +, Linux +, MCP, MCTS, CCENT
A.S Network Administration -
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billscott92787
Member Posts: 933
cleanwithit wrote: »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?
Yeah 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