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subnetting into another subnet
arcolino
Member Posts: 3 ■□□□□□□□□□
in CCNA & CCENT
Hello all, I got this subnetting problem from the net, I got it right but I have had folks say it should be 8 not 16, what do you guys think, and is there a way to proof out the answer weather 8 or 16?
how many subnets can be gained by subnetting 172.17.32.0/23 into a /27 mask, and how many usable host addresses will there be per subnet?
a 8 subnets, 31 hosts
b 8 subnets, 32 hosts
c 16 subnets, 30 hosts
d a class b address cant be subnetted into the fourth octet.
I picked c this is my math
27 -23 = 4
2 to the 4th power = 16 so thats 16 subnets
27 to binary 24 + 3 (1's) = 27 leaves 5(0's) which are the host
2 to the 5th power = 32 - 2 = 30 which is 30 hosts
btw does anybody have a quick way to do 2 to the power of and or increments?
how many subnets can be gained by subnetting 172.17.32.0/23 into a /27 mask, and how many usable host addresses will there be per subnet?
a 8 subnets, 31 hosts
b 8 subnets, 32 hosts
c 16 subnets, 30 hosts
d a class b address cant be subnetted into the fourth octet.
I picked c this is my math
27 -23 = 4
2 to the 4th power = 16 so thats 16 subnets
27 to binary 24 + 3 (1's) = 27 leaves 5(0's) which are the host
2 to the 5th power = 32 - 2 = 30 which is 30 hosts
btw does anybody have a quick way to do 2 to the power of and or increments?
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dynamik
Banned Posts: 12,312 ■■■■■■■■■□
You're right.
Just memorize a table up to 12 for the powers of two. It only takes a few minutes, and it makes everything from there on out so much easier. Also, the previous bit is half, and the next bit is double. If you remember five is 32 and ten is 1024, you should be easily able to double or halve your way to the desired number in a couple of steps.
Welcome to the forums
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Optionsblackngold4877
Member Posts: 19 ■□□□□□□□□□
Remember that a /27 address is always 30 usable hosts and you have the answer easily.
/24: 256 - 2 = 254 usable
/25: 128 - 2 = 126 usable
/26: 64 - 2 = 62 usable
/27: 32 - 2 = 30 usable
/28: 16 - 2 = 14 usable
/29: 8 - 2 = 6 usable
/30: 4 -2 = 2 usable (These are for the WAN links.) -
Optionsdynamik
Banned Posts: 12,312 ■■■■■■■■■□
blackngold4877 wrote: »/30: 4 -2 = 2 usable (These are for the WAN links.)
Don't forget about small businesses; usually one is sufficient for their needs. -
Optionsarcolino
Member Posts: 3 ■□□□□□□□□□
hmm, so is 16 subnets correct? and can the answer be proofed out? vs the 8? I have a senior network person saying its 8 subnets not 16?You're right.
Just memorize a table up to 12 for the powers of two. It only takes a few minutes, and it makes everything from there on out so much easier. Also, the previous bit is half, and the next bit is double. If you remember five is 32 and ten is 1024, you should be easily able to double or halve your way to the desired number in a couple of steps.
Welcome to the forums -
Optionssandman748
Member Posts: 104
It is definately 16. 172.17.32.0/23 ranges from 172.17.32.0 - 172.17.33.255 . If you go to a /27 mask you have 5 bits for hosts (32 hosts) . If you have to, just count up in increments of 32 to get each subnet address. You'll have 16 subnets.
0 32 64 96 128 160 192 224
twiceWorking on CCIE Collaboration:
Written Exam Completed June 2015 ~ 100 hrs of study
Lab Exam Scheduled for Dec 2015 -
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gravyjoe
Member Posts: 260
It is 16. That one almost tricked me. It's so easy to go by the rules of the /24 (255.255.255.0) mask. If it was 172.17.32.0/24, then it would be 8 subnets. Since the mask has one less bit in it (/23), it allows for more hosts. In this case, this allows for 172.17.32.x and 172.17.33.x addresses.
172.17.32.0/23
172.17.32.0 - 172.17.33.255
172.17.32.0 network address
172.17.32.1 - 172.17.33.254 usable addresses
172.17.33.255 broadcast addressThe biggest risk in life is not taking one.
