a particularly annoying subnetting question
usgrant
Registered Users Posts: 2 ■□□□□□□□□□
in CCNA & CCENT
I was grinding away on Subnettingquestions.com like a champ, and then i got this one:
Question: How many subnets and hosts per subnet can you get from the network 172.23.0.0/23?
Answer: 128 subnets and 510 hosts
I cant even begin to follow the math on this one. I would really appreciate anyone helping me on this one. I gound away at this one for an hour!
Question: How many subnets and hosts per subnet can you get from the network 172.23.0.0/23?
Answer: 128 subnets and 510 hosts
I cant even begin to follow the math on this one. I would really appreciate anyone helping me on this one. I gound away at this one for an hour!
Comments

IRONMONKUS Member Posts: 143 ■■■□□□□□□□So, you are given an IP and a subnet mask of 172.23.0.0/23
The easiest way that I do it is I break it down like this.
1. To find how many subnets, you need to know the difference between classful and classless. The classful ranges are A = 1126 with a subnet mask of /8 (127 is in here, but it's loopback), B = 128191 with a subnet mask of /16, and a C = 192223 with a subnet mask of /24. They have given you a classless, so find the difference. Well, 172 is in a Class B range, so the difference of /23 and /16 is 7.
 So, I set up my decimal to binary chart of 128 64 32 16 8 4 2 1, I guess it's a per bit representation of an octet. Don't qoute me... Anyway count over from right to left starting at 2 (because binary is multiples of 2) 7 times and you get 128. So you have 128 subnets. If you're good at math, you can do the 2 to the nth, nth being 7. I just like the graph.
2. For how many hosts, you use the leftover bits. You know there is 32 bits. So, the difference of /32 and the given /23 is 9.
 This time, your chart needs to be bigger. Binary is multiples of 2, so just go out two more times
512 256 128 64 32 16 8 4 2 1. Count over 9 times from right to left, starting at 2 and you get 512. Remember that for hosts, the math is 2 to the nth  2. nth being 9. Minus 2, because you have a network IP and a broadcast IP. So 512  2 = 510.
I hope this helped and actually didn't confuse you even further... 
Jon_Cisco Member Posts: 1,772 ■■■■■■■■□□9 network bits 2, 4, 8, 16, 32, 64, 128, 256 next is 512 then subtract two for the network and broadcast. now for the 7 subnet bits 2, 4, 8, 16, 32, 64, 128 For questions like this you need to know your classes to understand the subnets. Just keep trying the more practice you have the quicker you become.

_Gonzalo_ Member Posts: 113Question: How many subnets and hosts per subnet can you get from the network 172.23.0.0/23?
Answer: 128 subnets and 510 hosts
The question is the issue. It should be:
Which is the maximum number of subnets for 172.23.0.0/23? And the maximum number of hosts?
You would better get used to this... Cisco exams have plenty of questions that are incorrectly worded like this!:) 
usgrant Registered Users Posts: 2 ■□□□□□□□□□Thank you very much for your answers. Expect to see more in the future as i begin my journey into networking....