How can I improve my subnetting method?

SurferdudeHB

Here's an example of how I do my subnetting, feel free to make suggestion, thank you.

Sub problem from New York Business Institute - NYBI
Q. What are the subnet mask, first available host address of subnet one (NOT subnet zero), and the maximum number of hosts per subnet for the network below? (You need to allow for maximum number of hosts. Also, you can use subnet zero and all-ones subnet. RFC 187

Network ID: 222 17 128 0
Subnets Required: 7

Subnet Mask: ?
1st Available Host ?
Address of Subnet 1: ?
Max # of hosts/subnet: ?

Fist thing I do is write down
128 192 224 240 248 252 254 255
01-126 128-191 192-223

Then work the problem in this fashion,

1) 128 64 32 16 8 4 2 1, need 7 subnets therefore borrow from the first 3 bits = 3 bits needed
2) Network ID: 222 17 128 0 is a class C therefore 255.255.255.0
255.255.255.000 0000 borrow from last octect .1110 0000
convert this into /27 or .224
3)Find increment, last bit on .1110 000 is 32
therefore 1st Available Host is 222.17.128.33
4) Max # of hosts/subnet: is 2n-2 = 2x5 -2 result is 32-2 = 30
30 hosts.

This method takes me about 1 minute. I know there's got to be a more efficient way to do this..

Chris_

SurferdudeHB wrote: »

3)Find increment, last bit on .1110 000 is 32
therefore 1st Available Host is 222.17.128.33

Nothing that wrong with your method, except the above - firstly subnetting a class C with a /27 mask gives you 8 subnets (I'm sure you are aware of this but you didn't explicitly state it) and your first available host on the first subnet (as your requirements state that subnet zero can be used) would be 222.17.128.1

SurferdudeHB wrote: »

This method takes me about 1 minute. I know there's got to be a more efficient way to do this..

1 minute is hardly the end of the world! but you can probably speed it up by not writing out the binary equivalents and just getting used to doing it in your head - I'm not ashamed to admit that I still count up (or down) the powers of 2 on my fingers!! got 10, only really need 8

ITdude

Always good to have 2 extra, as spares.

instant000

SurferdudeHB wrote: »

Here's an example of how I do my subnetting, feel free to make suggestion, thank you.

Sub problem from New York Business Institute - NYBI
Q. What are the subnet mask, first available host address of subnet one (NOT subnet zero), and the maximum number of hosts per subnet for the network below? (You need to allow for maximum number of hosts. Also, you can use subnet zero and all-ones subnet. RFC 187

Network ID: 222 17 128 0
Subnets Required: 7

Subnet Mask: ?
1st Available Host ?
Address of Subnet 1: ?
Max # of hosts/subnet: ?

Fist thing I do is write down
128 192 224 240 248 252 254 255
01-126 128-191 192-223

Then work the problem in this fashion,

1) 128 64 32 16 8 4 2 1, need 7 subnets therefore borrow from the first 3 bits = 3 bits needed
2) Network ID: 222 17 128 0 is a class C therefore 255.255.255.0
255.255.255.000 0000 borrow from last octect .1110 0000
convert this into /27 or .224
3)Find increment, last bit on .1110 000 is 32
therefore 1st Available Host is 222.17.128.33
4) Max # of hosts/subnet: is 2n-2 = 2x5 -2 result is 32-2 = 30
30 hosts.

This method takes me about 1 minute. I know there's got to be a more efficient way to do this..

I would have done it like this (but it doesn't really matter)

Class A, B, or C? Class C
You need an even number of networks, so you divide by next largest even number from 7 (eDIT: incorrect. you need an exponential number of 2 networks, so, you use 2^1=2, 2^2 = 4, and 2^3 = 8, so you have 8 subnets to meet your requirements.)

256/8 = 32 (count networks in groups of 32)
block size of 32

0, 32, ... (first available host in the 1 network is ending in .33)
address of subnet 1 is ending in .32

256 - 32 = 224 (boom, subnet mask = 255.255.255.224)


Solution:
Subnet Mask: 255.255.255.224
1st Available Host in subnet 1: 222.17.128.33
Address of Subnet 1: 222.17.128.32/27

I hope this helps.

But, do it whatever way works for you.

o1n

reading Chapter 'Subnetting ' on the book 'Sybex CCNA study guide'