nav[aria-label="Primary Navigation"] { padding: 0; & ul { list-style: none; width: 100%; display: flex; flex-direction: row; justify-content: start; align-items: start; gap: 30px; padding: 0; & li { margin: 0; } & ul li { list-style: none; } } }

magic number for broadcasts issue

SephStorm

Hi, trying the odom/cisco magic bumber interesting octet method of subnetting. i am trying to calculate the broadcast addr for 172.22.241.224 /23 (254)

the formula says magic number (120 is the a multiple of2 which gets you 240, closest to 241..) + ip address bit which is 241... that is 360-1 359... even i know you cant go over 255... what am i doing wrong?

Find more posts tagged with

SAVE $250 on Your CISCO Boot Camp — Use Code "EXAM26"

Exclusively for TechExams members for Infosec Boot Camps starting before April 30, 2026

View CISCO Boot Camps

Comments

m3zilla

Do you know how to get the range of IP for a given network? If so, the broadcast is the last IP in the range.

SephStorm

the range will be 1 above and 1 below, but i cant have a 300 as my broadcast address... calculator says the broadcast should be 241, but i dont know where my math or process is wrong

SephStorm

it looks like i had originally miscalculated the magic number as 120 and was using it in my calculations.

But I would still like to know how we are supposed to know these multiples in less than 30 seconds...

Ltat42a

/23....you are subnetting a Class B address in the third octet. /23 = 255.255.254.0. Your increments are 2 in the third octet. Start at zero, 172.22.0.0, now go up two in the third octet until you go past 172.22.241.0. 172.22.242.0 will be the next subnet.
172.22.241.224 is in the network (subnet):

172.22.240.0 <Network
172.22.240.1 <First available address
172.22.241.254 <Last available address
172.22.241.255 <Broadcast

172.22.242.0 <Next subnet

hth

SephStorm

its annoying because (I) cant calculate those multiples that fast. i've done a few more, and i can do the full chart, but im spending so much time trying to figure out what the multiples of 2 or 16 or w/e are... sorry if im repeating the same question... idk. i'm practicing, but i dont seem to be getting much faster...

Reck_

Hi,

it will be 172.22.241.255 to be exact
your interesting/magic/increment # will be 2 and will be applied to the 3rd octet

so it means, you will start from....

172.22.0.0 _______1st network
172.22.2.0
172.22.4.0
172.22.6.0
.
.
.
.
172.22.240.0

Network (121st Network)
172.22.240.1

First usable IP address
172.22.241.254---Last usable IP address
172.22.241.255---Broadcast Address

172.22.242.0

Next subnet

A no memorization tip to get your interesting/magic/increment # is 256-NSM(New Subnet Mask)
A question may show a DDN (dotted decimal notation) or CIDR (/) represenation for subnet mask
so be sure you know how to convert your CIDR to DDN (or vice-versa)

in your example,we get 2 because...

256-NSM(New Subnet Mask)= interesting/magic/increment #
256-254 = 2

Also, I get a grip on the .240.0 subnet because counting by 2's is easy. If your 3rd octet is
241, it would turn out that you will jump right from 240 then 242,244 and so on.

Question is, on what octet you will start counting from 0...

If your NSM will fall on the range of...

9-16 : apply (start counting) your interesting/magic/increment # in the SECOND OCTET
17-24 : apply (start counting) your interesting/magic/increment # in the THIRD OCTET
25-32 : apply (start counting) your interesting/magic/increment # in the FOURTH OCTET

Setting a sample question similar to your example taken from subnettingquestions.com

Question: What is the broadcast address of the network 192.168.146.128/26?

What do we know.

1. It is Class C address with a DSM(Default Subnet Mask) of 24
2. The NSM (New Subnet Mask) is /26
3. My interesting/magic/increment # is 64 (/26 = 192)
256-192=64
4. I'll apply my interesting/magic/increment # in the 4th octet (/26 is within 25-32 range),
meaning I'll start counting by 64 starting from 0 on the FOURTH OCTET right ?

plotting down....

192.168.146.0

1st network
192.168.146.64

192.168.146.128

3rd network
192.168.146.129---- First usable IP address
192.168.146.190

Last usable IP address
192.168.146.191----Broadcast Address

OUR ANSWER

192.168.146.192 Next Subnet

Good luck.

192.168.146.128.JPG

CodeBlox

Think of it this way - The multiplier is the place value of the least significant subnet bit in your special octet.

So your multiplier will always be one of the following: 128,64,32,16,8,4,2,1

boredgamelad

SephStorm wrote: »

its annoying because (I) cant calculate those multiples that fast. i've done a few more, and i can do the full chart, but im spending so much time trying to figure out what the multiples of 2 or 16 or w/e are...

Honestly, it sounds like you're doing too much math. Sounds counterintuitive, but you are. Memorize your multiples of 32 or even better your multiples of 16 and you can start doing this stuff in your head. You shouldn't really be having to do that much calculation to get where you want to go.

If you're spending time adding and subtracting by hand, or guessing how to get where you're trying to go, you aren't working efficiently

atorven

Don't focus on speed, focus on getting your method 100%, the speed will come in time. If in doubt, write your multiples down on paper.

goldenlight

Its important to find a method you are most comfortable with. I use the Magic Number Trick to find THE Network and Broadcast Address. Everything between are Host. Downloading the Subnet calculator also help me better understand how the numbers progress. Similar to a odometer in a car

I usually make a quick chart. So I can answer all my subnetting questions. Not an expert yet, but someday I'm sure I will be one in time with practice.

128 64 32 16 8 4 2 1

128 192 224 240 248 252 254 255

then my powers of 2^n 2,4,8,16,32.. essentially doubling each number to get my sub nets and I'm good to go.

JasonIT

+1 for goldenlight. I usually make the same chart, but space mine out so that it corresponds correctly.

128 192 224 240 248 252 254 255
128 64 32 16 8 4 2 1

so as seen above, if your last octet is .248 (255.255.255.24

, then your networks are:
192.168.1.0
192.168.1.8
192.168.1.16
192.168.1.24

So, in the chart above the .248 corresponds with 8

hope this helps

**edit I guess you can't space them out correctly**

Ltat42a

JasonIT wrote: »

+1 for goldenlight. I usually make the same chart, but space mine out so that it corresponds correctly.

You mean like this??

Futura

Please don;t worry about subnetting too much, I used to find it tricky and now i find it very easy. Its totally logical. Please stay in the last octet for a while.

I just did /30's for ages increments of 4. 2usable adresses.
then I moved on to /28's Increments of 16, 14 addresses usable.

Forget the rest for now till you can sing about them in the shower, then. add a 1 number to 28 = 29. move along the binary scale to /29's you know that a /28 was 16 increments. so guess what. 29 is the next one being increments of 8, now the next one you already know /30 again. now move backwards. You know that /28 was 16 increments so guess what. /27 is 32. and /26 is 64 and 25 is 128, and guess what 24 is.

Once you have the last octet down move into the third octet. This will be very easy then.

I'm not being patronising, I had the same issue, Learn it slowly. Forget magic calculator etc, it just adds to the confusion.

I install cisco networks and create vlans for sites every day. And I still use this method.

Summary, learn that 240 = /28 = 16 increments and move backwards and forwards from there.

I also had the same issue using GNS3. I spent more time trying to get to the problems of software instead of learning cisco.

Hope this helps