Subnetting question...

Hi guys,

I'm going through subnettingquestions.com and doing quite well. There is, however, one question that is stumping me:

Question: What is the broadcast address of the network 172.31.70.208/28?

Answer: 172.31.70.223

Could someone relate to me the steps I take to get this answer?

I understand it is class B with 4096/4094 Subnets (not sure whether to "n-2" or not) and 14 hosts.

Thanks,
Prophet

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Comments

BubbaJ

You have 4 bits to work with so 2^4=16. The 1st address (20

is the network address. The 16th address (223) will be the broadcast address.

1 - 208
2 - 209
3 - 210
4 - 211
5 - 212
6 - 213
7 - 214
8 - 215
9 - 216
10 - 217
11 - 218
12 - 219
13 - 220
14 - 221
15 - 222
16 - 223

wildfire

Ok. the easiest way to break these things down is to get back to the good old binary..

so

172.31.70.208
10101100.00011111.01000100.11010000

/28 = 255.255.255.240

11111111.11111111.11111111.11110000

Now draw a line between the last 1 and the first 0

this gives you the last 4 0000's to mess with

0000 = the network address
0001 = The first host
1110 = The last host
1111 = Broadcast addy

so for the last octet

1101111 = 223

then the trick is to build up using math in you head, you I know that a 28 bit mask means I have 4 bits on the end so thats a range of 16, (14 useable hosts)
Once you master the binary and come to know the numbers that are associated with the mask it all becomes clear.

-prophet-

Wow!

Thanks guys, very helpful!

I tried it out and yup, it works great. I guess it all comes down to practice...
I'm sure I'll be dreaming 0 and 1 tonight and for the next month ahead.

Cheers,
Prophet

leilani

I am sure everyone has an easy way to do things....so here is mine.....
First always write this table out

128 064 032 016 008 004 002 001
128 192 224 240 248 252 254 255

/28 = to 255.255.255.240

If you look at the table...right above 240 is 16 which is the number of subnets of couse you minus 2.

Now you count by 16 until you reach your your subnet(I am doing this long way so you can hopefully understand)
16
32
48
64
80
96
112
128
144
160
196
192
208
224
(These are your 14 subnets)

Your .208 in between 208 and 224. Remember the first address is network and the last address is broadcast. So the last address before the new subnet is 223. There first address before the previous subnet is 208.

Anyway...thats how I do mine.

Danman32

wildfire wrote:

1101111 = 223

You missed a bit. Should be:

1101 1111 = 223.

The funny thing is, I kept making the same mistake on the calculator, coming up with an answer of 111 decimal until I discovered my missing bit. Then I got 223 as expected.

Another thing you can do to find the broadcast is: If finding the next subnet network address is easier for you in a particular example, subtract one for the broadcast of the previous subnet possibility.

binarysoul

I thought I should use an existing thread to ask questions about subnetting.

1. How do I know whether on the CCNA exam they just want the number of subnets/hosts or 'USABLE' subnets/hosts? Here is an example.

If mask 255.255.255.128 were used with a Class B network, how many subnets could exist, with how many hosts per subnet, respectively?

a. 256 and 256
b. 254 and 254
c. 510 and 126
d. 512 and 128

The answer is C. But do I make the assumption that Cisco always wants the 'usable' hosts/subnets?

EdTheLad

If this question is on the exam it will be phrased like so.

If mask 255.255.255.128 were used with a Class B network with subnet-zero disabled, how many subnets could exist, with how many hosts per subnet, respectively?

a. 256 and 256
b. 254 and 254
c. 510 and 126
d. 512 and 128

Anyway with subnet zero enabled it would still be C, i'm sure you meant
d to be 512 and 126 as you will always lose 2 addresses from the host side due to subnet and broadcast address.

marlon23

You are using such complicated methods...

Here is mine:

/28 means 4 subneting bits means 240 as fourth octet (from table in my head)
256-240=16 (magic number)
becouse it is network address : 208+16-1=223 finish.

You dont need to count nothing else, becouse you wasn't asked to. anyway you have now enought info to do so quickly.
so for example another subnets: just +-16 and you have them.

binarysoul

Your network can use class B network 130.1.0.0. What subnet masks meet the requirments that you plan to allow at most 50 subnets , with at most 200 hosts per subnet?

This is an example I'm reading in the book, it doesn't give a clear answer. I believe the answer should be 255.255.252.0. What do you think?

Danman32

Well the closest power of 2 that 50 will fit is 64. That's 6 bits.
A six bit subnet mask would be 252.

so 255.255.255.0 would work. But there's quite a lot of leeway in the criteria, since you only need 8 bits to accomodate the 200 hosts, which means the subnet can go as far as 8 bits.

binarysoul

How much calculation am supposed to know with respect to calculating valid # of hosts and subnets, e.g. 2 to the power 21?

I can manage to do upto 2 to the power of 16 in my head, but not beyond

. Of course I can use the board/marker in the exam room to get beyond that, but I want to know the extent of calculations on CCNA.

-prophet-

Hey binarysoul, check out what I wrote down before the exam started. The powers of 2 aren't a problem if you just make a note before you start.

http://www.techexams.net/forums/viewtopic.php?t=16380

Cheers,
Prophet

Danman32

You won't have to go much farther than 2 octets, so 2 ^ 16 should do you fine. It's great you can do that in your head.

For those of us that had been around in the PC/XP days, you might remember some of those powers beyond 8 bits with respect to memory amounts. Remember 8K was 8192 bytes? how about 16K? That was 16,384.
I remember 65536 and 32768 as memory map limits with the 6502 procesor used in the Atari computers such as the 800. Man I miss that machine.

binarysoul

Thanks everybody for all the help. Here's a bit of a method I use to quickly do the math for calculting hosts and subnets for upto 2 to the power of 16 or 2^16.

2^ 5 32
2^ 6 64
2^ 7 128
2^8 256
2^ 9 512
2^10 1024
2^11 2048
2^12 4096
2^ 13 8192
2^ 14 16384
2 ^ 15 32768
2^ 16 65536

**********The trick is to focus on the last two digits. **************

For example, if you don't have the time to calculate 8192, just remember:

2^13= 2^8 * 2^5 (so 2^(5+

, or 2^13, elementry Algebra)

Now you know
2^8=256
2^5=32

When u multiply the last digits (2*6), you get 12. Now the last digit being 2, corresponds to last digit in 8192.

There is a bit of thinking involved and it's not a perfect science, but just a way to make calculation easy. For example, the last digits of 32 and 512 is also 2, but you should at least think that 2^13 can't be that a small number

What about 512? Well, that's 2^4 * 2^5 or 16*32. Multiply the last digit (2*6, you get 12 which is the last digits of 512.

What about 65,536? Well, that's 256*256, i.e (2^8 * 2^

. Again, multiply the last digits (6*6), you get 36 which matches the last two digits in 65536.

What do you think?

binarysoul

Sorry folks, it regarded 8 and ) as a smiley face. Here it's again

2^ 5 32
2^ 6 64
2^ 7 128
2^8 256
2^ 9 512
2^10 1024
2^11 2048
2^12 4096
2^ 13 8192
2^ 14 16384
2 ^ 15 32768
2^ 16 65536

**********The trick is to focus on the last two digits. **************

For example, if you don't have the time to calculate 8192, just remember:

2^13= 2^8 * 2^5 (so 2^(5+8 ), or 2^13, elementry Algebra)

Now you know
2^8=256
2^5=32

When u multiply the last digits (2*6), you get 12. Now the last digit being 2, corresponds to last digit in 8192.

There is a bit of thinking involved and it's not a perfect science, but just a way to make calculation easy. For example, the last digits of 32 and 512 is also 2, but you should at least think that 2^13 can't be that a small number.

What about 512? Well, that's 2^4 * 2^5 or 16*32. Multiply the last digit (2*6, you get 12 which is the last digits of 512.

What about 65,536? Well, that's 256*256, i.e (2^8 * 2^8 ). Again, multiply the last digits (6*6), you get 36 which matches the last two digits in 65536.

What do you think?

zom

ed_the_lad wrote:

If this question is on the exam it will be phrased like so.

If mask 255.255.255.128 were used with a Class B network with subnet-zero disabled, how many subnets could exist, with how many hosts per subnet, respectively?

a. 256 and 256
b. 254 and 254
c. 510 and 126
d. 512 and 128

Anyway with subnet zero enabled it would still be C, i'm sure you meant
d to be 512 and 126 as you will always lose 2 addresses from the host side due to subnet and broadcast address.

may i know how to do the subnet from the back like this question above? i just couldn't get it understand....sry for the noob question

binarysoul

First, you need to ignore the first two octets as it's a class B address or all 1's. Now, you're left with 16 bits, al zeros. Correct?

Network.network.host.host

Now, of that 16 bits, 9 are borrowed for subnetting e.g. 255.128.

So 2^9 is 512. You subtruct 2 and it's 510 subnets

You have 7 bits left for hosts which is 128, subtruct 2, and you have 126.

PS. 2^9 is same as 2^5 * 2^4, or 32*16.