Subnetting

what is the easyest way and quick to workout the broadcast of an IP Address
here is example

Given the following IP address 10.2.3.4/24, what is the broadcast address on this network?
A 10.2.3.256
B 10.2.3.198
C 10.2.3.0
D 10.2.3.255

here the correct answer is D but it tookme while to workout does anyone know quick way?

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Comments

malcybood

faisal79 wrote:

what is the easyest way and quick to workout the broadcast of an IP Address
here is example

Given the following IP address 10.2.3.4/24, what is the broadcast address on this network?
A 10.2.3.256
B 10.2.3.198
C 10.2.3.0
D 10.2.3.255

here the correct answer is D but it tookme while to workout does anyone know quick way?

My warped way would be to do the following

/24 = 3 octets (8 bits each) so the network address is 10.2.3.0 the broadcast is 10.2.3.255

Do you have the sybex book? It explains it well in there

markzab

Yeah, that one came to me before I even finished the question.

Check out the sybex book ch3 or go into the FAQ. There are a lot of great threads in the FAQ explaining subnetting.

JBeam

I looked at the last octet of the IP for a 255 and got D in about 5 secs. I know that the broadcast is always .255 so thats how I came up with it, If there was another .255 option then you would have to spend a little more time on it.

jason2713

JBeam wrote:

I looked at the last octet of the IP for a 255 and got D in about 5 secs. I know that the broadcast is always .255 so thats how I came up with it, If there was another .255 option then you would have to spend a little more time on it.

255 is not always the broad cast address, be careful with that, because they will get you with that on the exam. make sure you know how to calculate first, last, broadcast, and network addresses.

Darthn3ss

well... 255 will always be a broad cast... but yeah, i don't think thats a safe assumption ^^

the 24 bit mas is kind of what gives the answer away though.

jason2713

Darthn3ss wrote:

well... 255 will always be a broad cast... but yeah, i don't think thats a safe assumption ^^

the 24 bit mas is kind of what gives the answer away though.

255 will not always be the broadcast, sorry to say, with an 18 bit subnet i believe 255 is an actual host address. I just remember doing a problem on a practice test asking which were valid broadcast addresses.

One of the choices was 172.31.128.255 /18 (which is a host address) another choice was 192.168.24.50 /30 (which ended up being a broad cast address)

don't get caught with that.

remyforbes777

Jason is right. 255 is not always the broadcast address at all. If you have 192.168.1.5 /28 and was asked what the broadcast address was for this it would not be .255 it would be .7. The broadcast depends on what the subnet mask is.

markzab

jason2713 wrote:

Darthn3ss wrote:

well... 255 will always be a broad cast... but yeah, i don't think thats a safe assumption ^^

the 24 bit mas is kind of what gives the answer away though.

255 will not always be the broadcast, sorry to say, with an 18 bit subnet i believe 255 is an actual host address. I just remember doing a problem on a practice test asking which were valid broadcast addresses.

One of the choices was 172.31.128.255 /18 (which is a host address) another choice was 192.168.24.50 /30 (which ended up being a broad cast address)

don't get caught with that.

Your kind of right on this but you didn't go far enough as well. In your example .255 WILL be the broadcast address, but the 3rd and 4th octet will be 191.255. So simply saying with an 18 bit mask that .255 wont be the broadcast isn't correct technically...because as you can see, .255 IS the broadcast.

Know what I'm saying?

Darthn3ss

you mis read what i said, that or i mis-typed that. i meant that .255 will always be a broadcast address, as in not for a host or network, not that it is always THE broadcast for a network...

jason2713

the way i do it, and maybe someone else has a better way is the following:

so you have a 28 bit SN mask, I'll put the last octet in decimal form and subtract 256 from it. so in this case 28 bits will be a 240 mask in the last octet. Subtract 256 and you got 16.

Then I just start from 0 and work my way up past my choices, start at 0 + 16 + 16 + 16 those will be your network numbers. your broadcast address for each network will be 1 less than the networks, the last will be 1 less than the broadcast, and finally the first will be 1 more than the network:

First Last Broadcast Network
1 14 15 0
17 30 31 16
33 46 47 32
49 62 63 48
65 78 79 64
81 94 95 80

and so on. my trick is get the network, then the broadcast, then first and last. Usually makes it easy for me. anyone do this different?

CCNA_2007

I'm a beginner but here's my own way of understanding it so far, and I could be off, so I'm hoping for feedback:

Whenever the subnet/host division happens within an octet (divides an octet anywhere within the octet), all octets to the right of the split octet are 255 for the broadcast address.

For example, the split between subnetting and host cannot ever happen in the first octet, so the only possibilities left are the 2nd, 3rd and 4th octets.

So, if the split happens in the 2nd octet, then the broadcast address will contain 255 in both the 3rd and 4th octets.

If the split happens in the 3rd octet, then the broadcast address will contain 255 only in the 4th octet.

And if the split happens in the 4th octet, then the broadcast address must always be less than 255.

Can you more experienced folks tell me if I'm on the right track with this. There are so many ways to look at this and I'm trying to find a ways that are true, thanks.

CCNA_2007

Apologies I can see that my own explanation above is *wrong*. Hope I didn't confuse anyone! The split octet can sometimes contain 255 as part of the broadcast address.