Do classful networks use broadcast?

I'm watching one of Andre's lectures and when he moves on from talking about classful networks to classless (6 min-isch) he mentaions you have to add +2 to the host count for broadcast & network, which makes sense.

However, it got me thinking; did the older style classful network NOT need a broadcast and network identifier? It's probably a stupid question, just can't wrap my head around it, but I want to know in order to make sense of things.

Thanks,

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rob42

No such thing as a stupid/hard/easy question; just questions to which you know the answer, and questions to which you don't...

Broadcast IP Addresses are always the last available IP Address associated with any given IP Network. This is because all the 'host bits' of the Address are set (= '1') for a 'broadcast' (meaning all the hosts on the Network will receive the IP Packet/s), regardless of classful or classless addressing. The Network ID is the IP Address with none of the 'host bits' set.

Two to the power of the number of 'host bits' - 2 will tell you how many hosts can be addressed.

You'll mostly see IP Addresses written in CIDR notation (classless) rather than and IP Address and a Mask.

So, 192.168.1.20/24 is the same as 192.168.1.20 255.255.255.0: both have 24 of the 32 bits used for the Network, leaving 8 bits for the hosts.

2^8 = 256 - 2 = 254; the number of hosts that can be addressed by the IP Protocol.

ccie14023

It's not a stupid question, it's a stupid to make people study classful addressing when it hasn't been used for eons. (A bit of a pet peeve of mine, I'm afraid.) They took ISDN and Token Ring off these exams, I'm not sure why you still have to study address classes. In addition to the answer above, I would give you the short answer of "yes". In other words, if you had a class C network, you did indeed have a broadcast address. However, given the size of these, class A and B networks were always subnetted, and class C often were, so the subnets would have their own broadcast address. Of course, the one exception to this would be class D addresses, which represent multicast groups and are arguably the only reason to mention classes any more.

rob42

ccie14023 wrote: »

...I'm not sure why you still have to study address classes.

I had this same debate with you in this post [ http://www.techexams.net/forums/ccna-wireless/123393-class-subnetting-question-newbie-one.html#post1060639 ] and I still don't know how you'd work out how many Subnets a network could address, without knowing about Classful Addressing.

ImYourOnlyDJ

Sounds like you have tunnel vision and aren't seeing the bigger picture (very common when studying CCNA). I'd recommend watching some videos on how ARP works and that should clarify it for you. Same for the rest of your studies if you don't get it at first try taking a step back and look at the bigger picture.

ccie14023

rob42 wrote: »

I had this same debate with you in this post [ http://www.techexams.net/forums/ccna-wireless/123393-class-subnetting-question-newbie-one.html#post1060639 ] and I still don't know how you'd work out how many Subnets a network could address, without knowing about Classful Addressing.

I know, I know, I'm like a broken record on this one. But I'm not sure I understand your question. Classful addresses have not been used since 1993. With the exception of class D they just don't exist any more. How do you work out how many subnets are in a network? You don't. Everything is basically a subnet.

Th3Ph3n0m

well,thats because you can tell whether its classful or classless by just looking at a givn ip address.

I mean,Class A -> IP Range 1-126 & default mask is 255.0.0.0 (or /

Class B -> IP Range 128-191 & default mask is 255.255.0.0 (or /16)
Class C -> IP Range 192 - 223 & default mask is 255.255.255.0 ( or /24)

Now, if you see an IP address, say, 10.1.1.2 and is using a mask other than the default /8, such as 10.1.1.2/24, it will be considered a classeless address because it's not using its default mask of /8.

Th3Ph3n0m

that was the reply to rob42 query why you must study classful addresses to subnet classless addresses...hope it clarifies.

networker050184

rob42 wrote: »

I had this same debate with you in this post [ http://www.techexams.net/forums/ccna-wireless/123393-class-subnetting-question-newbie-one.html#post1060639 ] and I still don't know how you'd work out how many Subnets a network could address, without knowing about Classful Addressing.

Yeah... That's something you'll need to know for the test only. You're never going to need to use that in the real world.

WastedHat

When people say classful addressing isnt necessary anymore do they mean the first octet address ranges? Are the /8 /16 /24 boundries still used in CIDR to seperate the network and subnetwork bits?

rob42

WastedHat wrote: »

When people say classful addressing isnt necessary anymore do they mean the first octet address ranges? Are the /8 /16 /24 boundries still used in CIDR to seperate the network and subnetwork bits?

/8 /16 /24 are the 'natural' boundaries for Class A, Class B and Class C Addresses and you'll see them in CIDR notation. The 'natural' boundaries are exactly one Octet apart.

The boundaries between the Network Bits and the Host Bits don't have to be one Octet apart, which is were Subnets and Supernets come in: you use bits from adjoining Octets by way of the Mask, to address the Network and Hosts.

E.g:

IP 63.24.89.21/18

IP - 63.24.89.21 in Binary ...........|0 0 1 1 1 1 1 1|0 0 0 1 1 0 0 0|0 1 0 1 1 0 0 1|0 0 0 1 0 1 0 1
Mask - 255.255.192.0 in Binary........|1 1 1 1 1 1 1 1|1 1 1 1 1 1 1 1|1 1 0 0 0 0 0 0|0 0 0 0 0 0 0 0
Logical AND...........................|0 0 1 1 1 1 1 1|0 0 0 1 1 0 0 0|0 1 0 0 0 0 0 0|0 0 0 0 0 0 0 0

With this example, the Network is being addressed using 18 bits: 8 from the 1st and 2nd Octets, and 2 from the 3rd Octet. The hosts are being address by 6 bits from the 3rd Octet and 8 bits from the 4th Octet.

Edit to add:
To work out the number of Subnets that can be addressed using pen and paper, you need to know where the 'natural' boundaries are (Classful Addressing), unless someone can tell me how you'd do this without knowing that.

Th3Ph3n0m

you wanna bring it rob??...ya common man bring it on...

Afaque

Hi rob42,i am new to networking, ,plz can you take your busy time to explain how you got binary numbers from dotted decimal ip addresses?..will be very useful to me and newbies.

rob42

Th3Ph3n0m wrote: »

you wanna bring it rob??...ya common man bring it on...

Excuse me, what's this?

rob42

Afaque wrote: »

Hi rob42,i am new to networking, ,plz can you take your busy time to explain how you got binary numbers from dotted decimal ip addresses?..will be very useful to me and newbies.

No problem, it's quite simple really.

Set out the decimal values that each binary position represents, like this...

128
64
32
16
8
4
2
1

Now, all you have to do is to put a one in the most significant place (left to right, most significant to least significant) where the position value is less than the decimal number you have, then take that value from your number and repeat the process. It sounds more complicated than it is, so as an example; dec 21

128
64
32
16
8
4
2
1

0
0
0
1

16 is the most significant place. 21 - 16 = 5. Repeat the process.

128
64
32
16
8
4
2
1

0
0
0
1
0
1

4 is the most significant place. 5 - 4 = 1. Repeat the process.

128
64
32
16
8
4
2
1

0
0
0
1
0
1
0
1

Done.

Now, 89.

128
64
32
16
8
4
2
1

0
1

89 - 64 = 25

128
64
32
16
8
4
2
1

0
1
0
1

25 - 16 = 9

128
64
32
16
8
4
2
1

0
1
0
1
1

9 - 8 = 1

128
64
32
16
8
4
2
1

0
1
0
1
1
0
0
1

Done.

You can also work backwards from 255. So, 255 is…

128
64
32
16
8
4
2
1

1
1
1
1
1
1
1
1

To work out say, 240. 255 - 240 = 15 (8+4+2+1)

128
64
32
16
8
4
2
1

1
1
1
1
0
0
0
0

With practice, you get to ‘know’ what the binary looks like, i.e I know that 192 is 11000000

Also, you should be able to ‘see’ that 63 is 00111111, because 64 would be 01000000.

Any help?

Afaque

Awesome rob42,believe me when my teacher was explaining this half of the class were scratching their heads,thanks for explaining it so crystal clearly.

Afaque

but when doing from backwards,i am getting correct values when i replace '1" with '0' in the most significant position.....why is that?

rob42

Sorry, maybe I confused you. All I meant was that it's sometimes quicker to work back from 255 down to whatever number you're starting from.

With 223, for example, it's quicker to take 223 away from 255 (32) and know that the answer will be to set all the bits other then the '32' position: 1 1 0 1 1 1 1 1, rather than going through the process that I've outlined.

Show me the example you're stuck on...