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08-16-2000 08:42 AM
08-16-2000 08:42 AM
Can someone please explain to me how HPUX determines a broadcast address? I am not an expert in networking, but I thought the last octet of a broadcast address was always ".255". What am I missing here?
Solved! Go to Solution.
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08-16-2000 08:47 AM
08-16-2000 08:47 AM
Re: broadcast address
http://www.docs.hp.com/dynaweb/hpux11/@Generic__CollectionView
http://docs.hp.com/hpux/communications/#doc
or HP_UX Sys Admin -part II - course material
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08-16-2000 08:47 AM
08-16-2000 08:47 AM
Re: broadcast address
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08-16-2000 08:53 AM
08-16-2000 08:53 AM
Re: broadcast address
depending on the "class" of your network, your broadcast is not necessarily .255
these days with CIDR, networks can be broken on boundries other than the traditional class A, B, or C
- a class C network is called a /24 network because the first 24 bits of the IP address forms the network address - in that case, the last 8 = 32 - 24 bits are set to all 1's or all 0's for the broadcast address (ie. 0 or 255). The netmask for this is 255.255.255.0
In otherwords, your netmask determines your broadcast address. Your broadcast is obtained by ORing the network (IP) address with (wo's compliment of your netmask - ie. 0.0.0.255 in the case of a class C/24.
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08-16-2000 09:33 AM
08-16-2000 09:33 AM
Re: broadcast address
Perhaps I should be more specific. If I assign the following:
IP address: 206.36.231.130
netmask: 255.255.255.224
Why do I get a broadcast address of:
206.36.231.159
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08-16-2000 09:47 AM
08-16-2000 09:47 AM
Solutionnetmask = ...224= 11100000 (binary)
compliment netmask = 00011111 (binary)
so:
IP Addr OR netmask compliment = 10000010
OR = 00011111
= 10011111 = 159
that is why your broadcast is 159.
You do not have a full class C network - you have only a /27 network which is a part of the class C (/24) network. In the scheme you have, there will be another network that starts from .160