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| Computers Forum Index » Computer - DCOM - Lans (Ethernet) » STP - bridge ID... |
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| Mark... |
 Posted: Wed Nov 04, 2009 6:15 am |
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Guest
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Hello
I'm reading the STP defined by 802.1d and can't understand one thing.
All bridges are is uniquely identified by 64-bit long IDs, that in fact is a
concatenation of a globally-unique 48-bit bridge MAC address and 16-bit
priority value. The MAC address is included to make the BridgeID unique,
it's clear.
What isn't clear to me is what happens when priority field values are the
same, how will the spanning protocol elect the root? MAC addresses aren't
necessary to be comparable in terms of arithmetic. Could anybody explain
this to me?
Thanks.
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Mark |
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| anoop... |
| Posted: Wed Nov 04, 2009 6:15 am |
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On Nov 3, 7:12 pm, "Mark" <mark_cruzNOTFORS... at (no spam) hotmail.com> wrote:
Quote: What isn't clear to me is what happens when priority field values are the
same, how will the spanning protocol elect the root? MAC addresses aren't
necessary to be comparable in terms of arithmetic. Could anybody explain
this to me?
MAC addresses are just 48-bit numbers. Why do you say that addresses
can't be compared? |
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| Chris Hills... |
| Posted: Wed Nov 04, 2009 6:11 pm |
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On 04/11/09 04:12, Mark wrote:
Quote: Hello
I'm reading the STP defined by 802.1d and can't understand one thing.
All bridges are is uniquely identified by 64-bit long IDs, that in fact
is a concatenation of a globally-unique 48-bit bridge MAC address and
16-bit priority value. The MAC address is included to make the BridgeID
unique, it's clear.
What isn't clear to me is what happens when priority field values are
the same, how will the spanning protocol elect the root? MAC addresses
aren't necessary to be comparable in terms of arithmetic. Could anybody
explain this to me?
If the path cost is equal, the bridge with the lowest ID is used. Logic
dictates that if the priority is equal then the bridge with the lowest
mac address is used.
For example, given two bridges:-
# Priority MAC
1 8192 01:ca:db:53:25:60
2 8192 04:53:91:4f:eb:f2
Bridge #1 will win the election.
The actual IDs are as follows ("/" is inserted between the priority and
mac for readability):-
0010000000000000/000000011100101011011011010100110010010101100000
0010000000000000/000001000101001110010001010011111110101111110010
Or in decimal:-
36029289712634200
36029986260646660 |
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| Rich Seifert... |
| Posted: Wed Nov 04, 2009 6:39 pm |
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On Nov 3, 7:12 pm, "Mark" <mark_cruzNOTFORS... at (no spam) hotmail.com> wrote:
Quote: Hello
I'm reading the STP defined by 802.1d and can't understand one thing.
All bridges are is uniquely identified by 64-bit long IDs, that in fact is a
concatenation of a globally-unique 48-bit bridge MAC address and 16-bit
priority value. The MAC address is included to make the BridgeID unique,
it's clear.
What isn't clear to me is what happens when priority field values are the
same, how will the spanning protocol elect the root? MAC addresses aren't
necessary to be comparable in terms of arithmetic. Could anybody explain
this to me?
You are correct that a MAC address is NOT a "number", i.e., it is a
unique bit pattern that ordinarily has no numerical significance.
However, for the purposes of electing a Root Bridge (and Designated
Bridges, in the event of equal path cost), the algorithm treats the
address as a number, and chooses the one with the numerically lowest
value.
(The phrase "Not a Number" always reminds me of that great 60s British
spy show, "The Prisoner". The imprisoned protagonist refuses to be
treated simply as "Number Six", lamenting that he is "not a number". I
once had a girlfriend so enamored of the show that she had a vanity
license plate "NOT A NBR".)
--
Rich Seifert Networks and Communications Consulting
21885 Bear Creek Way
(408) 395-5700 Los Gatos, CA 95033
(408) 228-0803 FAX
Send replies to: usenet at richseifert dot com |
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