| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
I get the following information from VPA and Monitor (?)
+-------------------------------------------------------------------+
| The following table gives a summary of the average amount of lock |
| traffic per second in the cluster. |
+-------------------------------------------------------------------+
Local Incoming Outgoing
Node ENQ/CVT/DEQ ENQ/CVT/DEQ ENQ/CVT/DEQ
-------- ----------- ----------- -----------
MELODY 3/ 3/ 3 2/ 3/ 2 1/ 2/ 1
PISCES 7/ 8/ 7 16/ 15/ 16 0/ 1/ 0
GUMDRP 20/ 33/ 20 0/ 0/ 0 15/ 11/ 15
FORNOW 0/ 0/ 0 0/ 0/ 0 0/ 1/ 0
I once tried to find out the actual traffic from any node to other nodes.
For instance, Melody has two incoming ENQs and 1 outgoing ENQs. Where are
they coming from and where is it going to? I tried to generalize this for
N nodes. I did not spend too much time on it. I gave up thinking that not
enough information is available to solve the problem. Did any one solve this
problem?
Swamy
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 840.1 | but I digress | ZFC::DERAMO | I voted on Super Tuesday! | Tue Mar 08 1988 22:53 | 9 |
That reminds me of a [a) similar b) totally unrelated] problem.
The NFL standings were in the newspaper two weeks into the
season, with the won-loss and points for-against totals for
each team. I also had a schedule so I knew who had played whom.
I never did figure out if this was enough information to derive
the score of each game.
Dan
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| 840.2 | CLT::GILBERT | Wed Mar 09 1988 12:30 | 8 | ||
Suppose we have 3 nodes: A, B, and C, with i arcs between A and B,
j arcs between A and C, and k arcs between B and C. We are given
i+j (arity of A), i+k (arity of B), and j+k (arity of C), and want
to determine i, j, and k. That gives 3 equations in 3 unknowns.
Suppose we have 4 nodes. We get 4 equations in 6 unknowns.
Suppose we have n nodes. We get n equations in n(n-1)/2 unknowns.
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| 840.3 | ZFC::DERAMO | Think of it as evolution in action. | Wed Mar 09 1988 12:59 | 4 | |
But the fact that the solutions are limited to nonnegative integers
gives more constraints than just "n equations in n(n-1)/2 unknowns."
So maybe some n=4 cases can be solved, too. (-:
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| 840.4 | TLE::BRETT | Wed Mar 09 1988 16:39 | 17 | ||
Consider the trivial case,
outgoing incoming
A C
B D
Now, if A and B each have one out-going link, and C and D each have
two incoming links, there is (obviously) no way of determining
whether (AC, BD) or (AD, BC) was the actual pattern.
Since you can't solve the simple case...
/Bevin
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