Thanks. Much appreciated.

- thomas

On 2013-09-05 13:43, Daniel Le Berre wrote:
The penalty for root IUs (explicitely installed IUs) or IUs to install is 1, 
the other penalties is a power of K
(K for the most recent IU, K**2 for the second most recent IU, etc) where K is 
the number of installed IUs on the current system + 1.

The relevant par of the code in p2 is here (line 73):
http://git.eclipse.org/c/equinox/rt.equinox.p2.git/tree/bundles/org.eclipse.equinox.p2.director/src/org/eclipse/equinox/internal/p2/director/OptimizationFunction.java

Daniel

Le 5 sept. 2013 à 13:31, Thomas Hallgren <[email protected]> a écrit :

Hi Daniel,

Thanks for an elaborate explanation. It all makes sense. Next question is of 
course, how is the magnitude of the two penalties (higher number of IU's versus 
older versions) determined?

- thomas

On 2013-09-05 10:48, Daniel Le Berre wrote:
Hi Thomas,

The resolver tries to install as few IUs as possible, because each installed IU 
is associated with a penalty.

Furthermore, there is also another rule which will prefer most recent versions 
to older ones.
It means that the penalty is smaller for most recent versions than for older 
versions.

So it means that when you use the most recent IUs, solutions with fewer 
installed IUs will be preferred to solutions with a higher number of IUs.

But since at the end we aggregate all penalties, there might be some 
compensation effects:
a solution installing many new recent IUs may be preferred to a solution 
installing a few older IUs.

Daniel

Le 4 sept. 2013 à 19:23, Thomas Hallgren <[email protected]> a écrit :

Confronted with two OK solutions, what would the p2 resolver do if one solution 
was resolved by brining in a higher number of IUs? If it prefers the one with a 
lower IU count, what does it do if that one satisfies some requirements with 
lower versions?

- thomas
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          Daniel Le Berre     mailto:[email protected]
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         Daniel Le Berre     mailto:[email protected]
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         http://www.cril.univ-artois.fr/~leberre






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