I have an optimization problem that I'm working on in which the optimizer's decision for one variable needs to constrain the available choices for another variable, and I'm wondering about the suitability of different ways of accomplishing this. I'll try to pare it down for demonstration purposes (in kotlin).
#PlanningEntity
class AuctionBid {
#PlanningId
lateinit var entity: String
#PlanningVariable(valueRangeProviderRefs = ["availOptions"])
var bidNow: Boolean? = null
#PlanningVariable(valueRangeProviderRefs = ["availOptions"])
var bidFinal: Boolean? = null
var forceBid: Boolean? = null
#ValueRangeProvider(id = "availOptions")
private fun availOptions(): List<Boolean> {
return if(this.forceBid != null) {
listOf(this.forceBid!!)
} else {
listOf(true,false)
}
}
}
In this scenario, we are trying to determine if we should bid on a given entity in a combinatorial auction (multiple entities being bid on in each round), given a hard budget and various cost/benefit tradeoffs. The bidNow choice is determined by the prices in the current round of the auction, and the bidFinal choice is determined by the anticipated/forecasted final prices in the auction. In order to avoid constantly changing decisions about what to bid on right now (switching bids is penalized in the auction rules), we want to constrain our choices on bidNow based on our decision for bidFinal, such that if we will want to bid on it at our anticipated future price, we will always choose to bid on it now. A state table seen below:
bidNow
bidFinal
possible
Y
Y
Y
Y
N
Y
N
N
Y
N
Y
N
As seen, we already have a method of constraining choices for situations where we know we don't want the product at all, or where we know we want it at any price. This is using the availOptions ValueRangeProvider, which only lets the optimizer choose a predetermined option if we know that is our only choice.
However, this constraint is more dynamic, as it results from a separate choice. If we are wanting to bid at the anticipated final price, our only choice for right now is to continue bidding in the current round.
I've thought of a few ways of doing this, as this is a classic Local Consistency Constraint Propagation example, but without knowing too much about optaplanner's internals, I don't really know the best way to do this. Some options:
Use a separate ValueRangeProvider for bidNow than bidFinal, and constrain the value range for bidNow to True if bidFinal is true. However, I don't know if optaplanner can even handle a dynamic value range that is determined by another planning variable's value, nor do I know if this is an efficient method.
Place a hard constraint (using ConstraintStreams, which is super cool BTW) on bidNow to always be True if bidFinal is True. I know this is feasible (it's my current solution), but it seems really slow.
Maybe try to use a Shadow variable which I've never used before, although I'm not sure this is a good fit because bidFinal==True implies bidNow==True, but bidFinal=False doesn't necessarily imply anything about bidNow.
I know I can always just try all the different options and see, but this problem is already big and unwieldy (the demo code is dramatically simplified), and I really would like some insight into how Optaplanner is working behind the scenes because I really like working with it but come across these this-way-or-that-way scenarios all the time and it's difficult to try them all. Is there any insight into modeling this constraint propagation problem? Or problems like this?
Both 1 and 2 should be possible; 2 being the straightforward way.
Another option would be to use filtered move selection to skip the moves that would set bidFinal to Y if bidNow is false (and the opposite corner case). The downside of this is that the filter needs to be applied to every move you use, which makes solver configuration verbose. And the moves will still be generated before they are thrown away, so you lose a bit of performance.
However, why not make bid an enum with three options: NOTHING, NOW_ONLY, BOTH. That way, you have the benefits of only having one planning variable and completely eliminating the original problem. Of course this is only suitable when the number of options is this low, which it may not be in your actual use case.
This last option is an example of designing your data model in a way to avoid corner cases. If they are impossible because the model precludes them, you do not need to worry about working around them later. This is always your best bet.
Related
Can we use two planning variables with (nullable = true) for each of them?
If so, how can we deal with them in the Drools rule file?
I know that when we use one planning variable we define it with (nullable = true) and then in the rule we use $planningVariable != null as in the "pas" example, I tried this and it worked well, but what about using two planning variables?
Can we apply this on the curriculumCourse? and if so, the over constrained data should appear in the output as unassigned for the two planning variables or appear in only one of them?
Yes, of course you can. But as usual, you 'll have to make sure your score constraints (= score rules) penalize/reward what you want to achieve.
For example on CurriculumCourse, I presume you 'd have a negative medium constraint that penalizes a Lecture if either room or period is null. If both are null, don't penalize it more, or you'll end up with a lot semi assigned entities. But despite that, you'll still probably end up with a few semi assigned entities, so to fix that:
Either do some post-processing to make all those not assigned at all (= both vars null) as a semi-assignment is useless.
Or add a hard constraint against semi assignments to avoid them entirely (even in intermediate solution states).
Additional solving efficiently can be gained from:
A ChangeMove selector that moves both room and period, as changing just one to/from null will never yield a better solution.
I successfully amended the nice CloudBalancing example to include the fact that I may only have a limited number of computers open at any given time (thanx optaplanner team - easy to do). I believe this is referred to as a bounded-space problem. It works dandy.
The processes come in groupwise, say 20 processes in a given order per group. I would like to amend the example to have optaplanner also change the order of these groups (not the processes within one group). I have therefore added a class ProcessGroup in the domain with a member List<Process>, the instances of ProcessGroup being stored in a List<ProcessGroup>. The desired optimisation would shuffle the members of this List, causing the instances of ProcessGroup to be placed at different indices of the List List<ProcessGroup>. The index of ProcessGroup should be ProcessGroup.index.
The documentation states that "if in doubt, the planning entity is the many side of the many-to-one relationsship." This would mean that ProcessGroup is the planning entity, the member index being a planning variable, getting assigned to (hopefully) different integers. After every new assignment of indices, I would have to resort the list List<ProcessGroup in ascending order of ProcessGroup.index. This seems very odd and cumbersome. Any better ideas?
Thank you in advance!
Philip.
The current design has a few disadvantages:
It requires 2 (genuine) entity classes (each with 1 planning variable): probably increases search space (= longer to solve, more difficult to find a good or even feasible solution) + it increases configuration complexity. Don't use multiple genuine entity classes if you can avoid it reasonably.
That Integer variable of GroupProcess need to be all different and somehow sequential. That smelled like a chained planning variable (see docs about chained variables and Vehicle Routing example), in which case the entire problem could be represented as a simple VRP with just 1 variable, but does that really apply here?
Train of thought: there's something off in this model:
ProcessGroup has in Integer variable: What does that Integer represent? Shouldn't that Integer variable be on Process instead? Are you ordering Processes or ProcessGroups? If it should be on Process instead, then both Process's variables can be replaced by a chained variable (like VRP) which will be far more efficient.
ProcessGroup has a list of Processes, but that a problem property: which means it doesn't change during planning. I suspect that's correct for your use case, but do assert it.
If none of the reasoning above applies (which would surprise me) than the original model might be valid nonetheless :)
I have a model implemented in OPL. I want to use this model to implement a local search in java. I want to initialize solutions with some heuristics and give these initial solutions to cplex find a better solution based on the model, but also I want to limit the search to a specific neighborhood. Any idea about how to do it?
Also, how can I limit the range of all variables? And what's the best: implement these heuristics and local search in own opl or in java or even C++?
Thanks in advance!
Just to add some related observations:
Re Ram's point 3: We have had a lot of success with approach b. In particular it is simple to add constraints to fix the some of the variables to values from a known solution, and then re-solve for the rest of the variables in the problem. More generally, you can add constraints to limit the values to be similar to a previous solution, like:
var >= previousValue - 1
var <= previousValue + 2
This is no use for binary variables of course, but for general integer or continuous variables can work well. This approach can be generalised for collections of variables:
sum(i in indexSet) var[i] >= (sum(i in indexSet) value[i])) - 2
sum(i in indexSet) var[i] <= (sum(i in indexSet) value[i])) + 2
This can work well for sets of binary variables. For an array of 100 binary variables of which maybe 10 had the value 1, we would be looking for a solution where at least 8 have the value 1, but not more than 12. Another variant is to limit something like the Hamming distance (assume that the vars are all binary here):
dvar int changed[indexSet] in 0..1;
forall(i in indexSet)
if (previousValue[i] <= 0.5)
changed[i] == (var[i] >= 0.5) // was zero before
else
changed[i] == (var[i] <= 0.5) // was one before
sum(i in indexSet) changed[i] <= 2;
Here we would be saying that out of an array of e.g. 100 binary variables, only a maximum of two would be allowed to have a different value from the previous solution.
Of course you can combine these ideas. For example, add simple constraints to fix a large part of the problem to previous values, while leaving some other variables to be re-solved, and then add constraints on some of the remaining free variables to limit the new solution to be near to the previous one. You will notice of course that these schemes get more complex to implement and maintain as we try to be more clever.
To make the local search work well you will need to think carefully about how you construct your local neighbourhoods - too small and there will be too little opportunity to make the improvements you seek, while if they are too large they take too long to solve, so you don't get to make so many improvement steps.
A related point is that each neighbourhood needs to be reasonably internally connected. We have done some experiments where we fixed the values of maybe 99% of the variables in a model and solved for the remaining 1%. When the 1% was clustered together in the model (e.g. all the allocation variables for a subset of resources) we got good results, while in comparison we got nowhere by just choosing 1% of the variables at random from anywhere in the model.
An often overlooked idea is to invert these same limits on the model, as a way of forcing some changes into the solution to achieve a degree of diversification. So you could add a constraint to force a specific value to be different from a previous solution, or ensure that at least two out of an array of 100 binary variables have a different value from the previous solution. We have used this approach to get a sort-of tabu search with a hybrid matheuristic model.
Finally, we have mainly done this in C++ and C#, but it would work perfectly well from Java. Not tried it much from OPL, but it should be fine too. The key for us was being able to traverse the problem structure and use problem knowledge to choose the sets of variables we freeze or relax - we just found that easier and faster to code in a language like C#, but then the modelling stuff is more difficult to write and maintain. We are maybe a bit "old-school" and like to have detailed fine-grained control of what we are doing, and find we need to create many more arrays and index sets in OPL to achieve what we want, while we can achieve the same effect with more intelligent loops etc without creating so many data structures in a language like C#.
Those are several questions. So here are some pointers and suggestions:
In Cplex, you give your model an initial solution with the use of IloOplCplexVectors()
Here's a good example in IBM's documentation of how to alter CPLEX's solution.
Within OPL, you can do the same. You basically set a series of values for your variables, and hand those over to CPLEX. (See this example.)
Limiting the search to a specific neighborhood: There is no easy way to respond without knowing the details. But there are two ways that people do this:
a. change the objective to favor that 'neighborhood' and make other areas unattractive.
b. Add constraints that weed out other neighborhoods from the search space.
Regarding limiting the range of variables in OPL, you can do it directly:
dvar int supply in minQty..maxQty;
Or for a whole array of decision variables, you can do something along the lines of:
range CreditsAllowed = 3..12;
dvar int credits[student] in CreditsAllowed;
Hope this helps you move forward.
Question after BIG edition :
I need to built a ranking using genetic algorithm, I have data like this :
P(a>b)=0.9
P(b>c)=0.7
P(c>d)=0.8
P(b>d)=0.3
now, lets interpret a,b,c,d as names of football teams, and P(x>y) is probability that x wins with y. We want to build ranking of teams, we lack some observations P(a>d),P(a>c) are missing due to lack of matches between a vs d and a vs c.
Goal is to find ordering of team names, which the best describes current situation in that four team league.
If we have only 4 teams than solution is straightforward, first we compute probabilities for all 4!=24 orderings of four teams, while ignoring missing values we have :
P(abcd)=P(a>b)P(b>c)P(c>d)P(b>d)
P(abdc)=P(a>b)P(b>c)(1-P(c>d))P(b>d)
...
P(dcba)=(1-P(a>b))(1-P(b>c))(1-P(c>d))(1-P(b>d))
and we choose the ranking with highest probability. I don't want to use any other fitness function.
My question :
As numbers of permutations of n elements is n! calculation of probabilities for all
orderings is impossible for large n (my n is about 40). I want to use genetic algorithm for that problem.
Mutation operator is simple switching of places of two (or more) elements of ranking.
But how to make crossover of two orderings ?
Could P(abcd) be interpreted as cost function of path 'abcd' in assymetric TSP problem but cost of travelling from x to y is different than cost of travelling from y to x, P(x>y)=1-P(y<x) ? There are so many crossover operators for TSP problem, but I think I have to design my own crossover operator, because my problem is slightly different from TSP. Do you have any ideas for solution or frame for conceptual analysis ?
The easiest way, on conceptual and implementation level, is to use crossover operator which make exchange of suborderings between two solutions :
CrossOver(ABcD,AcDB) = AcBD
for random subset of elements (in this case 'a,b,d' in capital letters) we copy and paste first subordering - sequence of elements 'a,b,d' to second ordering.
Edition : asymetric TSP could be turned into symmetric TSP, but with forbidden suborderings, which make GA approach unsuitable.
It's definitely an interesting problem, and it seems most of the answers and comments have focused on the semantic aspects of the problem (i.e., the meaning of the fitness function, etc.).
I'll chip in some information about the syntactic elements -- how do you do crossover and/or mutation in ways that make sense. Obviously, as you noted with the parallel to the TSP, you have a permutation problem. So if you want to use a GA, the natural representation of candidate solutions is simply an ordered list of your points, careful to avoid repitition -- that is, a permutation.
TSP is one such permutation problem, and there are a number of crossover operators (e.g., Edge Assembly Crossover) that you can take from TSP algorithms and use directly. However, I think you'll have problems with that approach. Basically, the problem is this: in TSP, the important quality of solutions is adjacency. That is, abcd has the same fitness as cdab, because it's the same tour, just starting and ending at a different city. In your example, absolute position is much more important that this notion of relative position. abcd means in a sense that a is the best point -- it's important that it came first in the list.
The key thing you have to do to get an effective crossover operator is to account for what the properties are in the parents that make them good, and try to extract and combine exactly those properties. Nick Radcliffe called this "respectful recombination" (note that paper is quite old, and the theory is now understood a bit differently, but the principle is sound). Taking a TSP-designed operator and applying it to your problem will end up producing offspring that try to conserve irrelevant information from the parents.
You ideally need an operator that attempts to preserve absolute position in the string. The best one I know of offhand is known as Cycle Crossover (CX). I'm missing a good reference off the top of my head, but I can point you to some code where I implemented it as part of my graduate work. The basic idea of CX is fairly complicated to describe, and much easier to see in action. Take the following two points:
abcdefgh
cfhgedba
Pick a starting point in parent 1 at random. For simplicity, I'll just start at position 0 with the "a".
Now drop straight down into parent 2, and observe the value there (in this case, "c").
Now search for "c" in parent 1. We find it at position 2.
Now drop straight down again, and observe the "h" in parent 2, position 2.
Again, search for this "h" in parent 1, found at position 7.
Drop straight down and observe the "a" in parent 2.
At this point note that if we search for "a" in parent one, we reach a position where we've already been. Continuing past that will just cycle. In fact, we call the sequence of positions we visited (0, 2, 7) a "cycle". Note that we can simply exchange the values at these positions between the parents as a group and both parents will retain the permutation property, because we have the same three values at each position in the cycle for both parents, just in different orders.
Make the swap of the positions included in the cycle.
Note that this is only one cycle. You then repeat this process starting from a new (unvisited) position each time until all positions have been included in a cycle. After the one iteration described in the above steps, you get the following strings (where an "X" denotes a position in the cycle where the values were swapped between the parents.
cbhdefga
afcgedbh
X X X
Just keep finding and swapping cycles until you're done.
The code I linked from my github account is going to be tightly bound to my own metaheuristics framework, but I think it's a reasonably easy task to pull the basic algorithm out from the code and adapt it for your own system.
Note that you can potentially gain quite a lot from doing something more customized to your particular domain. I think something like CX will make a better black box algorithm than something based on a TSP operator, but black boxes are usually a last resort. Other people's suggestions might lead you to a better overall algorithm.
I've worked on a somewhat similar ranking problem and followed a technique similar to what I describe below. Does this work for you:
Assume the unknown value of an object diverges from your estimate via some distribution, say, the normal distribution. Interpret your ranking statements such as a > b, 0.9 as the statement "The value a lies at the 90% percentile of the distribution centered on b".
For every statement:
def realArrival = calculate a's location on a distribution centered on b
def arrivalGap = | realArrival - expectedArrival |
def fitness = Σ arrivalGap
Fitness function is MIN(fitness)
FWIW, my problem was actually a bin-packing problem, where the equivalent of your "rank" statements were user-provided rankings (1, 2, 3, etc.). So not quite TSP, but NP-Hard. OTOH, bin-packing has a pseudo-polynomial solution proportional to accepted error, which is what I eventually used. I'm not quite sure that would work with your probabilistic ranking statements.
What an interesting problem! If I understand it, what you're really asking is:
"Given a weighted, directed graph, with each edge-weight in the graph representing the probability that the arc is drawn in the correct direction, return the complete sequence of nodes with maximum probability of being a topological sort of the graph."
So if your graph has N edges, there are 2^N graphs of varying likelihood, with some orderings appearing in more than one graph.
I don't know if this will help (very brief Google searches did not enlighten me, but maybe you'll have more success with more perseverance) but my thoughts are that looking for "topological sort" in conjunction with any of "probabilistic", "random", "noise," or "error" (because the edge weights can be considered as a reliability factor) might be helpful.
I strongly question your assertion, in your example, that P(a>c) is not needed, though. You know your application space best, but it seems to me that specifying P(a>c) = 0.99 will give a different fitness for f(abc) than specifying P(a>c) = 0.01.
You might want to throw in "Bayesian" as well, since you might be able to start to infer values for (in your example) P(a>c) given your conditions and hypothetical solutions. The problem is, "topological sort" and "bayesian" is going to give you a whole bunch of hits related to markov chains and markov decision problems, which may or may not be helpful.
What I want to do is implement submission scoring for a site with users voting on the content, much like in e.g. reddit (see the 'hot' function in http://code.reddit.com/browser/sql/functions.sql). Edit: Ultimately I want to be able to retrieve an arbitrarily filtered list of arbitrary length of submissions ranked according to their score.
My submission model currently keeps track of up and down vote totals. Currently, when a user votes I create and save a related Vote object and then use F() expressions to update the Submission object's voting totals. The problem is that I want to update the score for the submission at the same time, but F() expressions are limited to only simple operations (it's missing support for log(), date_part(), sign() etc.)
From my limited experience with Django I can see 5 options here:
extend F() somehow (haven't looked at the code yet) to support the missing SQL functions; this is my preferred option and seems to fit within the Django framework the best
define a scoring function (much like reddit's 'hot' function) in my database, and have Django use the value of that function for the value of the score field; as far as I can tell, #2 is not possible
wrap my two step voting process in a suitably isolated transaction so that I can calculate the voting totals in Python and then update the Submission's voting totals without fear that another vote against the submission could be added/changed in the meantime; I'm hesitant to take this route because it seems overly complex - what is a "suitably isolated transaction" in this case anyway?
use raw SQL; I would prefer to avoid this entirely -- what's the point of an ORM if I have to revert to SQL for such a common use case as this! (Note that this coming from somebody who loves sprocs, but is using Django for ease of development.)
(edit: added this after further discussion) compute the score using an extra select parameter containing a call to my function; this would work but impose unnecessary load on the DB (would be forced to calculate the score for every submission ever made every time the query ran; caching could help here, but it still seems like a bit of lame workaround)
Before I embark on this mission to extend F() (which I'm not sure is even possible), am I about to reinvent the wheel? Is there a more standard way to do this? It seems like such a common use case and yet in an hour of searching I have yet to find a common solution...
EDIT: There is another option: set the default value of the field in the database script to be an expression containing my function. This is not as flexible as #1, but probably the quickest and cleanest approach to solving the problem (although my initial investigation into extending F() looks promising).
Why can't you just denormalize the score and reconstruct it with the Vote objects every once and a while?
If you can't do that, it is very easy to make a 'property' function that acts as an object attribute for scoring.
#property
def score(self):
... calculate score from Vote objects ...
return score
I've never used F() on a property like this, but it's Python, so I bet it works.
If you are using django-voting (which I recommend), you can put #3 in the manager's record_vote function since that's how all vote transactions take place.