This uppaal model showing in the image is part of a top system.
It's used to check the condition var==1 hold for at least 10 time units. The range of the integer variable var is [0, 20000].
I use query E<>condition.hold to get the trace, but can't get result within minutes. If I change the range of var to [0, 1000], uppaal return result within seconds.
The question is:
Do I use uppaal in a right way?
Is uppaal suitable for this kind of model-checking? or any other options?
Thanks for any help
Due to the problem domain that we deal with, the model need to support integer and float data type. And now I understood that Uppaal treat clock as symbolic representation but other variables as concrete value, so maybe it's not possible to meet our requirement. And I am going to try nuxmv, which seems OK because SMT solver is used. I'm not sure whether nuxmv is suitable for timed automata modeling. Have a try.
Related
Suppose we're given some sort of graph where the feasible region of our optimization problem is given. For example: here is an image
How would I go on about constructing these constraints in an integer optimization problem? Anyone got any tips? Thanks!
Mate, I agree with the others that you should be a little more specific than that paint-ish picture ;). In particular you are neither specifying any objective/objective direction nor are you giving any context, what about this graph should be integer-variable related, except for the existence of disjunctive feasible sets, which may be modeled by MIP-techniques. It seems like your problem is formalization of what you conceptualized. However, in case you are just being lazy and are just interested in modelling disjunctive regions, you should be looking into disjunctive programming techniques, such as "big-M" (Note: big-M reformulations can be problematic). You should be aiming at some convex-hull reformulation if you can attain one (fairly easily).
Back to your picture, it is quite clear that you have a problem in two real dimensions (let's say in R^2), where the constraints bounding the feasible set are linear (the lines making up the feasible polygons).
So you know that you have two dimensions and need two real continuous variables, say x[1] and x[2], to formulate each of your linear constraints (a[i,1]*x[1]+a[i,2]<=rhs[i] for some index i corresponding to the number of lines in your graph). Additionally your variables seem to be constrained to the first orthant so x[1]>=0 and x[2]>=0 should hold. Now, to add disjunctions you want some constraints that only hold when a certain condition is true. Therefore, you can add two binary decision variables, say y[1],y[2] and an additional constraint y[1]+y[2]=1, to tell that only one set of constraints can be active at the same time. You should be able to implement this with the help of big-M by reformulating the constraints as follows:
If you bound things from above with your line:
a[i,1]*x[1]+a[i,2]-rhs[i]<=M*(1-y[1]) if i corresponds to the one polygon,
a[i,1]*x[1]+a[i,2]-rhs[i]<=M*(1-y[2]) if i corresponds to the other polygon,
and if your line bounds things from below:
-M*(1-y[1])<=-a[i,1]*x[1]-a[i,2]+rhs[i] if i corresponds to the one polygon,
-M*(1-y[1])<=-a[i,1]*x[1]-a[i,2]+rhs[i] if i corresponds to the other polygon.
It is important that M is sufficiently large, but not too large to cause numerical issues.
That being said, I am by no means an expert on these disjunctive programming techniques, so feel free to chime in, add corrections or make things clearer.
Also, a more elaborate question typically yields more elaborate and satisfying answers ;) If you had gone to the effort of making up a true small example problem you likely would have gotten a full formulation of your problem or even an executable piece of code in no time.
I'm taking a course on coursera that uses minizinc. In one of the assignments, I was spinning my wheels forever because my model was not performing well enough on a hidden test case. I finally solved it by changing the following types of accesses in my model
from
constraint sum(neg1,neg2 in party where neg1 < neg2)(joint[neg1,neg2]) >= m;
to
constraint sum(i,j in 1..u where i < j)(joint[party[i],party[j]]) >= m;
I dont know what I'm missing, but why would these two perform any differently from eachother? It seems like they should perform similarly with the former being maybe slightly faster, but the performance difference was dramatic. I'm guessing there is some sort of optimization that the former misses out on? Or, am I really missing something and do those lines actually result in different behavior? My intention is to sum the strength of every element in raid.
Misc. Details:
party is an array of enum vars
party's index set is 1..real_u
every element in party should be unique except for a dummy variable.
solver was Gecode
verification of my model was done on a coursera server so I don't know what optimization level their compiler used.
edit: Since minizinc(mz) is a declarative language, I'm realizing that "array accesses" in mz don't necessarily have a direct corollary in an imperative language. However, to me, these two lines mean the same thing semantically. So I guess my question is more "Why are the above lines different semantically in mz?"
edit2: I had to change the example in question, I was toting the line of violating coursera's honor code.
The difference stems from the way in which the where-clause "a < b" is evaluated. When "a" and "b" are parameters, then the compiler can already exclude the irrelevant parts of the sum during compilation. If "a" or "b" is a variable, then this can usually not be decided during compile time and the solver will receive a more complex constraint.
In this case the solver would have gotten a sum over "array[int] of var opt int", meaning that some variables in an array might not actually be present. For most solvers this is rewritten to a sum where every variable is multiplied by a boolean variable, which is true iff the variable is present. You can understand how this is less efficient than an normal sum without multiplications.
I have an optimization problem where I want to minimize for the total cost of a system, so I write an objective function that is the sum of my different costs. The problem includes using one of three machines each one with different cost at a different threshold of usage. I define each machine (model.Machine#) as a binary variable and declare the parameters of each machine cost model.Cost#). I am trying to get the cost to be able to minimize it but when I write the constraint:
model.Cost1*model.Machine1 + model.Cost2*model.Machine2 + model.Cost3*model.Machine3 == model.MachineCost
Where I also write:
model.Machine1 + model.Machine2 + model.Machine3 == 1
Gurobi is telling me that it can't handle an quadratic function referring to the first constraint mentioned above. However it is parameters multiplied by binary variables there isn't anything quadratic.
I know the question is vague and part of a larger problem but I hope you can understand what I am referring to and help me!
Thank you so much for your assistance!
What is model.MachineCost? Is it an Expression component with some kind of quadratic expression stored inside of it?
If not, can you start commenting out things in your model until you get down to a minimal working example (that causes this error) and post that? Otherwise, we can't be sure that there are not other quadratic pieces of the model that you are not showing.
Is there a function in Julia that is similar to the solver function in Excel where I can provide and equation, and it will solve for the unknown variable? If not, does anybody know the math behind Excel's solver function?
I am not expecting anybody to solve the equation, but if it helps:
Price = (Earnings_1/(1+r)^1)+(Earnings_2/(1+r)^2)++(Earnings_3/(1+r)^3)+(Earnings_4/(1+r)^4)+(Earnings_5/(1+r)^5)+(((Earnings_5)(RiskFreeRate))/((1+r)^5)(1-RiskFreeRate))
The known variables are: Price, All Earnings, and RiskFreeRate. I am just trying to figure out how to solve for r.
Write this instead as an expression f(r) = 0 by subtracting Price over to the other side. Now it's a rootfinding problem. If you only have one variable you're solving for (looks to be the case), then Roots.jl is a good choice.
fzero(f, a::Real, b::Real)
will search for a solution between a and b for example, and the docs have more choices for algorithms when you don't know a range to start with and only give an initial condition for example.
In addition, KINSOL in Sundials.jl is good when you know you're starting close to a multidimensional root. For multidimensional and needing some robustness to the initial condition, I'd recommend using NLsolve.jl.
There's nothing out of the box no. Root finding is a science in itself.
Luckily for you, your function has an analytic first derivative with respect to r. That means that you can use Newton Raphson, which will be extremely stable for your function.
I'm sure you're aware your function behaves badly around r = -1.
I am receiving as input a "map" represented by strings, where certain nodes of the map have significance (s). For example:
---s--
--s---
s---s-
s---s-
-----s
My question is, what reasonable options are there for representing this input as an object.
The only option that really comes to mind is:
(1) Each position translated to node with up,down,left,right pointers. The whole object contains a pointer to top right node.
This seems like just a graph representation specific to this problem.
Thanks for the help.
Additionally, if there are common terms for this type of input, please let me know
Well, it depends a lot on what you need to delegate to those objects. OOP is basically about asking objects to perform things in order to solve a given problem, so it is hard to tell without knowing what you need to accomplish.
The solution you mention can be a valid one, as can also be having a matrix (in this case of 6x5) where you store in each matrix cell an object representing the node (just as an example, I used both approaches once to model the Conway's game of life). If you could give some more information on what you need to do with the object representation of your map then a better design can be discussed.
HTH