I am trying to calculate a log with base 2 in JAGS, but can't find a way to implement this.
In the documentation I can't find a way to do this, and I am hoping I am missing something,
or that someone knows a workaround.
Thanks in advance for any help,
Benny
Log base 2 (or the binary logarithm) can be calculated with this trick here (link to wikipedia). As an example in R using the natural log:
log_2_result <- log(15, base = 2)
log_2_trick <- log(15) / log(2)
identical(log_2_result, log_2_trick)
[1] TRUE
JAGS has the log function, so you could use a similar approach to above (for log_2_trick). An important thing to note, however, is that because log is a link function in JAGS you can only input a scalar into it.
Related
I am using python's MIP module for optimization. I have set up a model with few parameters. I would want to limit the number of solutions and add stop timer if I don't find any solution in given time. I have added these parameters as given below:
m = Model(name='opt', sense=MAXIMIZE, solver_name=CBC)
m.optimize(max_solutions=1, max_seconds= 300)
somehow none of them seem working to me. it does not even stop looking for a solution after given time and it returns 2 solution sometimes even if I want to limit it to 1. Is there something I am missing in syntax?
One more thing, Gurobi has an option to add multiple variable using add_Vars parameter. Is there anything similar available in MIP too?
Thanks.
Yeah I have been doing some tests myself (with the Python-MIP solver) and seen some similar issues. Apparently it's still quite new and many improvements have been implemented recently or are yet to be developed. I will post from what I've learned:
regarding max_seconds: There's been at least one (closed) issue on the official repo related to using max_seconds parameter and CBC.
regarding max_solutions: If you are using version 1.6.2 or before, here's an explanation for that: until 1.6.1, the m.optimize(max_solutions=1) wasn't setting the maximum solution parameter to CBC. In that case you should try with the following lines (or just update to current version):
m.max_solutions = 1
m.optimize()
If the former don't help for the max_seconds and max_solutions parameters, I guess you'd better post your question as an issue at the library's repo to get answers and support from the project contributors.
Adding multiple variables, similar to gurobipy's Model.addVars() method: Yes, you can do it as follows
p = {(i, j): m.add_var(var_type=CONTINUOUS, lb=0, name="p[%d,%d]" % (i, j))
for i in Set_i for j in Set_j}
In this example, we are adding a variable p_ij and specifying it's continuous
(vs. binary or integer), has lower bound 0, and the sets where the indexes run. Set_i and Set_j are Python lists. See the documentation here for a more detailed explanation on how to use it. Similarly, you can create indexed constraints using the add_constr method, similar to Gurobi's Model.AddConstrs() method.
I have an implementation in Gurobi in python. My problem has different choices of selecting parameters to reach the optimal result. Now I need all solutions which reach the optimal value of result. How can I get them ? I know the blow code which just returns one solution.
if m.status == GRB.Status.OPTIMAL:
for v in m.getVars():
print (v.varname, v.x)
This can be achieved using the SolutionPool. You can specify how many solutions you want Gurobi to return.
I am trying to find the global maxima of a function and have found this package in j. However, after reading through the examples in demo I did not know how to use it yet. Anyone who can give an example about how to find the optimized parameters in a multi-parameter function, such as the function written below? I appreciate what you will do.
given that epsilon are data points given, omega and alpha have the range between 0 and 1.
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 working on a project, which is based on optix. I need to use progressive photon mapping, hence I am trying to use the Progressive Photon Mapping from the samples, but the transparency material is not implemented.
I've googled a lot and also tried to understand other samples that contains transparency material (e.g. Glass, Tutorial, whitted). At last, I got the solution as follows;
Find the hit point (intersection point) (h below)
Generate another ray from that point
use the color of the new generated points
By following you can also find the code of that part, by I do not understand why I get black color(.0f, .0f, 0.f) for the new generated ray (part 3 above).
optix::Ray ray( h, t, rtpass_ray_type, scene_epsilon );
HitPRD refr_prd;
refr_prd.ray_depth = hit_prd.ray_depth+1;
refr_prd.importance = importance;
rtTrace( top_object, ray, refr_prd );
result += (1.0f - reflection) * refraction_color * refr_prd.attenuation;
Any idea will be appreciated.
Please note that refr_prd.attenuation should contains some colors, after using function rtTrace(). I've mentioned reflection and reflaction_color to help you better understand the procedure. You can simply ignore them.
There are a number of methods to diagnose your problem.
Isolate the contribution of the refracted ray, by removing any contribution of the reflection ray.
Make sure you have a miss program. HitPRD::attenuation needs to be written to by all of your closest hit programs and your miss programs. If you suspect the miss program is being called set your miss color to something obviously bad ([1,0,1] is my favorite).
Use rtPrintf in combination with rtContextSetPrintLaunchIndex or setPrintLaunchIndex to print out the individual values of the product to see which term is zero from a given pixel. If you don't restrict the output to a given launch index you will get too much output. You probably also want to print out the depth as well.