optimization

问题 The function lmer in the lme4 package uses by default bobyqa from the minqa package as optimization algorithm. According to the following post https://stat.ethz.ch/pipermail/r-sig-mixed-models/2013q1/020075.html, it is possible to use also the other optimization algorirthms in the minqa package How can one use uobyqa or newuoa as optimization algorithm for lmer ? library(lme4) fm1 <- lmer(Reaction ~ Days + (Days | Subject), sleepstudy, control=lmerControl(optimizer="bobyqa")) 回答1: You can't

问题 The function lmer in the lme4 package uses by default bobyqa from the minqa package as optimization algorithm. According to the following post https://stat.ethz.ch/pipermail/r-sig-mixed-models/2013q1/020075.html, it is possible to use also the other optimization algorirthms in the minqa package How can one use uobyqa or newuoa as optimization algorithm for lmer ? library(lme4) fm1 <- lmer(Reaction ~ Days + (Days | Subject), sleepstudy, control=lmerControl(optimizer="bobyqa")) 回答1: You can't

问题 I have the next LP problem Maximize 1000 x1 + 500 x2 - 500 x5 - 250 x6 Subject To c1: x1 + x2 - x3 - x4 = 0 c2: - x3 + x5 = 0 c3: - x4 + x6 = 0 With these Bounds 0 <= x1 <= 10 0 <= x2 <= 15 0 <= x5 <= 15 0 <= x6 <= 5 By solving this problem with Cplex dual algorithm I get an optimal solution of 6250. But checking the reduced costs of the variables I get the next results Variable value reduced cost 1 10.0 500.0 1 0.0 -0.0 2 5.0 -0.0 3 5.0 -0.0 4 5.0 -0.0 5 5.0 250.0 Is it possible to have a

问题 I have the next LP problem Maximize 1000 x1 + 500 x2 - 500 x5 - 250 x6 Subject To c1: x1 + x2 - x3 - x4 = 0 c2: - x3 + x5 = 0 c3: - x4 + x6 = 0 With these Bounds 0 <= x1 <= 10 0 <= x2 <= 15 0 <= x5 <= 15 0 <= x6 <= 5 By solving this problem with Cplex dual algorithm I get an optimal solution of 6250. But checking the reduced costs of the variables I get the next results Variable value reduced cost 1 10.0 500.0 1 0.0 -0.0 2 5.0 -0.0 3 5.0 -0.0 4 5.0 -0.0 5 5.0 250.0 Is it possible to have a

问题 So, basically the problem is that there is a series of hill points. we can only move from one hill point to the next greater hill point. Please see the below image to get an idea. On above image black points are hill tops, each hill top has some spices with some taste value. we can move from one hill top to another tasting spices in them. Please See below image to know valid movements. green lines gives valid movements . So, Now we will be given a query with 2 integers b,c and for each query

问题 I need to check a short string for matches with a list of substrings. Currently, I do this like shown below (working code on ideone) bool ContainsMyWords(const std::wstring& input) { if (std::wstring::npos != input.find(L"white")) return true; if (std::wstring::npos != input.find(L"black")) return true; if (std::wstring::npos != input.find(L"green")) return true; // ... return false; } int main() { std::wstring input1 = L"any text goes here"; std::wstring input2 = L"any text goes here black";

问题 I've found some techniques in other SO answers, but apparently I've been unable to convince SBCL to do inline fixnum arithmetic: (declaim (optimize (speed 2) (safety 1))) (declaim (ftype (function (fixnum fixnum) double-float) fixnumtest) (inline fixnumtest)) (defun fixnumtest (i j) (declare (type fixnum i j)) (let* ((n (the fixnum (+ i j))) (n+1 (the fixnum (1+ n)))) (declare (type fixnum n n+1)) (/ 1.0d0 (the fixnum (* n n+1)) ) ) ) (defun main () (format t "~11,9F~%" (fixnumtest 2 3)) )

问题 I've found some techniques in other SO answers, but apparently I've been unable to convince SBCL to do inline fixnum arithmetic: (declaim (optimize (speed 2) (safety 1))) (declaim (ftype (function (fixnum fixnum) double-float) fixnumtest) (inline fixnumtest)) (defun fixnumtest (i j) (declare (type fixnum i j)) (let* ((n (the fixnum (+ i j))) (n+1 (the fixnum (1+ n)))) (declare (type fixnum n n+1)) (/ 1.0d0 (the fixnum (* n n+1)) ) ) ) (defun main () (format t "~11,9F~%" (fixnumtest 2 3)) )

问题 I have to modify a Dataset<Row> according to some rules that are in a List<Row> . I want to iterate over the Datset<Row> columns using Dataset.withColumn(...) as seen in the next example: (import necesary libraries...) SparkSession spark = SparkSession .builder() .appName("appname") .config("spark.some.config.option", "some-value") .getOrCreate(); Dataset<Row> dfToModify = spark.read().table("TableToModify"); List<Row> ListListWithInfo = new ArrayList<>(Arrays.asList()); ListWithInfo.add(0

问题 I'm using gekko for the first time to do optimization with python. I don't have a lot of experience with python, but I know the basics. I get error code -13 when I run the optimization: #import Gekko optimization package from gekko import gekko import math #create gekko model m = gekko() #constants pi = math.pi #initialize needed variables frictionLoss = m.Var() empirical = m.Var(value=1) widthInlet = m.Var(value=1) #in meters heightInlet = m.Var(value=1) #in meters diameterOutlet = m.Var