具有等式和不等式约束的 R 优化

但是，有一个专门用于解决此类问题的软件包，即 Rsolnp.您可以通过以下方式使用它:#指定你的功能opt_func <- 函数(x){10 - 5*x[1] + 2 * x[2] - x[3]}#指定等式函数.数字 15(函数等于)#被指定为附加参数等于 <- 函数(x){x[1] + x[2] + x[3]}#the optimiser - 默认最小化solnp(c(5,5,5), #starting values (random - 显然需要是正数并且总和为15)opt_func, #优化函数eqfun=equal, #equality 函数eqB=15, #等式约束LB=c(0,0,0), #参数的下限，即大于零UB=c(100,100,100)) #参数上限(我只是随机选了100个)输出:>解决方案(c(5,5,5)，+ opt_func,+ eqfun=相等，+ eqB=15,+ LB=c(0,0,0),+ UB=c(100,100,100))Iter: 1 fn: -65.0000 Pars: 14.99999993134 0.00000002235 0.00000004632Iter: 2 fn: -65.0000 Pars: 14.999999973563 0.000000005745 0.000000020692解决方案-->2次迭代完成$pars[1] 1.500000e+01 5.745236e-09 2.069192e-08$收敛[1] 0$值[1] -10 -65 -65$拉格朗日[,1][1,] -5$黑森州[,1] [,2] [,3][1] 121313076 121313076 121313076[2] 121313076 121313076 121313076[3] 121313076 121313076 121313076$ineqx0空值$nfuneval[1] 126$outer.iter[1] 2$elapsed0.1770101秒的时间差$vscale[1] 6.5e+01 1.0e-08 1.0e+00 1.0e+00 1.0e+00所以得到的最优值是:$pars[1] 1.500000e+01 5.745236e-09 2.069192e-08这意味着第一个参数是 15，其余的零和零.这确实是您函数中的全局最小值，因为 x2 正在添加到函数中，并且 5 * x1 对结果的(负面)影响比 x3 大得多(负面).15, 0, 0的选择是函数根据约束的解和全局最小值.这个功能很好用！I am trying to find the local minimum of a function, and the parameters have a fixed sum. For example,Fx = 10 - 5x1 + 2x2 - x3and the conditions are as follows,x1 + x2 + x3 = 15(x1,x2,x3) >= 0Where the sum of x1, x2, and x3 have a known value, and they are all greater than zero. In R, it would look something like this,Fx = function(x) {10 - (5*x[1] + 2*x[2] + x[3])}opt = optim(c(1,1,1), Fx, method = "L-BFGS-B", lower=c(0,0,0), upper=c(15,15,15))I also tried to use inequalities with constrOptim to force the sum to be fixed. I still think this may be a plausible work around, but I was unable to make it work. This is a simplified example of the real problem, but any help would be very appreciated. 解决方案 On this occasion optim will not work obviously because you have equality constraints. constrOptim will not work either for the same reason (I tried converting the equality to two inequalities i.e. greater and less than 15 but this didn't work with constrOptim).However, there is a package dedicated to this kind of problem and that is Rsolnp.You use it the following way:#specify your functionopt_func <- function(x) { 10 - 5*x[1] + 2 * x[2] - x[3]}#specify the equality function. The number 15 (to which the function is equal)#is specified as an additional argumentequal <- function(x) { x[1] + x[2] + x[3]}#the optimiser - minimises by defaultsolnp(c(5,5,5), #starting values (random - obviously need to be positive and sum to 15) opt_func, #function to optimise eqfun=equal, #equality function eqB=15, #the equality constraint LB=c(0,0,0), #lower bound for parameters i.e. greater than zero UB=c(100,100,100)) #upper bound for parameters (I just chose 100 randomly)Output:> solnp(c(5,5,5),+ opt_func,+ eqfun=equal,+ eqB=15,+ LB=c(0,0,0),+ UB=c(100,100,100))Iter: 1 fn: -65.0000 Pars: 14.99999993134 0.00000002235 0.00000004632Iter: 2 fn: -65.0000 Pars: 14.999999973563 0.000000005745 0.000000020692solnp--> Completed in 2 iterations$pars[1] 1.500000e+01 5.745236e-09 2.069192e-08$convergence[1] 0$values[1] -10 -65 -65$lagrange [,1][1,] -5$hessian [,1] [,2] [,3][1,] 121313076 121313076 121313076[2,] 121313076 121313076 121313076[3,] 121313076 121313076 121313076$ineqx0NULL$nfuneval[1] 126$outer.iter[1] 2$elapsedTime difference of 0.1770101 secs$vscale[1] 6.5e+01 1.0e-08 1.0e+00 1.0e+00 1.0e+00So the resulting optimal values are:$pars[1] 1.500000e+01 5.745236e-09 2.069192e-08which means that the first parameter is 15 and the rest zero and zero. This is indeed the global minimum in your function since the x2 is adding to the function and 5 * x1 has a much greater (negative) influence than x3 on the outcome. The choice of 15, 0, 0 is the solution and the global minimum to the function according to the constraints.The function worked great! 这篇关于具有等式和不等式约束的 R 优化的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！