Question

我正在使用NLoptr包来解决使用该程序的9变量成本函数的优化问题：

function(x){return( list( "objective" = 0.0404*x[1]^2 + 4.4823*x[1] + 0.4762+0.024*x[2]^2 + 3.9767*x[2] + 0.3737+0.0246*x[3]^2 + 3.6992*x[3] + 0.9425+0.0214*x[4]^2 + 3.5896*x[4] + 0.7615+0.0266*x[5]^2 + 3.8197*x[5] + 0.2799+0.0262*x[6]^2 + 3.7884*x[6] + 0.307+0.0362*x[7]^2 + 4.4927*x[7] + 0.1549+0.0344*x[8]^2 + 4.4066*x[8] - 0.2472+0.0241*x[9]^2 + 4.227*x[9],"gradient" = c(2*0.0404*x[1]+4.4823, 2*0.024*x[2]+3.9767, 2*0.0246*x[3], 2*0.0214*x[4]+3.5896, 2*0.0266*x[5]+3.8197,2*0.0262*x[6]+3.7884,2*0.0362*x[7]+4.4927, 2*0.0344*x[8]+4.4066, 2*0.0241*x[9]+4.227)))}

function( x ) {
    constr <- c(x[1] + x[2]+ x[3] + x[4]+x[5]+x[6]+x[7]+x[8]+x[9]-Balance)
    grad <- c(1,1,1,1,1,1,1,1,1)
    return( list( "constraints"=constr, "jacobian"=grad ) )
}

lb<-c(50,50,50,50,50,50,50,50,50)
ub<-c(0,0,0,0,0,0,0,0)
x_0<-c(25,25,25,25,25,25,25,25,25)

local_opts <- list( "algorithm" = "NLOPT_LD_MMA","xtol_rel" = 1.0e-9 )
opts <- list( "algorithm" = "NLOPT_LD_AUGLAG","xtol_rel" = 1.0e-9,"maxeval" = 10000, "local_opts" = local_opts )

res <- nloptr(x0=x_0, eval_f=eval_f,lb=lb,ub=ub,eval_g_eq=eval_g_eq,opts=opts)

代码工作正常，但问题是我需要在168h的时间内解决这个优化，并且每个时间步骤的下限和上限必须不同。有没有人以前实现过这个？

BR

Answer 1

我强烈建议您为此使用OSQP。您可以download it from CRAN。您可以在手册中找到更新问题向量的示例。我在这里重写了它：

library(Matrix)

# Define problem data in the form
# minimize      (1/2) x' P x + q' x
# subject to    l <= A x <= u
#
P <- Matrix(c(11., 0., 0., 0.), 2, 2, sparse = TRUE)
q <- c(3., 4.)
A <- Matrix(c(-1., 0., -1., 2., 3., 0., -1., -3., 5., 4.), 5, 2, sparse = TRUE)
u <- c(0., 0., -15., 100., 80)
l <- rep_len(-Inf, 5)
settings <- osqpSettings(verbose = FALSE)
model <- osqp(P, q, A, l, u, settings)

# Solve
res <- model$Solve()

# Get solution
x_opt <- res$x

# Define new vector
q_new <- c(10., 20.)

# Update model and solve again
model$Update(q = q_new)
res <- model$Solve()

# Get new solution
x_opt_new <- res$x

Disclamer：我是OSQP的作者之一。

更新优化问题中的下限和上限

1 个答案: