使用 NLOPT/Gurobi 求解混合约束优化
Using NLOPT/Gurobi for solving mixed constraint optimization
我目前正在做一个项目,我想使用 R 和 NLOPT 包(或 Gurobi)来解决以下优化问题:
找到最小 ||y-y_h||_L^2 使得 x = Ay_h, y >= 0,其中 x, y 是 给定的大小为16*1的向量,A = 16*24的矩阵也给定了
我的尝试:
R码
nrow=16;
ncol = 24;
lambda = matrix(sample.int(100, size = ncol*nrow, replace = T),nrow,ncol);
lambda = lambda - diag(lambda)*diag(x=1, nrow, ncol);
y = rpois(ncol,lambda) + rtruncnorm(ncol,0,1,mean = 0, sd = 1);
x = matrix (0, nrow, 1);
x_A1 = y[1]+y[2]+y[3];
x_A2 = y[4]+y[7]+y[3];
x_B1 = y[4]+y[5]+y[6];
x_B2 = y[11]+y[1];
x_C1 = y[7]+y[8]+y[9];
x_C2 = y[2]+y[5]+y[12];
x_D1 = y[10]+y[11]+y[12];
x_D2 = y[3]+y[6]+y[9];
x_E1 = y[13]+y[14]+y[15];
x_E2 = y[18]+y[19]+y[23];
x_F1 = y[20]+y[21]+y[19];
x_F2 = y[22]+y[16]+y[13];
x_G1 = y[23]+y[22]+y[24];
x_G2 = y[14]+y[17]+y[20];
x_H1 = y[16]+y[17]+y[18];
x_H2 = y[15]+y[21]+y[24];
d <- c(x_A1, x_A2,x_B1, x_B2,x_C1, x_C2,x_D1, x_D2,x_E1,
x_E2,x_F1, x_F2,x_G1, x_G2,x_H1, x_H2)
x <- matrix(d, nrow, byrow=TRUE)
A = matrix(c(1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, #x_A^1
0,0,0,1,0,0,1,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0, #x_A^2
0,0,0,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, #x_B^1
1,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0, #x_B^2
0,0,0,0,0,0,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, #x_C^1
0,1,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0, #x_C^2
0,0,0,0,0,0,0,0,0,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0, #x_D^1
0,0,1,0,0,1,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, #x_D^2
0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,0,0,0,0,0,0,0,0,0, #x_E^1
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,0,0,0,1,0, #x_E^2
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,0,0,0, #x_F^1
0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,1,0,0,0,0,0,1,0,0, #x_F^2
0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,1,0,0,1,0,0,0,0, #x_G^2
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1, #x_G^1
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,0,0,0,0,0,0, #x_H^1
0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,1), #x_H^2
nrow, ncol, byrow= TRUE)
试了两个代码解决问题: min ||y - y_h||_L^2 where x= Ay_h, y>= 0 其中 x、y、A 都在上面给出。
# f(x) = ||yhat-y||_L2
eval_f <- function( yhat ) {
return( list( "objective" = norm((mean(yhat-y))^2, type = "2")))
}
# inequality constraint
eval_g_ineq <- function( yhat ) {
constr <- c(0 - yhat)
return( list( "constraints"=constr ))
}
# equalities constraint
eval_g_eq <- function( yhat ) {
constr <- c( x-A%*%yhat )
return( list( "constraints"=constr ))
}
x0 <- y
#lower bound of control variable
lb <- c(matrix (0, ncol, 1))
local_opts <- list( "algorithm" = "NLOPT_LD_MMA",
"xtol_rel" = 1.0e-7 )
opts <- list( "algorithm" = "NLOPT_LD_AUGLAG",
"xtol_rel" = 1.0e-7,
"maxeval" = 1000,
"local_opts" = local_opts )
res <- nloptr( x0=x0,
eval_f=eval_f,
eval_grad_f = NULL,
lb=lb,
eval_g_ineq = eval_g_ineq,
eval_g_eq=eval_g_eq,
opts=opts)
print(res)
古罗比码:
**#model <- list()
#model$B <- A
#model$obj <- norm((y-yhat)^2, type = "2")
#model$modelsense <- "min"
#model$rhs <- c(x,0)
#model$sense <- c('=', '>=')
#model$vtype <- 'C'
#result <- gurobi(model, params)
#print('Solution:')
#print(result$objval)
#print(result$yhat)**
我的问题:首先,当我运行上面的R代码时,它一直给我这样的信息:
is.nloptr(ret) 错误:
objective 梯度中的元素数量错误
另外: 警告信息:
在 is.na(f0$gradient) 中:
is.na() 应用于 'NULL'
类型的非(列表或向量)
我尽量避免计算梯度,因为我没有任何关于 y 的密度函数的信息。谁能帮我解决上面的错误?
对于 Gurobi 代码,我收到此消息:错误:is(model$A, "matrix") ||是(模型$A,"sparseMatrix")||是(模型$A,....不是真的
但是我的矩阵A输入正确,这个错误是什么意思?
我几天前才开始使用 nloptr。这个问题已经是老问题了,但我还是会回答的。当您将 'nloptr' 与 'NLOPT_LD_AUGLAG' 算法一起使用时,'LD' 代表局部和使用梯度...因此您需要选择中间带有 'LN' 的其他内容。例如,'NLOPT_LN_COBYLA' 应该可以在没有渐变的情况下正常工作。
其实你可以看看nloptr包手册。
我目前正在做一个项目,我想使用 R 和 NLOPT 包(或 Gurobi)来解决以下优化问题:
找到最小 ||y-y_h||_L^2 使得 x = Ay_h, y >= 0,其中 x, y 是 给定的大小为16*1的向量,A = 16*24的矩阵也给定了
我的尝试:
R码
nrow=16;
ncol = 24;
lambda = matrix(sample.int(100, size = ncol*nrow, replace = T),nrow,ncol);
lambda = lambda - diag(lambda)*diag(x=1, nrow, ncol);
y = rpois(ncol,lambda) + rtruncnorm(ncol,0,1,mean = 0, sd = 1);
x = matrix (0, nrow, 1);
x_A1 = y[1]+y[2]+y[3];
x_A2 = y[4]+y[7]+y[3];
x_B1 = y[4]+y[5]+y[6];
x_B2 = y[11]+y[1];
x_C1 = y[7]+y[8]+y[9];
x_C2 = y[2]+y[5]+y[12];
x_D1 = y[10]+y[11]+y[12];
x_D2 = y[3]+y[6]+y[9];
x_E1 = y[13]+y[14]+y[15];
x_E2 = y[18]+y[19]+y[23];
x_F1 = y[20]+y[21]+y[19];
x_F2 = y[22]+y[16]+y[13];
x_G1 = y[23]+y[22]+y[24];
x_G2 = y[14]+y[17]+y[20];
x_H1 = y[16]+y[17]+y[18];
x_H2 = y[15]+y[21]+y[24];
d <- c(x_A1, x_A2,x_B1, x_B2,x_C1, x_C2,x_D1, x_D2,x_E1,
x_E2,x_F1, x_F2,x_G1, x_G2,x_H1, x_H2)
x <- matrix(d, nrow, byrow=TRUE)
A = matrix(c(1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, #x_A^1
0,0,0,1,0,0,1,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0, #x_A^2
0,0,0,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, #x_B^1
1,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0, #x_B^2
0,0,0,0,0,0,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, #x_C^1
0,1,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0, #x_C^2
0,0,0,0,0,0,0,0,0,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0, #x_D^1
0,0,1,0,0,1,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, #x_D^2
0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,0,0,0,0,0,0,0,0,0, #x_E^1
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,0,0,0,1,0, #x_E^2
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,0,0,0, #x_F^1
0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,1,0,0,0,0,0,1,0,0, #x_F^2
0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,1,0,0,1,0,0,0,0, #x_G^2
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1, #x_G^1
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,0,0,0,0,0,0, #x_H^1
0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,1), #x_H^2
nrow, ncol, byrow= TRUE)
试了两个代码解决问题: min ||y - y_h||_L^2 where x= Ay_h, y>= 0 其中 x、y、A 都在上面给出。
# f(x) = ||yhat-y||_L2
eval_f <- function( yhat ) {
return( list( "objective" = norm((mean(yhat-y))^2, type = "2")))
}
# inequality constraint
eval_g_ineq <- function( yhat ) {
constr <- c(0 - yhat)
return( list( "constraints"=constr ))
}
# equalities constraint
eval_g_eq <- function( yhat ) {
constr <- c( x-A%*%yhat )
return( list( "constraints"=constr ))
}
x0 <- y
#lower bound of control variable
lb <- c(matrix (0, ncol, 1))
local_opts <- list( "algorithm" = "NLOPT_LD_MMA",
"xtol_rel" = 1.0e-7 )
opts <- list( "algorithm" = "NLOPT_LD_AUGLAG",
"xtol_rel" = 1.0e-7,
"maxeval" = 1000,
"local_opts" = local_opts )
res <- nloptr( x0=x0,
eval_f=eval_f,
eval_grad_f = NULL,
lb=lb,
eval_g_ineq = eval_g_ineq,
eval_g_eq=eval_g_eq,
opts=opts)
print(res)
古罗比码:
**#model <- list()
#model$B <- A
#model$obj <- norm((y-yhat)^2, type = "2")
#model$modelsense <- "min"
#model$rhs <- c(x,0)
#model$sense <- c('=', '>=')
#model$vtype <- 'C'
#result <- gurobi(model, params)
#print('Solution:')
#print(result$objval)
#print(result$yhat)**
我的问题:首先,当我运行上面的R代码时,它一直给我这样的信息: is.nloptr(ret) 错误: objective 梯度中的元素数量错误 另外: 警告信息: 在 is.na(f0$gradient) 中: is.na() 应用于 'NULL'
类型的非(列表或向量)我尽量避免计算梯度,因为我没有任何关于 y 的密度函数的信息。谁能帮我解决上面的错误?
对于 Gurobi 代码,我收到此消息:错误:is(model$A, "matrix") ||是(模型$A,"sparseMatrix")||是(模型$A,....不是真的
但是我的矩阵A输入正确,这个错误是什么意思?
我几天前才开始使用 nloptr。这个问题已经是老问题了,但我还是会回答的。当您将 'nloptr' 与 'NLOPT_LD_AUGLAG' 算法一起使用时,'LD' 代表局部和使用梯度...因此您需要选择中间带有 'LN' 的其他内容。例如,'NLOPT_LN_COBYLA' 应该可以在没有渐变的情况下正常工作。 其实你可以看看nloptr包手册。