我目前工作的一個項目,我想使用R和NLOPT包(或Gurobi)解決以下優化問題:使用NLOPT/Gurobi求解混合約束優化
查找分鐘|| Y-這樣x = Ay_h,y> = 0,其中x,y是給出的大小爲16 * 1的向量,A = 16 * 24矩陣也給出。
我嘗試:
R代碼裏面
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)
試過兩個代碼來解決問題:分鐘||ý - y_h || _L^2,其中x = Ay_h, y> = 0其中x,y,A全部在上面給出。
#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)
Gurobi代碼:
**#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)**
我的問題:首先,當我跑將R代碼上面,它不斷給我這個消息: 錯誤在is.nloptr(ret): 錯誤的目標梯度元素數 另外:警告消息: is.na(f0 $梯度): is.na()適用於類型爲'NULL'的非(列表或向量)
我試圖避免計算梯度,因爲我沒有任何關於y的密度函數。任何人都可以請幫我解決上面的錯誤?
對於Gurobi代碼,我收到了這條消息:Error:is(model $ A,「matrix」)||是(型號$ A,「sparseMatrix」)||是(型號$ A,......是不是真
但我的矩陣A是正確的輸入,所以這個錯誤是什麼意思?
請不要在這裏發佈代碼的圖片,包括一個[可重現的例子](http:// stackoverflow .COM /問題/ 5963269 /如何對化妝一個偉大-R - 可重複的例子),我們可以將樣本輸入數據複製/粘貼到R中運行。 – MrFlick
@MrFlick:我修好了。非常感謝,因爲我是新來的。但是學習LATEX在這個論壇上無法正常工作令人討厭; p請你幫我解決這個問題,因爲我已經花了7個小時,並且仍然無法看到我如何避免使用漸變,但仍然使得「NLOPT」代碼工作:( – user177196
這裏沒有人想幫我修復上面的代碼??) – user177196