问题描述
我正在研究对预测进行排名的方法,并找到了这篇论文 A novel ranking procedure for forecasting approaches using Data Envelopment Analysis,我一直在研究这篇文章并设置我的数据,但我似乎无法在 R 中复制他们的 LP 公式?
这是我所指的公式:LP Formulation
这是一个示例,我正在尝试使用上述公式作为参考复制他们的结果。数据基于上述文章中的“表 2. 示例数据的对数值”。
library(lpSolve)
library(nonparaeff)
DMU = c("FOR01","FOR02","FOR03","FOR04","FOR05")
log.data = matrix(data = as.numeric(c("1.794","1.575","3.576","2.228","2.106","6.628","2.399","1.871","6.354","2.619","1.983","5.849","2.559","1.541","5.676")),ncol = 3,byrow = TRUE)
colnames(log.data) = c("M1","M2","M3")
rownames(log.data) = DMU
THETA = c(-1,-1,0)
add.to.one = c(1,1,1) # Constraint so each lambda adds up to one.
f.obj = c(1)
f.con = cbind(THETA,rbind(t(log.data),add.to.one))
f.dir = c("<=","<=","=")
f.rhs = c(1.794,1.575,3.576,1)
lp2(direction = "min",f.obj,f.con,f.dir,f.rhs,free.var = c(1))
我正在使用包 lpsolve 和 nonparaeff; nonparaeff 扩展了 lp()
函数,因此它可以处理自由变量。
然而,在文章中,他们最终将 theta 设为 0 或 1,第一个 lambda 返回为 1。所以我一定是做错了什么。
我是否正确应用了第四个约束('add.to.one')?此外,lp()
已经假定每个变量 >= 为零,但它是否假定其他任何变量?
我是否正确地将公式转换为 R?我是否正确使用了 lp2
函数?
我在这里查看了其他类似的 lp 问题,但我没有看到很多关于自由变量的问题。但如果您不这么认为,请将我链接到其他问题。
提前致谢。
解决方法
lpSolve 使用起来有点麻烦。但是,使用称为变量拆分的技术来实现自由变量应该不会太困难。 IE。用 xplus-xmin 用 xplus,xmin>=0 替换自由变量 x。不是两者都可以是非零的(那么基矩阵将是奇异的)。
我认为这是正确的更新代码:
library(lpSolve)
DMU = c("FOR01","FOR02","FOR03","FOR04","FOR05")
log.data = matrix(data = as.numeric(c("1.794","1.575","3.576","2.228","2.106","6.628","2.399","1.871","6.354","2.619","1.983","5.849","2.559","1.541","5.676")),ncol = 3,byrow = TRUE)
colnames(log.data) = c("M1","M2","M3")
rownames(log.data) = DMU
theta = c(-1,-1,0)
add.to.one = c(1,1,1) # Constraint so each lambda adds up to one.
f.obj = c(0,-1)
f.con = cbind(rbind(t(log.data),add.to.one),theta,-theta)
f.dir = c("<=","<=","=")
f.rhs = c(1.794,1.575,3.576,1)
r <- lp(direction = "min",f.obj,f.con,f.dir,f.rhs)
最好使用功能更强大的工具,例如 CVXR。