求解线性规划中R-误差的应用

这是我的R代码：

#install.packages("linprog")
library(linprog)


roi=c(0.04,0.02,0.03,0.11,0.03,0.2,0.17,0.24,0.03,0.04,0.04)

bvet=c(100000,0,0,0,0,0,0,0,0,0,0)

Amat=rbind(

c(1,1,1,1,1,1,1,1,1,1,1),  
c(0.96,-0.04,-0.04,-0.04,-0.04,-0.04,-0.04,-0.04,-0.04,-0.04,-0.04),
  c(-0.02,0.98,-0.02,-0.02,-0.02,-0.02,-0.02,-0.02,-0.02,-0.02,-0.02),
  c(-0.03,-0.03,0.97,-0.03,-0.03,-0.03,-0.03,-0.03,-0.03,-0.03,-0.03),
  c(-0.11,-0.11,-0.11,0.89,-0.11,-0.11,-0.11,-0.11,-0.11,-0.11,-0.11),
  c(-0.03,-0.03,-0.03,-0.03,0.97,-0.03,-0.03,-0.03,-0.03,-0.03,-0.03),
  c(-0.2,-0.2,-0.2,-0.2,-0.2,0.8,-0.2,-0.2,-0.2,-0.2,-0.2),
  c(-0.17,-0.17,-0.17,-0.17,-0.17,-0.17,0.83,-0.17,-0.17,-0.17,-0.17),
  c(-0.24,-0.24,-0.24,-0.24,-0.24,-0.24,-0.24,0.83,-0.24,-0.24,-0.24),
  c(-0.03,-0.03,-0.03,-0.03,-0.03,-0.03,-0.03,0.97,-0.03,-0.03,-0.03),
  c(-0.04,-0.04,-0.04,-0.04,-0.04,-0.04,-0.04,-0.04,-0.04,0.96,-0.04),
  c(-0.04,-0.04,-0.04,-0.04,-0.04,-0.04,-0.04,-0.04,-0.04,-0.04,0.96)
)

LP=solveLP(roi,bvet,Amat,TRUE)

我得到这个错误：

Error in solveLP(roi, bvet, Amat, TRUE) : 
  Matrix A must have as many rows as constraints (=elements of vector b) and as many columns as variables (=elements of vector c). 

---

我检查了我的代码多次，同样的问题仍然存在。

用

bvet

的元素数检查

Amat

的行数。

Amat

中有12行，但

bvet

和

roi

中都有11个条目。这不可行。但在“Amat”中，我能使用不同数量的约束吗？谢谢Rhertel…它起作用了！！