Question

我试着运行教科书中给出的例子＆＃34;使用R＆＃34建模和求解线性规划;可以找到教科书的链接here

我尝试复制第75页第3.8章中的确切代码，但似乎打印出不同的结果

Assignment01 <- function(c){
n <- dim(c)[1]
coef <- as.vector(t(c))
rhs <- rep(1,2*n)

Amatrix <- matrix(0, 2*n, n*n)

for(i in 1:n){
    for(j in 1:n){
        Amatrix[i, n*(i-1)+j] <- 1
    }
}

for(i in 1:n){
    for(j in 1:n){
        Amatrix[n+i, n*(j-1)+i] <- 1
    }
}

signs <- rep("==", 2*n)
var_type <- rep("B", 2*n)
library(Rglpk)

solution <- Rglpk_solve_LP(obj=coef, mat=Amatrix, dir=signs, types=var_type, rhs=rhs, max=TRUE)
return(solution)

}

Assignment02 <- function(c){
n <- dim(c)[1]
coef <- c(rep(0, n*n),1)
rhs <- c(rep(1, 2*n), rep(0,n))
Amatrix <- matrix(0, 3*n, n*n + 1)

for(i in 1:n){
    for(j in 1:n){
        Amatrix[i, n*(i-1)+j] <- 1
    }
}

for(i in 1:n){
    for(j in 1:n){
        Amatrix[n+i, n*(j-1)+i] <- 1
    }
}

for(i in 1:n){
    for(j in 1:n){
        Amatrix[2*n+1, n*(j-1)+i] <- c[j,i]
    }
}

for(i in 1:n){
    Amatrix[2*n+1, n*n + 1] <- -1

}

signs <- c(rep("==", 2*n),rep(">=",n))

var_type <- c(rep("B", n*n), "C")

library(Rglpk)

solutionPL <- Rglpk_solve_LP(obj=coef, mat=Amatrix, dir=signs, types=var_type, rhs=rhs, max=TRUE)
return (solutionPL) 
}


set.seed(1)
c <- matrix(sample(10:100, 25),100,100)
solAss01 <- Assignment01(c)
m.01 <- matrix(solAss01$solution[1:25],5,5, byrow=TRUE)
solAss02 <- Assignment02(c)
m.02 <- matrix(solAss02$solution[1:25],5,5,byrow=TRUE)
print(m.01)
print(m.02)

并获得此输出

应该是

Answer 1

此代码无法编译：

<?xml version="1.0" encoding="UTF-8" standalone="no"?>
<svg width="36px" height="36px" viewBox="0 0 36 36" version="1.1" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink">
<defs></defs>
<g id="States" stroke="none" stroke-width="1" fill="none" fill-rule="evenodd">
    <g id="Buttons-&amp;-States-(Documents-Steps)" transform="translate(-104.000000, -138.000000)">
        <g id="Close-Button" transform="translate(104.000000, 138.000000)">
            <g>
                <circle id="circle" fill="#E9EEF2" cx="18" cy="18" r="18"></circle>
                <polygon id="X" fill="#9FADB5" transform="translate(17.868000, 17.808000) rotate(-45.000000) translate(-17.868000, -17.808000) " points="16.968 23.616 18.768 23.616 18.768 18.672 23.736 18.672 23.736 16.944 18.768 16.944 18.768 12 16.968 12 16.968 16.944 12 16.944 12 18.672 16.968 18.672"></polygon>
            </g>
        </g>
    </g>
</g>

我相信你会for(j in 1:){与x(i,j)混淆。 c(i,j)值为零，每行和每列只有一个x。

另请注意，1 c(i,j)的每个j具有相同的值：

> c[1:5,1:5]
     [,1] [,2] [,3] [,4] [,5]
[1,]   34   34   34   34   34
[2,]   43   43   43   43   43
[3,]   60   60   60   60   60
[4,]   89   89   89   89   89
[5,]   27   27   27   27   27

从书籍示例中最大化R语言中的最小质量的赋值错误

1 个答案: