constrained optimization in R setting up constraints - r

Question

I have been trying to solve a constrained optimization problem in R using constrOptim() (my first time) but am struggling to set up the constraints for my problem.
The problem is pretty straight forward and i can set up the function ok but am a bit at a loss about passing the constraints in.
e.g. problem i've defined is (am going to start with N fixed at 1000 say so i just want to solve for X ultimately i'd like to choose both N and X that max profit):
so i can set up the function as:
fun <- function(x, N, a, c, s) { ## a profit function
x1 <- x[1]
x2 <- x[2]
x3 <- x[3]
a1 <- a[1]
a2 <- a[2]
a3 <- a[3]
c1 <- c[1]
c2 <- c[2]
c3 <- c[3]
s1 <- s[1]
s2 <- s[2]
s3 <- s[3]
((N*x1*a1*s1)-(N*x1*c1))+((N*x2*a2*s2)-(N*x2*c2))+((N*x3*a3*s3)-(N*x3*c3))
}
The constraints i need to implement are that:
x1>=0.03
x1<=0.7
x2>=0.03
x2<=0.7
x3>=0.03
x2<=0.7
x1+x2+x3=1
The X here represents buckets into which i need to optimally allocate N, so x1=pecent of N to place in bucket 1 etc. with each bucket having at least 3% but no more than 70%.
Any help much appreciated...
e.g. here is an example i used to test the function does what i want:
fun <- function(x, N, a, c, s) { ## profit function
x1 <- x[1]
x2 <- x[2]
x3 <- x[3]
a1 <- a[1]
a2 <- a[2]
a3 <- a[3]
c1 <- c[1]
c2 <- c[2]
c3 <- c[3]
s1 <- s[1]
s2 <- s[2]
s3 <- s[3]
((N*x1*a1*s1)-(N*x1*c1))+((N*x2*a2*s2)-(N*x2*c2))+((N*x3*a3*s3)-(N*x3*c3))
};
x <-matrix(c(0.5,0.25,0.25));
a <-matrix(c(0.2,0.15,0.1));
s <-matrix(c(100,75,50));
c <-matrix(c(10,8,7));
N <- 1000;
fun(x,N,a,c,s);

Answer 1

You can use The lpSolveAPI package.
## problem constants
a <- c(0.2, 0.15, 0.1)
s <- c(100, 75, 50)
c <- c(10, 8, 7)
N <- 1000
## Problem formulation
# x1 >= 0.03
# x1 <= 0.7
# x2 >= 0.03
# x2 <= 0.7
# x3 >= 0.03
# x1 +x2 + x3 = 1
#N*(c1- a1*s1)* x1 + (a2*s2 - c2)* x2 + (a3*s3- c3)* x3
library(lpSolveAPI)
my.lp <- make.lp(6, 3)
The best way to build a model in lp solve is columnwise;
#constraints by columns
set.column(my.lp, 1, c(1, 1, 0, 0, 1, 1))
set.column(my.lp, 2, c(0, 0, 1, 1, 0, 1))
set.column(my.lp, 3, c(0, 0, 0, 0, 1, 1))
#the objective function ,since we need to max I set negtive max(f) = -min(f)
set.objfn (my.lp, -N*c(c[1]- a[1]*s[1], a[2]*s[2] - c[2],a[3]*s[3]- c[3]))
set.rhs(my.lp, c(rep(c(0.03,0.7),2),0.03,1))
#constraint types
set.constr.type(my.lp, c(rep(c(">=","<="), 2),">=","="))
take a look at my model
my.lp
Model name:
Model name:
C1 C2 C3
Minimize 10000 -3250 2000
R1 1 0 0 >= 0.03
R2 1 0 0 <= 0.7
R3 0 1 0 >= 0.03
R4 0 1 0 <= 0.7
R5 1 0 1 >= 0.03
R6 1 1 1 = 1
Kind Std Std Std
Type Real Real Real
Upper Inf Inf Inf
Lower 0 0 0
solve(my.lp)
[1] 0 ## sucess :)
get.objective(my.lp)
[1] -1435
get.constraints(my.lp)
[1] 0.70 0.70 0.03 0.03 0.97 1.00
## the decisions variables
get.variables(my.lp)
[1] 0.03 0.70 0.27

Answer 2

Hi Just in case of use to anyone i also found an answer as below:
First of all, your objective function can be written a lot more concisely using vector operations:
> my_obj_coeffs <- function(N,a,c,s) N*(a*s-c)
> fun <- function(x,N,a,c,s) sum(my_obj_coeffs(N,a,c,s) * x)
You're trying to solve a linear program, so you can use solve it using the simplex algorithm. There's a lightweight implementation of it in the 'boot' package.
> library(boot)
> solution <- function(obj) simplex(obj, diag(3), rep(0.7,3), diag(3), rep(0.03,3), rep(1,3), 1, maxi=TRUE)
Then for the example parameters you used, you can call that solution function:
> a <- c(0.2,0.15,0.1)
> s <- c(100,75,50)
> c <- c(10,8,7)
> N <- 1000
> solution(my_obj_coeffs(N,a,c,s))
Linear Programming Results
Call : simplex(a = obj(N, a, s, c), A1 = diag(3), b1 = rep(0.7, 3),
A2 = diag(3), b2 = rep(0.03, 3), A3 = matrix(1, 1, 3), b3 = 1,
maxi = TRUE)
Maximization Problem with Objective Function Coefficients
[,1]
[1,] 10000
[2,] 3250
[3,] -2000
attr(,"names")
[1] "x1" "x2" "x3"
Optimal solution has the following values
x1 x2 x3
0.70 0.27 0.03
The optimal value of the objective function is 7817.5.