CODEWITHSUNDEEP
roptimizationlinear-programming
Qwo
Qwo

Reputation: 15

Optimization function applied to table of values in R

`values <-    matrix(c(0.174,0.349,1.075,3.1424,0.173,0.346,1.038,3.114,0.171,0.343,1.03,3.09,0.17,0.34,1.02,3.06),ncol=4)    `

I am attempting to maximize the total value for the dataset taking only one value from each row, and with associated costs for each column

subject to:

  1. One value column used per row.

  2. cost of each use of column 1 is 4

    cost of each use of column 2 is 3

    cost of each use of column 3 is 2

    cost of each use of column 4 is 1

    total cost <= 11

These are stand in values for a larger dataset. I need to be able to apply it directly to all the rows of a dataset.

I have been trying to use the lpSolve package, with no success.

`f.obj <- values
f.con <- c(4,3,2,1)
f.dir <- "<="
f.rhs <- 11

lp("max", f.obj, f.con, f.dir, f.rhs)`

I am getting a solution of "0"

I do not know how to model this in a way that chooses one value per row and then uses a different value in calculating the constraints.

Upvotes: 0

Views: 175

Answers (2)

Erwin Kalvelagen
Erwin Kalvelagen

Reputation: 16772

Looks like the problem is as follows:

  1. We have a matrix a[i,j] with values, and a vector c[j] with costs.

  2. We want to select one value for each row such that:

    a. total cost <= 11

    b. total value is maximized

  3. To develop a mathematical model, we introduce binary variables x[i,j] ∈ {0,1}. With this, we can write:

    max sum((i,j), a[i,j]*x[i,j])
subject to 
     sum((i,j), c[j]*x[i,j]) <= 11  
     sum(j, x[i,j]) = 1   ∀i
     x[i,j] ∈ {0,1}

  4. Implement in R. I use here CVXR.

     #
 # data
 # A : values
 # C : cost
 #

 A <- matrix(c(0.174,0.349,1.075,3.1424,0.173,0.346,1.038,3.114,0.171,0.343,1.03,3.09,0.17,0.34,1.02,3.06),ncol=4)

 C <- c(4,3,2,1)

 maxcost <- 11

 #
 # form a matrix cmat[i,j] indicating the cost of element i,j
 #    
 cmat <- matrix(C,nrow=dim(A)[1],ncol=dim(A)[2],byrow=T)

 #
 # problem:
 # pick one value from each row
 # such that total value of selected cells is maximized
 # and cost of selected cells is limited to maxcost
 #
 # model:
 # min sum((i,j), a[i,j]*x[i,j])
 # subject to
 #     sum((i,j), c[j]*x[i,j]) <= maxcost
 #     sum(j,x[i,j]) = 1    ∀i
 #     x[i,j] ∈ {0,1}
 #
 #

 library(CVXR)

 x = Variable(dim(A), name="x", boolean=T)

 p <- Problem(Maximize(sum_entries(A*x)),
          constraints=list(
            sum_entries(cmat*x) <= maxcost,
            sum_entries(x,axis=1) == 1
            ))

 res <- solve(p,verbose=T)
 res$status
 res$value
 res$getValue(x)*A

The output looks like:

> res$status
[1] "optimal"

> res$value
[1] 4.7304

> res$getValue(x)*A
       [,1] [,2]  [,3] [,4]
[1,] 0.0000    0 0.000 0.17
[2,] 0.0000    0 0.343 0.00
[3,] 1.0750    0 0.000 0.00
[4,] 3.1424    0 0.000 0.00

The description in the original post is not very precise. For instance, I assumed that we need to select precisely one cell from each row. If we just want "select at most one cell from each row", then replace

sum(j, x[i,j]) = 1   ∀i

by

sum(j, x[i,j]) <= 1   ∀i

Upvotes: 1

Oliver
Oliver

Reputation: 8592

As mentioned by Steve, the lpSolve package expects a single objective function not a matrix. You could reformulate as maximize(sum(RowSums(values*xij)) given constraint

Eg, change the matrix to a vector, and change the problem to a integer optimization problem

obj <- as.vector(values)
f.con <- rep(f.con, each = 4)
r <- lp('max', obj, matrix(f.con, nrow = 1), f.dir, f.rhs, int.vec  = seq_along(obj))

#' Success: the objective function is 9.899925

Upvotes: 0

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