Simulation/Optimization Package in R for tuning weights to achieve maximum allocation for groups

Question

I am looking to identify the simulation package in R to identify the perfect weights, that enables me allocate my datapoints into the maximum bucket.

Basically, i want to tune my weights in a such a way the achieve my goal.

Below is the example.

  Score1,Score2,Score3,Final,Group
0.87,0.73,0.41,0.63,"60-100"
0.82,0.73,0.85,0.796,"70-80"
0.82,0.37,0.85,0.652,"60-65"
0.58,0.95,0.42,0.664,"60-65"
1,1,0.9,0.96,"90-100"

Weight1,Weight2,Weight3
0.2,0.4,0.4

Final Score= Score1*Weight1+ Score2*Weight2+Score3*Weight3

The sum of my weights is 1. W1+W2+W3=1

i want to tune my weights in such a way that most of my cases lie into the "90-100" bucket. I know there won't be a perfect combination, but want to capture the maximum cases. I am currently trying to do the same in excel manually, using Pivot, but want to know if there is any package in R, that helps me to achieve my goal.

THe group allocation "70-80" "80-90" is something i have made in excel, using if else condition.

R Pivot Result:

"60-100",1
"60-65",2
"70-80",1
"90-100",1

Would appreciate if someone can help me to for the same.

Thanks,

score 1 · Answer 1 · answered Jun 13 '18 at 18:30

Here's an approach that tries to get all the final scores as close as possible to 0.9 using a nested optimisation approach.

Here's your original data:

# Original data
df <- read.table(text = "Score1, Score2, Score3 
0.87,0.73,0.41 
0.82,0.73,0.85 
0.82,0.37,0.85 
0.58,0.95,0.42 
1,1,0.9", header = TRUE, sep = ",")

This is the cost function for the first weight.

# Outer cost function
cost_outer <- function(w1){
  # Run nested optimisation
  res <- optimise(cost_nested, lower = 0, upper = 1 - w1, w1 = w1)

  # Spit second weight into a global variable
  res_outer <<- res$minimum

  # Return the cost function value
  res$objective
}

This is the cost function for the second weight.

# Nested cost function
cost_nested <- function(w2, w1){
  # Calculate final weight
  w <- c(w1, w2, 1 - w2 -w1)

  # Distance from desired interval
  res <- 0.9 - rowSums(w*df) 

  # Zero if negative distance, square distance otherwise
  res <- sum(ifelse(res < 0, 0, res^2))
}

Next, I run the optimisation.

# Repackage weights
weight <- c(optimise(cost_outer, lower = 0, upper = 1)$minimum, res_outer)
weight <- c(weight, 1 - sum(weight))

Finally, I show the results.

# Final scores
cbind(df, Final = rowSums(weight * df))


#   Score1 Score2 Score3     Final
# 1   0.87   0.73   0.41 0.7615286
# 2   0.82   0.73   0.85 0.8229626
# 3   0.82   0.37   0.85 0.8267400
# 4   0.58   0.95   0.42 0.8666164
# 5   1.00   1.00   0.90 0.9225343

Notice, however, that this code gets the final scores as close as possible to the interval, which is different from getting the most scores in that interval. That can be achieved by switching out the nested cost function with something like:

# Nested cost function
cost_nested <- function(w2, w1){
  # Calculate final weight
  w <- c(w1, w2, 1 - w2 -w1)

  # Number of instances in desired interval
  res <- sum(rowSums(w*df) < 0.9)
}

Thanks Lyngbakr.. Can you also help me on how to extract the values for W1,W2 and W3? The W1,W2 and W3 are the constant values that are applied on the entire value. by constant i mean, fixed numbers, rather than dynamic number (changing datapoint per datapoint). — Jay, Jun 13 '18 at 18:43
I checked the weight.. it is 0.33 for W1, W2 and W3... can't it vary in such a way to give maximum score, apart from being 0.33 for each weight. This would actually be similar to linear algebra, where we say maximum value for ax+by+cz=1, will be when x=y=z=1/3.. — Jay, Jun 13 '18 at 18:47
@Jay I don't know which weights you're referring to, but when I run it I get `9.999339e-01 2.525061e-05 4.085635e-05` for the latter case and `2.253008e-01 4.217429e-05 7.746570e-01` for the former case. In the latter case, I think there is no solution (for the given data set) that can put more than one record in the chosen range. — Dan, Jun 14 '18 at 11:17

Erwin Kalvelagen · Answer 2 · 2018-06-14T15:25:30.640

This can be formulated as a Mixed Integer Programming (MIP) problem. The mathematical model can look like:

The binary variable δ_i indicates if final weight F_i is inside the interval [0.9,1]. M is "large" value (if all your data is between 0 and 1 we can choose M=1). a_i,j is your data.

The objective function and all constraints are linear, so we can use standard MIP solvers to solve this problem. MIP solvers for R are readily available.

PS in the example groups overlap. That does not make much sense to me. I think if we have "90-100" we should not also have "60-100".

PS2. If all data is between 0 and 1, we can simplify the sandwich equation a bit: we can drop the right part.

For the small example data set I get:

----     56 PARAMETER a  

            j1          j2          j3

i1       0.870       0.730       0.410
i2       0.820       0.730       0.850
i3       0.820       0.370       0.850
i4       0.580       0.950       0.420
i5       1.000       1.000       0.900


----     56 VARIABLE w.L  weights

j1 0.135,    j2 0.865


----     56 VARIABLE f.L  final scores

i1 0.749,    i2 0.742,    i3 0.431,    i4 0.900,    i5 1.000


----     56 VARIABLE delta.L  selected

i4 1.000,    i5 1.000


----     56 VARIABLE z.L                   =        2.000  objective

(zeros are not printed)

Simulation/Optimization Package in R for tuning weights to achieve maximum allocation for groups

2 Answers2