lpSolve - find maximum value - formula with multiplications

Question

I'm looking for a way to use lpSolve in a similar way to how I use it succesfully in Excel. I've calculated elasticity for various product. Based on whether a product is elasticity or not elastic I want to give an advice on which price to ask.

I have the following values:

current.price = 15
sales.lastmonth = 50
elasticity = -1.5

And I want to optimize the sales.prediction by changing the suggested.price

Formula: sales.prediction = (sales.lastmonth - ((abs(elasticity)(suggested.price-current.price))(sales.lastmonth/current.price)))*suggested.price

I've tried the following:

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

objective.fn <- c(sales.prediction) # determine what the objective is

# constrain sales.lastmonth and current.price
const.mat <- matrix(c(1,1),ncol=1,byrow=T)
const.dir <- c("=","=")
const.rhs <- c(sales.lastmonth, current.price)

lp("max",objective.fn,const.mat,const.dir,const.rhs,compute.sens=TRUE)

Any suggestions?

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EDIT - V2: Based on the comment below, I would like to add the constraint that the suggested.price should be at max 25% higher of 25% lower than the the suggested price I have in my mind without the optimization. For example, let's say I am already thinking about lowering the price to 12.5. Is that possible as well?

Can you write out your constraints? These values of yours seem to be constant, this is not constrained opzimization? — user2974951, Jun 21 '22 at 10:51
The objective function shown is quadratic, not linear, in suggested price so linear programming is not the right approach. — G. Grothendieck, Jun 21 '22 at 12:26
@G.Grothendieck Do you have a suggestion for a better way to optimzie the price? — Frank Beke, Jun 21 '22 at 13:04

user2974951 · Answer 1 · 2022-06-21T12:15:26.257

1

If I understand correctly you are simply trying to optimize (maximize) this function, there are no constraints since your variables are constant. If this is true then you can do

optimize(
  f=function(x){
    sales.lastmonth-((abs(elasticity)*(x-current.price))*(sales.lastmonth/current.price))*x
  },
  interval=c(-100,100),
  maximum=T
)

$maximum
[1] 7.5

$objective
[1] 331.25

maximum at 7.5.

Edit: a simple hack to limit the optimization is to use the interval argument, for your edit interval=c(current.price*0.75,current.price*1.25).

edited Jun 21 '22 at 12:15

answered Jun 21 '22 at 10:59

user2974951

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I've altered the interval argument, is that possible as well? – Frank Beke Jun 21 '22 at 13:11
@FrankBeke If by that you mean that you want to find an optimal value in a custom interval, say [3,6], then yes, you can input this in `interval`. – user2974951 Jun 21 '22 at 13:13
Yes, I already figured it out. Thanks! – Frank Beke Jun 21 '22 at 13:16

lpSolve - find maximum value - formula with multiplications

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1 Answers1