CODEWITHSUNDEEP
r
Parseltongue
Parseltongue

Reputation: 11657

Using R to solve a series of simple equations

I'm having trouble understanding the solve command in R, even after reading some basic examples and documentation.

Let's say I have the following formula describing the growth rate of an economy:

Y = K^α * (AL)^1−α

Let's set α = 1/3. Let's read in a dummy data-frame

data <- read.table(text = "
Economy Year y k L
A 1960 100 1 1000
A 2000 1350 27 2000
B 1960 20 2 4000
B 2000 1350 54 8000
", header = TRUE)

Now, how would I use the solve command (or something more appropriate) to calculate the value of "A" in each of these rows? Something like this doesn't work obviously:

with(data, solve(y, k^(1/3), L^(2/3))

Upvotes: 0

Views: 84

Answers (1)

Karthik Arumugham
Karthik Arumugham

Reputation: 1350

I used the ompr Optimization Modelling Package. You need to rearrange the equation to make it work.

library(ROI)
library(ROI.plugin.glpk)
library(ompr)
library(ompr.roi)
library(dplyr)

f <- function(y, k, L, alpha){
    MIPModel() %>%
    add_variable(A) %>%
    set_objective(0) %>%
    add_constraint((((y/(k^alpha))^(1/((1-alpha))))/L) == A) %>%
    solve_model(with_ROI(solver = "glpk")) %>% 
    get_solution(A)
}

apply(data[,3:5], 1, function(x) f(y=x['y'],k=x['k'],L=x['L'], alpha=0.3))
#[1] 0.71968567 0.35984284 0.17992142 0.08996071
apply(data[,3:5], 1, function(x) f(y=x['y'],k=x['k'],L=x['L'], alpha=1/3))
#[1] 1.00000000 4.77297077 0.01581139 0.84375000

This might be an overkill for this example. But is scalable.

Upvotes: 1

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