How to Use optim Function in R (2 Examples)

You can use the optim function in R for general-purpose optimizations.

This function uses the following basic syntax:

optim(par, fn, data, ...)

where:

par: Initial values for the parameters to be optimized over
fn: A function to be minimized or maximized
data: The name of the object in R that contains the data

The following examples show how to use this function in the following scenarios:

1. Find coefficients for a linear regression model.

2. Find coefficients for a quadratic regression model.

Let’s jump in!

Example 1: Find Coefficients for Linear Regression Model

The following code shows how to use the optim() function to find the coefficients for a linear regression model by minimizing the residual sum of squares:

#create data frame
df <- data.frame(x=c(1, 3, 3, 5, 6, 7, 9, 12),
                 y=c(4, 5, 8, 6, 9, 10, 13, 17))

#define function to minimize residual sum of squares
min_residuals <- function(data, par) {
                   with(data, sum((par[1] + par[2] * x - y)^2))
}

#find coefficients of linear regression model
optim(par=c(0, 1), fn=min_residuals, data=df)

$par
[1] 2.318592 1.162012

$value
[1] 11.15084

$counts
function gradient 
      79       NA 

$convergence
[1] 0

$message
NULL

Using the values returned under $par, we can write the following fitted linear regression model:

y = 2.318 + 1.162x

We can verify this is correct by using the built-in lm() function in R to calculate the regression coefficients:

#find coefficients of linear regression model using lm() function
lm(y ~ x, data=df)

Call:
lm(formula = y ~ x, data = df)

Coefficients:
(Intercept)            x  
      2.318        1.162

These coefficient values match the ones we calculated using the optim() function.

Example 2: Find Coefficients for Quadratic Regression Model

The following code shows how to use the optim() function to find the coefficients for a quadratic regression model by minimizing the residual sum of squares:

#create data frame
df <- data.frame(x=c(6, 9, 12, 14, 30, 35, 40, 47, 51, 55, 60),
                 y=c(14, 28, 50, 70, 89, 94, 90, 75, 59, 44, 27))

#define function to minimize residual sum of squares
min_residuals <- function(data, par) {
                   with(data, sum((par[1] + par[2]*x + par[3]*x^2 - y)^2))
}

#find coefficients of quadratic regression model
optim(par=c(0, 0, 0), fn=min_residuals, data=df)

$par
[1] -18.261320   6.744531  -0.101201

$value
[1] 309.3412

$counts
function gradient 
     218       NA 

$convergence
[1] 0

$message
NULL

Using the values returned under $par, we can write the following fitted quadratic regression model:

y = -18.261 + 6.744x – 0.101x²

We can verify this is correct by using the built-in lm() function in R:

#create data frame
df <- data.frame(x=c(6, 9, 12, 14, 30, 35, 40, 47, 51, 55, 60),
                 y=c(14, 28, 50, 70, 89, 94, 90, 75, 59, 44, 27))

#create a new variable for x^2
df$x2 <- df$x^2

#fit quadratic regression model
quadraticModel <- lm(y ~ x + x2, data=df)

#display coefficients of quadratic regression model
summary(quadraticModel)$coef

               Estimate  Std. Error    t value     Pr(>|t|)
(Intercept) -18.2536400 6.185069026  -2.951243 1.839072e-02
x             6.7443581 0.485515334  13.891133 6.978849e-07
x2           -0.1011996 0.007460089 -13.565470 8.378822e-07

These coefficient values match the ones we calculated using the optim() function.