Computational Tools for Economics and Management

Exercise number one

First version

Consider the numbers n₁ and n₂. Then say whether both numbers are small and the first is larger than the second.

Conditions:

• n₁ < 1
• n₂ < 1
• n₁ > n₂

If the result is TRUE: 1.00 FALSE: 0.00

Second version

Consider the numbers n₁ and n₂. Then say whether both numbers are large and n₁ is larger than n₂.

Conditions:

• n₁ > 1
• n₂ > 1
• n₁ > n₂

If the result is TRUE: 1.00 FALSE: 0.00

Exercise number two

Draw the line and the parabola y = f₁, y = f₂ over the interval [n₁, n₂] and find the ordinate of the lowest point belonging to the graphs.

• f <- function(x) f write first the parabola
• g <- function(x) f than the line
• curve(f, n₁, n₂)
• curve(g, n₁, n₂, add = T, col = 2)

Analyse the graph, there can be two alternatives:

• First case: The lowest point coincides with the minimum of the parabola between point a and b.
• Second case: The lowest point coincides with the minimum of the line between point a and b.
• Third case: From the graph, the lowest y corresponds to x = ...

optimize(f, c(a, b))
optimize(g, c(a, b))

Ordinate of the lowest point: $objective

Exercise number three

Consider the expression a + b. Which of the following statements holds?

• The last recycled number was ... or The first recycled number was ...
• There was no need to recycle any argument

Recycling: When vectors have different lengths, in an operation some arguments are replicated to perform the computation

• a <- vector a
• b <- vector b
• a + b (shows the results)
• a (shows the numbers in a)
• b (shows the numbers in b)

Check which was the last or first recycled number:

• ! 1 - 11!

Exercise number four

Let x <- and y <-. Consider the functions g(x) = and h(x) =. Plot both the exponential function and the line functions and the points defined by x and y on the interval [a, b]. Which of the following sentence is true?

• The curve is mostly below the line and ... points are below the graph of the exponential.
• The curve is mostly above the line and ... points are below the graph of the exponential.

• x <- point
• y <- function
• g <- function(x) exp(...)
• h <- function(x) line
• curve(g, a, b)
• curve(h, a, b, add = T, col = 2)
• analyse the graph
• points(x, y)

Exercise number five

First version: vector

Type and press enter. Then immediately store in a vector normal pseudo-random numbers with mean and standard deviation. Compute the sum of components of the vector, starting from position a.

• set.seed(...) v <- rnorm(100, m, sd)
• sum(v[a:((a+b)-1)])
• generates a vector: rnorm(...)

Second version: matrix

Type and press enter. Then immediately create a matrix, filled by columns with (standard) random uniform numbers. Let be the product of the matrix times the vector (n₁, n₂, n₃). Finally, provide the sum of the components of u.

• set.seed(...) Z <- matrix(runif(r*c), r, c, byrow = F)
• v <- c(n₁, n₂, n₃)
• u <- Z %*% v
• sum(u) → answer
• To create a matrix by rows instead of by columns, use the argument byrow = T
• generates random uniform numbers runif(...)

Exercise number six

First version: maximize

Maximize, if possible, the following function over the interval [a, b]: f₁(x) =. Write the objective function at the optimum or -999 if there is no solution.

• f <- function(x) func
• curve(f, a, b)
• maxlooking at the graph you can see x = ... point can be found in an interval
• f(...) optimize(f, c(a, b), maximum = T)

Second version: minimize

Minimize, if possible, the following function over the interval [a, b]: f₁(x) =. Write the objective function at the optimum or -999 if there is no solution.

• f <- function(x) func
• curve(f, a, b)
• looking at the graph you can see x = ... min point can be found in an interval
• f(...) optimize(f, c(a, b))
• ! 2 - 11! or: value of min and max on the x-axis | value on the y-axis $minimum $maximum $objective

Exercise number seven

Adds lines to a previously drawn plot: abline

First version

First case

In the smartphone market, the number of customers (in billions) of Samsung and Apple can be modeled by the functions:

• f(x) = x/100 + 0.17e^0.079x
• g(x) = 0.32 + 0.024x

Where x > 0 denotes year 2000 + x.

· When is Samsung reaching one billion customers for the first time?

• f <- function(x) x/100 + 0.17 * exp(0.079 * x)
• h <- function(x) f(x) - 1
• x refers to years; we must consider only positive values
• curve(h(x), 0, 100)
• since...
• curve(h(x), 0, 50)
• curve(h(x), 0, ...)
• abline(h = 0, col = 2)
• grid(col = 3)
• shown in graph
• uniroot(h, c(0, ...)) interval
• → $root correct result

Second case

In the smartphone market, the number of customers (in billions) of Motorola and Apple can be modeled by the functions:

• f(x) = 0.21e^0.088x
• g(x) = 0.3 + 0.024x

Where x denotes year 2000 + x.

· When are the two competitors going to have the same number of customers?

• f <- function(x) 0.21 * exp(0.088 * x)
• g <- function(x) 0.3 + 0.024 * x
• h <- function(x) f(x) - g(x)
• curve(h, 0, 50)
• curve(h, 0, ...)
• abline(h = 0, col = 2)
• uniroot(h, c(10, 15))
• → $root correct result

Second version:

First case: Find the rightmost root of...