EC255-PS4-Key

EC 255 PS4 Key

The Investment Club on campus received a donation of $100,000 in 1995 under the condition that students manage the money as part of the college’s endowment. The current value of the portfolio is $2 million. Assuming continuous compounding and that the club has invested for 30 full years, what is the annual return for the Investment Club over this time? How many times has the investment doubled over that time period?

\[ r = \frac{ln(FV/PV)}{t} \\ doubles = \frac{ln(FV/PV)}{ln(2)} \]

PV <- 100000 
FV <- 2000000
t <- 30

r <- round(log(FV/PV)/t,4) # Calculation for Annual Rate of Return

double <- round(log(FV/PV)/log(2),2) # Equation to determine the number of doubles

The annual rate of return for the investment club portfolio is 0.0999. The investment has doubled 4.32 times.

Using the rule of 70, approximately how long would it take for an investment to double with a 5% return?

tm <- 70/5

It will take the investment approximately 14 years for the investment to double.

You are able to secure $100,000 in retirement savings by your 30th Birthday. You would like that sum to double six times in the next 30 years. Use this information to answer the following questions.

PV = 100000 # Present Value
mult = 64 # 2^6Value of 1 doubling six times

FV1 = mult*PV

rate <- round(log(mult)/30,3) # Rate required for the investment to double 6 times in 30 years.

What is the value of your investment after 30 years if it doubles 6 times? 6.4^{6}
What interest rate would be required for your investment to double 6 times over 30 years? 0.139

There is a link to the data set ”C02.txt” available on Moodle. Use this link to answer the following questions.

carbon <- read.table("https://raw.githubusercontent.com/tjdevine/EC255/refs/heads/main/co2.txt",header=TRUE)

names(carbon)

## [1] "year" "CO2"

plot(carbon$CO2 ~ carbon$year, type = "l", main = "CO2 over time")

ln_co2 <- log(carbon$CO2)

plot(ln_co2 ~ carbon$year,type="l",main = "Log_CO2 over time")

# 66 Years

start_carbon <- carbon$CO2[1]
end_carbon <- carbon$CO2[66]

rate <- log(end_carbon/start_carbon)/66 # Find c.c. rate

Read the data into R and create a line plot of CO2 Level against year. Comment on the shape of the relationship.
Create a new variable that is the log transformation of CO2. Plot the log transformed CO2 level against year.
Assuming continuous compounding, what is the annual percentage change in carbon emissions?

The annual percentage change in carbon emissions is 0.0044772

The following plot and table represent the natural log transformations of CO2 concentrations over time. Use both to answer the following questions.

Q5_a <- exp(5.966403)

Q5_b <- log(400)  
  
start_lnco2 <- 5.966403
end_lnco2 <- 6.042823

# years = 14

Q5_c <- round((end_lnco2 - start_lnco2)/14)

What was the concentration of CO2 in 2010? 390.0999545
It was recommended to try and keep CO2 concentrations below 400 PPM. What year did CO2 concentrations climb above 400 for the first time? The log transformed carbon values in 2010 are5.9914645. Carbon concentrations above that value are above 400 PPM. In 2015, carbon concentrations cross that threshold.
Assuming CO2 concentrations are growing at a continuous rate (identical to continuous compounding), what is the average annual percentage increase in CO2 from 2010 to 2023?

Revenue and costs for a firm are the following: R(Q) = 15 Q C(Q) = 10 + Q2 Determine the quantity that maximizes profits.

There are a couple ways to show this. You can plot it out to visualize the maximum. Or, you can use calculus or the optim function.

\[\begin{align*} \pi = -Q^{2} + 15Q - 10 \end{align*}\]

q <- seq(from = 0, to = 15, by = 1)
pi <- -q^2 + 15*q - 10


plot(pi ~ q, type = "l")

q_max <- -15/(-2)
pi_max <- -q_max^2 + 15*q_max - 10
q_max

## [1] 7.5

pi_max

## [1] 46.25

pi_fun <- function(x) {
  -x^{2} + 15*x - 10
}



optimize(pi_fun,maximum = TRUE, interval = c(0,25))

## $maximum
## [1] 7.5
## 
## $objective
## [1] 46.25

The quantity that maximizes profits is 7.5 with a profit level of 46.25