Instructions

Main instructions. Select one public firm from any country that has a relatively poor credit rating like B or C according to Moodys and propose at least three alternatives to improve its credit rating for the following years. Further instructions/recommendations. Your whole analysis and alternatives have to rely on the Merton credit risk model. Make sure you have the required data in order to estimate your model. Taking Moodys as a reference is better because you could use Table 24.1 from Hull to justify that your proposals actually change the probability of default in such a way that Moodys would have to improve its credit rating. Your alternatives should represent changes in the capital structure of the firm: change the debt, change the equity, and change the debt and the equity. You may need to incorporate proposals for more than one period, like 1 year, 5 years and 10 years, according to the available data and the firms' debt structure. Deliverables. You have to submit one report as a Jupyter notebook created in Deepnote and one video. The Jupyter notebook should have loaded all the required data that you use and your R code. You have to clearly explain both methodology and results (intermediate and final). You are free to propose the rest of the structure as long as it is clear, easy to follow, and with a comprehensive set of explanations and interpretations. Please submit your video not as a file, but as a YouTube link so you can delete it and avoid further distributions if necessary. The video is special because you have to assume that you are presenting your results to the actual CEO of the firm. The video must have the following stucture: (1) Who are you and what is your main objective; (2) briefly show the actual situation of the firm; (3) show your proposals and demonstrate these are reasonable to tackle your main objective; (4) a brief conclusion and farewell. Please note that the video is not about reading your detailed report, it is about the four parts above. Marks. Report: 65%. Video: 35%. Mechanics. Let's use Deepnote to develop this assignment. Follow these steps: (1) In your Deepnote dashboard select create new team; (2) in the Free Pro plan for teachers, students or community organizers select learn more; (3) select Select education plan; (4) type your team name, type your udem email, and click continue. Once you have your own team, click on new project, and select R 4.0 as the environment. Finally, invite your team members using their udem email, and add me as well martin.lozano@udem.edu. If you want me to see something you can write a comment with @martin.lozano@udem.edu and I will receive a notification. I may also add comments to your notebook. Bonus. If the report and video is good enough you may consider to send it to the actual CEO. This may have different outcomes. Probably the CEO ignores you (I am afraid this may be the most likely outcome). Probably the CEO minimize your approach because you rely in one model and in limited set of information, so he or she may ignore the potential of someone who can produce a credit risk report without even knowing the firm from the inside. Probably the CEO reply with compliments. Probably the CEO would be interested in knowing more about you and your professional services. Some of your colleagues have done their PEF (under my supervision) based on a research/consulting approach with very nice results, the last one was for Axtel.

Introduction

In this assignment I will analyze the credit rating of a public company with a low rating such as B or C according to Moody's and propose three alternatives to improve its credit rating in the following years. In this report I decided to analyze the company Alsea, which has a credit rating of B1, this indicates that this company lacks the characteristics of a desirable investment and is subject to a high credit risk. First let’s talk about Credit risk, this refers to the risk of loss that someone or some institution is exposed to, it may occur from the failure of any party to avoid the terms and conditions of any financial contract like for example the failure to make required payments on loans. One of the principal objectives of the risk management area is to ensure that it understands, measures and monitors the different risks that the bank or company faces and the possible emerging risks that could appear in a future. In order to assess the credit risk associated with the different financial proposals, the risk management division of the institution first needs to assess a variety of risks relating to the borrower and the industry.

The Merton model

The Merton model is useful when we are interested to evaluate the credit risk of public firms. In short, it evaluates how likely is that the value of the firm’s assets fall in the future below a certain threshold represented by the firm’s debt. By doing that, the model is able to estimate a probability of default among other results. The model can be used as a tool to propose changes (for example) in the balance sheet to reduce the credit risk exposure of a firm. In this case I will us the merton model to find three alternatives that the company could use to increase its credit rating by decreasing its probability of default.

# Merton model install.packages("knitr") install.packages("Sim.DiffProc") install.packages("bazar") install.packages("scatterplot3d") library(knitr) library(Sim.DiffProc) library(bazar) library(scatterplot3d)

# List of known parameters. E0 <- 82.882851 # Value of the equity as today. se <- 1.388978 # Stock returns volatility. rf <- 0.04 # Risk-free rate. TT <- 1 # Maturity. D <- 75.249007 # Value of the debt. The Bloomberg function DDIS is useful.

To estimate the probability of default, it is necessary to have the data of these 5 variables, which were recovered from Grupo BMV in which it shows us the financial information of ALSEA, the first value we need is the value of the equity as today which showed us a value of 82,882,851, in order to interpret these values in the graphs it was decided to represent the numbers in millions so that the values were smaller, the second value captured was the Stock returns volatility this value was obtained from Investing, from which I used the historical volatility of March 4 with a period of 30 days, then calculated the standard deviation with the following calculation: square root (daily return - the average return) + 2, the daily volatility was obtained with this calculation, and to convert it to annual, that result was multiplied by the square root of 250, which gave me the result of 138.8979, the Risk free rate and the value of the debt was also obtain from Grupo BMV, and for the maturity parameter I consider a time frame of 1 year.

With these five parameters: E0 = 82.882851, σE = 1.388978, rf = 0.04, TT = 1, D = 75.249007 we can lead to an estimate of the value of assets today V0 and the volatility of assets σV. With these seven parameters we can estimate the probability of default (pd) among other results.

# We need equations 24.3 and 24.4 to estimate V0 and sv. eq24.3 <- function(V0, sv) { ((V0 * pnorm((log(V0 / D) + (rf + sv^2 / 2) * TT) / (sv * sqrt(TT))) - D * exp(-rf * TT) * pnorm(((log(V0 / D) + (rf + sv^2 / 2) * TT) / (sv * sqrt(TT))) - sv * sqrt(TT)) - E0)) } eq24.4 <- function(V0, sv) { ((pnorm((log(V0 / D) + (rf + sv^2 / 2) * TT) / (sv * sqrt(TT))) * sv * V0 - se * E0)) } # Footnote 10 indicates that we should minimize F(x,y)ˆ2+G(x,y)ˆ2 min.footnote10 <- function(x) (eq24.3(x[1], x[2]))^2 + (eq24.4(x[1], x[2]))^2 # The minimization leads to the values of V0 and sv. V0_sv <- optim(c(1, 1), min.footnote10) # Define the values as parameters. V0 <- V0_sv$par[1] sv <- V0_sv$par[2] # Calculate the probability of default as a function. PD <- function(V0, sv) { pnorm(-(((log(V0 / D) + (rf + sv^2 / 2) * TT) / (sv * sqrt(TT))) - sv * sqrt(TT))) } # Calculate the probability of default given the previous parameters. pd <- PD(V0, sv) V0 sv pd

# Gather the results. ans <- data.frame(pd, V0, sv) kable(ans, caption = "Main model results.")

As this table shows us, the estimate of the value of assets is 26.74795 (V0), the volatility of assets is 0.3306842 (sv), and the probability of default is 0.5406482 (pd), this shows us a greater than 50% chance of default which indicates a bad credit rating

# Inspect the role of d and N(d). d1 <- (log(V0 / D) + (rf + sv^2 / 2) * TT) / (sv * sqrt(TT)) d2 <- d1 - sv * sqrt(TT) Nd1 <- pnorm(d1) Nd2 <- pnorm(d2) ans <- data.frame(d1, Nd1, d2, Nd2) kable(ans, caption = "Inspect the role of d and N(d).")

xseq <- seq(-4, 4, 0.01) densities <- dnorm(xseq, 0, 1) # Standard normal distribution. # Graphical evaluation of N(-d_2). plot(xseq, densities, xlab = "d values (d_2 is dotted line)", ylab = "Density", type = "l", lwd = 2, cex = 2) abline(h = 0) polygon(c(xseq[xseq >= d2], d2), c(densities[xseq >= d2], 0), col = "red") polygon(c(xseq[xseq < d2], d2), c(densities[xseq < d2], 0), col = "blue") legend("topleft", legend = c("N(d_2)=0.624836 is represented in blue and N(-d_2)=0.375164 in red. N(d_2)+N(-d_2)=1"), bg = "white", bty = "n", cex = 0.9) abline(v = d2, lty = 2, lwd = 2)

this histogram shows us Nd2 in blue, which is the probability that the company does not default with a probability of 62.48%, and in red it shows us the N-d2 which is the probability of defaults, with a probability of 37.52%

# Evaluate how good is the minimization. min.footnote10(x = c(V0, sv))

# Fix sv and evaluate different values of V0. V0.seq <- seq(from = 4, to = 300, length.out=50) sv.rep <- rep(sv, 50) # Function to be evaluated. FG <- function(V0, sv) { (eq24.3(V0, sv))^2 + (eq24.4(V0, sv))^2 } # Apply the function with fixed sv and different V0 values. FG.V0 <- mapply(FG, V0.seq, sv.rep)

# Plot the results. plot(V0.seq, FG.V0, type = "l", xlab = expression(paste(V[0])), ylab = expression(paste(F(x,y)^2+G(x,y)^2)), cex.lab = 0.8, lwd = 3) abline(v = V0, lty = 2) abline(h = FG(V0, sv), lty = 2) points(V0, FG(V0, sv), pch = 1, cex = 3, col = "red", lwd = 2)

Here we fix the volatility of the assets and change the value of the assets at time zero

# Now fix V0, and evaluate different sv values. sv.seq <- seq(0, 2, length.out = 50) V0.rep <- rep(V0, 50) # Evaluate the function with these parameters. FG.sv <- mapply(FG, V0.rep, sv.seq) # Plot the results. plot(sv.seq, FG.sv, type = "l", xlab = expression(paste(sigma[V])), ylab = expression(paste(F(x, y)^2 + G(x, y)^2)), cex.lab = 0.8, lwd = 3) abline(v = sv, lty = 2) abline(h = FG(V0, sv), lty = 2) points(sv, FG(V0, sv), pch = 1, cex = 3, col = "red", lwd = 2)

Here we fix the value of the assets at time zero and change the volatility of the assets.

# outer evaluates the function for each combination of the vectors. z <- outer(V0.seq, sv.seq, FG) # z is now 50x50 matrix. persp(V0.seq, sv.seq, z, col = "springgreen", shade = 0, xlab = "V0", ylab = "sv", zlab = "Probability of default")

# The nlevels is high to emphasize the minimum value in blue. contour(V0.seq, sv.seq, z, xlab = "V0", ylab = "sv", lwd = 2, nlevels = 300) points(V0, sv, pch = 19, col = "blue", cex = 2)

# Other results in the Merton's model. market_value_debt <- V0 - E0 pv_promised_payment_debt <- D * exp(-rf * TT) Expected_Loss <- (pv_promised_payment_debt - market_value_debt) / pv_promised_payment_debt recovery_rate <- 1 - (Expected_Loss / pd) ans <- data.frame(market_value_debt, pv_promised_payment_debt, Expected_Loss, recovery_rate, pd) kable(t(ans), caption = "Other results in the Merton's model.")

spread <- function(E0, se, rf, TT, D) { eq24.3 <- function(V0, sv) { ((V0 * pnorm((log(V0 / D) + (rf + sv^2 / 2) * TT) / (sv * sqrt(TT))) - D * exp(-rf * TT) * pnorm(((log(V0 / D) + (rf + sv^2 / 2) * TT) / (sv * sqrt(TT))) - sv * sqrt(TT)) - E0)) } eq24.4 <- function(V0, sv) { ((pnorm((log(V0 / D) + (rf + sv^2 / 2) * TT) / (sv * sqrt(TT))) * sv * V0 - se * E0)) } min.footnote10 <- function(x) (eq24.3(x[1], x[2]))^2 + (eq24.4(x[1], x[2]))^2 V0_sv <- optim(c(1, 1), min.footnote10) V0 <- V0_sv$par[1] sv <- V0_sv$par[2] pd <- pnorm(-(((log(V0 / D) + (rf + sv^2 / 2) * TT) / (sv * sqrt(TT))) - sv * sqrt(TT))) market_value_debt <- V0 - E0 pv_promised_payment_debt <- D * exp(-rf * TT) Expected_Loss <- (pv_promised_payment_debt - market_value_debt) / pv_promised_payment_debt recovery_rate <- 1 - (Expected_Loss / pd) spread <- pd * (1-recovery_rate) spread} spread(E0 = 82.882851, se = 1.388978, rf = 0.04, TT = 1, D = 75.249007)

x.T <- seq(0.5, 10, 0.5) spread.T <- mapply(spread, E0 = 82.882851, se = 1.388978, rf = 0.04, x.T, D = 75.249007) data.frame(x.T, spread.T)

Inside view of the Merton’s model

# Merton assumes V follows a geometric brownian motion process. V0.sim <- function(i) { # the argument i is not used here. GBM(N = 365, T = 1, t0 = 0, x0 = V0, theta = 0.05, sigma = sv) } set.seed(3) # Reproducibility paths <- sapply(1:10, V0.sim) # Create 10 paths for V. # Plot results. matplot(paths, type ="l", col = "black", lwd = 1, lty = 1, ylab = expression(paste(V[t])), xlab = "Time in days (0 to T)") abline(h = D, col = "red", lwd = 2) points(1, V0, pch = 19, cex = 1.5, col = "blue")

# Which path went lower than D? which(colSums(paths < 75.249007) > 0)

# There might be an easier way to select these cases. matplot(paths[,which(colSums(paths < 75.249007) > 0)], type = "l", col = c("green", "blue", "red"), lwd = 2, lty = 1, ylab = expression(paste(V[t])), xlab = "Time in days (0 to T)") abline(h = D, col = "red", lwd = 2) points(1, V0, pch = 19, cex = 1.5, col = "blue")

These 5 paths show us the cases in which the assets of the company are not high enough to pay the debt at maturity, so this simulation shows us a 40% probability of default. In this case the 40% probability of default got very close to 0.375164 but even so it is a little far from this probability. This is because the number of simulations is small. Example below show how these values tend to converge as we increase the number of simulations from 100 to 100,000.

sum(paths[366,] < 75.249007) / 10

set.seed(3) # Reproducibility paths <- sapply(1:100, V0.sim) # Create 100 paths. # Plot results. matplot(paths, type = "l", col = "black", lwd = 1, lty = 1, ylab = expression(paste(V[t])), xlab = "Time in days (0 to T)") abline(h = D, col = "red", lwd = 2) points(1, V0, pch = 19, cex = 1.5, col = "blue")

sum(paths[366,] < 75.249007) / 100

With this graph we can now see how by performing more simulations (in this case 100 simulations) we get closer to the probability of default.

# These are the 35 default cases. matplot(paths[,which(paths[366,] < 75.249007)], type = "l", col = "black", lwd = 1, lty = 1, ylab = expression(paste(V[t])), xlab = "Time in days (0 to T)") abline(h = D, col = "red", lwd = 2) points(1, V0, pch = 19, cex = 1.5, col = "blue")

# V_t < 75.249007 almost_default <- which(colSums(paths < 75.249007) > 0) # V_T < 75.249007 default <- which((paths[366,] < 75.249007)) # There should be an easier way to do this: matplot(paths[,almost_default[which(almost_default %in% default == FALSE)]], type = "l", col = c(1:11), lwd = 1, lty = 1, ylab = expression(paste(V[t])), xlab = "Time in days (0 to T)") abline(h = D, col = "red", lwd = 2) points(1, V0, pch = 19, cex = 1.5, col = "blue")

set.seed(3) # Reproducibility paths <- sapply(1:100, V0.sim) # Create 100 paths. # Select those paths with all 366 values lower or equal than V0. select.paths <- which(colSums(paths <= V0) == 366) new.paths <- paths[,select.paths] # Plot. matplot(new.paths, type = "l", col = c(1:100), lwd = 1, lty = 1, ylab = expression(paste(V[t])), xlab = "Time in days (0 to T)") abline(h = D, col = "red", lwd = 2) points(1, V0, pch = 19, cex = 1.5, col = "blue")

# Probability sum(new.paths < 75.249007) / length(new.paths)

set.seed(3) # Reproducibility paths <- sapply(1:10000, V0.sim) # Create 10,000 paths. # Select those paths with all 366 values lower or equal than V0. select.paths <- which(colSums(paths <= V0) == 366) new.paths <- paths[,select.paths] # Plot. matplot(new.paths, type = "l", col = c(1:10000), lwd = 1, lty = 1, ylab = expression(paste(V[t])), xlab = "Time in days (0 to T)") abline(h = D, col = "red", lwd = 2) points(1, V0, pch = 19, cex = 1.5, col = "blue")

# Probability sum(new.paths < 75.249007) / length(new.paths)

set.seed(13) # 100,000 values of VT at once. # Here, I added a TT value twice. This was not a mistake, because in the tutorial this VT <- exp(rnorm(100000, log(V0) + (0.04 - (sv^2) / 2)*TT, sv*sqrt(TT))) # Plot results. # You may need to change the 100 if colors do not show properly. Sometimes this value h <- hist(VT, 100, plot = FALSE) ccat <- cut(h$breaks, c(-Inf, D, Inf)) # You may also need to change the xlim if a big red square appears as changes in matur plot(h, main = NULL, col = c("red", "blue")[ccat], xlab = "Value of the assets at maturity", xlim = c(4, 1000)) legend("topright", legend = c("The red area is 37.438%, and represents the simulated probability of default. The probability of default of the model is N(-d_2)=37.5164%. As we can see, both are very close."), col = c("red", "blue"), pch = c(19), bg = "white", bty = "n", cex = 0.8)

sum(VT < 75.249007) / 100000

With this graph we can now see how by performing more simulations (in this case 100,000 simulations) we get much closer to the probability of default. Now both probabilities are very close.

ET <- pmax(VT - D, 0) # payoff function of a typical call option. plot(sort(VT), sort(ET), type = "l", lwd = 4, xlim = c(5, 200), ylim = c(0, 100), xlab = "Simulated assets at maturity (V_T)", ylab ="Simulated equity at maturity (E_T)") abline(v = 10, lty = 2) legend("right", legend = c("E_T=max(V_T-D,0). Given the simulation, E_T=0 happens in 37.438% of the cases. According to the model, E_T=0 happens in 37.5164% of the cases."), bg = "white", bty = "n", cex = 0.7)

# Black-Scholes call. E0.theo <- function(V0, D, rf, TT, sv) { d1 <- (log(V0 / D) + (rf + sv^2 / 2) * TT) / (sv * sqrt(TT)) d2 <- d1 - sv * sqrt(TT) c <- V0 * pnorm(d1) - D * exp(-rf * TT) * pnorm(d2) } V0 <- 144.16388321583 # Assets. D <- 75.249007 # Debt. rf <- 0.04 # Risk-free rate. TT <- 1 # Maturity. sv <- 0.898983106997233 # Volatility of assets. (E0.theo(V0, D, rf, TT, sv))

Then, we can argue that the market value of the firm is slightly overrated since the theoretical value of the firm is 82.87659 compared with 82.882851. This approach opens the possibility to value firms. It also opens the possibility to compare the book value in the Balance sheet of the total assets versus the estimate V0. This will allow us to understand the difference between book value and market value of the assets, not only of the equity. We can even implement a similar analysis about the market value of the debt just as we did before.

Probability of default as a function of some parameters

pd <- function(E0, se, rf, TT, D) { eq24.3 <- function(V0, sv) { ((V0 * pnorm((log(V0 / D) + (rf + sv^2 / 2) * TT) / (sv * sqrt(TT))) - D * exp(-rf * TT) * pnorm(((log(V0 / D) + (rf + sv^2 / 2) * TT) / (sv * sqrt(TT))) - sv * sqrt(TT)) - E0)) } eq24.4 <- function(V0, sv) { ((pnorm((log(V0 / D) + (rf + sv^2 / 2) * TT) / (sv * sqrt(TT))) * sv * V0 - se * E0)) } min.footnote10 <- function(x) (eq24.3(x[1], x[2]))^2 + (eq24.4(x[1], x[2]))^2 V0_sv <- optim(c(1, 1), min.footnote10) V0 <- V0_sv$par[1] sv <- V0_sv$par[2] pd <- pnorm(-(((log(V0 / D) + (rf + sv^2 / 2) * TT) / (sv * sqrt(TT))) - sv * sqrt(TT))) pd }

# Evaluate the case of the textbook example. pd1 <- pd(82.882851, 1.388978, 0.04, 1, 75.249007) pd1

l = 50 # Vectors of length 50. # Create vectors of the parameters. x.E <- seq(from = 1, to = 200, length.out = l) x.rf <- seq(0, 4, length.out = l) x.D <- seq(1, 80, length.out = l) x.T <- seq(0, 20, length.out = l) x.se <- seq(0, 3, length.out = l) # Evaluate the pd at different values. pd.E <- mapply(pd, x.E, se, rf, TT, D) pd.rf <- mapply(pd, E0, se, x.rf, TT, D) pd.D <- mapply(pd, E0, se, rf, TT, x.D) pd.T <- mapply(pd, E0, se, rf, x.T, D) pd.se <- mapply(pd, E0, x.se, rf, TT, D) pd.E # vemos que hay en pd.E

plot(x.E, pd.E, type = "l", ylab = "Probability of default", xlab = "Equity at time zero", lwd = 3, ylim = c(0, 0.7)) lines(x.E, pd.E) abline(v = E0, lty = 2) abline(h = pd1, lty = 2) points(E0, pd1, pch = 1, cex = 3, col = "red", lwd = 2)

plot(x.rf, pd.rf, type = "l", ylab = "Probability of default", xlab = "Risk-free rate", lwd = 3) abline(h = 0, lty = 2) abline(v = rf, lty = 2) abline(h = pd1, lty = 2) points(rf, pd1, pch = 1, cex = 3, col = "red", lwd = 2)

plot(x.D, pd.D, type = "l", ylab = "Probability of default", xlab = "Debt", lwd = 3, ylim = c(0, 0.5)) abline(h = 0, lty = 2) abline(v = D, lty = 2) abline(h = pd1, lty = 2) points(D, pd1, pch = 1, cex = 3, col = "red", lwd = 2)

plot(x.T, pd.T, type = "l", ylab = "Probability of default", xlab = "Maturity", lwd = 3) abline(h = 0, lty = 2) abline(v = TT, lty = 2) abline(h = pd1, lty = 2) points(TT, pd1, pch = 1, cex = 3, col = "red", lwd = 2)

plot(x.se, pd.se, type = "l", ylab = "Probability of default", xlab = "Standard deviation of equity", lwd = 3) abline(h = 0, lty = 2) abline(v = se, lty = 2) abline(h = pd1, lty = 2) points(se, pd1, pch = 1, cex = 3, col = "red", lwd = 2)

ED <- mapply(pd, x.E, x.se, rf, TT, D) p.ED <- scatterplot3d(x.E, x.se, ED, pch = 16, type = "h", color = 1:50, angle = 100, xlab = "Equity at time zero", ylab = "Volatility of equity", zlab = "Probability of default") p.ED$points3d(E0, se, pd1, type = "h", col = "red", pch = 20, cex = 3)

# Create the empty matrix. p_E_se <- matrix(0, nrow = l, ncol = l) # Fill the empty matrix with probability of default values. for(i in 1 : l){ # Is there an easier way to do this? for(j in 1 : l){ p_E_se[i,j] <- mapply(pd, x.E[i], x.se[j], rf, TT, D) } }

# Plot results. p.ED1 <- persp(x.E, x.se, p_E_se, zlab = "pd", xlab = "Equity at time zero", ylab = "Volatility of equity", theta = 60, phi = 10, expand = 0.5, col = "orange", shade = 0.2, ticktype = "detailed") # Add the original pd value as in the textbook example. points(trans3d(E0, se, pd1, p.ED1), cex = 2, pch = 19, col = "blue")

contour(x.E, x.se, p_E_se, xlab = "Equity at time zero", ylab = "Volatility of equity", lwd = 2) points(E0, se, pch = 19, col = "blue", cex = 2)

GoT: capital structure.

D.seq <- seq(from = 0, to = 100, by = 0.1) # Evaluate pd function: E0 changes from 1 to 5; debt goes from 0 to 13. E1 <- mapply(pd, 62, se, rf, TT, D.seq) E2 <- mapply(pd, 72, se, rf, TT, D.seq) E3 <- mapply(pd, 82, se, rf, TT, D.seq) E4 <- mapply(pd, 92, se, rf, TT, D.seq) E5 <- mapply(pd, 102, se, rf, TT, D.seq)

# We use these vectors for colors and legends in the following plots. colors <- c("green", "purple", "black", "blue", "red") leg <- c("E0=62", "E0=72", "E0=82 (initial value)", "E0=92", "E0=102") # Plot. plot(D.seq, E1, type = "l", col = "green", lwd = 3, xlab = "Debt value. Initially, D=75.249007", ylab = "Probability of default. Initially PD=37.5164%") lines(D.seq, E2, col = "purple", lwd = 3) lines(D.seq, E3, col = "black", lwd = 3) lines(D.seq, E4, col = "blue", lwd = 3) lines(D.seq, E5, col = "red", lwd = 3) abline(v = D, lty = 2) abline(h = pd1, lty = 2) legend("bottomright", legend = leg, lwd = rep(3, 5), col = colors, bg = "white") points(D, pd1, pch = 19, cex = 2)

plot(D.seq, E1, type = "l", col = "green", lwd = 3, xlab = "Debt value. Initially, D=75.25", ylab = "Probability of default. Initially PD=37.5164%") lines(D.seq, E2, col = "purple", lwd = 3) lines(D.seq, E3, col = "black", lwd = 3) lines(D.seq, E4, col = "blue", lwd = 3) lines(D.seq, E5, col = "red", lwd = 3) abline(v = D, lty = 2) abline(v = 55, lty = 2) abline(h = pd1, lty = 2) abline(h = pd(E0, se, rf, TT, 55), lty = 2) abline(h = pd(102, se, rf, TT, D), lty = 2) legend("bottomright", legend = leg, lwd = rep(3, 5), col = colors, bg = "white") points(D, pd1, pch = 19, cex = 2) points(D, pd(102, se, rf, TT, D), pch = 19, cex = 2, col = "red") points(55, pd(82, se, rf, TT, 55), pch = 19, cex = 2, col = "yellow")

plot(D.seq, E1, type = "l", col = "green", lwd = 3, xlab = "Debt value. Initially, D=2", ylab = "Probability of default. Initially PD=7.13%", xlim = c(0, 6), ylim = c(0, 0.08)) lines(D.seq, E2, col = "purple", lwd = 3) lines(D.seq, E3, col = "black", lwd = 3) lines(D.seq, E4, col = "blue", lwd = 3) lines(D.seq, E5, col = "red", lwd = 3) abline(v = 2, lty = 2) abline(v = 1, lty = 2) abline(h = pd(2, se, rf, TT, 2), lty = 2) abline(h = pd(2, se, rf, TT, 1), lty = 2) abline(h = pd(E0, se, rf, TT, 2), lty = 2) legend("bottomright", legend = leg2, lwd = rep(3, 5), col = colors, bg = "white", cex = 0.6) points(2, pd(2, se, rf, TT, 2), pch = 19, cex = 2, col = "purple") points(1, pd(2, se, rf, TT, 1), pch = 19, cex = 2, col = "yellow") points(2, pd(E0, se, rf, TT, 2), pch = 19, cex = 2)

T1 <- mapply(pd, E0, se, rf, 0.5, D.seq) T2 <- mapply(pd, E0, se, rf, 1, D.seq) T3 <- mapply(pd, E0, se, rf, 1.5, D.seq) T4 <- mapply(pd, E0, se, rf, 2, D.seq) T5 <- mapply(pd, E0, se, rf, 2.5, D.seq)

leg3 <- c("T=2.5", "T=2", "T=1.5", "T=1", "T=0.5") plot(D.seq, T5, type = "l", col = "green", lwd = 3, xlab = "Debt value. Initially, D=6", ylab = "Probability of default. Initially PD=20.33%") lines(D.seq, T4, col = "purple", lwd = 3) lines(D.seq, T3, col = "black", lwd = 3) lines(D.seq, T2, col = "blue", lwd = 3) lines(D.seq, T1, col = "red", lwd = 3) abline(v = 6, lty = 2) abline(h = pd(E0, se, rf, 1.5, 6), lty = 2) legend("bottomright", legend = leg3, lwd = rep(3, 5), col = colors, bg = "white") points(6, pd(E0, se, rf, 1.5, 6), pch = 19, cex = 2)

plot(D.seq, T5, type = "l", col = "green", lwd = 3, xlab = "Debt value. Initially, D=6", ylab = "Probability of default. Initially PD=20.33%") lines(D.seq, T4, col = "purple", lwd = 3) lines(D.seq, T3, col = "black", lwd = 3) lines(D.seq, T2, col = "blue", lwd = 3) lines(D.seq, T1, col = "red", lwd = 3) abline(v = 6, lty = 2) abline(v = 8, lty = 2) abline(h = pd(E0, se, rf, 1.5, 6), lty = 2) abline(h = pd(E0, se, rf, 2, 6), lty = 2) abline(h = pd(E0, se, rf, 1.5, 6), lty = 2) abline(h = pd(E0, se, rf, 1.5, 8), lty = 2) legend("bottomright", legend = leg3, lwd = rep(3, 5), col = colors, bg = "white") points(6, pd(E0, se, rf, 1.5, 6), pch = 19, cex = 2) points(8, pd(E0, se, rf, 1.5, 8), pch = 19, cex = 2, col = "yellow") points(6, pd(E0, se, rf, 2, 6), pch = 19, cex = 2, col = "purple")

Conclusion

The main goal of this work was to find 3 alternatives that the company ALSEA could do to improve its credit rating, to achieve this, first I had to analyze the financial information of the company to calculate the probability of default that the company had, by doing this I used many simulators to calculate the number of cases in which the company defaulted so I could have the probability that this could happen. To find the alternatives that the company could use, I carried out a simulation in which I modified the company's capital and the company's debt by increasing and decreasing it so that I could see if the probability of default increased or decreased. With this information I was able to realize that it was better for the company to reduce its debt instead of increasing its capital, so that this would make it less likely to default. After obtaining this alternative, I looked for another option by modifying the maturity of the company, with this information I was able to conclude that it was better for the company to reduce its maturity to have less probability of default. So we can conclude that the alternatives for the company would be to reduce its debt and maturity.

https://youtu.be/mNRkHFKa3cg