source('../Chapter12/utils.R')
source('../Chapter3/utils.R')

sigma = 0.18
u = exp(sigma)
d = 1/u

r = 0.10

S0 = 100 ## the inital value of the house

R = 1 + r ## total return over the next year

q = (R - d)/(u - d)
print(c(u, d, R, q))

n_years = 15

delta_t = 1 ## yearly steps

## Build the value lattice:
##
SSpot = make_spot_table(S0, u, d, n_years)
print('Spot value of the house=')
print(round(SSpot, 1))

## Build the value of the loan (the mortgage balance i.e. what you still have to pay):
##
LA = c(0.9*S0) ## finance 90% of the loan initially
A = annuity_payment(LA[length(LA)], n_years, r) ## what we will pay each year
print(sprintf('Yearly payment amount A= %f', A))
for( ye in 1:n_years ){
    new_loan_value = LA[length(LA)] * R - A ## the loan value increases due to interest and then we pay off an amount A
    LA = c(LA, new_loan_value)
}
print("Value of loan='")
print(data.frame(start_of_year=seq(1, 16), loan_amount=round(LA, 2)))

##
## Working backwards lets evaluate the option to default:
##
n_steps = dim(SSpot)[2] # = number of columns = number of time units + one column for the initial spot value
n_rows = n_steps

V = matrix(NA, nrow=n_rows, ncol=n_steps)
for( jj in seq(n_rows, 1, -1) ){
    for( ii in 1:jj ){
        if( jj == n_steps ){ ## the final column/timestep
            V[ii, jj] = max(LA[jj] - 1.05 * SSpot[ii, jj], 0) ## amount saved if we default
        }else{
            pv_option = (q*V[ii, jj+1] + (1-q) *V[ii+1, jj+1])/R 
            amt_saved_if_default = LA[jj] - 1.05 * SSpot[ii, jj]
            V[ii, jj] = max( c(pv_option, amt_saved_if_default) )
        }
    }
}
print(round(V, 2))
print(sprintf('Put value= %f', V[1, 1]))