source('utils.R')

r = 0.1

R = 1 + r ## one-year discount factor

u = 1.2 ## the original example
p_up = 0.75

d = 0.9
p_down = 0.27

q = (R - d)/(u - d)

S0 = 400 ## current spot price

## Get the grid of gold spot prices:
##
n_years = 10
SSpot = make_spot_table(S0, u, d, n_years)
print('Spot gold price=')
print(round(SSpot, 1))

## Solve for the_value of the mine (in terms of K):
##
n_steps = n_years+1 ## = number of columns = number of years + the initial spot_value
n_rows = n_steps

K = matrix(NA, nrow=n_rows, ncol=n_steps)
for( jj in seq(n_steps, 1, -1) ){
    for( ii in 1:jj ){
        if( jj == n_steps ){ ## the final column/timestep
            K[ii, jj] = 0
        }else{
            if( jj == (n_steps-1) ){ ## the one before last column/timestep
	    	g_hat = q*SSpot[ii, jj+1] + (1-q)*SSpot[ii+1, jj+1]
                g = (SSpot[ii, jj] + g_hat)/2
                K[ii, jj] = (g/R)^2 / 2000
            }else{
                K_right = K[ii, jj+1] ## the node just to the right of this spot
                K_prime = K[ii+1, jj+1] ## the node just to the right and below this spot
                K_hat = q*K_right + (1-q)*K_prime
	    	g_hat = q*SSpot[ii, jj+1] + (1-q)*SSpot[ii+1, jj+1]
                g = (SSpot[ii, jj] + g_hat)/2
                K[ii, jj] = (g - K_hat)^2 / ( 2000 * R^2 ) + K_hat/R ## moving R will include the discount factor into the profit
            }
        }
    }
}
print('K=')
print(round(K, 1))

x0 = 50000 ## initial amount of gold in the mine
V0 = K[1, 1] * x0
print(sprintf('Value of the lease= %f (MM)', V0/1e6) )