Hello, I am trying to numerically solve the dynamic problem of optimal natural resource depletion (aka “eating the cake”) with InfiniteOpt.jl.

The script below works fine, but for the time horizon (i.e. the upper bound of the domain of the infinite dimension “time”) I need to specify it from the analytical resolution that I already have.

Is there a way to let the time horizon “free”, i.e. let it to be endogenously retrieved in the optimisation ?

[Side question] Is is possible to use an objective version that is itself an integral computed with `quadgk`

(like `welfare_auto`

?). I am having errors with it, so I had to compute it manually…

```
# "Eating the cake" model of depletion of natural resources
cd(@__DIR__)
using Pkg
Pkg.activate(".")
# Pkg.add(["QuadGK", "InfiniteOpt", "Ipopt", "Plots"]) # only once to install packages
using QuadGK, InfiniteOpt, Ipopt, Plots
# Parameters
k=100; a = 10 # Parameters of the demand function
S₀ = 1000 # Initial stock
ρ = 0.01 # Discount rate
Sτ = 0 # Final stocks
test = sqrt(2*S₀*a/ρ) # this comes from the analytical resolution that I already have !!
T = test # time horizon
# Demand function at each point in time
demand_price(r;k=k,a=a) = k*exp(-a*r)
# OBJECTIVE FUNCTION
# Welfare (utility) at each point in time
welfare_auto(r;k=k,a=a) = quadgk(q -> demand_price(q,k=k,a=a), 0, r, rtol=1e-4)[1] # quadgk first output is the value of the integral, second value is an estimated error
welfare_manual(r;k=k,a=a) = - k*(exp(-a*r) - 1) / a
discount(t; ρ=ρ) = exp(-ρ*t) # discount function
# Equation of motion of the state variable
var_stocks(r) = -r
opt = Ipopt.Optimizer # desired solver
ns = 1_000; # number of points in the time grid
m = InfiniteModel(opt)
@infinite_parameter(m, t in [0, T], num_supports = ns)
@variable(m, S, Infinite(t)) ## state variables
@variable(m, r, Infinite(t)) ## control variables
@objective(m, Max, integral(welfare_manual(r), t, weight_func = discount))
@constraint(m, S(0) == S₀)
@constraint(m, S(test) == Sτ)
@constraint(m, r >= 0)
@constraint(m, c1, deriv(S, t) == var_stocks(r))
optimize!(m)
termination_status(m)
r_opt = value(r)
S_opt = value(S)
ts = supports(t)
opt_obj = objective_value(m)
ix = 2:(length(ts)-1) # index for plotting
plot(ts[ix], S_opt[ix], lab = "S: resource stock")
plot(ts[ix], r_opt[ix], lab = "r: resource extraction")
plot(ts[ix], demand_price.(r_opt[ix]), lab = "p: price")
# For checking...
price_manual(t;k=k,ρ=ρ,T=T) = k*exp(ρ*(t-T))
plot!(ts[ix], price_manual.(ts[ix]), lab = "p: price (manual")
```