%matplotlib inline
from modsim import *
import pandas as pd
import matplotlib.pyplot as plt

plt.style.use('ggplot')

filename = 'data/World_population_estimates.html'
tables = pd.read_html(filename, header=0, index_col=0, decimal='M')
table2 = tables[2]
table2.columns = ['census', 'prb', 'un', 'maddison',
                  'hyde', 'tanton', 'biraben', 'mj',
                  'thomlinson', 'durand', 'clark']

census = table2.census / 1e9
t0 = census.index[0]
p0 = census[t0]
t_end = census.index[-1]
system = System(t0=t0,
                t_end=t_end,
                p0=p0,
                alpha=0.025,
                beta=-0.0018)


def run_simulation(system, update_func):
    """Run a model.

    Adds TimeSeries to `system` as `results`.

    system: System object
    update_func: function that computes the population next year
    """
    results = Series([])
    results[system.t0] = system.p0
    for t in linrange(system.t0, system.t_end):
        results[t + 1] = update_func(results[t], t, system)
    system.results = results


def update_func2(pop, t, system):
    """Update population based on a quadratic model.

    pop: current population in billions
    t: what year it is
    system: system object with model parameters
    """
    net_growth = system.alpha * pop + system.beta * pop**2
    return pop + net_growth

What happens if we start with an initial population above the carrying capacity, like 20 billion? The the model with initial populations between 1 and 20 billion, and plot the results on the same axes.

ax = census.plot()
for system.p0 in range(1,20):
    temp_sys = system.copy()
    run_simulation(temp_sys, update_func2)
    temp_sys.results.plot(ax=ax)

Suppose there are two banks across the street from each other, The First Geometric Bank (FGB) and Exponential Savings and Loan (ESL). They offer the same interest rate on checking accounts, 3%, but at FGB, they compute and pay interest at the end of each year, and at ESL they compound interest continuously.

If you deposit $p_0$ dollars at FGB at the beginning of Year 0, the balanace of your account at the end of Year $n$ is

$ x_n = p_0 (1 + α)^n $

where $α = 0.03$. At ESL, your balance at any time $t$ would be

$ x(t) = p_0 exp(α t) $

If you deposit \$1000 at each back at the beginning of Year 0, how much would you have in each account after 10 years?

Is there an interest rate FGB could pay so that your balance at the end of each year would be the same at both banks? What is it?

Hint: modsim provides a function called exp, which is a wrapper for the NumPy function exp.

import sympy as sp
import numpy as np

p0, a, n, t = sp.symbols('p0 a n t')
f = sp.Function('f')
fgb = p0 * (1 + a) ** t
esl = p0 * sp.exp(a * t)
fgbamnt = fgb.subs(a, 0.03).subs(p0, 1000).subs(t,10)
eslamnt = esl.subs(a, 0.03).subs(p0, 1000).subs(t, 10)
fgbamnt,eslamnt

Because both equations are fundamentally different, α for FGB would have to change every year to match ESL.

Suppose a new bank opens called the Polynomial Credit Union (PCU). In order to compete with First Geometric Bank and Exponential Savings and Loan, PCU offers a parabolic savings account where the balance is a polynomial function of time:

$ x(t) = p_0 + β_1 t + β_2 t^2 $

As a special deal, they offer an account with $β_1 = 30$ and $β_2 = 0.5$, with those parameters guaranteed for life. Suppose you deposit \$1000 at all three banks at the beginning of Year 0. How much would you have in each account at the end of Year 10? How about Year 20? And Year 100?

b1, b2 = sp.symbols('b1 b2')
pcu = p0 + b1 * t + b2 * t ** 2
pcuyrs = pcu.subs(p0, 1000).subs(b1, 30).subs(b2, 0.5)
fgbyrs = fgb.subs(a, 0.03).subs(p0, 1000)
eslyrs = esl.subs(a, 0.03).subs(p0, 1000)
yrs = 10, 20, 100

amnts = pd.DataFrame({'PCU': [pcuyrs.subs(t, y) for y in yrs], 'FGB': [
                     fgbyrs.subs(t, y) for y in yrs], 'ESL': [eslyrs.subs(t, y) for y in yrs]})
amnts = amnts.set_index(pd.Series(yrs))
print(amnts)

                  ESL               FGB               PCU
10   1349.85880757600  1343.91637934412  1350.00000000000
20   1822.11880039051  1806.11123466941  1800.00000000000
100  20085.5369231877  19218.6319808563  9000.00000000000