#!/usr/bin/env python
# coding: utf-8

# # DiscreteDP Example: Automobile Replacement

# **Daisuke Oyama**
# 
# *Faculty of Economics, University of Tokyo*

# We study the finite-state version of the automobile replacement problem as considered in
# Rust (1996, Section 4.2.2).

# * J. Rust, "Numerical Dynamic Programming in Economics",
#   <i>Handbook of Computational Economics</i>, Volume 1, 619-729, 1996.

# In[1]:


get_ipython().run_line_magic('matplotlib', 'inline')


# In[2]:


from __future__ import division, print_function
import numpy as np
import itertools
import scipy.optimize
import matplotlib.pyplot as plt
import pandas as pd
from quantecon.markov import DiscreteDP


# ## Setup

# In[3]:


lambd = 0.5  # Exponential distribution parameter
c = 200  # (Constant) marginal cost of maintainance
net_price = 10**5  # Replacement cost


# In[4]:


n = 100  # Number of states; s = 0, ..., n-1: level of utilization of the asset
m = 2  # Number of actions; 0: keep, 1: replace


# In[5]:


# Reward array
R = np.empty((n, m))
R[:, 0] = -c * np.arange(n)  # Costs for maintainance
R[:, 1] = -net_price - c * 0  # Costs for replacement


# In[6]:


# Transition probability array
# For each state s, s' distributes over
# s, s+1, ..., min{s+supp_size-1, n-1} if a = 0
# 0, 1, ..., supp_size-1 if a = 1
# according to the (discretized and truncated) exponential distribution
# with parameter lambd
supp_size = 12
probs = np.empty(supp_size)
probs[0] = 1 - np.exp(-lambd * 0.5)
for j in range(1, supp_size-1):
    probs[j] = np.exp(-lambd * (j - 0.5)) - np.exp(-lambd * (j + 0.5))
probs[supp_size-1] = 1 - np.sum(probs[:-1])

Q = np.zeros((n, m, n))

# a = 0
for i in range(n-supp_size):
    Q[i, 0, i:i+supp_size] = probs
for k in range(supp_size):
    Q[n-supp_size+k, 0, n-supp_size+k:] = probs[:supp_size-k]/probs[:supp_size-k].sum()

# a = 1
for i in range(n):
    Q[i, 1, :supp_size] = probs


# In[7]:


# Discount factor
beta = 0.95


# ## Continuous-state benchmark

# Let us compute the value function of the continuous-state version
# as described in equations (2.22) and (2.23) in Section 2.3.

# In[8]:


def f(x, s):
    return (c/(1-beta)) * \
        ((x-s) - (beta/(lambd*(1-beta))) * (1 - np.exp(-lambd*(1-beta)*(x-s))))


# The optimal stopping boundary $\gamma$ for the contiuous-state version, given by (2.23):

# In[9]:


gamma = scipy.optimize.brentq(lambda x: f(x, 0) - net_price, 0, 100)
print(gamma)


# The value function for the continuous-state version, given by (2.24):

# In[10]:


def value_func_cont_time(s):
    return -c*gamma/(1-beta) + (s < gamma) * f(gamma, s)


# In[11]:


v_cont = value_func_cont_time(np.arange(n))


# ## Solving the problem with `DiscreteDP`

# Construct a `DiscreteDP` instance for the disrete-state version:

# In[12]:


ddp = DiscreteDP(R, Q, beta)


# Let us solve the decision problem by
# 
# (0) value iteration,  
# (1) value iteration with span-based termination
# (equivalent to modified policy iteration with step $k = 0$),  
# (2) policy iteration,  
# (3) modified policy iteration.

# Following Rust (1996), we set:
# 
# * $\varepsilon = 1164$ (for value iteration and modified policy iteration),
# * $v^0 \equiv 0$,
# * the number of iteration for iterative policy evaluation $k = 20$.

# In[13]:


v_init = np.zeros(ddp.num_states)
epsilon = 1164
methods = ['vi', 'mpi', 'pi', 'mpi']
labels = ['Value iteration', 'Value iteration with span-based termination',
          'Policy iteration', 'Modified policy iteration']
results = {}

for i in range(4):
    k = 20 if labels[i] == 'Modified policy iteration' else 0
    results[labels[i]] = \
        ddp.solve(method=methods[i], v_init=v_init, epsilon=epsilon, k=k)


# In[14]:


columns = [
    'Iterations', 'Time (second)', r'$\lVert v - v_{\mathrm{pi}} \rVert$',
    r'$\overline{b} - \underline{b}$', r'$\lVert v - T(v)\rVert$'
]
df = pd.DataFrame(index=labels, columns=columns)


# The numbers of iterations:

# In[15]:


for label in labels:
    print(results[label].num_iter, '\t' + '(' + label + ')')
    df[columns[0]].loc[label] = results[label].num_iter


# Policy iteration gives the optimal policy:

# In[16]:


print(results['Policy iteration'].sigma)


# Takes action 1 ("replace") if and only if $s \geq \bar{\gamma}$, where $\bar{\gamma}$ is equal to:

# In[17]:


(1-results['Policy iteration'].sigma).sum()


# Check that the other methods gave the correct answer:

# In[18]:


for result in results.values():
    if result != results['Policy iteration']:
        print(np.array_equal(result.sigma, results['Policy iteration'].sigma))


# The deviations of the returned value function from the continuous-state benchmark:

# In[19]:


diffs_cont = {}
for label in labels:
    diffs_cont[label] = np.abs(results[label].v - v_cont).max()
    print(diffs_cont[label], '\t' + '(' + label + ')')


# In[20]:


label = 'Policy iteration'
fig, ax = plt.subplots(figsize=(8,5))
ax.plot(-v_cont, label='Continuous-state')
ax.plot(-results[label].v, label=label)
ax.set_title('Comparison of discrete vs. continuous value functions')
ax.ticklabel_format(style='sci', axis='y', scilimits=(0,0))
ax.set_xlabel('State')
ax.set_ylabel(r'Value $\times\ (-1)$')
plt.legend(loc=4)
plt.show()


# In the following we try to reproduce Table 14.1 in Rust (1996), p.660,
# although the precise definitions and procedures there are not very clear.

# The maximum absolute differences of $v$ from that by policy iteration:

# In[21]:


for label in labels:
    diff_pi = \
        np.abs(results[label].v - results['Policy iteration'].v).max()
    print(diff_pi, '\t' + '(' + label + ')')
    df[columns[2]].loc[label] = diff_pi


# Compute $\lVert v - T(v)\rVert$:

# In[22]:


for label in labels:
    v = results[label].v
    diff_max = \
        np.abs(v - ddp.bellman_operator(v)).max()
    print(diff_max, '\t' + '(' + label + ')')
    df[columns[4]].loc[label] = diff_max


# Next we compute $\overline{b} - \underline{b}$
# for the three methods other than policy iteration, where
# $I$ is the number of iterations required to fulfill the termination condition, and
# $$
# \begin{aligned}
# \underline{b} &= \frac{\beta}{1-\beta} \min\left[T(v^{I-1}) - v^{I-1}\right], \\\\
# \overline{b}  &= \frac{\beta}{1-\beta} \max\left[T(v^{I-1}) - v^{I-1}\right].
# \end{aligned}
# $$

# In[23]:


for i in range(4):
    if labels[i] != 'Policy iteration':
        k = 20 if labels[i] == 'Modified policy iteration' else 0
        res = ddp.solve(method=methods[i], v_init=v_init, k=k,
                        max_iter=results[labels[i]].num_iter-1)
        diff = ddp.bellman_operator(res.v) - res.v
        diff_span = (diff.max() - diff.min()) * ddp.beta / (1 - ddp.beta)
        print(diff_span, '\t' + '(' + labels[i] + ')')
        df[columns[3]].loc[labels[i]] = diff_span


# For policy iteration, while it does not seem really relevant,
# we compute $\overline{b} - \underline{b}$ with the returned value of $v$
# in place of $v^{I-1}$:

# In[24]:


label = 'Policy iteration'
v = results[label].v
diff = ddp.bellman_operator(v) - v
diff_span = (diff.max() - diff.min()) * ddp.beta / (1 - ddp.beta)
print(diff_span, '\t' + '(' + label + ')')
df[columns[3]].loc[label] = diff_span


# Last, time each algorithm:

# In[25]:


for i in range(4):
    k = 20 if labels[i] == 'Modified policy iteration' else 0
    print(labels[i])
    t = get_ipython().run_line_magic('timeit', '-o ddp.solve(method=methods[i], v_init=v_init, epsilon=epsilon, k=k)')
    df[columns[1]].loc[labels[i]] = t.best


# In[26]:


df


# ### Notes

# It appears that our value iteration with span-based termination is different in some details
# from the corresponding algorithm (successive approximation with error bounds) in Rust.
# In returing the value function, our algorithm returns
# $T(v^{I-1}) + (\overline{b} + \underline{b})/2$,
# while Rust's seems to return $v^{I-1} + (\overline{b} + \underline{b})/2$.
# In fact:

# In[27]:


i = 1
k = 0
res = ddp.solve(method=methods[i], v_init=v_init, k=k,
                max_iter=results[labels[i]].num_iter-1)
diff = ddp.bellman_operator(res.v) - res.v
v = res.v + (diff.max() + diff.min()) * ddp.beta / (1 - ddp.beta) / 2


# $\lVert v - v_{\mathrm{pi}}\rVert$:

# In[28]:


np.abs(v - results['Policy iteration'].v).max()


# $\lVert v - T(v)\rVert$:

# In[29]:


np.abs(v - ddp.bellman_operator(v)).max()


# Compare the Table in Rust.

# ## Convergence of trajectories

# Let us plot the convergence of $v^i$ for the four algorithms;
# see also Figure 14.2 in Rust.

# ### Value iteration

# In[30]:


label = 'Value iteration'
iters = [2, 20, 40, 80]
v = np.zeros(ddp.num_states)

fig, ax = plt.subplots(figsize=(8,5))
for i in range(iters[-1]):
    v = ddp.bellman_operator(v)
    if i+1 in iters:
        ax.plot(-v, label='Iteration {0}'.format(i+1))
ax.plot(-results['Policy iteration'].v, label='Fixed Point')

ax.ticklabel_format(style='sci', axis='y', scilimits=(0,0))
ax.set_ylim(0, 2.4e5)
ax.set_yticks([0.4e5 * i for i in range(7)])
ax.set_title(label)
ax.set_xlabel('State')
ax.set_ylabel(r'Value $\times\ (-1)$')
plt.legend(loc=(0.7, 0.2))
plt.show()


# ### Value iteration with span-based termination

# In[31]:


label = 'Value iteration with span-based termination'
iters = [1, 10, 15, 20]
v = np.zeros(ddp.num_states)

fig, ax = plt.subplots(figsize=(8,5))
for i in range(iters[-1]):
    u = ddp.bellman_operator(v)
    if i+1 in iters:
        diff = u - v
        w = u + ((diff.max() + diff.min()) / 2) * ddp.beta / (1 - ddp.beta)
        ax.plot(-w, label='Iteration {0}'.format(i+1))
    v = u
ax.plot(-results['Policy iteration'].v, label='Fixed Point')

ax.ticklabel_format(style='sci', axis='y', scilimits=(0,0))
ax.set_ylim(1.0e5, 2.4e5)
ax.set_yticks([1.0e5+0.2e5 * i for i in range(8)])
ax.set_title(label)
ax.set_xlabel('State')
ax.set_ylabel(r'Value $\times\ (-1)$')
plt.legend(loc=(0.7, 0.2))
plt.show()


# ### Policy iteration

# In[32]:


label = 'Policy iteration'
iters = [1, 2, 3]
v_init = np.zeros(ddp.num_states)

fig, ax = plt.subplots(figsize=(8,5))
sigma = ddp.compute_greedy(v_init)
for i in range(iters[-1]):
    # Policy evaluation
    v_sigma = ddp.evaluate_policy(sigma)
    if i+1 in iters:
        ax.plot(-v_sigma, label='Iteration {0}'.format(i+1))
    # Policy improvement
    new_sigma = ddp.compute_greedy(v_sigma)
    sigma = new_sigma
ax.plot(-results['Policy iteration'].v, label='Fixed Point')

ax.ticklabel_format(style='sci', axis='y', scilimits=(0,0))
ax.set_ylim(1e5, 4.2e5)
ax.set_yticks([1e5 + 0.4e5 * i for i in range(9)])
ax.set_title(label)
ax.set_xlabel('State')
ax.set_ylabel(r'Value $\times\ (-1)$')
plt.legend(loc=4)
plt.show()


# ### Modified policy iteration

# In[33]:


label = 'Modified policy iteration'
iters = [1, 2, 3, 4]
v = np.zeros(ddp.num_states)
k = 20 #- 1

fig, ax = plt.subplots(figsize=(8,5))
for i in range(iters[-1]):
    # Policy improvement
    sigma = ddp.compute_greedy(v)
    u = ddp.bellman_operator(v)
    if i == results[label].num_iter-1:
        diff = u - v
        break
    # Partial policy evaluation with k=20 iterations
    for j in range(k):
        u = ddp.T_sigma(sigma)(u)
    v = u
    if i+1 in iters:
        ax.plot(-v, label='Iteration {0}'.format(i+1))
ax.plot(-results['Policy iteration'].v, label='Fixed Point')

ax.ticklabel_format(style='sci', axis='y', scilimits=(0,0))
ax.set_ylim(0, 2.8e5)
ax.set_yticks([0.4e5 * i for i in range(8)])
ax.set_title(label)
ax.set_xlabel('State')
ax.set_ylabel(r'Value $\times\ (-1)$')
plt.legend(loc=4)
plt.show()


# ## Increasing the discount factor

# Let us consider the case with a discount factor closer to $1$, $\beta = 0.9999$.

# In[34]:


ddp.beta = 0.9999


# In[35]:


v_init = np.zeros(ddp.num_states)
epsilon = 1164
ddp.max_iter = 10**5 * 2

results_9999 = {}

for i in range(4):
    k = 20 if labels[i] == 'Modified policy iteration' else 0
    results_9999[labels[i]] = \
        ddp.solve(method=methods[i], v_init=v_init, epsilon=epsilon, k=k)


# In[36]:


df_9999 = pd.DataFrame(index=labels, columns=columns)


# The numbers of iterations:

# In[37]:


for label in labels:
    print(results_9999[label].num_iter, '\t' + '(' + label + ')')
    df_9999[columns[0]].loc[label] = results_9999[label].num_iter


# Policy iteration gives the optimal policy:

# In[38]:


print(results_9999['Policy iteration'].sigma)


# Takes action 1 ("replace") if and only if $s \geq \bar{\gamma}$, where $\bar{\gamma}$ is equal to:

# In[39]:


(1-results_9999['Policy iteration'].sigma).sum()


# Check that the other methods gave the correct answer:

# In[40]:


for result in results_9999.values():
    if result != results_9999['Policy iteration']:
        print(np.array_equal(result.sigma, results_9999['Policy iteration'].sigma))


# $\lVert v - v_{\mathrm{pi}}\rVert$:

# In[41]:


for label in labels:
    diff_pi = \
        np.abs(results_9999[label].v - results_9999['Policy iteration'].v).max()
    print(diff_pi, '\t' + '(' + label + ')')
    df_9999[columns[2]].loc[label] = diff_pi


# $\lVert v - T(v)\rVert$:

# In[42]:


for label in labels:
    v = results_9999[label].v
    diff_max = \
        np.abs(v - ddp.bellman_operator(v)).max()
    print(diff_max, '\t' + '(' + label + ')')
    df_9999[columns[4]].loc[label] = diff_max


# $\overline{b} - \underline{b}$:

# In[43]:


for i in range(4):
    if labels[i] != 'Policy iteration':
        k = 20 if labels[i] == 'Modified policy iteration' else 0
        res = ddp.solve(method=methods[i], v_init=v_init, k=k,
                        max_iter=results_9999[labels[i]].num_iter-1)
        diff = ddp.bellman_operator(res.v) - res.v
        diff_span = (diff.max() - diff.min()) * ddp.beta / (1 - ddp.beta)
        print(diff_span, '\t' + '(' + labels[i] + ')')
        df_9999[columns[3]].loc[labels[i]] = diff_span


# For policy iteration:

# In[44]:


label = 'Policy iteration'
v = results_9999[label].v
diff = ddp.bellman_operator(v) - v
diff_span = (diff.max() - diff.min()) * ddp.beta / (1 - ddp.beta)
print(diff_span, '\t' + '(' + label + ')')
df_9999[columns[3]].loc[label] = diff_span


# In[45]:


for i in range(4):
    k = 20 if labels[i] == 'Modified policy iteration' else 0
    print(labels[i])
    t = get_ipython().run_line_magic('timeit', '-o ddp.solve(method=methods[i], v_init=v_init, epsilon=epsilon, k=k)')
    df_9999[columns[1]].loc[labels[i]] = t.best


# In[46]:


df_9999


# In[47]:


df_time = pd.DataFrame(index=labels)
df_time[r'$\beta = 0.95$'] = df[columns[1]]
df_time[r'$\beta = 0.9999$'] = df_9999[columns[1]]
second_max = df_time[r'$\beta = 0.9999$'][1:].max()

plt.rcParams['axes.color_cycle'] = ['g', 'b']
for xlim in [None, (0, second_max*1.2)]:
    ax = df_time.ix[reversed(labels)][df_time.columns[::-1]].plot(
        kind='barh', legend='reverse', xlim=xlim, figsize=(8,5)
    )
    ax.set_xlabel('Time (second)')


# In[48]:


import platform
print(platform.platform())


# In[49]:


import sys
print(sys.version)


# In[50]:


print(np.__version__)


# In[ ]: