Daisuke Oyama
Faculty of Economics, University of Tokyo
From Miranda and Fackler, Applied Computational Economics and Finance, 2002, Section 7.6.6
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%matplotlib inline
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import numpy as np
import scipy.sparse as sp
import matplotlib.pyplot as plt
import quantecon as qe
from quantecon.markov import DiscreteDP, backward_induction, sa_indices
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emax = 8 # Energy carrying capacity
n = emax + 1 # Number of states, 0, ..., emax
m = 3 # Number of areas (actions), 0, ..., m-1
e = [2, 4, 5] # Energy offerings
p = [1.0, 0.7, 0.8] # Survival probabilities
q = [0.5, 0.8, 0.7] # Success probabilities
T = 10 # Time horizon
# Terminal values
v_term = np.ones(n)
v_term[0]= 0
We follow the state-action pairs formulation approach.
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L = n * m # Number of feasible state-action pairs
s_indices, a_indices = sa_indices(n, m)
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# Reward vector
R = np.zeros(L)
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# Transition probability array
Q = sp.lil_matrix((L, n))
it = np.nditer((s_indices, a_indices), flags=['c_index'])
for s, k in it:
i = it.index
if s == 0:
Q[i, 0] = 1
elif s == 1:
Q[i, np.minimum(emax, s-1+e[k])] = p[k] * q[k]
Q[i, 0] = 1 - p[k] * q[k]
else:
Q[i, np.minimum(emax, s-1+e[k])] = p[k] * q[k]
Q[i, s-1] = p[k] * (1 - q[k])
Q[i, 0] = 1 - p[k]
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# Discount factor
beta = 1
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# Create a DiscreteDP
ddp = DiscreteDP(R, Q, beta, s_indices, a_indices)
Let us use the backward_induction routine to solve our finite-horizon problem.
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vs, sigmas = backward_induction(ddp, T, v_term)
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fig, axes = plt.subplots(1, 2, figsize=(12, 4))
ts = [0, 5]
for i, t in enumerate(ts):
axes[i].bar(np.arange(n), vs[t], align='center', width=1, edgecolor='k')
axes[i].set_xlim(0-0.5, emax+0.5)
axes[i].set_ylim(0, 1)
axes[i].set_xlabel('Stock of Energy')
axes[i].set_ylabel('Probability')
axes[i].set_title('Survival Probability, Period {t}'.format(t=t))
plt.show()
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