n = 500
pmin = -1 # minimum log price
pmax = 1 # maximum log price
Value = BasisSpline(n, pmin, pmax, labels=['logprice'], l=['value'])
print(Value)
# In the last expression, by passing the option `l` with a one-element list we are telling the ```BasisSpline``` class that we a single function named "value". On creation, the function will be set by default to $V(p)=0$ for all values of $p$, which conveniently corresponds to the terminal condition of this problem.
# ## Finding the critical exercise prices
#
# We are going to find the prices recursively, starting form a option in the expiration date. Notice that the solution to this problem is trivial: since next-period value is zero, the exercise price is $K$. Either way, we can find it numerically by calling the ```zero``` method on the `f` object.
# In[7]:
pcrit[0] = f.zero(0.0)
# Next, for each possible price shock, we compute next period log-price by adding the shock to current log-prices (the nodes of the Value object). Then, we use each next-period price to compute the expected value of an option with one-period to maturity (save the values in ```v```). We update the value function to reflect the new time-to-maturity and use ```broyden``` to solve for the critical value. We repeat this procedure until we reach the $T=300$ horizon.
# In[8]:
for t in range(T):
v = np.zeros((1, n))
for k in range(m):
pnext = Value.nodes + e[k]
v += w[k] * np.maximum(K - np.exp(pnext), delta * Value(pnext))
Value[:] = v
pcrit[t + 1] = f.broyden(pcrit[t])
# ### Print Critical Exercise Price 300 Periods to Expiration
pmax = 1 # maximum log price
Value = BasisSpline(n, pmin, pmax,
labels=['logprice'], l=['value'])
print(Value)
# In the last expression, by passing the option `l` with a one-element list we are telling the ```BasisSpline``` class that we a single function named "value". On creation, the function will be set by default to $V(p)=0$ for all values of $p$, which conveniently corresponds to the terminal condition of this problem.
# ## Finding the critical exercise prices
#
# We are going to find the prices recursively, starting form a option in the expiration date. Notice that the solution to this problem is trivial: since next-period value is zero, the exercise price is $K$. Either way, we can find it numerically by calling the ```zero``` method on the `f` object.
# In[7]:
pcrit[0] = f.zero(0.0)
# Next, for each possible price shock, we compute next period log-price by adding the shock to current log-prices (the nodes of the Value object). Then, we use each next-period price to compute the expected value of an option with one-period to maturity (save the values in ```v```). We update the value function to reflect the new time-to-maturity and use ```broyden``` to solve for the critical value. We repeat this procedure until we reach the $T=300$ horizon.
# In[8]:
for t in range(T):
v = np.zeros((1, n))
for k in range(m):
pnext = Value.nodes + e[k]
v += w[k] * np.maximum(K - np.exp(pnext), delta * Value(pnext))
Value[:] = v
pcrit[t + 1] = f.broyden(pcrit[t])
print('q1 = ', x[0], '\nq2 = ', x[1])
''' Example page 43 '''
# numbers don't match those of Table 3.1, but CompEcon2014' broyden gives same answer as this code
example(43)
opts = {'maxit': 30, 'all_x': True, 'print': True}
g = NLP(lambda x: sqrt(x), 0.5)
f = NLP(lambda x: (x - sqrt(x), 1 - 0.5 / sqrt(x)), 0.5)
x_fp = g.fixpoint(**opts)
x_br = f.broyden(**opts)
x_nw = f.newton(**opts)
''' Example page 51 '''
example(51)
f = MCP(lambda x: (1.01 - (1 - x)**2, 2 * (1 - x)), 0, np.inf)
x = f.zero(0.0) # ssmooth transform by default
print('Using', f.opts.transform, 'transformation, x = ', x)
x = f.zero(0.0, transform='minmax')
print('Using', f.opts.transform, 'transformation, x = ', x,
'FAILED TO CONVERGE')
# ==============================================================
''' Exercise 3.1 '''
exercise('3.1')
def newtroot(c):
x = c / 2
for it in range(150):
fx = x**2 - c
if abs(fx) < 1.e-9:
'linewidth': 1.0,
'markersize': 6,
'markerfacecolor': 'red',
'markeredgecolor': 'red'
}
contour_options = {'levels': [0.0], 'colors': '0.25', 'linewidths': 0.5}
Q1, Q2 = np.meshgrid(q1, q2)
Z0 = np.reshape(z[0], (n, n), order='F')
Z1 = np.reshape(z[1], (n, n), order='F')
methods = ['newton', 'broyden']
cournot_problem.opts['maxit', 'maxsteps', 'all_x'] = 10, 0, True
qmin, qmax = 0.1, 1.3
fig = plt.figure()
for it in range(2):
x = cournot_problem.zero(method=methods[it])
demo.subplot(1, 2, it + 1, methods[it].capitalize() + "'s method", '$q_1$',
'$q_2$', [qmin, qmax], [qmin, qmax])
plt.contour(Q1, Q2, Z0, **contour_options)
plt.contour(Q1, Q2, Z1, **contour_options)
plt.plot(*cournot_problem.x_sequence, **steps_options)
demo.text(0.85, qmax, '$\pi_1 = 0$', 'left', 'top')
demo.text(qmax, 0.55, '$\pi_2 = 0$', 'right', 'center')
plt.show()
demo.savefig([fig])
contour_options = {'levels': [0.0],
'colors': '0.25',
'linewidths': 0.5}
Q1, Q2 = np.meshgrid(q1, q2)
Z0 = np.reshape(z[0], (n,n), order='F')
Z1 = np.reshape(z[1], (n,n), order='F')
methods = ['newton', 'broyden']
cournot_problem.opts['all_x'] = True
fig = plt.figure()
for it in range(2):
cournot_problem.zero(x0=q, method=methods[it])
demo.subplot(1, 2, it + 1, methods[it].capitalize() + "'s method",
'$x$', '$y$', [min(q1), max(q1)], [min(q2), max(q2)])
plt.contour(Q1, Q2, Z0, **contour_options)
plt.contour(Q1, Q2, Z1, **contour_options)
plt.plot(*cournot_problem.x_sequence, **steps_options)
ax = plt.gca()
ax.set_xticks([0.3, 0.7, 1.1])
ax.set_yticks([0.4, 0.8, 1.2])
demo.text(0.7, 0.45, '$f_1(x,y) = 0$', 'left', 'top',fs=12)
demo.text(0.3, 0.55, '$f_2(x,y) = 0$', 'left', 'center',fs=12)
plt.show()
demo.savefig([fig])
'markersize': 6,
'markerfacecolor': 'red',
'markeredgecolor': 'red'
}
contour_options = {'levels': [0.0], 'colors': '0.25', 'linewidths': 0.5}
Q1, Q2 = np.meshgrid(q1, q2)
Z0 = np.reshape(z[0], (n, n), order='F')
Z1 = np.reshape(z[1], (n, n), order='F')
methods = ['newton', 'broyden']
cournot_problem.opts['all_x'] = True
fig = plt.figure()
for it in range(2):
cournot_problem.zero(x0=q, method=methods[it])
demo.subplot(1, 2, it + 1, methods[it].capitalize() + "'s method", '$x$',
'$y$', [min(q1), max(q1)], [min(q2), max(q2)])
plt.contour(Q1, Q2, Z0, **contour_options)
plt.contour(Q1, Q2, Z1, **contour_options)
plt.plot(*cournot_problem.x_sequence, **steps_options)
ax = plt.gca()
ax.set_xticks([0.3, 0.7, 1.1])
ax.set_yticks([0.4, 0.8, 1.2])
demo.text(0.7, 0.45, '$f_1(x,y) = 0$', 'left', 'top', fs=12)
demo.text(0.3, 0.55, '$f_2(x,y) = 0$', 'left', 'center', fs=12)
plt.show()
demo.savefig([fig])
'markerfacecolor': 'red',
'markeredgecolor': 'red'}
contour_options = {'levels': [0.0],
'colors': '0.25',
'linewidths': 0.5}
Q1, Q2 = np.meshgrid(q1, q2)
Z0 = np.reshape(z[0], (n,n), order='F')
Z1 = np.reshape(z[1], (n,n), order='F')
methods = ['newton', 'broyden']
cournot_problem.opts['maxit', 'maxsteps', 'all_x'] = 10, 0, True
qmin, qmax = 0.1, 1.3
fig = plt.figure()
for it in range(2):
x = cournot_problem.zero(method=methods[it])
demo.subplot(1, 2, it + 1, methods[it].capitalize() + "'s method",
'$q_1$', '$q_2$', [qmin, qmax], [qmin, qmax])
plt.contour(Q1, Q2, Z0, **contour_options)
plt.contour(Q1, Q2, Z1, **contour_options)
plt.plot(*cournot_problem.x_sequence, **steps_options)
demo.text(0.85, qmax, '$\pi_1 = 0$', 'left', 'top')
demo.text(qmax, 0.55, '$\pi_2 = 0$', 'right', 'center')
plt.show()
demo.savefig([fig])
