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Solving a simple DP problem (student stuff) .[embedded content] In case you want to solve the problem yourself, here’s a Python script yours truly made (assuming the utility function is of the natural log kind and with some explicit numerical parameterization): import numpy as np import matplotlib.pyplot as plt # Parameters beta = 0.9 # Discount factor r = 0.1 # Depreciation rate T = 3 # Time horizon k0 = 4.0 # Initial capital # Production function: f(k) = k^0.5 def f(k): return k**0.5 # Utility function: u(c) = ln(c), ensuring c > 0 def u(c): return np.log(c) if c > 0 else -np.inf # Discretize capital grid k_vals = np.linspace(0.1, 10, 100) # Initialize value function and policy storage V = [{} for _ in range(T + 1)] policy = [{} for
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Lars Pålsson Syll considers the following as important: Economics
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Solving a simple DP problem (student stuff) .[embedded content] In case you want to solve the problem yourself, here’s a Python script yours truly made (assuming the utility function is of the natural log kind and with some explicit numerical parameterization): import numpy as np import matplotlib.pyplot as plt # Parameters beta = 0.9 # Discount factor r = 0.1 # Depreciation rate T = 3 # Time horizon k0 = 4.0 # Initial capital # Production function: f(k) = k^0.5 def f(k): return k**0.5 # Utility function: u(c) = ln(c), ensuring c > 0 def u(c): return np.log(c) if c > 0 else -np.inf # Discretize capital grid k_vals = np.linspace(0.1, 10, 100) # Initialize value function and policy storage V = [{} for _ in range(T + 1)] policy = [{} for
Topics:
Lars Pålsson Syll considers the following as important: Economics
This could be interesting, too:
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Solving a simple DP problem (student stuff)
.
In case you want to solve the problem yourself, here’s a Python script yours truly made (assuming the utility function is of the natural log kind and with some explicit numerical parameterization):
import numpy as np
import matplotlib.pyplot as plt
# Parameters
beta = 0.9 # Discount factor
r = 0.1 # Depreciation rate
T = 3 # Time horizon
k0 = 4.0 # Initial capital
# Production function: f(k) = k^0.5
def f(k):
return k**0.5
# Utility function: u(c) = ln(c), ensuring c > 0
def u(c):
return np.log(c) if c > 0 else -np.inf
# Discretize capital grid
k_vals = np.linspace(0.1, 10, 100)
# Initialize value function and policy storage
V = [{} for _ in range(T + 1)]
policy = [{} for _ in range(T)]
# Terminal condition at T
for k in k_vals:
cT = f(k) + (1 - r) * k # Consume all resources at the last period
V[T][k] = u(cT) if cT > 0 else -np.inf # Ensure feasibility
# Backward induction
for t in range(T - 1, -1, -1): # t = 2, 1, 0
for k in k_vals:
best_c = 0
best_v = -np.inf
c_vals = np.linspace(0.1, f(k) + (1 - r) * k, 100) # Consumption choices
for c in c_vals:
k_next = f(k) + (1 - r) * k - c # Capital transition
if k_next < 0:
continue # Skip infeasible values
# Interpolate next period's value function
k_next_values = np.array(list(V[t + 1].keys()))
V_next_values = np.array(list(V[t + 1].values()))
V_next = np.interp(k_next, k_next_values, V_next_values)
value = u(c) + beta * V_next # Bellman equation
if value > best_v:
best_v = value
best_c = c
V[t][k] = best_v
policy[t][k] = best_c
# Compute optimal paths starting from k0
opt_c = []
opt_k = [k0]
for t in range(T):
k_current = opt_k[-1]
# Interpolate the optimal policy for k_current
k_policy_values = np.array(list(policy[t].keys()))
c_policy_values = np.array(list(policy[t].values()))
c_t = np.interp(k_current, k_policy_values, c_policy_values)
opt_c.append(c_t)
opt_k.append(f(k_current) + (1 - r) * k_current - c_t)
# Print results
print("Optimal Consumption Path:", opt_c)
print("Optimal Capital Path:", opt_k)
# Plot results
plt.figure(figsize=(10, 5))
plt.plot(range(T+1), opt_k, marker='o', label='Capital')
plt.plot(range(T), opt_c, marker='s', label='Consumption')
plt.xlabel("Time")
plt.ylabel("Value")
plt.legend()
plt.title("Optimal Capital and Consumption Paths")
plt.grid()
plt.show()