This lecture and Cass-Koopmans Competitive Equilibrium describe a model that Tjalling Koopmans [Koopmans, 1965] and David Cass [Cass, 1965] used to analyze optimal growth.
The model can be viewed as an extension of the model of Robert Solow described in an earlier lecture but adapted to make the saving rate be a choice.
(Solow assumed a constant saving rate determined outside the model.)
We describe two versions of the model, one in this lecture and the other in Cass-Koopmans Competitive Equilibrium.
Together, the two lectures illustrate what is, in fact, a more general connection between a planned economy and a decentralized economy organized as a competitive equilibrium.
This lecture is devoted to the planned economy version.
In the planned economy, there are
Instead there is a dictator that tells people
The lecture uses important ideas including
Let’s start with some standard imports:
import matplotlib.pyplot as plt
plt.rcParams["figure.figsize"] = (11, 5) #set default figure size
from numba import njit, float64
from numba.experimental import jitclass
import numpy as np
Time is discrete and takes values $ t = 0, 1 , \ldots, T $ where $ T $ is finite.
(We’ll eventually study a limiting case in which $ T = + \infty $)
A single good can either be consumed or invested in physical capital.
The consumption good is not durable and depreciates completely if not consumed immediately.
The capital good is durable but depreciates.
We let $ C_t $ be the total consumption of a nondurable consumption good at time $ t $.
Let $ K_t $ be the stock of physical capital at time $ t $.
Let $ \vec{C} $ = $ \{C_0,\dots, C_T\} $ and $ \vec{K} $ = $ \{K_0,\dots,K_{T+1}\} $.
We use a concept of a representative consumer to be thought of as follows.
There is a unit mass of identical consumers indexed by $ \omega \in [0,1] $.
Consumption of consumer $ \omega $ is $ c(\omega) $.
Aggregate consumption is
$$ C = \int_0^1 c(\omega) d \omega $$Consider a welfare problem that chooses an allocation $ \{c(\omega)\} $ across consumers to maximize
$$ \int_0^1 u(c(\omega)) d \omega $$where $ u(\cdot) $ is a concave utility function with $ u' >0, u'' < 0 $ and maximization is subject to
$$ C = \int_0^1 c(\omega) d \omega . \tag{40.1} $$
Form a Lagrangian $ L = \int_0^1 u(c(\omega)) d \omega + \lambda [C - \int_0^1 c(\omega) d \omega ] $.
Differentiate under the integral signs with respect to each $ \omega $ to obtain the first-order necessary conditions
$$ u'(c(\omega)) = \lambda. $$These conditions imply that $ c(\omega) $ equals a constant $ c $ that is independent of $ \omega $.
To find $ c $, use feasibility constraint (40.1) to conclude that
$$ c(\omega) = c = C. $$This line of argument indicates the special aggregation theory that lies beneath outcomes in which a representative consumer consumes amount $ C $.
It appears often in aggregate economics.
We shall use this aggregation theory here and also in this lecture Cass-Koopmans Competitive Equilibrium.
A representative household is endowed with one unit of labor at each $ t $ and likes the consumption good at each $ t $.
The representative household inelastically supplies a single unit of labor $ N_t $ at each $ t $, so that $ N_t =1 \text{ for all } t \in \{0, 1, \ldots, T\} $.
The representative household has preferences over consumption bundles ordered by the utility functional:
$$ U(\vec{C}) = \sum_{t=0}^{T} \beta^t \frac{C_t^{1-\gamma}}{1-\gamma} \tag{40.2} $$
where $ \beta \in (0,1) $ is a discount factor and $ \gamma >0 $ governs the curvature of the one-period utility function.
Larger $ \gamma $’s imply more curvature.
Note that
$$ u(C_t) = \frac{C_t^{1-\gamma}}{1-\gamma} \tag{40.3} $$
satisfies $ u'>0,u''<0 $.
$ u' > 0 $ asserts that the consumer prefers more to less.
$ u''< 0 $ asserts that marginal utility declines with increases in $ C_t $.
We assume that $ K_0 > 0 $ is an exogenous initial capital stock.
There is an economy-wide production function
$$ F(K_t,N_t) = A K_t^{\alpha}N_t^{1-\alpha} \tag{40.4} $$
with $ 0 < \alpha<1 $, $ A > 0 $.
A feasible allocation $ \vec{C}, \vec{K} $ satisfies
$$ C_t + K_{t+1} \leq F(K_t,N_t) + (1-\delta) K_t \quad \text{for all } t \in \{0, 1, \ldots, T\} \tag{40.5} $$
where $ \delta \in (0,1) $ is a depreciation rate of capital.
A planner chooses an allocation $ \{\vec{C},\vec{K}\} $ to maximize (40.2) subject to (40.5).
Let $ \vec{\mu}=\{\mu_0,\dots,\mu_T\} $ be a sequence of nonnegative Lagrange multipliers.
To find an optimal allocation, form a Lagrangian
$$ \mathcal{L}(\vec{C} ,\vec{K} ,\vec{\mu} ) = \sum_{t=0}^T \beta^t\left\{ u(C_t)+ \mu_t \left(F(K_t,1) + (1-\delta) K_t- C_t - K_{t+1} \right)\right\} \tag{40.6} $$
and pose the following min-max problem:
$$ \min_{\vec{\mu}} \max_{\vec{C},\vec{K}} \mathcal{L}(\vec{C},\vec{K},\vec{\mu} ) \tag{40.7} $$
Before computing first-order conditions, we present some handy formulas.
The following technicalities will help us.
Notice that
$$ F(K_t,N_t) = A K_t^\alpha N_t^{1-\alpha} = N_t A\left(\frac{K_t}{N_t}\right)^\alpha $$Define the output per-capita production function
$$ \frac{F(K_t,N_t)}{N_t} \equiv f\left(\frac{K_t}{N_t}\right) = A\left(\frac{K_t}{N_t}\right)^\alpha $$whose argument is capital per-capita.
It is useful to recall the following calculations for the marginal product of capital
$$ \begin{aligned} \frac{\partial F(K_t,N_t)}{\partial K_t} & = \frac{\partial N_t f\left( \frac{K_t}{N_t}\right)}{\partial K_t} \\ &= N_t f'\left(\frac{K_t}{N_t}\right)\frac{1}{N_t} \quad \text{(Chain rule)} \\ &= f'\left.\left(\frac{K_t}{N_t}\right)\right|_{N_t=1} \\ &= f'(K_t) \end{aligned} \tag{40.8} $$
and the marginal product of labor
$$ \begin{aligned} \frac{\partial F(K_t,N_t)}{\partial N_t} &= \frac{\partial N_t f\left( \frac{K_t}{N_t}\right)}{\partial N_t} \quad \text{(Product rule)} \\ &= f\left(\frac{K_t}{N_t}\right){+} N_t f'\left(\frac{K_t}{N_t}\right) \frac{-K_t}{N_t^2} \quad \text{(Chain rule)} \\ &= f\left(\frac{K_t}{N_t}\right){-}\frac{K_t}{N_t}f'\left.\left(\frac{K_t}{N_t}\right)\right|_{N_t=1} \\ &= f(K_t) - f'(K_t) K_t \end{aligned} $$(Here we are using that $ N_t = 1 $ for all $ t $, so that $ K_t = \frac{K_t}{N_t} $.)
We now compute first-order necessary conditions for extremization of Lagrangian (40.6):
$$ C_t: \qquad u'(C_t)-\mu_t=0 \qquad \text{for all} \quad t= 0,1,\dots,T \tag{40.9} $$
$$ K_t: \qquad \beta \mu_t\left[(1-\delta)+f'(K_t)\right] - \mu_{t-1}=0 \qquad \text{for all } \quad t=1,2,\dots,T \tag{40.10} $$
$$ \mu_t:\qquad F(K_t,1)+ (1-\delta) K_t - C_t - K_{t+1}=0 \qquad \text{for all } \quad t=0,1,\dots,T \tag{40.11} $$
$$ K_{T+1}: \qquad -\mu_T \leq 0, \ \leq 0 \text{ if } K_{T+1}=0; \ =0 \text{ if } K_{T+1}>0 \tag{40.12} $$
In computing (40.10) we recognize that $ K_t $ appears in both the time $ t $ and time $ t-1 $ feasibility constraints (40.5).
Restrictions (40.12) come from differentiating with respect to $ K_{T+1} $ and applying the following Karush-Kuhn-Tucker condition (KKT) (see Karush-Kuhn-Tucker conditions):
$$ \mu_T K_{T+1}=0 \tag{40.13} $$
Combining (40.9) and (40.10) gives
$$ \beta u'\left(C_t\right)\left[(1-\delta)+f'\left(K_t\right)\right]-u'\left(C_{t-1}\right)=0 \quad \text{ for all } t=1,2,\dots, T+1 $$which can be rearranged to become
$$ \beta u'\left(C_{t+1}\right)\left[(1-\delta)+f'\left(K_{t+1}\right)\right]= u'\left(C_{t}\right) \quad \text{ for all } t=0,1,\dots, T \tag{40.14} $$
Applying the inverse marginal utility of consumption function on both sides of the above equation gives
$$ C_{t+1} =u'^{-1}\left(\left(\frac{\beta}{u'(C_t)}[f'(K_{t+1}) +(1-\delta)]\right)^{-1}\right) $$which for our utility function (40.3) becomes the consumption Euler equation
$$ \begin{aligned} C_{t+1} =\left(\beta C_t^{\gamma}[f'(K_{t+1}) + (1-\delta)]\right)^{1/\gamma} %\notag\\= C_t\left(\beta [f'(K_{t+1}) + %(1-\delta)]\right)^{1/\gamma} \end{aligned} $$which we can combine with the feasibility constraint (40.5) to get
$$ \begin{aligned} C_{t+1} & = C_t\left(\beta [f'(F(K_t,1)+ (1-\delta) K_t - C_t) + (1-\delta)]\right)^{1/\gamma} \\ K_{t+1} & = F(K_t,1)+ (1-\delta) K_t - C_t . \end{aligned} $$This is a pair of non-linear first-order difference equations that map $ C_t, K_t $ into $ C_{t+1}, K_{t+1} $ and that an optimal sequence $ \vec C , \vec K $ must satisfy.
It must also satisfy the initial condition that $ K_0 $ is given and $ K_{T+1} = 0 $.
Below we define a jitclass that stores parameters and functions
that define our economy.
planning_data = [
('γ', float64), # Coefficient of relative risk aversion
('β', float64), # Discount factor
('δ', float64), # Depreciation rate on capital
('α', float64), # Return to capital per capita
('A', float64) # Technology
]
@jitclass(planning_data)
class PlanningProblem():
def __init__(self, γ=2, β=0.95, δ=0.02, α=0.33, A=1):
self.γ, self.β = γ, β
self.δ, self.α, self.A = δ, α, A
def u(self, c):
'''
Utility function
ASIDE: If you have a utility function that is hard to solve by hand
you can use automatic or symbolic differentiation
See https://github.com/HIPS/autograd
'''
γ = self.γ
return c ** (1 - γ) / (1 - γ) if γ!= 1 else np.log(c)
def u_prime(self, c):
'Derivative of utility'
γ = self.γ
return c ** (-γ)
def u_prime_inv(self, c):
'Inverse of derivative of utility'
γ = self.γ
return c ** (-1 / γ)
def f(self, k):
'Production function'
α, A = self.α, self.A
return A * k ** α
def f_prime(self, k):
'Derivative of production function'
α, A = self.α, self.A
return α * A * k ** (α - 1)
def f_prime_inv(self, k):
'Inverse of derivative of production function'
α, A = self.α, self.A
return (k / (A * α)) ** (1 / (α - 1))
def next_k_c(self, k, c):
''''
Given the current capital Kt and an arbitrary feasible
consumption choice Ct, computes Kt+1 by state transition law
and optimal Ct+1 by Euler equation.
'''
β, δ = self.β, self.δ
u_prime, u_prime_inv = self.u_prime, self.u_prime_inv
f, f_prime = self.f, self.f_prime
k_next = f(k) + (1 - δ) * k - c
c_next = u_prime_inv(u_prime(c) / (β * (f_prime(k_next) + (1 - δ))))
return k_next, c_next
We can construct an economy with the Python code:
pp = PlanningProblem()
We use shooting to compute an optimal allocation $ \vec{C}, \vec{K} $ and an associated Lagrange multiplier sequence $ \vec{\mu} $.
First-order necessary conditions (40.9), (40.10), and (40.11) for the planning problem form a system of difference equations with two boundary conditions:
We have no initial condition for the Lagrange multiplier $ \mu_0 $.
If we did, our job would be easy:
However, we woujld not be assured that the Kuhn-Tucker condition (40.13) would be satisfied.
Furthermore, we don’t have an initial condition for $ \mu_0 $.
So this won’t work.
Indeed, part of our task is to compute the optimal value of $ \mu_0 $.
To compute $ \mu_0 $ and the other objects we want, a simple modification of the above procedure will work.
It is called the shooting algorithm.
It is an instance of a guess and verify algorithm that consists of the following steps:
The following Python code implements the shooting algorithm for the planning problem.
(Actually, we modified the preceding algorithm slightly by starting with a guess for $ c_0 $ instead of $ \mu_0 $ in the following code.)
@njit
def shooting(pp, c0, k0, T=10):
'''
Given the initial condition of capital k0 and an initial guess
of consumption c0, computes the whole paths of c and k
using the state transition law and Euler equation for T periods.
'''
if c0 > pp.f(k0):
print("initial consumption is not feasible")
return None
# initialize vectors of c and k
c_vec = np.empty(T+1)
k_vec = np.empty(T+2)
c_vec[0] = c0
k_vec[0] = k0
for t in range(T):
k_vec[t+1], c_vec[t+1] = pp.next_k_c(k_vec[t], c_vec[t])
k_vec[T+1] = pp.f(k_vec[T]) + (1 - pp.δ) * k_vec[T] - c_vec[T]
return c_vec, k_vec
We’ll start with an incorrect guess.
paths = shooting(pp, 0.2, 0.3, T=10)
fig, axs = plt.subplots(1, 2, figsize=(14, 5))
colors = ['blue', 'red']
titles = ['Consumption', 'Capital']
ylabels = ['$c_t$', '$k_t$']
T = paths[0].size - 1
for i in range(2):
axs[i].plot(paths[i], c=colors[i])
axs[i].set(xlabel='t', ylabel=ylabels[i], title=titles[i])
axs[1].scatter(T+1, 0, s=80)
axs[1].axvline(T+1, color='k', ls='--', lw=1)
plt.show()
Evidently, our initial guess for $ \mu_0 $ is too high, so initial consumption too low.
We know this because we miss our $ K_{T+1}=0 $ target on the high side.
Now we automate things with a search-for-a-good $ \mu_0 $ algorithm that stops when we hit the target $ K_{t+1} = 0 $.
We use a bisection method.
We make an initial guess for $ C_0 $ (we can eliminate $ \mu_0 $ because $ C_0 $ is an exact function of $ \mu_0 $).
We know that the lowest $ C_0 $ can ever be is $ 0 $ and that the largest it can be is initial output $ f(K_0) $.
Guess $ C_0 $ and shoot forward to $ T+1 $.
If $ K_{T+1}>0 $, we take it to be our new lower bound on $ C_0 $.
If $ K_{T+1}<0 $, we take it to be our new upper bound.
Make a new guess for $ C_0 $ that is halfway between our new upper and lower bounds.
Shoot forward again, iterating on these steps until we converge.
When $ K_{T+1} $ gets close enough to $ 0 $ (i.e., within an error tolerance bounds), we stop.
@njit
def bisection(pp, c0, k0, T=10, tol=1e-4, max_iter=500, k_ter=0, verbose=True):
# initial boundaries for guess c0
c0_upper = pp.f(k0)
c0_lower = 0
i = 0
while True:
c_vec, k_vec = shooting(pp, c0, k0, T)
error = k_vec[-1] - k_ter
# check if the terminal condition is satisfied
if np.abs(error) < tol:
if verbose:
print('Converged successfully on iteration ', i+1)
return c_vec, k_vec
i += 1
if i == max_iter:
if verbose:
print('Convergence failed.')
return c_vec, k_vec
# if iteration continues, updates boundaries and guess of c0
if error > 0:
c0_lower = c0
else:
c0_upper = c0
c0 = (c0_lower + c0_upper) / 2
def plot_paths(pp, c0, k0, T_arr, k_ter=0, k_ss=None, axs=None):
if axs is None:
fix, axs = plt.subplots(1, 3, figsize=(16, 4))
ylabels = ['$c_t$', '$k_t$', '$\mu_t$']
titles = ['Consumption', 'Capital', 'Lagrange Multiplier']
c_paths = []
k_paths = []
for T in T_arr:
c_vec, k_vec = bisection(pp, c0, k0, T, k_ter=k_ter, verbose=False)
c_paths.append(c_vec)
k_paths.append(k_vec)
μ_vec = pp.u_prime(c_vec)
paths = [c_vec, k_vec, μ_vec]
for i in range(3):
axs[i].plot(paths[i])
axs[i].set(xlabel='t', ylabel=ylabels[i], title=titles[i])
# Plot steady state value of capital
if k_ss is not None:
axs[1].axhline(k_ss, c='k', ls='--', lw=1)
axs[1].axvline(T+1, c='k', ls='--', lw=1)
axs[1].scatter(T+1, paths[1][-1], s=80)
return c_paths, k_paths
Now we can solve the model and plot the paths of consumption, capital, and Lagrange multiplier.
plot_paths(pp, 0.3, 0.3, [10]);
When $ T \rightarrow +\infty $, the optimal allocation converges to steady state values of $ C_t $ and $ K_t $.
It is instructive to set $ K_0 $ equal to the $ \lim_{T \rightarrow + \infty } K_t $, which we’ll call steady state capital.
In a steady state $ K_{t+1} = K_t=\bar{K} $ for all very large $ t $.
Evalauating feasibility constraint (40.5) at $ \bar K $ gives
$$ f(\bar{K})-\delta \bar{K} = \bar{C} \tag{40.15} $$
Substituting $ K_t = \bar K $ and $ C_t=\bar C $ for all $ t $ into (40.14) gives
$$ 1=\beta \frac{u'(\bar{C})}{u'(\bar{C})}[f'(\bar{K})+(1-\delta)] $$Defining $ \beta = \frac{1}{1+\rho} $, and cancelling gives
$$ 1+\rho = 1[f'(\bar{K}) + (1-\delta)] $$Simplifying gives
$$ f'(\bar{K}) = \rho +\delta $$and
$$ \bar{K} = f'^{-1}(\rho+\delta) $$For production function (40.4), this becomes
$$ \alpha \bar{K}^{\alpha-1} = \rho + \delta $$As an example, after setting $ \alpha= .33 $, $ \rho = 1/\beta-1 =1/(19/20)-1 = 20/19-19/19 = 1/19 $, $ \delta = 1/50 $, we get
$$ \bar{K} = \left(\frac{\frac{33}{100}}{\frac{1}{50}+\frac{1}{19}}\right)^{\frac{67}{100}} \approx 9.57583 $$Let’s verify this with Python and then use this steady state $ \bar K $ as our initial capital stock $ K_0 $.
ρ = 1 / pp.β - 1
k_ss = pp.f_prime_inv(ρ+pp.δ)
print(f'steady state for capital is: {k_ss}')
Now we plot
plot_paths(pp, 0.3, k_ss, [150], k_ss=k_ss);
Evidently, with a large value of $ T $, $ K_t $ stays near $ K_0 $ until $ t $ approaches $ T $ closely.
Let’s see what the planner does when we set $ K_0 $ below $ \bar K $.
plot_paths(pp, 0.3, k_ss/3, [150], k_ss=k_ss);
Notice how the planner pushes capital toward the steady state, stays near there for a while, then pushes $ K_t $ toward the terminal value $ K_{T+1} =0 $ when $ t $ closely approaches $ T $.
The following graphs compare optimal outcomes as we vary $ T $.
plot_paths(pp, 0.3, k_ss/3, [150, 75, 50, 25], k_ss=k_ss);
The following calculation indicates that when $ T $ is very large, the optimal capital stock stays close to its steady state value most of the time.
plot_paths(pp, 0.3, k_ss/3, [250, 150, 50, 25], k_ss=k_ss);
In the above graphs, different colors are associated with different horizons $ T $.
Notice that as the horizon increases, the planner keeps $ K_t $ closer to the steady state value $ \bar K $ for longer.
This pattern reflects a turnpike property of the steady state.
A rule of thumb for the planner is
The planner accomplishes this by adjusting the saving rate $ \frac{f(K_t) - C_t}{f(K_t)} $ over time.
Let’s calculate and plot the saving rate.
@njit
def saving_rate(pp, c_path, k_path):
'Given paths of c and k, computes the path of saving rate.'
production = pp.f(k_path[:-1])
return (production - c_path) / production
def plot_saving_rate(pp, c0, k0, T_arr, k_ter=0, k_ss=None, s_ss=None):
fix, axs = plt.subplots(2, 2, figsize=(12, 9))
c_paths, k_paths = plot_paths(pp, c0, k0, T_arr, k_ter=k_ter, k_ss=k_ss, axs=axs.flatten())
for i, T in enumerate(T_arr):
s_path = saving_rate(pp, c_paths[i], k_paths[i])
axs[1, 1].plot(s_path)
axs[1, 1].set(xlabel='t', ylabel='$s_t$', title='Saving rate')
if s_ss is not None:
axs[1, 1].hlines(s_ss, 0, np.max(T_arr), linestyle='--')
plot_saving_rate(pp, 0.3, k_ss/3, [250, 150, 75, 50], k_ss=k_ss)
We want to set $ T = +\infty $.
The appropriate thing to do is to replace terminal condition (40.12) with
$$ \lim_{T \rightarrow +\infty} \beta^T u'(C_T) K_{T+1} = 0 , $$a condition that will be satisfied by a path that converges to an optimal steady state.
We can approximate the optimal path by starting from an arbitrary initial $ K_0 $ and shooting towards the optimal steady state $ K $ at a large but finite $ T+1 $.
In the following code, we do this for a large $ T $ and plot consumption, capital, and the saving rate.
We know that in the steady state that the saving rate is constant and that $ \bar s= \frac{f(\bar K)-\bar C}{f(\bar K)} $.
From (40.15) the steady state saving rate equals
$$ \bar s =\frac{ \delta \bar{K}}{f(\bar K)} $$The steady state saving rate $ \bar S = \bar s f(\bar K) $ is the amount required to offset capital depreciation each period.
We first study optimal capital paths that start below the steady state.
# steady state of saving rate
s_ss = pp.δ * k_ss / pp.f(k_ss)
plot_saving_rate(pp, 0.3, k_ss/3, [130], k_ter=k_ss, k_ss=k_ss, s_ss=s_ss)
Since $ K_0<\bar K $, $ f'(K_0)>\rho +\delta $.
The planner chooses a positive saving rate that is higher than the steady state saving rate.
Note that $ f''(K)<0 $, so as $ K $ rises, $ f'(K) $ declines.
The planner slowly lowers the saving rate until reaching a steady state in which $ f'(K)=\rho +\delta $.
plot_saving_rate(pp, 0.3, k_ss*1.5, [130], k_ter=k_ss, k_ss=k_ss, s_ss=s_ss)
In Cass-Koopmans Competitive Equilibrium, we study a decentralized version of an economy with exactly the same technology and preference structure as deployed here.
In that lecture, we replace the planner of this lecture with Adam Smith’s invisible hand.
In place of quantity choices made by the planner, there are market prices that are set by a deus ex machina from outside the model, a so-called invisible hand.
Equilibrium market prices must reconcile distinct decisions that are made independently by a representative household and a representative firm.
The relationship between a command economy like the one studied in this lecture and a market economy like that studied in Cass-Koopmans Competitive Equilibrium is a foundational topic in general equilibrium theory and welfare economics.