15
Kuhn -Tucker conditions
There is a systematic approach to inequality-constrained
maximisation—a.k.a. concave programming or nonlinear programming.
Consider a version of the consumer problem in which
1
quasilinear utility x12 + 14 x2 is maximised subject to
x1 +x2 = 1. Mechanically applying the Lagrange multiplier/common slopes technique produces
5
3
x1
2
1
-4
-3
x2
-2
-1
0
p1x1 + p2x2 − m ≤ 0
−x1 ≤ 0
−x2 ≤ 0
The aspect I emphasise is that the constraints are inequalities. We have seen how the Lagrange multiplier
method for dealing with one equality constraint extends naturally to the case of several constraints. If
we wish to maximise f (x) subject to g(x) = b and
h(x) = c, we work with the Lagrangian
4
-5
To be explicit the full set of restrictions for the 2 good
consumer problem is not p1x1 + p2x2 = m but
1
2
Negative quantity!
The tangency solution violates an unspoken economic
requirement x2 ≥ 0.
L(x, λ, µ) = f (x) − λ [g(x) − b] − µ [h(x) − c]
with a multiplier for each constraint. (See Dixit 24ff)
The inequality constrained optimisation problem (SH
682) is:
• Maximise the objective function f (x) where f :
Rn → R
• subject to the constraints gj (x) ≤ 0 where gj :
Rn → R and j = 1, ..., m.
Terminology:
The set of points satisfying the
constraints is called the constraint set, admissible set
or feasible set. If at the optimum x∗, gj (x∗) = 0 then
the j-th constraint is binding ; if not, it is slack. If
at least one constraint is binding then x∗ is on the
boundary of the feasible set; if none are binding x∗ is
an interior point.
Example 15.1 In the consumer problems we have seen
the budget constraint is binding—all income is spent—
because the consumer is never satiated. The constraint −x2 ≤ 0 may be binding as in the situation
pictured at the beginning of this section.
The method of Kuhn-Tucker multipliers is a variation
on the Lagrange multiplier method. If all the constraints are binding then the Lagrange method will
produce the same results as KT.
The Kuhn-Tucker approach involves forming the Lagrangean in more or less the usual way
L = f (x) −
m
X
λj gj (x)
j=1
with the same derivative with respect to the choice
variables conditions
∂L
= 0, i = 1, ..., n or ∇L = 0.
∂xi
However the further conditions specify the interaction between the multipliers and the constraints. The
complementary slackness conditions state
λj ≥ 0 f or all i
λj = 0 whenever gj (x∗) < 0
If the constraint is slack the corresponding multiplier
is zero.
Solving this assortment of equalities and inequalities
in the n + m unknowns (choice variables and multipliers) is messier than the Lagrange method for equality
constrained problems.
c = 2 : x∗ = 0
c = 0 : x∗ = 0
To see how it works, consider a transparent case
c = −1 : x∗ = −1
Example 15.2 Maximise the (strictly concave) function y = 1 − x2 subject to x ≤ c. The optimal x∗ can
either be interior (< c) or on the boundary (= c) of
the feasible set—which will depend on the value of c.
The pictures show c = 2, 0, −1.
To do the Kuhn-Tucker analysis, form the Lagrangean
L = 1 − x2 − λ(x − c).
The K-T conditions are
∂L
= −2x − λ = 0 ⇒ λ∗ = −2x∗
∂x
λ ≥ 0
λ = 0 if (x∗ − c) < 0
∂ (1−x2)
Remark 15.1 The conditions ∂L
=
−λ = 0
∂x
∂x
and λ ≥ 0 imply that the derivative at the maximum
cannot be negative. It is obvious that the derivative
cannot be negative at a maximum because a reduction
in x (this is always feasible) would then raise the value
of the objective.
Remark 15.2 Often the Kuhn-Tucker conditions are
used not for finding a solution but for providing information about a solution. For example in the general
problem of maximising a strictly concave function subject to x ≤ c, the conditions imply that at a maximum
the slope cannot be negative.
Now for the three examples, c = −1, 0 and 2.
• c = −1 : there are 2 possibilities: x∗ = −1 or
x∗ < −1. The latter is impossible for it would
imply that λ∗ = 0 and hence x∗ = 0, a contradiction. So x∗ = −1.
• c = 0 : there are 2 possibilities: x∗ = 0 or x∗ < 0.
As before the latter is impossible. All the conditions are satisfied when x∗ = 0.
• c = 2 : there are 2 possibilities: x∗ = 2 or x∗ < 2.
The former is impossible for it makes −2x∗ and
hence λ∗ negative.
16
Kuhn -Tucker theorem
There are lots of propositions linking the Kuhn-Tucker
conditions to the existence of a maximum. The conditions can be interpreted as necessary conditions for
a maximum (compare the treatment of Lagrange multipliers in 8.2). Or, making strong assumptions about
f and gj , as sufficient conditions. That line is taken
in the next theorem.
Theorem 16.1 ( Kuhn-Tucker sufficiency) Consider the
inequality constrained optimisation problem with concave objective and convex constraints: i.e. to maximise f (x) (where f : Rn → R) subject to the constraints gj (x) ≤ 0 where gj : Rn → R and j =
1, ..., m. Define L = f (x) −
m
P
j=1
λj gj (x) and let x∗
be a feasible point. Suppose we can find numbers
λj such that ∇L(x∗) = 0, λj ≥ 0 f or all i and
λj = 0 whenever gj (x∗) < 0.
Then x∗ solves the maximisation problem
Proof. Since f is concave the supporting hyperplane
theorem takes the form
f (x) ≤ f (x∗) + ∇f (x∗)(x − x∗).
Using ∇L(x∗) = 0, we can write this as
f (x) ≤ f (x∗) +
X
λj ∇gj (x∗)(x − x∗).
The aim is to show that the sum term on the right is
not positive.
The multipliers associated with slack constraints will
be zero so we need only attend to the binding constraints gj (x∗) = 0. In such cases, since gj is convex
we have
0 ≥ gj (x) ≥ 0 + ∇gj (x∗)(x − x∗).
Because the λj are nonnegative,
is not positive—as required.
P
λj ∇gj (x∗)(x−x∗)
Remark 16.1 Like Lagrange multipliers these KuhnTucker multipliers can be interpreted as measures of
the sensitivity of the maximum value to changes in the
constraint (10.2) but we won’t go into the details. See
SH 696.
Remark 16.2 This theorem can be extended to apply
to quasi-concave objective functions. Dixit 97ff discusses the extension.
17
Quasi-linear utility again
Return to the quasi-linear utility case and now incorporate all the inequality constraints and include prices
and income
Maximise u(x)
s.t. p1x1 + p2x2 − m
−x1
−x2
The Lagrangean L is
1
x2
= 1 + αx2, α > 0
≤ 0,
≤ 0,
≤ 0.
1
x 2 +αx
1
2 −λ0(p1x1 +p2x2 −m)−λ1 (−x1 )−λ2 (−x2 )
The Kuhn-Tucker conditions are
∂L
∂x1
∂L
∂x2
λ0, λ1, λ2
λ0(p1x1 + p2x2 − m)
λi(−xi)
1 − 12
= x1 − λ0p1 + λ1 = 0
2
= α − λ0p2 + λ2 = 0
≥ 0
= 0
= 0, i = 1, 2.
1. Because the objective function is strictly increasing
in x1 and x2 the budget constraint is binding —so λ0 >
0.
2. The constraint −x1 ≤ 0, cannot bind for then
1
1 x− 2
2 1
would be infinitely large. So λ1 = 0.
3. The other constraint may or may not bind. Putting
this information about the budget constraint and λ1 =
0 into the Kuhn-Tucker conditions:
∂L
1 −1
= x1 2 − λ0p1 = 0,
∂x1
2
∂L
= α − λ0p2 + λ2 = 0,
∂x2
(p1x1 + p2x2 − m) = 0,
λ2(−x2) = 0.
Consider the possibility x2 = 0 : from the budget
constraint we get
m
x1 =
p1
and so
Ã
!− 1
2
Ã
!1
∂L
1 m
=
∂x1
2 p1
1
1
λ0 =
2 p1m
− λ0p1 = 0
2
It is reasonable that the consumer always consume
some of the first good because marginal utility w.r.t.
it approaches infinity as x1 → 0 while marginal utility
w.r.t. the other good is constant at α.
∂L = 0
Putting this value into ∂x
2
Ã
1
∂L
1
=α−
∂x2
2 p1m
!1
2
p2 + λ2 = 0
But as λ2 ≥ 0 it must be the case that when x2 = 0,
α satisfies
Ã
1 p22
α≤
2 p1m
!1
2
So small values of α produce a corner solution.
The interior solution x1, x2 > 0 is associated with
larger values of α and corresponds to the case λ1 =
λ2 = 0.)
18
Dynamic optimisation
In dynamic optimisation a time-path is chosen. Simple dynamic optimisation problems can be treated by
the same methods as the static optimisation problem. However dynamic problems have special features
which often suggest a different treatment.
Example 18.1 Consider a simple T -period problem where
a given stock b0 is consumer over T periods (formally
a variation on the consumer problem with logarithmic
utility)
max U(x) =
x
T
X
δ t−1 ln xt
• One difference between static and dynamic optimisation is that dynamic equations appear naturally in the latter.
t=1
s.t.
T
X
xt = b0 (f ixed)
(*)
• A second difference is that multiple constraints
(one for each time period) are routine.
t=1
Form the Lagrangean
L(x, λ) =
T
X
t=1
δ t−1 ln xt − λ(
T
X
t=1
xt − b0)
and go through the usual steps to obtain a solution
that involves
xt = δxt−1 for t = 2, ..., T
This kind of dynamic equation is called a difference
equation and is characteristic of the discrete time formulation. If the problem were formulated in continuous time a differential equation would appear at this
point. In more complicated problems diagonalising
methods are used for investigating the properties of
the solution.
Example 18.2 (continues Ex. 18.1) In complicated problems it is usually convenient to specify a ‘budget constraint’ for each time period. Thus the constraint (*)
would appear as:
bt = bt−1 − xt, t = 1, ..., T ; b0 f ixed
(**)
This law of motion describes how the available chocolate stock evolves: the bars of chocolate left over at
the end of period t equals the bars available at the end
of t − 1 less what has been consumed in period t. (*)
collapses these dynamic equations into one constraint
T
P
t=1
xt = b0, eliminating the b1, b2, ..., bT .
The Lagrange method extends to multiple constraints
by introducing a multiplier for each constraint. Thus
here
L(x, λ) =
T
X
t=1
δ t−1 ln xt −
T
X
t=1
λt(xt − bt − bt−1).
There are 2T equations to solve: T of the form Lt = 0
and T making up (**).
Just as there is a sequence {x1, ..., xT } there is a sequence of multipliers {λ1, ..., λT }. The usual algebra
produces conditions like
λ
xt
= δ t−1 .
xt−1
λt
We already know that xt = δxt−1 and it turns out
that λt is the same for all time periods.
• A third difference between static and dynamic
optimisation is the existence of specialised techniques for treating the latter—including (Pontryagin’s) maximum principle and dynamic programming.
18.1
Maximum principle
The maximum principle is widely used in Macroeconomics, usually in its continuous time form. I will go
through a discrete time version to suggest where the
coninuous time forms come from.
A fairly general formulation covering the chocolate
stock example and extensions to include production
and investment involves the choice variables c(1), ..., c(T );
these symbols are easier on the eye than c1 etc.
The notation reflects the terminology of control theory. There is a state variable s governed by an equation of motion or state equation. The problem is to
choose a control variable sequence c to maximise a
value function. This may involve one or both of the
state variable and the control variable.
max V (s, c) =
c
T
X
t=1
v(s(t), c(t))
s.t. s(t + 1) − s(t) = f (s(t), c(t))
(***)
for t = 1, ..., T and with s(1) and s(T + 1) fixed at
s1 and sT +1 respectively. (Other end conditions are
possible.)
The Lagrangian is
L =
T
X
v(s(t), c(t))
t=1
T
X
−
t=1
λ(t) [s(t + 1) − s(t) − f (s(t), c(t))]
These conditions can be obtained as first order conditions involving a new function H (Hamiltonian) defined for all t by
H(s(t), c(t), λ(t)) ≡ v(s(t), c(t)) + λ(t)f (s(t), c(t).
Differentiating w.r.t. c(t) and s(t)
∂H
= 0, t = 1, ..., T
∂c(t)
∂H
, t = 2, ..., T
λ(t) − λ(t − 1) = −
∂s(t)
18.1.1
In continuous time
In optimal control the λ’s are called co-state variables.
Differentiating w.r.t. c(t) and s(t) (writing partial
derivatives using subscripts) the first order conditions
are
In the more usual continuous time formulation the
problem is to choose the time path of consumption
c(t) to maximise
V
vc(t) + λ(t)fc(t) = 0, t = 1, ..., T
vs(t) − λ(t − 1) + λ(t) + λ(t)fs(t) = 0, t = 2, ..., T
=
ZT
v(s(t), c(t))dt
o
s.t
ds
= f (s(t), c(t))
dt
The first order conditions for a maximum are conditions on the partial derivatives of H,
H(t, s(t), c(t), λ(t)) = v(t, s(t), c(t))+λ(t)f (t, s(t), c(t))
The first order conditions are
∂H
= 0
∂c
dλ
∂H
= −
.
dt
∂s
Example 18.3 Logarithmic chocolate in continuous time.
Choose a consumption path x(t) to maximise
U(x(t)) =
ZT
ln x(t)e−ρtdt
0
subject to (writing k for the stock)
.
k = −x
k(0) = given
k(T ) = f ree
In this case the chocolate stock is the state variable—its
derivative appears in the constraint—and consumption
is the control variable. The choice of labels may not
seem very natural—you control the chocolate stock by
consuming chocolate. In this example the state variable does not appear in the objective function.
The Hamiltonian is
H(t, k(t), x(t), λ(t)) = ln x(t)e−ρt − λ(t)x(t)
The first order conditions are
.
∂H
= −x(t)
k =
∂λ
.
∂H
λ = −
=0
∂k
∂H
e−ρt
=
− λ(t) = 0
∂x
x(t)
.
The second condition λ= 0 is so simple because k
does not appear in the Hamiltonian; it implies that
λ(t) is constant So from the third condition
x(t) ∝ e−ρt
The time path of consumption is exponentially declining—
and so is the chocolate stock.
THE END