Download Some solutions to a first course in optimization and more Exercises Econometrics and Mathematical Economics in PDF only on Docsity!
Paolo Pin pin@unive.it http://venus.unive.it/pin
August, 2005
Rangaraian K. Sundaran, 1996, “A First Course in
Optimization Theory”, Cambridge University Press:
Solutions of (some) exercises
Appendix C (pp. 330-347): Structures on Vector Spaces.
‖x + y‖^2 − ‖x − y‖^2
< x + y, x + y > − < x − y, x − y >
< x, x > +2 < x, y > + < y, y > −(< x, x > − 2 < x, y > + < y, y >)
4 < x, y >
= < x, y > 2
‖x + y‖^2 + ‖x − y‖^2 = ‖x + y‖^2 − ‖x − y‖^2 + 2‖x − y‖^2 Polarization Identity =⇒ = 4 < x, y > +2‖x − y‖^2 = 2 < x, x > +2 < y, y > = 2
‖x‖^2 + ‖y‖^2
- Let’s take e.g. R^2 , ~x = (1, 0) and ~y = (0, 2):
‖x + y‖^21 + ‖x − y‖^21 = 3^2 + 3^2 = 18 6 = 10 = 2 · (1^2 + 2^2 ) = 2
‖x‖^21 + ‖y‖^21
- a) {xn} → x in (V, d 1 ) if and only if:
n^ lim→∞ d^1 (xn, x) = 0 =⇒^ nlim→∞ d^2 (xn, x)^ ≤^ nlim→∞ c^1 ·^ d^1 (xn, x) =^ c^1 ·^ nlim→∞ d^1 (xn, x) = 0
since d 2 (xn, x) ≥ 0 , ∀ n: limn→∞ d 2 (xn, x) = 0 ;
b) A is open in (V, d 1 ) if and only if:
∀ x ∈ A, ∃ r > 0 s.t. B 1 (x, r) = {y ∈ V | d 1 (x, y) < r} ⊂ A
=⇒ ∀ x ∈ A, ∃
r c 2
0 s.t. B 2 (x,
r c 2 ) = {y ∈ V | c 2 · d 2 (x, y) ≤ r} ⊆ B 1 (x, r) ⊂ A
the reverse of (a) and (b) is equivalent. 2
- Consider R with the usual metric d(x, y) ≡ |x − y| and with the other one defined as δ(x, y) ≡ max{d(x, y), 1 };
δ is a metric, to prove triangle inequality consider that:
δ(x, z) = max{d(x, z), 1 } ≤ max{d(x, y) + d(x, z), 1 }
≤ max{d(x, y) + d(x, z), 1 + d(x, y), 1 + d(y, z), 2 } = max{d(x, y), 1 } + max{d(x, z), 1 } = δ(x, y) + δ(y, z) ;
∀ Bδ (x, r) ⊂ R, Bd(x, r) ⊆ Bδ (x, r), similarly ∀ Bd(x, r) ⊂ R, Bδ (x, min{r, 12 }) ⊆ Bd(x, r), so they generate the same open sets;
nevertheless 6 ∃ c ∈ R+ such that d(x, y) < c · δ(x, y), ∀ x, y ∈ R, suppose by absurd it exists, then d(0, c) = c < c · δ(x, y) is impossible since δ(x, y) ≤ 1 , ∀ x, y ∈ R. 2
- d∞(~x, ~y) = max{|x 1 − y 1 |, |x 2 − y 2 |} is equivalent to any
dp(~x, ~y) =
|x 1 − y 1 |p^ + |x 2 − y 2 |p
) (^1) p , with p ∈ N, p ≤ 1 , because:
d∞ ≤ dp ∧ dp ≤
d∞
all the metrics are equivalent by symmetry and transitivity (see exercise 7 ). 2
- If d 1 and d 2 are equivalent in V , ∃ b 1 , b 2 ∈ R such that ∀ x, y ∈ V :
neither for V itself, so that if V is uncountable, there exists no countable subset of V whose closure is V. 2
- The correct statement of Corollary C.23 (page 340) is:
A function f : V 1 → V 2 is continuous if and only if, ∀ open set U 2 ⊆ V 2 , f −^1 (U 2 ) = {x ∈ V 1 |f (x) ∈ U 2 } is an open set in V 1.
(The inverse function of every open set is an open set.)
Since every subset of (R, ρ) is open, every function (R, ρ) → (R, d) and (R, ρ) → (R, ρ) is continuous, that is F = G = {f | f : R → R}; clearly not every function (R, d) → (R, d) is continuous (here continuity has its standard meaning), then H ⊂ F = G and H 6 = F. 2
- See exercise 11 for the correct statement of Corollary C.23 (page 340);
the open sets in (Rn, d∞) and (Rn, d 2 ) coincide, because:
∀ Bd x,r∞ = {y ∈ Rn^ | max{|yi − xi|}i∈{ 1 ,...n} ≤ r}
∃ B x,rd^2 = {y ∈ Rn^ | (
∑^ n
i=
(yi − xi)^2 ) (^12) ≤ r} ⊆ B x,rd∞
and
∀ B x,rd^2 = {y ∈ Rn^ | (
∑^ n
i=
(yi − xi)^2 ) (^12) ≤ r}
∃ q ≡
r^2 n
s.t. Bd x,q∞ = {y ∈ Rn^ | max{|yi − xi|}i∈{ 1 ,...n} ≤ q} ⊆ B x,rd^2 ;
hence also the continuous functions in (Rn, d∞) → (R, d) and the continuous func- tions in (Rn, d 2 ) → (R, d) coincide. 2
- [The p-adic valuation here is not well defined: if r = 0, a = 0 and any n would do; if r = 1 there are no a, b, n such that 1 = p−n
a b
, a is not a multiple of p.
For a survey: http://en.wikipedia.org/wiki/P-adic_number ]
- Since V with the discrete topology is metrizable from (V, ρ), topological com- pactness is equivalent to compactness in the sequential sense (page 343), hence exercise 9 proves that the compact subsets in (V, ρ) are all and only the finite ones. 2
As seen in exercise 11 , every such function is continuous. 2
d[(x, y), (x′, y′)] =
d 1 (x, y)^2 + d 2 (x′, y′)^2
positivity and symmetry are straightforward from positivity of d 1 and d 2 ;
triangle inequality:
d[(x, z), (x′^ , z′)] =
d 1 (x, z)^2 + d 2 (x′, z′)^2
(d 1 (x, y) + d 1 (y, z))^2 + d 2 (x′, z′)^2
(d 1 (x, y) + d 1 (y, z))^2 + (d 2 (x′, y′) + d 2 (y′, z′))^2
d 1 (x, y)^2 + 2d 1 (x, y)d 1 (y, z) + d 1 (y, z)^2 + d 2 (x′, y′)^2 + 2d 2 (x′, y′)d 2 (y′, z′) + d 2 (y′, z′)^2
[
d 1 (x, y)^2 + d 2 (x′, y′)^2 + d 1 (y, z)^2 + d 2 (y′, z′)^2 + 2
d 1 (x, y)d 1 (y, z) + d 2 (x′, y′)d 2 (y′, z′)
)] 12
[
d 1 (x, y)^2 + d 2 (x′, y′)^2 + d 1 (y, z)^2 + d 2 (y′, z′)^2 +...
... + 2
d 1 (x, y)d 1 (y, z) + d 1 (x, y)d 2 (y′, z′) + d 2 (x′, y′)d 1 (y, z) + d 2 (x′, y′)d 2 (y′, z′)
)]^12
[
d 1 (x, y)^2 + d 2 (x′, y′)^2 + d 1 (y, z)^2 + d 2 (y′, z′)^2 + 2
d 1 (x, y) + d 2 (x′, y′)
d 1 (y, z) + d 2 (y′, z′)
)] 12
since
a + b + 2
ab =
a +
b
=
d 1 (x, y)^2 + d 2 (x′, y′)^2
d 1 (y, z)^2 + d 2 (y′, z′)^2
= d[(x, y), (x′^ , y′)] + d[(y, z), (y′^ , z′)]. 2
- The base of a metric space (X, d 1 ) is the one of its balls B(x, r), every ball in (X, d 1 ) × (Y, d 2 ) = (X × Y, (d^21 + d^22 )
1 (^2) ) is of the form:
B((x∗, y∗), r) = {(x, y) ∈ X × Y | (d 1 (x, x∗)^2 + d 2 (y, y∗)^2 ) (^12) ≤ r} ⊆ {(x, y) ∈ X × Y | max{d 1 (x, x∗), d 2 (y, y∗)} ≤ r} = B(x∗, r) × B(y∗, r)
for every open set in (X, d 1 ) × (Y, d 2 ) there is a couple of open sets, one in (X, d 1 ) and one in (Y, d 2 ), whose product contains them, moreover
Chapter 3 (pp. 90-99): Existence of Solutions.
- As a counterexample consider f : (0, 1) → R such that f (x) = (^1) x :
sup (1,0)
f (x) = limx→ 0
x
- It can be shown (e.g. by construction) that in R (in any completely ordered set) sup and inf of any finite set coincide respectively with max and min. If D ⊂ Rn^ is finite also f (D) = {x ∈ R | ∃ ~x ∈ D s.t. f (~x) = x} is.
The result is implied by the Weierstrass theorem because every finite set A is com- pact, i.e. every sequence in A have a constant (hence converging) subsequence. 2
- a) Consider x¯ = max{D}, which exists because D is compact subset of R, f (¯x) is the desired maximum, because 6 ∃x ∈ D such that x ≥ ¯x, =⇒ 6 ∃x ∈ D such that f (x) ≥ f (¯x);
b) consider D ≡ {(x, y) ∈ R^2 + | x + y = 1} ∈ R^2 , D is the closed segment from (0, 1) to (1, 0), hence it is compact, there are no two points in D which are ordered by the ≥ partial ordering, hence every function D → R is nondecreasing, consider:
f (x, y) =
x if^ x^6 = 0 0 if x = 0
max f on D is limx→ 0 f (x, y) = limx→ (^0 1) x = +∞. 2
- A finite set is compact, because every sequence have a constant (hence converg- ing) subsequence. Every function from a finite set is continuous, because we are dealing with the discrete topology, and every subset is open (see exercise 11 of Appendix C for the correct statement of Corollary C.23 (page 340)). We can take a subset A of cardinality k in Rn^ and construct a one-to-one function assigning to every element of A a different element in R.
A compact and convex subset A ⊂ Rn^ cannot have a finite number k ≥ 2 of elements (because otherwise
a∈A a k is different from every^ a^ ∈^ A^ but should be in A). Suppose A is convex and ∃ a, b ∈ A such that f (a) 6 = f (b), then fa,b = f (λa + (1 − λ)b) is a restriction of f : A → R but can be also considered as a function f[a,b] : [0, 1] → R. By continuity every element of [f (a), f (b)] ⊂ R must be in the codomain of f[a,b]
(the rigorous proof is long, see Th. 1.69, page 60), which is then not finite, so that neither the codomain of f is. 2
- Consider sup(f ), it cannot be less than 1 , if it is 1 , then max(f ) = f (0). Suppose sup(f ) > 1 , then, by continuity, we can construct a sequence {xn ∈ R+ | f (xn) = sup(f ) − sup( nf^ )−^1 , moreover, since limx→∞ f (x) = 0, ∃x s.t.¯ ∀ x > x, f ¯ (x) < 1. {xn} is then limited in the compact set [0, x¯], hence it must converge to xˆ. f (ˆx) is however sup(f ) and then f (ˆx) = max(f ).
f (x) = e−x, restricted to R+, satisfies the conditions but has no minimum. 2
- Continuity (see exercise 11 of Appendix C for the correct statement of Corollary C.23 (page 340)) is invariant under composition, i.e. the composition of continuous functions is still continuous. Suppose g : V 1 → V 2 and f : V 2 → V 3 are continuous, if A ⊂ V 3 is open then, by continuity of f , also f −^1 (A) ⊂ V 2 is, and, by continuity of g, also g−^1 (f −^1 (A)) ⊂ V 1 is. g−^1 ◦ f −^1 : V 3 → V 1 is the inverse function of f ◦ g : V 1 → V 3 , which is then also continuous.
We are in the conditions of the Weierstrass theorem. 2
g(x) =
− 1 x ≤ 0 x x ∈ (0, 1) 1 x ≥ 1
, f (x) = e−|x|
arg max(f ) = 0, max(f ) = e^0 = 1 and also sup(f ◦ g) = 1, which is however never attained. 2
min ~p · ~x subject to ~x ∈ D ≡ {~y ∈ Rn + | u(~y) ≥ u¯}
s.t. u : Rn + → R is continuous, ~p ≫ ~ 0 a) ~p · ~x is a linear, and hence continuous function of ~x, there are two possibilities: (i) u is nondecreasing in ~x, (ii) u is not; (i) D is not compact, we can however consider a utility U and define WU ≡ {~y ∈ Rn + | u(~y) ≤ U }, by continuity of u, WU is compact, for U large enough D ∩ WU is not empty and the minimum can be found in it;
~y(k · ψ~) increases linearly with k and then also U (~y(k · ψ~)) increases with k,
max
U (~y(k · ψ~))
is not bounded above;
(⇐=) ∀ ψ~ ∈ RN^ such that ~p · ψ~ ≤ 0 there are two possibilities: either
∑N
i=1 ψizis^ = 0^ ∀^ s^ ∈ {^1 ,... S}, or ∃ s ∈ { 1 ,... S} such that
∑N
i=1 ψizis^ <^0 ; in the first case the function of k U (~y(k · ~φ)) : R → R is constant, then a maximum exists, in the second case ~y(φ~) ≥ 0 impose a convex constraint on the feasible set, by Weierstrass theorem a solution exists. 2
max W
u 1 (x 1 , h(x)),... un(xn, h(x))
with h(x) ≥ 0 , (~x, x) ∈ [0, w]n+1, w − x =
∑^ n
i=
xi
sufficient condition for continuity is that W , ~u and h are continuous, the problem is moreover well defined only if ∃ x ∈ [0, w] such that h(x) ≥ 0 ;
the simplex ∆w ≡ {(~x, x) ∈ [0, w]n+1^ | x +
∑n i=1 xi^ =^ w}^ is closed and bounded, if h(x) is continuous H = {(~x, x) ∈ [0, w]n+1^ | h(x) ≥ 0 } is closed because the inequality is not strict, then the feasible set ∆w ∩ H is closed and bounded because intersection of closed and bounded sets. 2
max π(x) = max x∈R+
xp(x) − c(x)
with p : R+ → R+ and c : R+ → R+ continuous, c(0) = 0 and p(·) decreasing;
a) ∃ x∗^ > 0 such that p(x∗) = 0, then, ∀ x > x∗, π(x) ≤ π(0) = 0, the solution can then be found in the compact set [0, x∗] ∈ R+;
b) now ∀ x > x∗, π(x) = π(0) = 0, as before;
c) consider c(x) = p 2 ¯ · x, now π(x) = ¯p 2 · x, so that the maximization problem is not bounded above. 2
- a)
max v(c(1), c(2)) subject to ~c(1) ≫ ~ 0 , ~c(2) ≫ ~ 0 , ~p(1) · ~c(1) + ~p(2) 1 + r
· ~c(2) ≤ W 0
b) we can consider, in R^2 n, the arrays ~c = (~c(1), ~c(2)) and ~p = (~p(1), ~p 1+(2)r ), the conditions
~c(1) ≫ ~ 0 , ~c(2) ≫ ~ 0 , ~p(1) · ~c(1) +
~p(2) 1 + r · ~c(2) ≤ W 0
become
~c ≫ ~ 0 , ~p · ~c ≤ W 0
we are in the conditions of example 3.6 (pages 92-93), and ~p ≫ ~ 0 if and only if ~p(1) ≫ ~ 0 and p~(2) ≫ ~ 0. 2
max
π(x 1 ) + π(x 2 ) + π(x 3 )
subject to
0 ≤ x 1 ≤ y 1 0 ≤ x 2 ≤ f (y 1 − x 1 ) 0 ≤ x 3 ≤ f (f (y 1 − x 1 ) − x 2 )
let’s call A the subset of R^3 + that satisfies the conditions, A is compact if it is closed and bounded (Th. 1.21, page 23). It is bounded because
A ⊆ [0, y 1 ] × [0, max f ([0, y 1 ])] × [0, max f ([0, max f ([0, y 1 ])])] ⊂ R^3 +
and maxima exist because of continuity. Consider (ˆx 1 , ˆx 2 , xˆ 3 ) ∈ Ac^ in R^3 +, then (if we reasonably suppose that f (0) = 0), becasue of continuity, at least one of the following holds: x ˆ 1 > y 1 , xˆ 2 > f (y 1 − ˆx 1 ) or xˆ 3 > f (f (y 1 − xˆ 1 ) − xˆ 2 ). If we take r = max{xˆ 1 − y 1 , xˆ 2 − f (y 1 − xˆ 1 ), xˆ 3 − f (f (y 1 − xˆ 1 ) + ˆx 2 )}, r > 0 and B((ˆx 1 , xˆ 2 , xˆ 3 ), r) ⊂ Ac. Ac^ is open =⇒ A is closed. 2
(− 25 a, − 15 a) and (− 15 a, − 25 a) are local minima ∀ a ∈ R;
d) when sin y 6 = 0, limits for x ±∞ are ±∞,
fx = sin y fy = x cos y
=⇒ (0, kπy) ∀ k ∈ Z = {... , − 2 , − 1 , 0 , 1 ,.. .}
f is null for critical points, adding any (ǫ, ǫ) to them (ǫ < π 2 ), f is positive, adding (ǫ, −ǫ) to them, f is negative, critical points are saddles;
e) limits for x or y to ±∞ are +∞, ∀ a ∈ R,
fx = 4 x^3 + 2xy^2 fy = 2 x^2 y − 1
=⇒ y =
2 x^2
=⇒ 4 x^3 = −
x
=⇒ impossible
there are no critical points;
f) limits for x or y to ±∞ are +∞,
fx = 4 x^3 − 3 x^2 fy = 4 y^3
since f (0, 0) = 0 and f ( 34 , 0) < 0 , ( 34 , 0) is a minimum and (0, 0) a saddle;
g) limits for x or y to ±∞ are 0,
fx = 1 −x (^2) +y 2 (1+x^2 +y^2 )^2 fy = (^) (1+−x^22 xy+y (^2) ) 2
since f (1, 0) = 12 and f (− 1 , 0) = − 12 , the previous is a maximum, the latter a minimum;
h) limits for x to ±∞ are +∞, limits for y to ±∞ depends on the sign of (x^2 − 1),
fx = x 3 8 + 2xy
fy = 2 y(x^2 − 1)
y = 0 =⇒ x = 2 ∨ x = 1 =⇒ y = ±
√ 7 4 ∨ x = − 1 =⇒ impossible
D^2 f (x, y) =
8 x
(^2) + 2y (^2 4) xy 4 xy 2 y(x^2 − 1)
D^2 f (2, 0) =
=⇒ saddle, D^2 f (1,
√ 7 4 ) =
4
√^7
=⇒ saddle,
D^2 f (1, −
√ 7 4 ) =
=⇒ saddle. 2
- The unconstrained function is given by the substitution y = 9 − x:
f (x) = 2 + 2x + 2(9 − x) − x^2 − (9 − x)^2 = − 2 x^2 + 18x − 61
which is a parabola with a global maximum in x = 92 =⇒ y = 92. 2
- a) x∗^ is a local maximum of f if ∃ǫ ∈ R+ such that ∀y ∈ B(x∗, ǫ), f (y) ≤ f (x∗), considering then that:
lim inf y→x∗^ f (x∗) − f (y) = lim inf y∈B(x∗^ ,ǫ)→x∗^
f (x∗) − f (y) ≥ 0
∀ y < x∗, x∗^ − y > 0 ∧ ∀ y > x∗, x∗^ − y < 0
− 1 · lim inf = − lim sup
we have the result;
b) a limit may not exist, while lim inf and lim sup always exist if the function is defined on all R;
c) if x∗^ is a strict local maximum the inequality for lim infy→x∗ is strict, so also the two inequalities to prove are. 2
- Consider f : R → R:
f (x) =
1 x ∈ Q 0 otherwise
where Q ⊂ R are the rational number; x = 0 is a local not strict maximum, f is not constant in x = 0. 2
- f is not constant null, otherwise also f ′^ would be constant null, since limy→∞ f (y) = 0, f has a global maximum ¯x, with f (¯x) > 0 and f ′(¯x) x is the only point for which f ′(x) = 0 =⇒ x = ¯x. 2
- a) φ′^ > 0 :
dφ ◦ f dxi
= φ′(f ) df dxi
df dxi
dφ ◦ f dxi
Df (~x∗) = ~ 0 =⇒ Dφ ◦ f (~x∗) = ~ 0
Chapter 5 (pp. 112-144): Equality Constraints.
- a)
L(x, y, λ) = x^2 − y^2 + λ(x^2 + y^2 − 1) Lx = 2x + 2λx = =⇒ λ = − 1 ∨ x = 0 Ly = − 2 y + 2λy = =⇒ λ = 1 ∨ y = 0
Lλ = x^2 + y^2 − 1 ==0⇒
x = 0 ⇒ y = ± 1 λ = 1 y = 0 ⇒ x = ± 1 λ = − 1
when λ = 1 we have a minimum, when λ = − 1 a maximum;
b)
f (x) = x^2 − (1 − x^2 ) = 2x^2 − 1 =⇒
min for x = 0 max for x = ±∞
the solution is different from (a) because the right substitution is y ≡
1 − x^2 , admissible only for x ∈ [− 1 , 1]. 2
- a) Substituting y ≡ 1 − x:
f (x) = x^3 − (1 − x)^3 = 3x^2 − 3 x + 1 =⇒ max for x = ±∞ b)
L(x, y, λ) = x^3 + y^3 + λ(x + y − 1) Lx = 3x^2 + λ ==0⇒ λ = − 3 x^2 Ly = 3y^2 + λ ==0⇒ λ = − 3 y^2
=⇒ x = ±y
Lλ = x + y − 1 x=y =⇒ x = y =
x = y = 12 is the unique local minimum, as can be checked in (a). 2
- a)
L(x, y, λ) = xy + λ(x^2 + y^2 − 2 a^2 )
Lx = y + 2λx ==0⇒ (y = 0 ∧ λ = 0) ∨ y = − 2 λx Ly = x + 2λy ==0⇒ (x = 0 ∧ λ = 0) ∨ x = − 2 λy
Lλ = x^2 + y^2 − 2 a^2
=⇒
x = 0 ⇒ y = ±
2 a y = 0 ⇒ x = ±
2 a y = − 2 λx ∧ x = − 2 λy
2 a) and (±
2 a, 0) are saddle points, because f (x, y) can be positive or neg- ative for any ball around them; y = − 2 λx ∧ x = − 2 λy imply: λ = ± 12 , x = ±y, and x = ±a, the sign of x does not matter, when x = y we have a maximum, when x = −y a minimum.
b) substitute xˆ ≡ (^1) x , ˆy ≡ (^1) y and aˆ ≡ (^1) a
L(ˆx, y, λ) = ˆx + ˆy + λ(ˆx^2 + ˆy^2 − ˆa^2 ) Lˆx = 1 + 2λxˆ ==0⇒ xˆ = −
λ
Lˆy = 1 + 2λˆy ==0⇒ yˆ = −
λ = ˆx
Lλ = ˆx^2 + ˆy^2 − ˆa^2
=⇒ xˆ = ˆy = ±
a
for xˆ and yˆ negative we have a minimum, otherwise a maximum.
c)
L(x, y, x, λ) = x + y + z + λ(x−^1 + y−^1 + z−^1 − 1) Lx = 1 − λx−^2 ==0⇒ x = ±
λ
... ==0⇒ y = ±
λ
... ==0⇒ z = ±
λ
we have a minimum when they are all negative (x = y = z = − 3 ) and a maximum when all positive (x = y = z = − 3 ).
d) substitute xy = 8 − (x + y)z = 8 − (5 − z)z:
f (z) = z(8 − (5 − z)z) = z^3 − 5 z^2 + 8z
fz = 3z^2 − 10 z + 8 ==0⇒ z =
4 3
as z → ±∞, f (z) → ±∞, fzz = 6z − 10 is negative for z = 43 (local maximum) and positive for z = 2 (local minimum),
- a) (x − 1)^3 = y^2 implies x ≥ 1 , f (x) = x^3 − 2 x^2 + 3x − 1 =⇒ fx = 3x^2 − 4 x + 3 which is always positive for x ≥ 1 , since f (x) is increasing, min f (x) = f (1) = 1 (y = 0);
b) the derivative of the constraint D((x − 1)^3 − y^2 ) is (3x^2 − 6 x + 3, − 2 y)′, which is (0, 0)′^ in (1, 0) and in any other point satisfying the constraint, hence the rank condition in the Theorem of Lagrange is violated. 2
- a)
L(~x, ~λ) = ~c′~x +
~x′D~x + ~λ′(A~x − ~b)
∑^ n
i=
cixi +
∑^ n
i=
( (^) ∑n
j=
xj Dji
xi +
∑^ m
i=
λi
(( (^) ∑n
j=
Aij xj
− bi
Lxi = ci + Diixi +
∑^ n
j=1,j 6 =i
xj Dji +
∑^ n
j=1,j 6 =i
xj Dij +
∑^ m
j=
λj Aji
= ci +
∑^ n
j=
xj Dij +
∑^ m
j=
λj Aji
Lλi =
( (^) ∑n
j=
Aij xj
− bi
b)...
- The constraint is the normalixation |~x| = 1, the system is:
f (~x) = ~x′A~x
=
∑^ n
i=
( (^) ∑n
j=
xj Aji
xi
fxi = 2
∑^ n
j=
xj Aji
x∗ i = −
∑n j=1,j 6 =i x ∗ j Aji Aii
x^ ~∗^ = −
0 A A^2111... A An 111 A 12 A 22 0...^
An 2 A 22 .. .
A 1 n Ann
A 2 n Ann...^0
· x~∗^ ≡ B · x~∗
such that | x~∗| = 1
for any eigenvectors of the square matrix B, its two normalization (one opposite of the other) are critical points of the problem;
since D^2 f (~x) = 2
A 11... An 1 .. .
A 1 n... Ann
,^ ∀^ ~x^ is constant,
all critical points are maxima, minima or saddles, according wether A is positive- definite, negative-definite or neither. 2
- a)
(^00 50 100 150 200 250 300 350 )
1
stock
days
The function s(t) quantifying the stock is periodic of period T = x dt dI ,
in this period the average stock is
R (^) T 0 s(t) T =^
x·T 2 T =^
x 2 , in the long run this will be the total average;
b)
L(x, n, λ) = Ch x 2
Lx = C 2 h + λn Ln = C 0 + λx
==0⇒ n = − Ch 2 λ
∧ x = −
C 0
λ
Lλ = nx − A ==0⇒
ChC 0 2 λ^2 = A =⇒ λ = ±
ChC 0 2 A
x and n negative have no meaning so λ must be negative. 2
The condition for equality constraint to suffice is that u(x 1 , x 2 ) is nondecreasing, this happens for α ≥ 1 ∧ β ≥ 1 ; if otherwise one of them , say α is less than 1 , the problem is unbounded at limx 1 → 0 xα 1 + xβ 2 = +∞. 2
