sundaram optimization solutions

8/14/2019 Sundaram Optimization Solutions

1/28

Paolo [email protected]

http://venus.unive.it/pin

August, 2005

Rangaraian K. Sundaran, 1996, A First Course inOptimization Theory, Cambridge University Press:

Solutions of (some) exercises

Appendix C (pp. 330-347): Structures on Vector Spaces.

1.

14

x+ y2 x y2 = 14

< x + y, x + y > < x y, x y >

= 1

4

< x, x >+2 < x, y >+ < y, y > (< x, x > 2< x, y >+ < y, y >)

=

1

4

4< x, y >

= < x, y >

2.

x+ y2 + x y2 = x + y2 x y2 + 2x y2

Polarization Identity= = 4< x, y >+2x y2

= 2< x, x >+2 < y, y >

= 2x2 + y2

3. Lets take e.g. R2, x= (1, 0) andy= (0, 2):

x+ y21+ x y21= 32 + 32 = 18 = 10 = 2 (12 + 22) = 2x21+ y21

1


2/28

4. a)

{xn

} x in (V, d1) if and only if:

limn d1(xn, x) = 0 = limn d2(xn, x) limn c1 d1(xn, x) =c1 limn d1(xn, x) = 0

sinced2(xn, x) 0, n: limn d2(xn, x) = 0 ;b)A is open in (V, d1) if and only if:

x A, r >0 s.t. B1(x, r) = {y V | d1(x, y)< r} A

= x A, r

c2>0 s.t. B2(x,

r

c2) = {y V | c2 d2(x, y) r} B1(x, r) A

the reverse of (a) and (b) is equivalent.

5. Consider R with the usual metric d(x, y) |x y|and with the other one defined as (x, y) max{d(x, y), 1};

is a metric, to prove triangle inequality consider that:

(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} =(x, y) + (y, z) ;

B(x, r) R, Bd(x, r) B(x, r),similarly Bd(x, r) R, B(x, min{r,12}) Bd(x, r),so they generate the same open sets;

nevertheless c R+ such that d(x, y)< c (x, y), x, y R,suppose by absurd it exists,thend(0, c) =c < c

(x, y) is impossible since (x, y)

1,

x, y

R.

6. d(x, y) = max{|x1 y1|, |x2 y2|} is equivalent to anydp(x, y) =

|x1 y1|p + |x2 y2|p

1p

, with p N, p 1, because:

d dp dp 12

d

all the metrics are equivalent by symmetry and transitivity (see exercise 7).

7. Ifd1 and d2 are equivalent in V,b1, b2 Rsuch that x, y V:

2


3/28

d2(x, y)b1

d1(x, y) b2 d2(x, y)

similarly ifd2 and d3 are equivalent in V,c1, c2 Rsuch thatx, y V:d3(x, y)

c1 d2(x, y) c2 d3(x, y)

hence (b1 c1), (b2 c2) Rsuch thatx, y V:d3(x, y)

b1 c1 d1(x, y) b2 c2 d3(x, y)

=

d1 and d3 are equivalent.

8.

(x, y) =

1 x =y0 x= y

Positivity and Simmetrycome from the definition;for Triangle inequality consider the exhaustive cases:

x= y = z = d(x, z) = 0 = 0 + 0 =d(x, y) + d(y, z)x= y

=z =

d(x, z) = 1 = 0 + 1 =d(x, y) + d(y, z)

x =y = z = symmetricx= z=y = d(x, z) = 0 1 + 1 =d(x, y) + d(y, z)

all different = d(x, z) = 1 1 + 1 =d(x, y) + d(y, z)

9. We generalize from Rto any V;every subset Xin(V, )is closed becauseXC is open (every subset is open in (V, ));

in (V, ) the only convergent sequences are those constant from a certain point on(not only constant sequencesas stated at page 337),consider a finite subset A ofV, every sequence in A have a constant (hence con-verging) subsequence,consider now an infinite subset B ofV, (by the Axiom of choice) we can constructa sequence with all different elements, where all subsequence will have all differentelements,= compact subsets in (V, ) are all and only the finite ones. 10. Since that every subset X in (V, ) is closed (exercise 9), then cl(X) =X, X V, and

X V, A X s.t. A =X and cl(A) =X

3


4/28

neither forVitself, so that ifV is uncountable, there exists no countable subset of

Vwhose closure is V.

11. The correct statement of Corollary C.23 (page 340) is:

A function f : V1 V2 is continuous if and only if, open setU2 V2, f1(U2) = {x V1|f(x) U2} is an open set inV1.

(The inverse function of every open set is an open set.)

Since every subset of (R, ) is open, every function (R, ) (R, d) and (R, )(R, ) is continuous, that is

F=

G=

{f

|f : R

R

};

clearly not every function (R, d) (R, d) is continuous (here continuity has itsstandard meaning), thenH F= G andH = F. 12. See exercise 11for the correct statement of Corollary C.23 (page 340);

the open sets in (Rn, d)and(Rn, d2)coincide, because:

Bdx,r = {y Rn | max{|yi xi|}i{1,...n} r}

B d2x,r = {y Rn | (n

i=1(yi xi)2) 12 r} Bdx,r

and

Bd2x,r = {y Rn | (n

i=1

(yi xi)2)12 r}

q

r2

n s.t. Bdx,q = {y Rn | max{|yi xi|}i{1,...n} q} Bd2x,r ;

hence also the continuous functions in(Rn, d) (R, d)and the continuous func-tions in (Rn, d2)

(R, d)coincide.

13. [The p-adic valuationhere is not well defined:ifr= 0, a = 0 and any n would do;

ifr= 1 there are no a, b, n such that 1 =pn

ab

, a is not a multiple ofp.

For a survey: http://en.wikipedia.org/wiki/P-adic_number]

14. Since Vwith the discrete topology is metrizable from (V, ), topological com-pactness is equivalent to compactness in the sequential sense (page 343), henceexercise 9 proves that the compact subsets in (V, ) are all and only the finiteones.

4


5/28

15. As seen in exercise 11, every such function is continuous.

16.

d[(x, y), (x, y)] =

d1(x, y)2 + d2(x

, y)2 1

2

positivity and symmetry are straightforward from positivity ofd1 and d2;

triangle inequality:

d[(x, z), (x, z)] =

d1(x, z)

2 + d2(x, z)2

12

(d1(x, y) + d1(y, z))2 + d2(x

, z)2 1

2

(d1(x, y) + d1(y, z))2 + (d2(x

, y) + d2(y, z))2 1

2

=

d1(x, y)2 + 2d1(x, y)d1(y, z) + d1(y, z)

2 + d2(x, y)2 + 2d2(x, y)d2(y, z) + d2(y, z)2

12

=

d1(x, y)2 + d2(x

, y)2 + d1(y, z)2 + d2(y, z)2 + 2

d1(x, y)d1(y, z) + d2(x, y)d2(y, z)

12

d1(x, y)2 + d2(x

, y)2 + d1(y, z)2 + d2(y, z)2 + . . .

. . .+ 2

d1(x, y)d1(y, z) + d1(x, y)d2(y, z) + d2(x, y)d1(y, z) + d2(x, y)d2(y, z)

12

=

d1(x, y)2 + d2(x

, y)2 + d1(y, z)2 + d2(y, z)2 + 2

d1(x, y) + d2(x, y)

d1(y, z) + d2(y

, z) 1

2

since

a+ b + 2

ab=

a +

b

=

d1(x, y)2 + d2(x

, y)2 1

2+

d1(y, z)2 + d2(y

, z)2 1

2

= d[(x, y), (x, y)] + d[(y, z), (y, z)] .

17. The base of a metric space (X, d1) is the one of its balls B (x, r),

every ball in (X, d1) (Y, d2) = (X Y, (d21+ d22) 12 ) is of the form:

B((x, y), r) = {(x, y) X Y | (d1(x, x)2 + d2(y, y)2)12 r}

{(x, y) X Y | max{d1(x, x), d2(y, y)} r}= B(x, r) B(y, r)

for every open set in (X, d1) (Y, d2)there is a couple of open sets, one in(X, d1)and one in (Y, d2), whose product contains them,moreover

5


6/28

B(x, r1) B(y, r2) = {(x, y) X Y | d1(x, x) r1 d2(y, y) r2} {(x, y) X Y | (d1(x, x)2 + d2(y, y)2)

12 r1+ r2}

= B((x, y), r1+ r2)

then for every couple of open sets in (X, d1)and (Y, d2), there is an open set in theproduct space containing their product,then the two topologies coincide.

18. [(a) is not hard only for fcompact = f sequentially compact,the other way round and (b) are really non trivial;

see e.g.:http://www-history.mcs.st-and.ac.uk/john/MT4522/Lectures/L22.html]

19. ftopological continuous = fsequentially continuous:considerxn x V, and an open Of(x) containing f(x) V,f1Of(x) contains xand all of the xn after a certain n0,then all of the f(xn), after the same n0, are in Of(x),fis sequelntially continuous;

fsequentially continuous = f topological continuous:non trivial. . .

20. a) 1) Clearly, [0, 1] ;

2) ifO1, O2 , O C1 andOC2 are finite,but then also their union OC1 OC2 = (O1 O2)C is,= O1 O2 ;

3) if{O}A , OC is finite A,but then also their intersection

A O

C = (

A O)

C is, =A O ;b) supposex [0, 1] with a countable base{Bi}iN,

sinceBCi is finitei N, iN BCi is countable,

since[0, 1] is uncountable, y=x, y [0, 1] such that yiN BCi ,= y Bi i N,{y}C contains x but is not contained in any element of its base.

6


7/28

Chapter 3 (pp. 90-99): Existence of Solutions.

1. As a counterexample consider f : (0, 1) Rsuch that f(x) = 1x :

sup(1,0)

f(x) =limx01

x= +.

2. It can be shown (e.g.by construction) that in R (in any completely ordered set)sup and infof any finite set coincide respectively with max and min.IfD 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 setA is com-pact, i.e. every sequence in A have a constant (hence converging) subsequence.

3. a) Consider x= max{D}, which exists becauseD is compact subset ofR,f(x)is the desired maximum, becausex D such that x x,= x D such that f(x) f(x);

b) considerD {(x, y) R2+| x + y= 1} R2,D is the closed segment from(0, 1) to (1, 0), hence it is compact,there are no two points inD which are ordered by thepartial ordering,hence every functionD R is nondecreasing,consider:

f(x, y) =

1

x if x = 0

0 if x= 0

max f onD is limx0 f(x, y) = limx0 1x = +. 4. 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 thediscrete topology, and every subset is open (see exercise 11of Appendix C for thecorrect statement of Corollary C.23 (page 340)).

We can take a subset A of cardinality kin Rn and construct a one-to-one functionassigning to every element ofA a different element in R.

A compact and convex subset A Rn cannot have a finite number k 2 ofelements (because otherwise

aA

ak

is different from every aA but should bein A).Suppose A is convex and a, b A such that f(a) = f(b), then f[a,b]() =f(a+ (1 )b) is a restriction of f : A R but can be also considered as afunctionf[a,b] : [0, 1] R.By continuity every element of [f(a), f(b)] R must be in the codomain off[a,b]

7


8/28

(the rigorous proof is long, see Th. 1.69, page 60), which is then not finite, so that

neither the codomain off is.

5. 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(f)1n , 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 x. f(x)ishoweversup(f)and then f(x) = max(f).

f(x) =ex, restricted to R+, satisfies the conditions but has no minimum.

6. Continuity(see exercise 11 of Appendix C for the correct statement of CorollaryC.23 (page 340)) is invariant under composition,i.e.the composition of continuousfunctions is still continuous.Suppose g : V1 V2 and f : V2 V3 are continuous, ifAV3 is open then, bycontinuity off, alsof1(A) V2is, and, by continuity ofg, alsog1(f1(A)) V1is.g1 f1 : V3 V1 is the inverse function off g : V1 V3, which is then alsocontinuous.

We are in the conditions of the Weierstrass theorem.

7.

g(x) =

1 x 0x x (0, 1)1 x 1

, f(x) =e|x|

arg max(f) = 0, max(f) =e0 = 1 and also sup(f g) = 1, which is however neverattained.

8.

min p x subject to x D {y Rn+| u(y) u}

s.t. u: Rn+ Ris continuous, p 0a) p xis a linear, and hence continuous function ofx,

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 defineWU {y Rn+| u(y) U},by continuity ofu,WU is compact,for U large enoughD WUis not empty and the minimum can be found in it;

8


9/28

(ii) for the components ofx where u is decreasing,

maximal utility may be at 0,soDmay be compact, if not we can proceed as in (i);

b) ifu is not continuous we could have no maximum even if it is nondecreasing(see exercise 3);if pi < 0, and u is nondecreasing inxi, the problem minimizes for xi +. 9.

max p g(x) w x s.t. x Rn+

with p >0, g continuous, w 0a solution is not guaranteed because Rn+ is not compact.

10.

F(p, w) = {(x, l) Rn+1+ | px w(H l), l H}F(p, w) compact = p 0 (by absurd):

ifpi 0, the constrain could be satisfied also at the limit xi +,Fwould not be compact;

p 0 = F(p, w) compact:

Fis closed because inequalities are not strict,l is limited in [0, H],

i {1, . . . n}, xi is surely limited in [0, w(Hl)pi ]. 11.

max U(y()) subject to { RN | p 0, y() 0}where y

s() =

s+ N

i=1

iz

is, U : RS

+R is continuous and strictly increasing

(y y in RS+ ifyi yii {1, . . . , S } and at least one inequality is strict).

Arbitrage: RN such that p 0 and Z 0.

A solution exists no arbitrage

(=) (by absurd: (A= B) (B= A) )suppose RN such that p 0 and Zt 0, then any multiple k of has the same property,

9


10/28

y(k

) increases linearly with k and then alsoU(y(k

)) increases with k ,

max

U(y(k )) is not bounded above;(=) RN such that p 0 there are two possibilities:

eitherN

i=1 izis = 0s {1, . . . S },ors {1, . . . S }such that Ni=1 izis < 0;in the first case the function ofk U(y(k )) : R Ris constant,then a maximum exists,in the second case y() 0 impose a convex constraint on the feasible set,by Weierstrass theorem a solution exists.

12.

max W

u1(x1, h(x)), . . . un(xn, h(x))

with h(x) 0, (x, x) [0, w]n+1, w x=n

i=1

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 theinequality is not strict,then the feasible set w His closed and bounded because intersection of closedand bounded sets.

13.

max (x) = maxxR+

xp(x) c(x)

with p : R+ R+ and c : R+ R+ continuous, c(0) = 0and p() decreasing;

a)x> 0 such that p(x) = 0, then, x > x, (x) (0) = 0,the solution can then be found in the compact set [0, x]

R+;

b) now x > x, (x) =(0) = 0, as before;

c) consider c(x) = p2 x, now(x) = p2 x, so that the maximization problem isnot bounded above.

14. 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) W0

10


11/28

b) we can consider, in R2n, the arraysc= (c(1), c(2)) and p= (p(1), p(2)1+r ), the

conditions

c(1) 0, c(2) 0, p(1) c(1) + p(2)1 + r

c(2) W0become

c 0, p c W0we are in the conditions of example 3.6 (pages 92-93), and p 0 if and only ifp(1) 0and p(2) 0. 15.

max

(x1) + (x2) + (x3)

subject to

0 x1 y10 x2 f(y1 x1)0 x3 f(f(y1 x1) x2)

lets callAthe subset ofR3+that satisfies the conditions, A is compact if it is closedand bounded (Th. 1.21, page 23).It is bounded because

A [0, y1] [0, max f([0, y1])] [0, max f([0, max f([0, y1])])] R3

+

and maxima exist because of continuity.Consider (x1,x2,x3) Ac in R3+, then (if we reasonably suppose that f(0) = 0),becasue of continuity, at least one of the following holds:x1 > y1, x2 > f(y1 x1) or x3 > f(f(y1 x1) x2).If we take r= max{x1 y1,x2 f(y1 x1),x3 f(f(y1 x1) + x2)}, r >0 andB((x1,x2,x3), r) Ac.Ac is open = A is closed.

11


12/28

Chapter 4 (pp. 100-111): Unconstrained Optima.

1. ConsiderD = [0, 1] and f : D R, f(x) =x:arg maxxDf(x) = 1, but Df(1) = 1 = 0. 2.

f(x) = 1 2x 3x2 = 0 for x= 1

1 + 3

3 =1

13

f(x) = 2 6x , f(1) = 4f(1

3

) =

4

1 is a local minimum and 13 a local maximum, they are not global becauselimx= + and limx+= . 3. The proof is analogous to the proof at page 107, reverting the inequalities.

4. a)

fx = 6x2 + y2 + 10x

fy = 2xy+ 2y

= y= 0 x= 0

D2f(x, y) = 12x+ 10 2y2y 2x+ 2

(0, 0) is a minimum;

b)

fx = e2x + 2e2x(x+ y2 + 2y) =e2x(1 + 2(x+ y2 + 2y))

fy = e2x(2y+ 2)

= y = 1

1 + 2(x+ 1 2) = 0 = x= 12

= f( 12

, 1) = 12

e

since limx f(x, 1) = 0 and limx+ f(x, 1) = +, the only critical point(12 , 1)is a global minimum;

c) the function is symmetric, limits for x or y to are +,a R:

fx = ay 2xy y2fy = ax 2xy x2

= (0, 0) (0, a) (a, 0) (2

5a, 1

5a) (1

5a, 2

5a)

D2f(x, y) =

2y a 2x 2y

a 2x 2y 2x

12


13/28

(

25a,

15a)and(

15a,

25a)are local minima

a

R;

d) when sin y= 0, limits for xare,

fx = sin yfy = x cos y

= (0,ky) k Z = {. . . , 2, 1, 0, 1, . . .}

fis null for critical points, adding any (, )to them ( < 2 ), f is positive,adding(, ) to them, f is negative,critical points are saddles;

e) limits for x or y toare +,a R,

fx = 4x3 + 2xy2

fy = 2x2y 1

= y = 1

2x2 = 4x3 = 1

x = impossible

there are no critical points;

f) limits for x or y toare +,

fx = 4x3 3x2

fy = 4y3

= (0, 0) ( 3

4, 0)

sincef(0, 0) = 0 and f(34 , 0)< 0, (34 , 0) is a minimum and (0, 0) a saddle;

g) limits for x or y toare 0,

fx = 1x2+y2(1+x2+y2)2

fy = 2xy(1+x2+y2)2

= (1, 0) (1, 0)

sincef(1, 0) = 12 and f(1, 0) = 12 , the previous is a maximum, the latter aminimum;

h) limits for x toare +,limits fory todepends on the sign of(x2 1),

fx = x3

8 + 2xy2 1

fy = 2y(x2 1)

=

y= 0 = x= 2

x= 1 = y = 74

x= 1 = impossible

D2f(x, y) =

38x

2 + 2y2 4xy4xy 2y(x2 1)

13


14/28

D2f(2, 0) = 32 00 0

= saddle, D2f(1,7

4 ) = 54

7

7 0 = saddle,D2f(1,

74 ) =

54

7

7 0

= saddle.

5. The unconstrained function is given by the substitution y= 9 x:

f(x) = 2 + 2x + 2(9 x) x2 (9 x)2 = 2x2 + 18x 61which is a parabola with a global maximum in x= 92 = y= 92 . 6. a)xis a local maximum off if R+such that y B(x, ), f(y) f(x),considering then that:

lim infyx

f(x) f(y) = lim inf yB(x,)x

f(x) f(y) 0

y < x, x y >0 y > x, x y 0 and f(x)x is the only point for which f(x) = 0 = x= x. 9. a) > 0:

d fdxi

=(f)df

dxi,

df

dxi= 0 = d f

dxi= 0

Df(x) = 0 = D f(x) = 0

14


15/28

b)

d2 fdxidxj

= d

dxj

(f)

df

dxi

=(f)

df

dxi

df

dxj+ (f)

d2f

dxidxj

df

dxi= 0 = d

2 fdxidxj

=(f) d2f

dxidxj

Df(x) = 0 = D2 f(x) =(f(x)) D2f(x)a negative definite matrix is still so if multiplied by a constant.

10. Let us check conditions on principal minors (see e.g. Hal R. Varian, 1992,

Microeconomic analysis, Norton, pp.500-501).

D2f(x) =

fx1x1 . . . f x1xn...

. . . ...

fx1xn . . . f xnxn

is positive definite if:1) fx1x1 >0,2) fx1x1 fx2x2 f2x1x2 >0,. . .

n)|D2

f(x)| >0;

D2g(x) =

gx1x1 . . . gx1xn...

. . . ...

gx1xn . . . gxnxn

=D2 f(x) =

fx1x1 . . . fx1xn...

. . . ...

fx1xn . . . fxnxn

is negative definite if:1)fx1x1 0,. . .n)

|D2

f(x)

|0 ifn is even;

the ith principal minor is a polinimial where all the elements have order i,it is easy to check that odd principal minors mantain the sign of its arguments,while even ones are always positive,henceD2f(x) positive definite = D2f(x) negative definite.

15


16/28

Chapter 5 (pp. 112-144): Equality Constraints.

1. a)

L(x,y ,) =x2 y2 + (x2 + y2 1)Lx = 2x+ 2x

=0= = 1 x= 0

Ly = 2y+ 2y =0= = 1 y= 0L = x

2 + y2 1 =0=

x= 0 y= 1 = 1y = 0 x= 1 = 1

when= 1we have a minimum, when = 1a maximum;b)

f(x) = x2 (1 x2) = 2x2 1 =

min for x= 0max for x=

the solution is different from (a) because the right substitution is y 1 x2,admissible only forx [1, 1]. 2. a) Substituting y 1 x:

f(x) = x3 (1 x)3 = 3x2 3x+ 1 = max for x = b)

L(x,y ,) =x3 + y3 + (x+ y 1)Lx= 3x

2 + =0= = 3x2

Ly = 3y2 +

=0= = 3y2

= x= y

L= x + y 1 x=y= x= y = 12

x= y = 12 is the unique local minimum, as can be checked in (a).

3. a)

L(x,y ,) =xy + (x2 + y2 2a2)Lx= y+ 2x

=0= (y= 0 = 0) y= 2x

Ly =x + 2y =0= (x= 0 = 0) x= 2y

L= x2 + y2 2a2 =0=

x= 0 y = 2ay= 0 x= 2a

y= 2x x= 2y

16


17/28

(0,

2a)and (

2a, 0)are saddle points, because f(x, y)can be positive or neg-

ative for any ball around them;y= 2x x= 2y imply:= 12 , x = y, and x= a,the sign ofx does not matter, when x= y we have a maximum, when x =y aminimum.

b) substitute x 1x

, y 1y

and a 1a

L(x ,y ,) = x + y+ (x2 + y2 a2)

Lx = 1 + 2x

=0

= x= 1

2

Ly = 1 + 2y =0= y = 1

2= x

L= x2 + y2 a2 =0= x= y=

2

2 a

for x and y negative we have a minimum, otherwise a maximum.

c)

L(x,y,x,) =x + y+ z+ (x1 + y1 + z1

1)

Lx= 1 x2 =0= x= . . .

=0= y=

. . . =0= z =

we have a minimum when they are all negative (x= y = z = 3) and a maximumwhen all positive (x= y = z = 3).

d) substitute xy = 8 (x + y)z = 8 (5 z)z:

f(z) =z(8 (5 z)z) =z3 5z2 + 8z

fz = 3z2 10z+ 8 =0= z= 5

25 243

=

243

as z , f(z) ,fzz = 6z 10 is negative for z = 43 (local maximum)and positive for z = 2 (local minimum),

17


18/28

whenz= 43 , x + y= 5

z = 113 and xy= 8

(x+ y)z= 289

= one is also 43 and the other 73 ,

whenz= 2, x + y= 5 z = 3and xy= 8 (x + y)z= 2= one is also 2 and the other is 1;

e) substituting y 16x

:

f(x) = x + 16x

= fx= 1 16x2=0

= x= 4whenx = 4and y = 14 we have a minimum, maxima are unbounded for limx0and limx;

f) substitutingz

6

x and y

2x:f(x) =x2 + 4x (6 x)2 = 16x 36 ,f(x)is a linear function with no maxima nor minima.

4. Actually a lemniscate is(x2+y2)2 =x2y2, while(x2y2)2 =x2+y2 identifiesonly the point (0, 0).

1 0.8 0.6 0.4 0.2 0 0.2 0.4 0.6 0.8 10.5

0

0.5

x

y

(x2+y

2)2x

2+y

2= 0

A lemniscate

In the lemniscate x + y maximizes, by simmetry, in the positive quadrant,in the point where the tangent of the explicit function y = f(x)is1;for x and y positive the explicit function becomes:

(x2 + y2)2 = x2 y2

x4

+ 2x2

y2

+ y4

x2

+ y2

= 0y4 + (2x2 + 1)y2 + x4 x2 = 0

y2 = 2x2 1 + 4x4 + 4x2 + 1 4x4 + 4x2

2

y= f(x) =

2x2 + 1 + 8x2 + 1

2

computingf(x)and finding where it is1 we find the arg max = (x, y),(x, y) will be the arg min.

18


19/28

5. a) (x

1)3 =y2 impliesx

1,

f(x) = x3 2x2 + 3x 1 = fx = 3x2 4x+ 3 which is always positive forx 1,sincef(x)is increasing, min f(x) = f(1) = 1 (y= 0);

b) the derivative of the constraint D((x 1)3 y2) is (3x2 6x+ 3, 2y),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.

6. a)

L(x,) = cx +

1

2 xDx +(Ax

b)

=

ni=1

cixi+1

2

ni=1

nj=1

xj Dji

xi+

mi=1

i

nj=1

Aijxj

bi

Lxi = ci+ Diixi+1

2

nj=1,j=i

xj Dji +1

2

nj=1,j=i

xj Dij+m

j=1

j Aji

= ci+n

j=1

xj Dij+m

j=1

j Aji

Li = n

j=1

Aijxj bib)...

7. The constraint is the normalixation|x| = 1, the system is:

f(x) = xAx

=n

i=1

nj=1

xj Aji

xi

fxi

= 2n

j=1

xj

Aji

xi = n

j=1,j=i xj Aji

Aii

x =

0 A21A11

. . . An1A11

A12A22

0 . . . An2A22

... ...

. . . ...

A1nAnn

A2nAnn

. . . 0

x B x

such that| x| = 1

19


20/28

for any eigenvectors of the square matrix B, its two normalization (one opposite

of the other) are critical points of the problem;

sinceD2f(x) = 2

A11 . . . An1...

. . . ...

A1n . . . Ann

, xis constant,

all critical points are maxima, minima or saddles, according wether A is positive-definite, negative-definite or neither.

8. a)

0 50 100 150 200 250 300 350 4000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

stock

days

The function s(t)quantifying the stock is periodic of period T =x dtdI

,

in this period the average stock is

RT

0

s(t)

T = xT

2T = x2 ,

in the long run this will be the total average;

b)

L(x,n,) = Chx

2+ C0n + (nx A)

Lx = Ch

2 + nLn = C0+ x

=0= n= Ch

2 x= C0

L= nx

A

=0=

ChC0

22

=A=

=

ChC0

2Ax and n negative have no meaning so must be negative.

9. The condition for equality constraint to suffice is that u(x1, x2)is nondecreasing,this happens for 1 1;if otherwise one of them , say is less than 1,the problem is unbounded at limx10 x1 + x

2 = +.

10.

20


21/28

min w1x1+ w2x2 s.t. (x1, x2) X {(x1, x2) R2+| x21+ x22 1}a)

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

x1

x2

X=x1

2+x

2

21

w1/ w

2

ifw1< w2 it is clear from the graph that (1, 0) is the cheapest point in X,similarly, ifw2< w1, the cheapest point is 0, 1,ifw1= w2 they both cost w1 = w2;

b) if nonnegativity constraints are ignored:

minx21+x

22

1

w1x1+ w2x2= lim(x1,x2)(,)

w1x1+ w2x2=

similarly for (x1, x2) (+, +) ifw1 andw2 are positive.

11. [x12 is not defined ifx


22/28

Chapter 6 (pp. 145-171): Inequality Constraints.

1. Since the function is increasing in both variables, we can search maxima on theboundary,substitutingy =

1 x2 (which means x [0, 1] = y [0, 1]):

f(x) = ln x+ln

1 x2 = ln x+ 12

ln(1x2) = fx= 1x

12

2x

1 x2 = 1 2x2(1 x2)x

=0=

x=

2

2 =

y=

2

2which is the argument of the maximum.

2. The function is increasing, we search maxima on the boundary,we can substitute to polar coordinates y x sin , z x cos :

f() =

x(pysin +pzcos ) = f =

x(pycos pzsin )

=0= sin

cos =

pypz

= yz

= pypz

economically speaking, marginal rate of substitution equals the price ratio,

sin = pyp2y+p

2z

cos = pzp2y+p

2z

y= x pyp2y+p

2z

z = x pzp2y+p

2z

(

x pyp2y+p

2z

,

x pzp2y+p

2z

)is a maximum and(x pyp2y+p

2z

, x pzp2y+p

2z

)is a minimum.

3. a) We can search maxima on the boundary determined by I:

max x1x2x3 s.t. x1 [0, 1], x2 [2, 4], x3= 4 x1 x2

max f(x1, x2) = 4x1x2 x21x2 x1x22 s.t. x1 [0, 1], x2 [2, 4]consider only x1,

f(x1) = 4x1x2 x21x2 x1x22 has a maximum for x1= 4x2x22

2x2= 2 12x2,

by simmetry a maximum is for x1= x2= 43 , which is however not feasible,

nevertheless, for x1 [0, 1] the maximum value for x2 [2, 4] is 2,we get then x1= 1 and x3= 1;

b)

22


23/28

max x1x2x3 s.t. x1 [0, 1], x2 [2, 4], x3= 6 x1 2x23

max f(x1, x2) = 2x1x2 13

x21x2 2

3x1x

22 s.t. x1 [0, 1], x2 [2, 4]

f(x1)has a maximum when x1= 2x2 23x22

23x2

= 3 x2,f(x2)when x2=

2x1 13x2243

x2= 32 14x1,

the two together give x1= 3 32 + 14x1 = x1= 2 and x2= 1, not feasible,nevertheless, for x1

[0, 1] the maximum value for x2

[2, 4] is again 2,

hencex1 is again 1 and x3= 1.

4. The argument of square-root must be positive, the function is increasing in allvariables, so we can use Lagrangean method:

L(x, ) =T

t=1

2t

xt+ T

t=1

xt 1

Lxi = 2i1x

12

i +

if = 0all xi are 0 (minimum);

otherwise, for i < j, i, j {1, . . . T }:

2i1x 1

2i = 2

j1x 1

2j =

xixj

12

=2j+1

2i+1 = 2ji = xi

xj= 22(ji)

the system becomes x1= x1, x2= 14x1, . . . xT =

122(T1)

x1,since they must sum to 1:

x1= 1/T1

i=0

4i , x2 = 1

4

/T1

i=0

4i , . . . xT = 1

4(T1)

/T1

i=0

4i

as T increases the sum quickly converges to 11 14

= 43 .

5. a)

min w1x1+ w2x2 s.t. x1x2= y2, x1 1, x2> 0

the feasible set is closed but unbounded as x1 +

b) substitutingx2= y2

x1we get:

23


24/28

f(x1) =w1x1+ w2y2

x1

f(x1) = w1 w2 y2

x21

=0= x1= +

w2w1

y since x1> 1

f(x1) = 2w2 y2

x31>0 since x1> 1

if

w2w1

y 1, this is the solution, and x2 =

w1w2

y,

otherwise1is, and x2= y2,

in both cases satisfying Kuhn-Tucker conditions:

L(x1, x2, 1, 2) = w1x1+ w2x2+ 1(y2 x1x2) + 2(x1 1)

forw2

w1y,

w1w2

y

Lx2 =w2 1x1 =0= 1=w2

y

w1w2

Lx1 =w1 1x2+ 2 =0= 2= 0

for (1, y2)

Lx2 =w2 1x1 =0= 1= w2Lx1 =w1 1x2+ 2 =0= 2= w2y2 w1 .

6.

max(x1,x2)R2+

f(x1, x2) = max(x1,x2)R2+

p1x121 +p2x

121 x

132 w1x1 w2x2

substitutex x121 and y x

132, the problem becomes

max(x,y)R2+ f(x, y) = max(x,y)R2+ x(p1+p2y) w1x2

w2y3

fx= p1+p2y 2w1x =0= x= p1+p2y2w1

fy =xp2 3w2y2 = p2p12w1

+ p22y

2w1 3w2y2 =0= y=

p222w1

p222w1

2+ 6 w2p2p1

w1

3w2

only the positive value will be admissible;for p1 = p2= 1 and w1= w2 = 2 we have:

24


25/28

y=14 +

116+ 6

6 =

97 124

= x= 9796

it is a maximum, because D2f(x, y) =

4 11 12y

is negative semidefinite for

every y 0,it is a global one for positive values because it is the only critical point.

7. a) Substitutingx x121 andy x2, considering that utility is increasing in both

variables, the problem is:

max(x,y)R2+ f(x, y) = x + x2

y such that 4x2

+ 5y = 100

substitutingy = 20 45x2 the problem becomes:

maxxR+

f(x) = x + 20x2 45

x4

fx= 1 + 40x 165

x3

with numerical methods it is possible to calculate that the only positive xfor whichfx= 0 is x

3.5480,where f(x)

128.5418,

sincefxx= 40 485x2 0, ue > 0, ul> 0

25


26/28

if is the amount of income spent in food, f= wlp

ande = (1)wlq

:

max uwl

p ,

(1 )wlq

, l

s.t. l [0, H], [0, 1], uf >0, ue > 0, ul


27/28

in principle Weierstrass theorem applies but not Kuhn-Tucker because min is con-

tinuous but not differentiable.The cheapest way to maximize min{x2, x3}is however when x2= x3, the problembecomes:

max x131 + x2 s.t. p1x1+ (p2+p3)x2 I

now also Kuhn-Tucker applies.

10. a)

max p

f(L+ l)

w1L

w2l s.t. l

0, f

C1 is concave =

fl < 0

b)

L(l, ) = p f(L+ l) w1L w2l+ lLl = p fl(L+ l) w2+

L = l

l= 0 and= w2 p fl(L+ l);

c)p f(L+l)w1Lw2lmaximizes once (by concavity) forp fl(L+l) =w2,when this happens for l 0, the maximum is (by chance) the Lagrangean point,when instead this happens for l >0, the maximum is not on the boundary.

11. a)

max pyx141 x

142p1(x1 K1) p2(x2 K2) s.t. x1 K1, x2 K2

L(x1, x2, 1, 2) = pyx

14

1 x

14

2p1(x1 K1) p2(x2 K2) + 1(K1+ x1) + 2(K2+ x2)Lx1 =

1

4pyx

34

1 x142p1x1+ 1x1

Lx2 = 1

4pyx

141 x

342 p2x2+ 2x2

L1 = K1+ x1

L2 = K2+ x2

b) the unbounded maximum off(x1, x2) =x141 x

142 x1 x2+ 4 is for:

27


28/28

fx1 = 14x

34

1 x142 1 = 0 = x

341 =

14x

142

fx2 = 14x

141 x

34

2 1 = 0 = x342 =

14x

141

= x1= x2 =

14

2=

1

16

bounds are respected,the firm sells most ofx1 and buys only

116 units ofx2 to produce

14 units ofy ;

c) is analogous to (b)since the problem is symmetric.

12. a)

max py(x1(x2+ x3)) wx

L(x, ) = py(x1(x2+ x3)) wx + xLx1 = py(x2+ x3) w1+ 1

Lxi i={2,3} = pyx1 wi+ iLi = xi

x1= w2 2py

= w3 3py

= 3 2= w3 w2

x2+ x3 = w1 1

py

b) [ w4 ??? ] x R3+ and R3 are not defined by the equations;

c) the problem can be solved considering the cheapest between x2 and x3,suppose it is x2, the problem becomes:

max pyx1x2

w1x1

w2x2

which has critical point( w2py

, w1py

)but its Hessian

0 pypy 0

is not negative semidef-

inite.The problem maximizes at (+, +) for any choice ofpy R+ and w R3+.

28

sundaram optimization solutions

Documents