Solution 1 A First Course in Optimization Theory. Rangarajan K. SundaramCambridge University Press (1996).pdf

Frederick Robinson

Math 368: Optimization

Homework Frederick Robinson 11 January 2010

Contents 1 Bo ok Problems 1.1 Problem 1.16 . . 1.2 Problem 1.32 . . 1.3 Problem 1.51 . . 1.4 Problem 1.52 . . 1.5 Problem 1.57 . . 1.6 Problem 1.63adf . 1.7 Problem 2.3 . . . 1.8 Problem 2.7 . . . 1.9 Problem 3.5 . . . 1.10 Pr P roblem 3.8 . . .

. . . . . . . . . .

1 1 2 2 3 4 5 6 6 7 7

2 Supplemental Problems 2.1 Problem 1.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Problem 3.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . .

8 8 8

1 1.1 1.1 1.1. 1.1.1 1

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Book ook Problems Prob Proble lem m 1.16 1.16 Ques Qu esti tion on

Find the supremum, infimum, maximum, and minimum of the set X set X in in each of the following cases:

1

Frederick Robinson


1. X = x [0, [0 , 1] x is irrational .

{ ∈ | } 2. X = {x | x = 1/n,n = /n,n = 1, 2, . . .}. 3. X = {x | x = 1 − 1/n,n = /n,n = 1, 2, . . .}. 4. X = {x ∈ [0, [0 , π] | sin x > 1/ 1 /2}. 1.1. 1.1.2 2

Answ Answer er

1. The suprem supremum um of X though it has no maxim maximum. um. Simila Similarly rly the X is 1, though infimum is 0 though it has no minimum. 2. The maximum maximum and supremum supremum are both 1. The infimum infimum is 0, but there there is no minimum. 3. The minimum minimum and infimum infimum are both 0. There There is no maximum, maximum, but the supremum is 1. 4. The infimum is π/6 5π/66 however there is no π/6 and the supremum is 5π/ minimum or maximum.

1.2 1.2 1.2. 1.2.1 1

Prob Proble lem m 1.32 1.32 Ques Qu esti tion on

Let A Let A = = 1, 1/2, 1/3, . . . , 1/ n , . . .

} ∪ {0}. Is A Is A closed? Is it compact?

{

1.2. 1.2.2 2

Answ Answer er

The set A is closed because it contains contains all of its limit points. points. It is compact since it is bounded (by 0, 1), and by the Heine-Borel theorem closed, bounded sets in Rn are compact.

1.3 1.3 1.3. 1.3.1 1


Let f Let f : R+

→ R be defined by  f ( f (x) =

0, x = 0 sin(1/x)), x = 0. x sin(1/x



Show that f that f is is continuous at 0. 2

Frederick Robinson

1.3. 1.3.2 2


Answ Answer er

We need to show that given some δ some δ there there is always some corresponding  so that each element x B 0 is mapped to f ( f (x) B δ f ( f (x) = B δ 0. This is easy though, for clearly setting  = δ/2 δ/2 yields the desired condition dition.. The fact fact that the the sin z [ 1, 1] implies that f ( sin(1/x)) f (x) = x sin(1/x [ x, x] for a given input x input x which is all we need.

∈

∈

∈ ∈ −

−

1.4 1.4 1.4. 1.4.1 1

∈


Let D be the unit square [0, [0 , 1] [0, [0, 1] in R2 . For (s, t) defined by 0, for all s all s [0, [0 , 1]. 1]. f ( f (s, 0) = 0,

×

∈ D, let f ( f (s, t) be

∈

and for t for t > 0,

2s

 f ( f (s, t) =  2 −0 t

2s

t

0, 2t s t s ,t 2 ( t, 1] s (t,

∈  ∈ ∈

 

Show that f that f is is separately continuous, but not jointly continuous.

1.4. 1.4.2 2

Answ Answer er

If t = 0 then f ( constant nt value valued d and therefo therefore re contin continuou uous. s. So set a f (s) is consta (0 , 1] we have t¯ (0, 2s t¯ 0 s , ¯ 2 t t¯ ¯ f ( f (s) = 2 2t¯s s , t 2 0 s ( t, t¯, 1]

∈

 

∈  ∈ ∈

−

 

In each of the intervals [0, [0 , t/2), function f ((s) is contint¯/2), (t/2 t¯/2, t¯) and (t, t¯, 1] the function f uous. In the first interv interval al it is a constant constant multiple multiple of a contin continuous uous function function (the identity function) and is therefore continuous. In the second interval it is the sum of two two continuous continuous functions functions and is therefore therefore continuous. continuous. On the third it is just constant valued and therefore continuous. Now we check that the function is continuous at t/2 t¯/2 and t¯ but it is easy ¯ ¯ to verify that we have lim s→t/ f (s) = lims→t/ f (s) since 2t¯ 2t = 2 2t¯ 2t = ¯ 2+ f ( ¯ 2− f ( 1. Simila Similarly rly we we verify verify that that lims→t¯+ f ( f (s) = lims→t¯− f ( f (s) by checking that 2¯ 2 t¯ t = 0.

−

−

3

Frederick Robinson


So we know that f (s, t) is separately continuous in s. Now we need to verify that it is separately continuous in t and that it is not jointly continuous. Towards this end fix an ¯s [0, 1]. We have

∈

2¯ s

 f (t) =  2 −0 t

s 2¯ t

0, 2t s¯ t s¯ ,t 2 s¯ (t, 1]

∈  ∈ ∈

 

s and f (t) is continuous since 2¯ts and 2 2¯ are continuous everywhere but at t 0 and 0 is continuous everywhere. Furthermore f (t) is defined to be 0 for t = 0. Lastly we must check that the left and right limit of f (t) are the same whenever it transitions between component functions s Thus, we check that lim t→s¯− f (t) = limt→s¯+ f (t) by computing 2 2¯ = 0, s¯ 2¯ 2¯ s s and check that lim t→2¯s− f (t) = limt→2¯s+ f (t) by computing 2¯s = 2 2¯s . Lastly, it remains to show that f is not jointly continuous. This is easy though. Consider the sequence defined by (sn , tn ) = (1/n, 1/n). The function evaluated on this sequence converges to 2 yet, the limit of this sequence is (0, 0) andf (0, 0) = 0. Hence, the function f is not jointly continuous.

−

− −

1.5

Problem 1.57

1.5.1

Question

Let f : R2

→ R be defined by f (0, 0) = 0, and for (x, y) = (0, 0). xy(x − y ) f (x, y) = 2

2

x2 + y 2

Show that the cross-partials ∂ 2 f (x, y)/∂x∂y and ∂ 2 f (x, y)/∂y∂x exist at all (x, y) R2 , but that these partials are not continuous at (0, 0). Show also that ∂ 2 f ∂ 2 f (0, 0) = (0, 0). ∂x∂y ∂y∂x

∈



1.5.2

Answer

We evaluate the cross-partials on (x, y) to yield ∂ 2 f (x, y) ∂ 2 f (x, y) x6 + 9x4 y2 9x2 y4 = = ∂x∂y ∂y∂x (x2 + y 2 )3

−

4

−y

6

Frederick Robinson


Furthermore, we observe that this value does not exist in the limit ( x, y) (0, 0) since limiting along the line y = 0 yields 1 whereas limiting along the line x = 0 yields 1.

→

−

1.6 1.6.1

Problem 1.63adf Question

Find the hessians D2 f of each of the following functions. Evaluate the hessians at the specified points, and examine if the hessian is positive definite, negative definite, positieve semidefinite, negative semidefinite, or indefinite: 2 (a) f : R2 R, f (x) = x 1 + x2 at x = (1, 1). 3 (d) f : R+ R, f (x) = x1 + x2 + x3 , at x = (2, 2, 2) 3 (f ) f : R+ R, f (x) = x 1 x2 + x2 x3 + x3 x1 at x = (1, 1, 1)

√ √ √

→ → →

1.6.2

√

Answer

(a) f : R2

→ R, f (x) = x + √ x 2 1

2

at x = (1, 1).

2

0

Positive definite1 (d) f : R3+

−

0 1/4



→ R, f (x) = √ x + √ x + √ x , at x = (2, 2, 2)

−  0 0

1

1

2

0

3/2 4x1

−

  − √  0 0 =  0 0

1

8

1 3/2 4x2

0

3

−

1

3/2 4x3

2

−

0 1 √ 8 2 0

−

0 0 1 √

8

Negative definite (f ) f : R3+ R, f (x) = x 1 x2 + x2 x3 + x3 x1 at x = (1, 1, 1)

→

0 1

1 1 0 1 1 1 0

Indefinite 1

Thanks to Simon Martin for a correction

5

 

2

 

Frederick Robinson

1.7 1.7.1


Problem 2.3 Question

Let = [0, 1]. Describe the set f ( ) in each of the following cases, and identify sup f ( ) and inf f ( ). In which cases does f attain its supremum? What about its infimum?

D

D

D

1. f (x) = 1 + x for all x

D

∈ D.

2. f (x) = 1 if x < 1/2, and f (x) = 2x otherwise. 3. f (x) = x, if x < 1, and f (1) = 2. 4. f (0) = 1, f (1) = 0, and f (x) = 3x for x (0, 1).

∈

1.7.2

Answer

1. f ( ) is just [1, 2]. The supremum is 2 and the infimum is 1. Both are attained.

D

2. Same as previous. 3. f ( ) is [0, 1) 2 . The infimum is 0 and it is attained. The supremum is 2 and it too is attained.

D

∪{ }

4. f ( ) is [0, 3). The infimum is 0 and is attained, whereas the supremum is 3 and is not attained.

D

1.8 1.8.1


Give an example of a set R and a continuous function f : that f attains its maximum, but not a minimum on .

D ⊂

1.8.2

D

D → R such

Answer

Let f be the identity function f (x) = x and = (4, 12]. The image of under f is just (4, 12] and clearly this has a maximum (12) but no minimum.

D

6

D

Frederick Robinson

1.9 1.9.1



R be continuous on R+ . Suppose that f also satisfies the Let f : R+ conditions that f (0) = 1 and lim x→∞ f (x) = 0. Show that f must have a maximum on R+ . What about a minimum?

→

1.9.2

Answer

Since limx→∞ f (x) = 0 for any choice of  > 0 there is some M such that f (x) <  x > M . So, pick  = 1. Since the interval [0, M ] is compact f must attain its maximum on this interval. Moreover, the maximum of the function on this interval is the maximum of the entire function, since by construction f (x) < 1 x > M and the maximum of f on [0, M ] is at least 1 (for after all we have in particular that f (0) = 1). The same cannot be said of a minimum. Consider the function given by f (x) = 1 for x [0, 1) and by f (x) = 1/x for x [1, ). Clearly this function is continuous, however it has no minimum since it is strictly decreasing (starting at x = 1).

∀

∀

∈

1.10

Problem 3.8

1.10.1

Question

∈ ∞

Use the Weierstrass Theorem to show that a solution exists to the expenditure minimization problem of subsection 2.3.2, as long as the utility function u is continuous on Rn+ and the price vector p satisfies p 0. What if one of these conditions fails?



1.10.2

Answer

Assuming that u is continuous we can show that there exists some compact subset of the constraint set which contains the solution to the expenditure minimization problem. In particular given some set x0 which produces the desired utility u ¯ the set of all vectors which cost at most p x is closed bounded and contains the solution to the expenditure minimization problem. So, the intersection of this set and the constraint set is a closed, bounded set which contains the answer to the expenditure minimization problem

·

7

Frederick Robinson


Since we have identified a compact set in which the minimum of a continuous function must be located we know that the minimum of the function over the whole domain must be attained (in particular in this subset.) If we don’t have continuity of u the compact subset of the constraint set which we identified earlier may not actually be bounded, closed. If we did not have p 0 we would not have had a closed constraint set. Either of these problems would have caused the proof to fail; it depends on pinning down the minimum in a set which lives in the intersection of a closed, and a closed and bounded set which must therefore itself be closed and bounded.



2 2.1 2.1.1

Supplemental Problems Problem 1.3 Question

(a) Show that the set S = (x, y) R2 : 1 < x < 1 is open. (b) Show that the set S = (x, y) R2 : x 1, y 0 is closed.

{

2.1.2

{

∈

∈

−

} ≤ ≤ }

Answer

(a) The set in question is open because, given any x S we know that there exists some  > 0 such that B x S and this is just the definition of openness. (b) The only limit points of this set are 1, and 0. These points are contained in the set, so the set is closed.

∈

⊂

2.2 2.2.1


Let u(x, y) = xy and the expenditure be E (x, y) = p1 x + p 2 y with p1 > 0 and p2 > 0. Fix a ¯u > 0. Let X (u ¯ ) = (x, y) R2+ : u(x, y) ¯ . The u objective is to show that E has a minimum on X (¯u). (Note that X (¯u) is not compact.) i. Find a point (x0 , y0 ) in X (¯u) and let E 0 = E (x0 , y0 ). ¯ ) be contained in the set (P, E 0 ) ii. Why must the minimum of E on X (u (using the notation of Section 2.3.1)?

{

∈

≥ } B

8

Frederick Robinson


iii. Why must E attain a minimum on X (¯u) ( p, E 0 )? iv. Using reasoning like for problem 3.4:5, explain why E attains a minimum on X (¯u).

∩ B

2.2.2

Answer

i. For the point (x0 , y0 ) = (1, ¯ u) we have u(1, ¯ u) = u¯. E 0 in this instance is just E 0 = p 1 + p2 u¯ ii. Since we have demonstrated a particular way to meet the goal utility of u¯ it follows that the expenditure minimizing way of meeting this goal costs this much or less. That is, since (x0 , y0 ) has the property u(x0 , y0 ) u ¯ we must have that E (x0 , y0 ) is an upper bound for E (xopt , yopt ). iii. X (¯u) and ( p, E 0 ) are both closed, so their intersection is also closed. Moreover the set ( p, E 0 ) is bounded. Thus, E attains a minimum on X (¯u) ( p, E 0 ) because this intersection is compact and E is continuous. iv. We have discovered that the minimum of E must lie in a compact subset of E . Thus, E being continuous implies that the minimum is always attained.

≥

B

B B

∩

9

Frederick Robinson



Contents 1 Book Problems 1.1 Problem 4.1 . . . . 1.2 Problem 4.2 . . . . 1.3 Problem 4.4 (a,b,g) 1.4 Problem 4.7 . . . . 1.5 Problem 5.1 . . . . 1.6 Problem 5.3d . . .

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1 1 2 3 6 6 8

2 Supplemental Problems 9 2.1 Problem 4.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 2.2 Problem 1.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1 1.1 1.1.1

Book Problems Problem 4.1 Question

Is Theorem 4.1 valid if x ∈ / int D? If yes, provide a proof. If not, a counterexample. ∗

1.1.2

Answer

Recall that Theorem 4.1 states:

1

Frederick Robinson


Suppose x∗ ∈ int D ⊂ Rn is a local maximum of f on D , i.e., there is r > 0 such that B (x∗ , r) ⊂ D and f (x∗ ) ≥ f (x) for all x ∈ B(x∗ , r). Suppose also that f is differentiable at x ∗ . Then Df (x∗ ) = 0 . The same result is true if, instead x∗ were a local minimum of f on D

Theorem 4.1 does not hold if we drop the assumption that x ∈ int D. For, consider the problem of trying to maximize f (x) subject to x ∈ [0, 5] = D. Clearly x = 5 has the property that x ∈ D is a local maximum of f . d However, we know that dx f = 1 for all x including our choice for x . So this modification of the theorem does not hold. ∗

∗

∗

1.2

Problem 4.2

1.2.1

Question

Find all the critical points (i.e., points where f (x) = 0) of the function f : R → R, defined as f (x) = x − x2 − x3 for x ∈ R. Which of these points can you identify as local maxima or minima using the second-order test? Are any of these global optima? 

1.2.2

Answer

The critical points are those where f (x) = 0 so we compute 

f (x) = x − x2 − x3 ⇒ f (x) = 1 − 2x − 3x2 

Hence if f (x) = 1 − 2x − 3x2 = 0 

we have x =

2±



4 − 4(−3) 1 ± 2 = 2(−3) −3

and the critical points are x = −1, x = 1/3 Now we apply the second order test. First we compute f (x) = 1 − 2x − 3x2 ⇒ f (x) = −2 − 6x. 



Substituting the critical points which we determined above we see that 



f (−1) = 4 f 2



1 = −4. 3

Frederick Robinson


So, x = −1 is a local minimum, whereas x = 1/3 is a local maximum. Neither of these are global extrema however as lim x f (x) = −∞ and limx f (x) = ∞. →∞

→−∞

1.3 1.3.1

Problem 4.4 (a,b,g) Question

Find and classify the critical points (local maximum, local minimum, neither) of each of the following functions. Are any of the local optima also global optima? 1. f (x, y) = 2x3 + xy 2 + 5x2 + y 2 2. f (x, y) = e 2x (x + y 2 + 2y) 3. f (x, y) =

1.3.2

x 1+x2 +y 2

Answer

1. To locate the critical points we compute the partial derivatives by ∂ 2x3 + xy 2 + 5x2 + y 2 = 6x2 + y 2 + 10x ∂x





∂ 2x3 + xy 2 + 5x2 + y 2 = 2xy + 2y ∂y and find those points at which





∂ ∂ f = f = 0. ∂x ∂y So, 0 = 2xy + 2y = 2y(x + 1) ⇒ y = 0 or x = −1 and 0 = 6x2 + y 2 + 10x ⇒ −y2 = x(6x + 10) Thus we have Case 1: y = 0 ⇒ x = 0 or x = −5/3 and Case 2: x = −1 ⇒ y = 2 or y = −2 So we conclude that the critical points are (0 , 0), (−5/3, 0), (−1, 2), (−1, −2), 3

Frederick Robinson


To determine which of these are local extrema we first compute the Hessian matrix as ∂ f ∂x∂x ∂ ∂y∂x f



∂ f ∂x∂y ∂ ∂y∂y f

  =

12x + 10 2y 2y 2x + 2



.

Evaluating this matrix at each critical point we get (0, 0) →



10 0 0 2

(−1, 2) →





2 4 4 0

= 20 (−5/3, 0) →





 

0 −10 = 40/3 0 −4/3

= −16 (−1, −2) →



2 −4 0 −4

= −16

So, since the corresponding Hessian matrix is positive definite (0, 0) is a local minimum. Similarly, since its corresponding Hessian is negative definite (−5/3, 0) is a local maximum. The other two critical points are indefinite, and so they are not local extrema. It is easy to check though, that these local extrema are not global extrema for in particular f (0, 0) = 0 > −125 = f (−5, 0) and f (−5/3, 0) = 125 < 18000 = f (20, 0) 27 2. Again we compute the partial derivatives in order to find the critical points ∂ 2x e (x + y 2 + 2y) = e 2x 1 + 2x + 4y + 2y2 ∂x ∂ 2x e (x + y 2 + 2y) = e 2x (2 + 2y) ∂y Since the critical point are those points with both partial derivatives zero and exponentiation always yields a nonzero result we must have 1 + 2x + 4y + 2y2 = 0 and 2 + 2y = 0. So, y = −1 and consequently 0 = 1 + 2x − 4 + 2 ⇒ x = 1/2.



  





Therefore we conclude that the only critical point is (1 /2, −1). To test this point we need to compute the Hessian matrix. We do this as



∂ ∂x∂x f ∂ f ∂y∂x

∂ ∂x∂y f ∂ f ∂y∂y

  =

4e2x (x + (1 + y)2 ) 4e2x (1 + y) 4e2x (1 + y) 2e2x

4



.

Frederick Robinson


Evaluating this matrix at the critical point we arrive at 4e2(1/2) ((1/2) + (1 + (−1))2 ) 4e2(1/2) (1 + (−1)) 4e2(1/2) (1 + (−1)) 2e2(1/2)



=



6e 0 0 2e





.

.

and since this matrix is positive definite we may conclude that the point (1, −1) is a local minimum. Furthermore it is apparent that this is a global minimum since any potential minimum needs to have y = −1 (since this of course minimizes y2 + 2y). Fixing this we evaluate lim x f and limx f yielding ∞ and 0. Since both of these exceed the value of the function at (1/2, −1) this point must be a global minimum. →∞

→−∞

3. Again here we begin by computing the partial derivatives

 

∂ ∂x ∂ ∂y

1 − x2 + y 2 x = 1 + x2 + y 2 (1 + x2 + y 2 )2

 

2xy x = − . 2 2 1 + x + y (1 + x2 + y 2 )2

The second expression is zero if either x or y is made zero. If x is zero however the first expression cannot be zero as well. So, the only case to consider is that in which y = 0. In this case we can make the first expression zero by letting x = 1. So, we conclude that (1, 0) is a critical point of the given function. The Hessian Matrix is just



∂ ∂x∂x ∂ ∂y∂x

∂ ∂x∂y ∂ ∂y∂y

so in this case we get



2(1) (1+1)3

0

 

2x(x2 −3(1+y 2 ))

=

0 − 2(1+1) (1+1) 3

−

(1+x2 +y 2 )3 2y (1−3x2 +y 2 ) (1+x2 +y 2 )3

  =

5

2 23

0

0 − 24

3

− −

2y (1−3x2 +y 2 ) (1+x2 +y 2 )3 2(x+x3 −3xy 2 ) (1+x2 +y 2 )3

  =

1 22

0

0 − 12

  

Frederick Robinson


Since the Hessian is negative definite we conclude that the point 1 , 0 is a local maximum. Moreover, since in the limit |(x, y)| → ∞ we have f (x, y) → 0 this local maximum is also the global maximum. For, were there a global maximum that is not also a local maximum it would have to occur in this limit. That is, suppose that there is some global maximum that does not occur in the limit |(x, y)| → ∞. Then there is some neighborhood of this maximum on which it is the maximum. But this is just the definition of a local maximum.

1.4 1.4.1


Suppose f : R → R has a local maximum at x that is not a strict local maximum. Does this imply that f is constant in some neighborhood of x? Prove your answer or provide a counterexample.

1.4.2

Answer

Consider the function f (x) =



sin (1/x) |x| > 0 1 x = 0

Clearly there is a maximum of this function at x = 0 since sin x is at most 1. However, this is not a strict local maximum, since in every neighborhood of 0 there is some point y in the function where f (y) = 1. (This follows by the density of Q ∈ R) Finally, the function f is not constant valued on any neighborhood of 0. Thus, we have constructed a counterexample.

1.5 1.5.1


Find the maximum and minimum of f (x, y) = x2 − y2 on the unit circle x2 + y 2 = 1 using the Lagrange multipliers method. Using the substitution 6

Frederick Robinson


y 2 = 1 − x 2 , solve the same problem as a single variable unconstrained problem. Do you get the same results? Why or why not?

1.5.2

Answer

First we solve this problem using Lagrange’s Theorem. We set up the Lagrangian by L(x, y) = f (x, y) + λ(x2 + y 2 − 1) = x 2 − y2 + λx2 + λy 2 − λ and then compute the derivatives with respect to x, y and λ, setting them equal to zero to yield the following three equations. 0 = 2x + 2λx = 2x(1 + λ)

(1)

0 = −2y + 2λy = −2y(1 − λ)

(2)

0 = x 2 + y 2 − 1

(3)

Observe that equation 1 may only be satisfied if λ = −1 or x = 0 Case 1: Assume λ = −1 . Then, we must have y = 0 to satisfy equation 2, and using this information we see that x = ±1 in equation 3. Case 2: Assume x = 0. Then, in order to satisfy equation 3 we must have y = ±1. We have now identified the critical points as (0 , 1), (0, −1), (1, 0), (−1, 0). Evaluating the function at these points we get f ((0, 1)) = −1,

f ((0, −1)) = −1

f ((1, 0)) = 1,

f ((−1, 0)) = 1

If we repeat the process using the substitution method we reduce the problem of optimizing f (x, y) = x 2 − y2 subject to x2 + y 2 = 1 to the easier problem of optimizing f (x) = x 2 − (1 − x2 ) without constraint. So, f (x) = x 2 − (1 − x2 ) = 2x2 − 1 

f (x) = 4x and f (x) = 0 ⇒ x = 0 Now, using the substitution equation y 2 = 1 − x 2 again we see that x = 0 ⇒ y = ±1. These are only two of the solutions which we got using 

7

Frederick Robinson


the Lagrangian technique however. We have already checked these points, observing earlier that f ((0, 1)) = −1,

f ((0, −1)) = −1

So, we see that we have only found the local minima, not the local maxima as we did with the lagrangian. The reason that we failed to find the other points is that, upon making the substitution we failed to constrain the problem correctly. The substitution makes little since for values of y 2 > 1 since, after all the unit circle contains no such points. Had we tried to solve the one variable optimization problem restricted to y ∈ [−1, 1] we would have found all of the solutions as we did with the Lagrangian technique.

1.6 1.6.1

Problem 5.3d Question

Find the maxima and minima of the following functions subject to the specified constraints: f (x,y,z ) = xyz subject to x + y + z = 5 and xy + xz + yz = 8

1.6.2

Answer

We set up the Lagrangian as usual by L(x,y,z,λ,µ) = xyz + λ1 (x + y + z − 5) + λ2 (xy + xz + yz − 8) Computing all of the partial derivatives and setting them equal to zero we see that ∂L = yz + λ1 + λ2 y + λ2 z = 0 (4) ∂x ∂L = xz + λ1 + λ2 x + λ2 z = 0 (5) ∂y ∂L = xy + λ1 + λ2 x + λ2 y = 0 (6) ∂z ∂L = x + y + z − 5 = 0 (7) ∂λ 1 ∂L = xy + xz + yz − 8 = 0 (8) ∂λ 2 8

Frederick Robinson


Now after some algebra (omitted) we see that the solutions are 16 4 4 4 7 λ1 = , λ2 = − , x = , y = , z = 9 3 3 3 3 16 4 4 7 4 λ1 = , λ2 = − , x = , y = , z = 9 3 3 3 3 16 4 7 4 4 λ1 = , λ2 = − , x = , y = , z = 9 3 3 3 3 λ1 = 4, λ2 = −2, x = 1, y = 2, z = 2 λ1 = 4, λ2 = −2, x = 2, y = 1, z = 2 λ1 = 4, λ2 = −2, x = 2, y = 2, z = 1 Now we just evaluate f (x,y,z ) = xyz at each of these points 4 4 7 112 f ( , , ) = 3 3 3 27 4 7 4 112 f ( , , ) = 3 3 3 27 7 4 4 112 f ( , , ) = 3 3 3 27 f (1, 2, 2) = 4 f (2, 1, 2) = 4 f (2, 2, 1) = 4 So, since 112 is greater than 4 each of the first three points is a local 27 maximum, and the last three are local minima. They are also globally maxima subject to the constraints since the constraints define a closed and bounded (and therefore compact) subset of R3

2

Supplemental Problems

2.1 2.1.1


Find all the critical points and classify them as local maximum, local minimum, or neither for the following functions 1. f (x,y,z ) = x 4 + x2 − 6xy + 3y2 + z 2 2. f (x,y,z ) = 3x − x3 − 2y2 + y 4 + z 3 − 3z 9

Frederick Robinson

2.1.2


Answer

1. We compute the partial derivatives to get ∂ 4 x + x2 − 6xy + 3y2 + z 2 = 2 x + 2x3 ∂x



   



∂ 4 x + x2 − 6xy + 3y2 + z 2 = 6y ∂y ∂ 4 x + x2 − 6xy + 3y2 + z 2 = 2z ∂z In order that a point be a critical point we must have each partial derivative zero. The only points which satisfy this criterion have z = 0 and y = 0. Finally the points must also have x + 2x3 = 0. The only real x which satisfy this last are x = 0 . Hence, the only critical point is (0, 0, 0).

 

So, in order to figure out whether this is a local minimum, local maximum, or neither we compute the Hessian.

 

∂ ∂x∂x ∂ ∂y∂x ∂ ∂z∂x

∂ ∂x∂y ∂ ∂y∂y ∂ ∂z∂y

∂ ∂x∂z ∂ ∂y∂z ∂ ∂z∂z

    =

2 + 12x2 0 0 0 6 0 0 0 2

So, evaluating at our critical point we just get

 

2 0 0 0 6 0 0 0 2

 

 

and since this matrix is positive definite we must be dealing with a local minimum. 2. We compute the partial derivatives to get ∂ 3x − x3 − 2y2 + y 4 + z 3 − 3z = 3 − 3x2 ∂x





∂ 3x − x3 − 2y2 + y 4 + z 3 − 3z = 4y −1 + y 2 ∂y



 

10



Frederick Robinson


∂ 3x − x3 − 2y2 + y 4 + z 3 − 3z = 3 −1 + z 2 ∂z So, to make each of these expressions zero we must have x = ±1, z = ±1 and y = 0 or y = ±1. That means that the critical points we have are in particular (1, −1, 1), (1, 0, 1), (1, 1, 1), (1, −1, −1), (1, 0, −1), (1, 1, −1), (−1, −1, 1), (−1, 0, 1), (−1, 1, 1), (−1, −1, −1), (−1, 0, −1), (−1, 1, −1)



 



0 0 −6x 2 0 0 −4 + 12y 0 0 6z

 

Next we find the hessian matrix

 

∂ ∂x∂x ∂ ∂y∂x ∂ ∂z∂x

∂ ∂x∂y ∂ ∂y∂y ∂ ∂z∂y

∂ ∂x∂z ∂ ∂y∂z ∂ ∂z∂z

    =

So we may now categorize the above critical points as follows (−1, 1, 1), (−1, −1, 1), Minimum (1, 1, −1), (1, −1, −1) Maximum Neither Others



Since, in order to be positive definite we must have x < 0, |y| > 1/3, and z > 0. Also, in order to be negative definite we must have x > 0, |y| > 1/3, z < 0.

2.2 2.2.1




A firm uses two inputs q 1 and q 2 to produce a single output Q, given by 2/5 1/5 the production function Q = kq 1 q 2 . Let P be the price of the output Q, p1 be the price of q 1 , and p2 be the price or q 2 . The profit is given b 2/5 π = P kq 1 − p1 q 1 − p2 q 2 . The inputs that maximize profit satisfy 0=

2P k 3/5 1/5 q q 2 − p1 5 1 −

and

P k 2/5 4/5 q q − p2 . 5 1 2 1. Show that these two equations can be used to determine the amounts of inputs q 1 and q 2 in terms of the price p1 , p2 , and P . Show that the relevant matrix has nonzero determinant. −

0=

11

Frederick Robinson


2. Write the matrix equation for the partial derivatives of q 1 and q 2 with respect to p 1 , p2 and P in terms of the variables 3. Solve for the matrix of partial derivatives of q 1 and q 2 in terms of p1 , p2 and P .

2.2.2

Answer

1. The two equations determine the amounts of inputs in terms of the prices by the implicit function theorem. The relevant matrix is



6P k −8/5 1/5 q 2 25 q 1 2P k −3/5 −4/5 q 2 25 q 1

−



2P k −3/5 −4/5 q 2 25 q 1 2/5 −9/5 −4P k 25 q 1 q 2

and we have

 

6P k −8/5 1/5 q 1 q 2 25 2P k −3/5 −4/5 q q 2 25 1 −

2P k −3/5 −4/5 q q 2 25 1 2/5 −9/5 −4P k q 1 q 2 25

 

24P 2 k 2 6/5 8/5 4P 2 k 2 6/5 8/5 = q q 2 − q q 2 252 1 252 1 −

−

−

−

20P 2 k 2 6/5 8/5 = q q 2 252 1 so the determinant is always nonzero for p 1 > 0, p 2 > 0 and the matrix is negative definite. −

−

2.

   0 0 = 0 0

   0 0 = 0 0

6P k −8/5 1/5 q 1 q 2 25 − 3/5 −4/5 2P k q q 2 25 1

2P k −3/5 −4/5 q q 2 25 1 2/5 −9/5 −4P k q 1 q 2 25



∂q 1 ∂p 1 ∂q 2 ∂p 1

∂q 1 ∂p 2 ∂q 2 ∂p 2



6P k −8/5 1/5 q 2 25 q 1 2P k −3/5 −4/5 q 2 25 q 1

2P k −3/5 −4/5 q 2 25 q 1 2/5 −9/5 −4P k 25 q 1 q 2



∂q 1 ∂p 1 ∂q 2 ∂p 1

∂q 1 ∂p 2 ∂q 2 ∂p 2

2P k −3/5 −4/5 q 2 25 q 1 2/5 −9/5 −4P k 25 q 1 q 2



∂q 1 ∂p 1 ∂q 2 ∂p 1

∂q 1 ∂p 2 ∂q 2 ∂p 2

−

−

∂g 1 ∂p 2 ∂g 2 ∂p 2





0 −1 + 0 −1







+

∂g 1 ∂p 1 ∂g 2 ∂p 1

3.

      0 0 = 0 0

6P k −8/5 1/5 q 2 25 q 1 2P k −3/5 −4/5 q 2 25 q 1

−

1 0 = 0 1

6P k −8/5 1/5 q 2 25 q 1 2P k −3/5 −4/5 q 2 25 q 1

−

12

2P k −3/5 −4/5 q 2 25 q 1 2/5 −9/5 −4P k 25 q 1 q 2



∂q 1 ∂p 1 ∂q 2 ∂p 1

0 −1 + 0 −1 ∂q 1 ∂p 2 ∂q 2 ∂p 2



Frederick Robinson


but since the matrix on the left is just the identity we have

 

6P k −8/5 1/5 q 1 q 2 25 2P k −3/5 −4/5 q q 2 25 1

−

6P k −8/5 1/5 q 1 q 2 25 2P k −3/5 −4/5 q 2 25 q 1

−

2P k −3/5 −4/5 q q 2 25 1 2/5 −9/5 −4P k q 1 q 2 25

2P k −3/5 −4/5 q q 2 25 1 2/5 −9/5 −4P k 25 q 1 q 2



1

−



8/5 1/5 −3q 1 q 2 −3/5 −4/5 q 1 q 2

6P k −8/5 1/5 q 1 q 2 25 − 3/5 −4/5 2P k q q 2 25 1

−

−

3/5

−

4/5

−

q 1 q 2 2/5 9/5 −2q 1 q 2

2P k −3/5 −4/5 q q 2 25 1 2/5 −9/5 −4P k q 1 q 2 25

−



13

=

25 = 2P k

and since

 

1

−

   

1

−

=

 

∂q 1 ∂p 1 ∂q 2 ∂p 1

∂q 1 ∂p 2 ∂q 2 ∂p 2

8/5 1/5

−

−3q 1 q 2 3/5 4/5 q 1 q 2 −

−

6/5

−

= 5q 1

−

3/5

−

4/5

−

q 1 q 2 2/5 9/5 −2q 1 q 2 −

  1

−

8/5

−

q 2

25 6/5 8/5 5q 1 q 2 2P k −





2/5 −9/5 −2q 1 q 2 −3/5 −4/5 −q 1 q 2

3/5

−

4/5

−

−q 1 q 2 8/5 1/5 −3q 1 q 2 −



Frederick Robinson



Contents 1 1.1 1.1.1

Book Problems Problem 6.7a Question

√

A consumer with a utility function given by u(x1 , x2 ) = x1 + x 1 x2 has an income of 100. The unit prices of x1 and x2 are 4 and 5 respectively. Compute the utility-maximizing commodity bundle, if consumption must be nonnegative. 1.1.2

Answer

We first note that the customer’s utility maximizing bundle must cause him to expend all his money because, suppose there were some maximizing bundle which did not cause him to spend all his money, then since the utility function is strictly increasing in both x1 and x2 he could construct a new bundle by increasing slightly his purchase of either, constructing an affordable set with a better utility. Contradiction. This established we reframe the problem in the following manner:

√ x + x x Subject to 4x + 5x = 100

Maximize

1

1

2

1

2

This problem can be solved easily, either by the Lagrangian method, or by simple reduction to a one dimensional problem. I will choose the first method, 1

Frederick Robinson


so we set up our Lagrangian by L =

√ x + x x + λ (4x (4x + 5x 5 x − 100) 1

1

2

1

2

Now we compute each partial derivative, and set them equal to zero 1 ∂L = 0 = √ + x2 + 4λ 4λ 2 x1 ∂x 1

∂L = 0 = x 1 + 5λ 5λ ∂x 2

∂L = 0 = −100 + 4x 4x1 + 5x 5 x2 ∂λ Solving this system we combine the first two equations to get 0=

1 √ 2 x

1

1 + x2 − 4 x1 5

Then we can solve the last equation for x 2 and plug this in to get 1 1 1 4 x2 = (100 − 4x1 ) ⇒ 0 = √ + (100 − 4x1 ) − x1 5 2 x1 5 5 Simplifying this is just



1 16x 16x1 = 5 40 + √ x1 or, better yet 16x 16x1 = 200 +



√ 5x

1

solving solving this numerically numerically we get 12.5881 ⇒ x2 = 9.92952 x1 = 12. Now, we should also check the boundary solutions, that is (0, (0, 20) and (25, 0) We see that (0, 20) = 0 f (25 (12.5881, 5881, 9.92952) = 128. 128.542 f (0, f (25,, 0) = 5 f (12. so indeed, our first solution is the utility maximizing bundle. 2

Frederick Robinson

2


Supp Supple leme men ntal tal Prob Proble lems ms

2.1 2.1 2.1. 2.1.1 1

Prob Proble lem m 5.1 Ques Questi tion on

Find the points satisfying the first order conditions for constrained extrema and then apply the second order test to determine whether they are local maxima or local minima. 1. f ( and g((x,y,z ) = 2x + 3y 3y + z = = 6. f (x,y,z ) = xyz xy z and g 2. f ( + y 2 f (x,y,z ) = 2x + y = 0 x + z = 2.1. 2.1.2 2

2

− z , g (x,y,z ) = x − 2y = 0, and g (x,y,z ) = 1

2

Answ Answer er

1. The Lagrangian Lagrangian in this instance is given given by = xyz + λ (2x (2x + 3y 3y + z − 6) L = xy z + We must therefore look for solutions to the following system 0 = yz + 2λ 2λ y z + 0 = xz + + 3λ 3λ 0 = xy + xy + λ 0 = 2x + 3y 3y + z − 6 Solving this system reveals the following solutions λ =

3 − 32 , z = = 3, x = , y = 1 2

= 0, x = 0, y = 2 λ = 0, z = = 0, x = 3, y = 0 λ = 0, z = Now we check second order conditions to determine whether these are local maxima or minima.

3

Frederick Robinson


The second order condition is checked by way of the Hessian matrix which, in this instance is given as

0 z y  z 0 x  y x 0

So for the critical points the corresponding hessians are

0 3

3 1 0 3/2 1 3/2 0

 0  0

0 2 0 0 2 0 0

 0  0

0 0 0 3 0 3 0

 

Unfortunat Unfortunately ely since these matrices matrices are indefinite indefinite the corresponding corresponding points must be saddles. 2. The Lagrangian Lagrangian in this case is given given by (x − 2y) + λ2 (x (x + z ) L = 2x + y2 − z 2 + λ1 (x Thus, solutions to the optimization problem must be solutions to the following system of equations 0 = 2 + λ1 + λ2 0 = 2y 2 y − 2λ1

0 = −2z + + λ2 0 = x − 2y 0 = x + x + z

We may immediately reduce the number of variables by using the last two equations to get 0 = 2 + λ1 + λ2 0 = x − 2λ1 0 = 2x + λ2

The first equation allows us to solve for λ1 in terms of λ λ 2. Employing this relation we get a system in two variables which is easily solved. λ1 =

−2 − λ ⇒ 0 = x − 2 (−2 − λ ) = x + 2λ x + 4 + 2λ 2

2

4

2

Frederick Robinson


So, x =

−4 − 2λ ⇒ − 83 = λ 2

2

Now, substituting throughout we get 4 2 x = , y = , z = 3 3

− 43 , λ = 23 , λ = − 83 1

2

Now we need only check second order conditions to determine whether this critical point is a local minimum or maximum. We compute the hessian matrix to be

0 0

0 2 0 0

0 0 −2

 

Since this is indefinite everywhere we know that our point is a saddle.

2.2 2.2.1


Consider the problem Maximize : f (x, y) = x + y Subject to g1 (x, y) = y − 2x + 1 ≥ 0, g2 (x, y) = x 2 − y ≥ 0, g3 (x, y) = 1 − y ≥ 0, g4 (x, y) = y ≥ 0. Carry out the following steps to show that the constraint qualification fails for this problem, although a maximum exists. 1. By drawing a a figure, show that the feasible set is a three sided (nonlinear) region with vertices at (0, 0), ( 12 , 0), and (1, 1). 2. Add the plot of several level curves of the objective function to your figure from part ?? and conclude geometrically that the point that maximizes f is at (1, 1). 3. Show that the constraint qualification fails at (1, 1). 5

Frederick Robinson 2.2.2


Answer 1.2

1.0

0.8

0.6

0.4

0.2

0.0

1.

0.2 0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.0

0.5

0.0 1.

1.0

0.5

0.0 1.0 0.5 0.0

2. 3. The constraint qualification fails here, because the first three constraints are effective and we get Dh1 (x , y ) = (−2, 1); Dh2 (x , y ) = ∗

6

∗

∗

∗

Frederick Robinson


(2x, −1) = (2, −1)); Dh3(x , y ) = (0, −1). So the number of effective constraints is not equal to the linearly independent derivatives since −(−2, 1) = (2, −1). ∗

2.3 2.3.1

∗


Maximize the revenue 1

1

1

π = p 1 y1 + p2 y2 = p 1x1 + p2 x1 x2 2

2

3

subject to a wealth constraint on the inputs w1 x1 + w2 x2 ≤ C > 0,

x ≥ 0 x2 ≤ 0.

1. Write down the constraint functions and the equations that must be satisfied for the Karush-Kuhn-Tucker Theorem 2. Take w 1 = w 2 = 2, p1 = p 2 = 1, and C = 8, and find explicit values of x1 and x 2 that attain the maximum. 2.3.2

Answer

1. We may construct the Lagrangian as /

/

/

1 2 1 2 1 3 L = p 1 x1 + p2x1 x2 + λ (w1x1 + w2 x2 − C ) + λ1 x1 + λ2 x2

So the critical points of the problem must be solutions to 1/3

∂L p1 p 2 x = 0 = √ + √ 2 + λw1 + λ1 2 x1 ∂x 1 x1 1 ∂L / / = 0 = p2 x11 2 x2 2 3 + λw2 + λ2 3 ∂x 2 −

λ ≥ 0,

(w1 x1 + w2 x2 − C ) ≥ 0,

λ (w1 x1 + w2 x2 − C ) = 0

λ1 ≥ 0,

x1 ≥ 0,

λ1 x1 = 0

λ2 ≥ 0,

x2 ≥ 0,

λ2 x2 = 0

7

Frederick Robinson


2. We first substitute the values provided into the above equations to get 1/3 1 x2 0 = √ + √ + 2λ + λ1 2 x1 x1

1 / / 0 = x11 2 x2 2 3 + 2λ + λ2 3 (2x1 + 2x2 − 8) ≥ 0, λ (2x1 + 2x2 − 8) = 0 −

λ ≥ 0,

λ1 ≥ 0,

x1 ≥ 0,

λ1 x1 = 0

λ2 ≥ 0,

x2 ≥ 0,

λ2 x2 = 0

Since the utility function is strictly increasing in both variables we should have equality in the case of the function (2 x1 + 2x2 − 8) ≥ 0 for otherwise we could increase one of the variables for a better outcome. Thus, x1 + x2 = 4. Similarly we may set λ = 0. So, after some algebra we attain solutions as x1 = 3.2966, x2 = 0.703403, λ1 = 0., λ2 = 0., λ =

8

−0.382601

Frederick Robinson


Homework Frederick Robinson 1 February 2010

Contents 1 Chapter 7

1.1 1.2 1.3 1.4

Problem Problem Problem Problem

1

1 . 2 . 15 20

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2 Chapter 8

5

2.1 Problem 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Problem 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 1.1 1.1.1

1 2 3 3 5 6

Chapter 7 Problem 1 Question

2 2 Define f : R2 R by f (x, y ) = ax + by + 2cxy + d . For what values of a, b, c, and d is f concave?

→

1.1.2

Answer

Instead of computing the answer directly from the definition concavity we will employ the result of Theorem 7.10 to check concavity from the derivatives of the function. This is a valid application of the Theorem since indeed the above function f is C 2 for any choice of a,b,c,d. Moreover, since the result 1

Frederick Robinson


holds for arbitrary open convex subsets of R2 we can use the information gained by this test to determine whether f is concave on all of R2 since in particular a function f is concave on R2 if and only if it is concave on every open convex subset thereof. So we compute the Hessian matrix corresponding to f as



∂ 2 f ∂x 2 ∂ 2 f ∂y∂x

∂ 2 f ∂x∂y ∂ 2 f ∂y 2

  =



2a 2c 2c 2b

.

Next we observe that the eigenvalues associated with this matrix are a + b a2 2ab + b2 + 4c2 and a + b + a2 2ab + b2 + 4c2 . So the Hessian is negative semidefinite and f is concave if and only if these are both nonpositive. In particular since the quantity under the square root is always positive it must be that a + b 0 and (a b)2 + 4c2 (a + b)2 c2 ab. So we have established necessary and sufficient conditions for f to be concave as desired.

√ −

≤

−

c2

1.2

−

√ −

≤

⇔ ≤

and a + b

≤ ab

≤0

Problem 2

1.2.1

Question

Let f : Rn++

→ R be defined by

f (x1 , . . . , xn ) = log(xα1

where α > 0. Is f concave? 1.2.2

α n

··· x )

Answer

First observe that we can express f in the following manner f (x1 , . . . , xn ) = log(xα1

α n

α

··· x ) = log((x ··· x ) ) = α log(x ··· x ) Since α > 0 we have log(x ··· x ) is concave if and only if α log(x ··· x ) is. Thus, it should be clear by inspection now that f is indeed concave. More formally however, let y, x ∈ R and λ ∈ (0, 1) λf (x) + (1 − λ)f (y ) = λ log(x x . . .) + (1 − λ) log(y y . . .) 1

1

n

1

n

1

n

n

n

1 2

(1−λ) (1−λ) y2

= log(xλ1 xλ2 . . . xλn y1

. . . yn(1

λ)

−

1 2

(1−λ) λ (1−λ) x2 y2 . . . xλn yn(1−λ) )

) = log(xλ1 y1

≥ log((λx + (1 − λ)y )(λx + (1 − λ)y ) . . . (λx + (1 − λ)y )) ≥ λx +(1 −λ)y . So log(x x . . .) is concave by definition. Since each x y 1

λ (1−λ) i i

1

i

2

2

1 2

i

2

n

n

Frederick Robinson

1.3 1.3.1


Problem 15 Question

Describe a set of conditions on the parameters p and I under which the budget set ( p, I ) of the utility-maximization problem of subsection 2.3.1 meets Slater’s condition.

B

1.3.2

Answer

Slater’s condition is met if there exists some possible point for which each of the constraints is slack. In the context of the particular problem presented in subsection 2.3.1 the (only) constraint is in particular

· ≤ I

p   x

That is, the dot product of the consumption bundle vector with the price vector must be less than or equal to the income. So, as long as I is nonzero Slater’s condition is met since the empty bundle always causes the price constraint to be slack.

1.4 1.4.1

Problem 20 Question

√

A firm produces an output y using two inputs x1 and x2 as y = x1 x2 . The firm is obligated to use at least one unit of x 1 in its production process. The input prices of x1 and x2 are given by w1 and w2 , respectively. Assume that the firm wishes to minimize the cost of producing y units of output. 1. Set up the firm’s cost-minimization problem. Is the feasible set closed? compact? convex? 2. Describe the Kuhn-Tucker first-order conditions. Are they sufficient for a solution? Why or why not? 3. Find a solution of the first-order conditions. What conditions on the parameters make x 1 = 1 a solution of the first order conditions? When is x1 = 1? ∗



∗

3



Answer

1. The firm wishes to minimize w1 x1 + w 2 x2 subject to x1 1 and y x1 x2 y2 x1 x2 for some constant y > 0. So, the feasible set is closed, as it contains its limit points of the form y 2 = x 1 x2 and x1 = 1. However, it is not bounded and is therefore not compact. The set is convex however since it is the intersection of two convex sets: x1 1 and x1 x2 .y 2

√ ⇔

≥

≤

≤

≥

≥

Now we check

√ x x is a concave function (b) {(x , x ) | x ≥ 1} is open and convex (a) f =

1

1 2

2

1

(c) x1 x2 = y 2 is concave

(d) There exists some x1 , x2 such that x1 x2

≥ y

2

So we can use the Kuhn-Tucker method. In particular we get the Lagrangian as L =

2

−w x − w x + λ (x x − y ) + λ (x − 1) 1 1

2 2

1

1 2

2

1

2. The first order Kuhn-Tucker conditions are l

Df (x ) + ∗



λi Dhi (x ) = 0 ∗

∗

i=1

and

l ∗

λ



≥ 0,

λ i hi (x ) = 0 . ∗

∗

i=1

They are sufficient since the constraints are all concave. They are necessary since the problem satisfies Slater’s condition. 3. We can solve the problem now by using these constraints. Solutions must satisfy the following equations

−w − w + λ + λ = 0 1

2

∗

∗

1

2

and 0 = λ 1 (x1 x2 ∗

∗

∗

2

− y ) + λ (x − 1) 4

∗

∗

2

1

Frederick Robinson


as well as the conditions Df (x) + λi Dhi (x) = 0

Thus the optimization problem is solved by x1 = max

2 2.1 2.1.1

   w2 ,1 w1

y

and

y2 x2 = x1

Chapter 8 Problem 4 Question

n R into R, where Let f 1 , . . . fl be functions mapping a1 , . . . , al be nonnegative numbers. Show that if for each i concave, then so is f , where f is defined by

D is convex. Let ∈ {1, . . . , l}, f is

D ⊂

i

l

f (x) =



ai f i (x),

i=1

x

∈ D.

Give an example to show that if each f i is only quasi-concave, then f need not be quasi-concave. 2.1.2

Answer

Fix some x, y and λ nonnegative we know that

∈ D

∈ (0, 1).

f i [λx + (1

Since each f i is concave and each ai is

− λ)y] ≥ λf (x) + (1 − λ)f (y) i

i

and so clearly f (x) inherits this property as l

f [λx + (1

− λ)y] =

 i=1

ai f i [λx + (1

− λ)y]

l

 ≥ i=1

ai (λf i (x) + (1

− λ)f (y)) = λf (x) + (1 − λ)f (y) i

5

Frederick Robinson


This is not true for merely quasi-concave functions however. We know for example that the functions f 1 (x, y ) = min x2 , (x

2

2

2

− 4) } and f (x, y) = min{y , (y − 4) } are both quasiconcave on [0, 4] × [0, 4] however the sum f + f is not qua{

2

1

2

siconcave since level sets of this function are not convex. This is easy to see if you consider that such level sets are symmetric about the lines y = 2 and x = 2 both and the points on a given level set which are furthest from (2 , 2) (the center of each level set by symmetry) occur along these lines.

2.2 2.2.1

Problem 8 Question 2

Show that the function f : concave if = R2+ , but not if

D ⊂ R → R defined by f (x, y) = xy is quasi D = R .

D

2.2.2

2

Answer

2 The function f : R R defined by f (x, y ) = xy is quasi-concave 2 if = R+ since given x and y in the first quadrant f (λx + (1 λ)y ) = λ2 x + 2λ(1 λ)xy + (1 λ)2 y 2 min(x2 , y 2 ) given any choice of λ. However if we cease to restrict our attention to the first quadrant this is no longer the case. Take the points (0 , 1) and (0, 1) with λ = .5. In this case 02 = 0 < min(12 , 12 ) so the function is not quasi-concave.

D

D⊂

−

→

−

−

≥

−

6

Frederick Robinson



Contents 1 Book Problems

1

1.1 Problem 9.2.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Problem 9.2.12 . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Supplemental Problems

3

2.1 Problem LP.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Problem LP.2 . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 1.1 1.1.1

1 2

Book Problems Problem 9.2.8 Question

Minimize 5x1 + 3x2 subject to 2x1 + 5x2 ≥ 10 3x1 + x2 ≥ 6 x1 + 7x2 ≥ 7 and x1 ≥ 0, x2 ≥ 0

1

3 5



Answer

We begin by observing that, since it contains the point (100 , 100) the feasible set is nonempty. Moreover we see that the objective function is bounded below by f (0, 0) = 0 since for a smaller value we would need negative x1 or x2 , but these are excluded from the feasible set. We know therefore that the optimum occurs, and moreover that it is at the “corners” and convex linear combinations of said. So, to compute the optimum we will evaluate the objective function at each of these corners. Towards this we compute each pairwise intersection of functions.

 20 18  35 4 

 7 3 

(0, 2)(5, 0) (0, 6)(2, 0)(0, 1)(7, 0)(0, 0) , , 9 9 4 4

, 13 13

Now we need to check which of these is actually in the feasible set. Doing so reveals that the only ones which are feasilbe are

 20 18  35 4  , 13 13

(0, 6)(7, 0) , 9 9

Evaluating the function at these points we get f

 20 18  ,

13 13

 

154 35 4 = ≈ 11.8462 f , 13 9 9

=

187 ≈ 20.7778 9

f (0, 6) = 18 f (7, 0) = 35 and by inspection we identify that the minimum of the function on the feasible set occurs as 20 18 154 = f , 13 13 13



1.2 1.2.1



Problem 9.2.12 Question

Mark each statement True of False. Justify each answer. 1. If a canonical linear programming problem does not have an optimal solution, then either the objective function is not bounded on the feasible set F or F is the empty set. 2. If x is an optimal solution of a canonical linear programming problem, then x is an extreme point of the feasible set. 2



Answer

1. True Theorem 6 state: “If the feasible set F is nonempty and if the objective function is bounded above on F , then the canonical linear programming problem has at least one optimal solution. Furthermore, at least one of the optimal solutions is an extreme point of F . ” This is merely the contrapositive of the statement in question. Thus the statement is true. 2. False This statement is not true. We know in particular that convex combinations of optimal solutions can be optimal, and that more than one extreme point may be optimal. Thus solutions along the “edge” connecting two optimal extreme solutions are also optimal, yet these points are not themselves extreme points. For a particular counterexample see Part 2.1.2.

2 2.1 2.1.1

Supplemental Problems Problem LP.1 Question

Show graphically that the following linear program does not have a unique solution: Maximize 30x + 40y Subject to : 3x + 4y ≤ 48 x + y ≤ 14 0 ≤ x, 0 ≤ y. 2.1.2

Answer

The following depict the feasible set, and the function evaluated on the feasible set respectively. Intuitively, each value on the “edge” from (0, 12) to (8, 6)

3

Frederick Robinson


maximizes the objective function since it is “highest” on each such value. 14

12

10

8

6

4

2

0 0

5

10

4

15

Frederick Robinson


15 10 5 0

40

200

0 0 5 10 15

2.2 2.2.1

Problem LP.2 Question

Show that the feasible set to the following problem permits the objective function to assume arbitrarily large values: Maximize 2x + y Subject to : x − y ≤ 3 −3x + y ≤ 1 0 ≤ x, 0 ≤ y. 2.2.2

Answer

I claim in particular that each point of the function g(x) = y = 2x − 1 is contained within the feasible set given x ≥ 1/2 and that the objective function tends to infinity along this funciton. Proof. Our the constraints on our feasible region may be rewritten as

x−y ≤3⇔y ≥x−3

− 3x + y ≤

5

1 ⇔ y ≤ 1 + 3x

Frederick Robinson


Now just compute (2x − 1) − (x − 3) = x + 2 which is greater than or equal to zero subject to x ≥ 1/2 as claimed. Similarly we compute (1 + 3x) − (2x − 1) = x + 2 which is again greater than zero given x ≥ 1/2. It remains only to verify that our function has x ≥ 0 and y ≥ 0 for x ≥ 1/2. That it satisfies this property for x is implied directly from our restricting x ≥ 1/2. Moreover since at x = 1/2 we get y = 0 and the y value is strictly increasing in x it is true as well that y ≥ 0. Now we just verify that our objective function tends to infinity along the function as claimed. Since as just demonstrated the portion of our function which has x ≥ 1/2 is within the feasible set this will suffice to demonstrate that the objective function is not bounded in the feasible set. For a given value of x we may evaluate the objective function on g(x) by 2x + y = 2x + 2x − 1 = 4x − 1 and clearly we have lim 4x − 1 = ∞ x→∞

so the objective function may assume arbitrarily large values on the feasible set as claimed.

6

Frederick Robinson



Contents 1 Book Problems

1

1.1 Problem 9.3.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Problem 9.3.12 . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Problem 9.3.14 . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Supplemental Problems

5

2.1 Problem L.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1

1 2 3 5

Book Problems

1.1 1.1.1


1. A solution is called a basic solution if m or fewer of the variables are nonzero 2. The basic feasible solutions correspond to the extreme points of the feasible region. 3. The bottom entry in the right column of a simplex tableau gives the maximum value of the objective function.

1



Answer

1. True . Page 30 states ”A solution to this system is called a basic solution if no more than m of the variables are nonzero” 2. True. Again, Page 30 says ”Geometrically, these basic feasible solutions correspond to the extreme points of the feasible set. ” 3. False. This is partially true. The entry in question gives the value of the objective function at the point in question which is sometimes the maximum value, but could be less if the algorithm has not yet been run.

1.2

Problem 9.3.12

1.2.1

Question

Maximize subject to

and x1 1.2.2

2x1 +5x2 +3x3 x1 +2x2 2x1 +4x3 x2 +x3

28 ≤ 16 ≤ 12 ≤

≥ 0, x2 ≥ 0.

Answer

In this case we have only resource contraints, so we can set up the simplex as described on page 6 of the notes.

x  1  2 

1

−2

x2

x3

2

s1 s2 s3

1

1

4 1

−5

−3

1 1

 28   16   12 

0

Now we can just perform the algorithm with this matrix as follows

x  1  2 0

1

x2

x3

s1 s2 s3

2 0 1

0 4 1

0 1 0 0

1 0 0 −2 −5 −3 0

0 0 1 0

 x  1 28   16  2 ⇒   0 12 

1

x2 x3

0 0 1 −2 0

0

2

−2

4 1 2

s1 s2 s3

1 0 0 0

0 1 0 0

−2

0 1 5

 4   16   12  60

Frederick Robinson

x  1 ⇒  00 0

1


x2 x3

s1

0 0 1 0

1

−2

s2 s3

0 8 −2 1 1 0 0 0 −2 2

−2

4 1 1

 x  1 4   8  0 ⇒   0 12  68

1

0

x2 x3 s 1

0 0 1 0

0 1 0 0

1 2

1

−4 1

4 3 2

s2

s3

1

−1

4 1

1

8

2 1

1

2

− 18 4

2

 6   1   11  70

So the solution (6, 11, 1) Maximizes the above problem giving in particular f (6, 11, 1) = 70. Moreover one can check that this solution does indeed satisfy all the constraints as desired.

1.3 1.3.1


Minimize subject to

and x1 1.3.2

2x1 +3x2 +3x3 x1 −2x2 2x2 +x3 2x1 −x2 +x3

≥ −8

15 ≤ 25 ≥

≥ 0, x2 ≥ 0, x3 ≥ 0.

Answer

Since in this case there are more than just resource constraints our matrix is slightly more complex. We need to rewrite the constraint involving a negative number as the opposite constraint on a positive value. Moreover, we must invert the sign of the minimization problem to make it a maximization problem before constructing the matrix.

x  −1   2  −1 1

2

x2

2 2

x3

1 −1 1 −3 −2 3 3

s1

s2

s3

r2

1 −1

1 1

−1

1

3

−1

  8 15   25   −48 

Frederick Robinson


Now we just apply the algorithm

x  −1  0  2  −1

x2

1

2 2

0 1 −1 1 −3 −2 3 3

2

x  − 1 ⇒  −

1 0 0 0 0

2

3 2

1

5 2

2

x  0  1 ⇒  0 0

1 0 0

0

0 1 1

s2

1

0

s1

1

0 1 0 0 0

s3

0

r2

0 1 0 0 0

  4  7 29   −36  −12

r2

1 15 0 2 2 1 7 −1 −1 0 3 3 4 1 − 2 37 2 2 3 5 −2 − 2 −1 2 − 37 2 0 −1 21 − 12 − 31 2

1 − 12 1 2 5 2

− 12

s1 s2

1 2 3 4

  8 15   25   −48 

r2

s3

s2

0

2

1 0 0 0 0

0

s1 2

x2 x3

1

s3

0 −1 −1 0 1 0 1 2 1 −2 2 −1 1 3 3 0 −2 0

1 0 0 0 0

0

s2

0 −1 0 0 1 −1 1 −1 0 0 0

x2 x3

1

s1

x2 x3

1 1

x  0 1 ⇒  0 0

x3

1

−4

0 0

0 0 1 0 0

s3

− 12

0

3

2 1

− 14

r2

1

1

15

2 1

2 65

4

4 37

3

4

2

−4

0 2

0

1

0

−1

−2

−55

4

Now we may remove the artificial variables

x  0 ⇒  10 0

1

x2 x3

1 0 0 0

1 2 3 4

− 14

0

s1 s2

0 0 1 0

4

s3

− 12

0 1

2 65

3

2 1

4 37

− 14 4

2

2

−1

15

4

−55

   

         

Frederick Robinson


x  0 ⇒  10

1

x2 x3

1 0 0 0

0

x  − ⇒  1

1 2

1 − 12 − 12

s1

s2

0

− 12

0 −1 −1 0 3 2 1 2 7 2 0 2

x 2 x3 s 1

1 1

1 0 0 0

2

1 2 1 2

0 1 0 0

s2

1

4 7 22 −33

   

15 2

7 37 2

s3

0

0 −1 −1 0 3 1 1 2 3 3 0 2 2

− 73 2

   

s3

So our optimum is just f (0, 4, 7) = 33.

2 2.1 2.1.1

Supplemental Problems Problem L.3 Question

Solve the linear program by the Simplex Algorithm Maximize subject to and x1 2.1.2

x1 +2x2 +3x3 x1 +2x2 +x3 = 36 2x1 +x2 +4x3 ≥ 12

≥ 0, x2 ≥ 0, x3 ≥ 0.

Answer

We begin by setting up the matrix as follows

x  1  2  −3 1

−1

x2

x3

s2

2 4

1 1

−1

−6

−2

1

−2

−3

r1 r2

1

5

1

 36   12   −48 

Frederick Robinson


Now we apply the simplex algorithm

x  1  2  −3

x2

x3

s2

2 4

1 1

0

x  −1  2 1

x2

x3 s2

−2

0 1 0 0

1

r1 r2

1 −1 0 0 −6 −2 1 −1 −2 −3 0 0 1

5

4 2 10

0 1 0 0

r1 r2

1

1 −1 0 −1 0 −3 0

−1

1 2 3

 x  0 36   12  ⇒   0 −48 

1

0 1 0 0

1 2

0

0

 x  −1 24   12  1 ⇒   0 −24  1

36

x2 x3

2

s2

r1 r2

1

1

2 1

2

1 0 0 0

1

4

−4

− 12

− 12

− 52

− 12

− 12 1

4 3 2 1 2

x2

x3 s 2 r 1 r 2

−2

0 1 0 0

2 0 4

1 0 0 0

1 1 1 3

−1

0 1 0

 30   3  −30  6

24 36 0 108

   

Now we may drop the artificial variables since they are no longer pivots and are zero

x  −1 1 1

2

x2

x3 s2

−2

0 1 0

2 4

1 0 0

24 36 108

 x  ⇒  0  0 2

1

x2 x3 s2

0 0 4

0 1 0

1 0 0

36 108

  

So we see that the solution to our problem is given by f (0, 0, 36) = 108

6

Frederick Robinson



1 1.1 1.1.1

Book Problems Problem 9.4.4 Question

State the dual problem of Exercise 12 in Section 9.3 1.1.2

Answer

Minimize Subject to

and y1

1.2 1.2.1

≥ 0, y2 ≥ 0, y3 ≥

28y1 +16y2 +12y3 y1 +2y2 2y1 +y3 4y2 +y3

2 ≥5 ≥3 ≥

0


Use the final tableau in the solution of Exercise 12 in Section 9.3 to solve its dual.

1



Answer

Recall that

  1  00

x1 x2 x3 s1

0

0 0 1 0

s2

s3

1

−1

1

0 1 0 0

2

4 1

1

−4 1

8

1 1

−8

4 3

1

2

4

2 1 2

2

 6   1  11  70

is the tableau of the final solution from last week. Thus, we see by inspection that the solution to the dual problem is just g

 3 1 

, , 2 = 70.

2 4

It is easy to check that these values do indeed satisfy the constraints from above.

1.3 1.3.1

Problem 9.4.10abc Question

Mark each statement True or False, and justify each answer. 1. The dual of the dual problem is the original problem. 2. If either the primal or the dual problem has an optimal solution, then they both do. 3. If the primal problem has an optimal solution, then the final tableau in the simplex method also gives the optimal solution of the dual problem. 1.3.2

Answer

1. True. In fact on Page 46 of the book it states explicitly “So the dual of the dual problem is the original primal problem.” 2. False Assuming that both feasible sets are nonempty this is just the Duality Theorem and therefore true. This is however false in the case that one of the feasible sets is empty.

2

Frederick Robinson


3. True The continuation of the Duality Theorem from Page 48 states “If either P or P is solved by the simplex method, then the solution of its dual is displayed in the bottom row of the final tableau in the columns associated with the slack variables. ” ∗

1.4

Problem 9.4.12

1.4.1

Question

Use the simplex method to solve the dual, and from this solve the original problem. Minimize 10x1 +14x2 subject to x1 +2x2 ≥ 3 2x1 +x2 ≥ 4 3x1 x2 ≥ 2 and x1 ≥ 0, x2 ≥ 0 1.4.2

Answer

First we observe that the dual to this problem is just the following Maximize subject to and y1

3y1 +4y2 +2y3 y1 +2y2 +3y3 2y1 +y2 +y3

≥ 0, y2 ≥ 0, y3 ≥

10 ≤ 14 ≤

0

So we can set up the tableau for the dual problem as

  

y1

y2

y3

s1 s2

1 2 −3

2 1 −4

3 1 −2

1 1

   10   = 14  

y1 y2 1

∼

2

0

y2

1

1 0 0

∼

0

  = 1

y1

1 0 0

2 3 2

−1

y3

s1

s2

3

1

0

2

1

2

1

2

−3

−3

11

5

3 2

3

3

3

y3

s1

s2

3

1

0 1 0

2

1

2

1

−2

−2

4

2

 5   6 

26

So, we see by inspection that the solution to the dual problem is g (6, 5, 0) = 26.

3

 5   9 

20

Frederick Robinson


Moreover, the solution to the dual of the dual problem — and therefore to the original problem is just f

 5 2  ,

3 3

= 26.

A quick check verifies that this does indeed satisfy the constraints to the original problem.

4

Frederick Robinson


Homework Frederick Robinson 2 March 2010

Contents 1 Problem 9.1

1

1.1 Question . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Problem 9.2

3


1

3 4 5

Problem 9.1

1.1

1 2

Question

1.

D(θ) =

 [0, 2θ]

D(θ) =

 [0, 1 − 2θ]

D(θ) =

 [0, 1 − 2θ]

for θ ∈ [0, 1/2) [0, 2 − 2θ] for θ ∈ [1/2, 1]

2. for θ ∈ [0, 1/2] [0, 2 − 2θ] for θ ∈ (1/2, 1]

3. for θ ∈ [0, 1/2) [0, 2 − 2θ] for θ ∈ [1/2, 1] 1

Frederick Robinson


4.

D(θ) = {0, θ} for θ ∈ [0, 1]

1.2

Answer

1. This is a closed correspondence since given arbitrary θ we have D(θ) closed. Moreover, since given arbitrary θ we have D(θ) bounded we may surmise that this is a compact correspondence. It is both upper and lower semicontinuous. To show that this is the case (upper-semicontinuous) we merely note that for every.  > 0 there exists δ > 0 such that θ ∈ B δ θ0 ∩ Θ ⇒ Φ(θ) ⊂ B  Φ(θ0 ). In particular θ = θ 0 ± δ ⇒ Φ(θ0 ± δ ) ⊂ B3 δ Φ(θ0 ). That is, given δ the condition is satisfied by taking  = 3δ ·

Now to establish lower semicontinuity we just need to show that for each  > 0 there exists δ > 0 such that θ ∈ Bδ θ0 ∩ Θ ⇒ Φ(θ0 ) ⊂ B Φ(θ). The inverse relation works here though. That is; given δ taking  = δ/3 gives us θ ∈ Bδ θ0 ∩ Θ ⇒ Φ(θ0) ⊂ Bδ/3 Φ(θ) as desired. Since the correspondence is both upper and lower semicontinuous it is continuous. 2. This is a closed correspondence since given arbitrary θ we have D(θ) closed. Moreover, since given arbitrary θ we have D(θ) bounded we may surmise that this is a compact correspondence. It is easy to see that it is lower semicontinuous at each point of the region [0, 1/2) ∪ (1/2, 1] since on these regions it is compact, linear. It therefore remains only to check the point { 1/2}. However it is lower semicontinuous here too since for  > 0 there exists δ > 0 such that θ ∈ B δ 1/2 ∩ Θ ⇒ Φ(1/2) ⊂ B  Φ(θ). In particular we may take any  > 0 since for any θ ∈ Bδ 1/2 ∩ Θ we have Φ(1/2) = {0} ⊂ B Φ(θ) as desired for any  > 0. The function is not upper semicontinuous since there does not exist δ > 0 such that θ ∈ Bδ 1/2 ∩ Θ ⇒ Φ(θ) ⊂ B Φ(1/2) for any . In particular θ = 1/2 ± δ ⇒ Φ(1/2 ± δ ) ⊂ B Φ(1/2) for every epsilon. That is, the continuity condition fails to be satisfied at θ = 1/2. Since the correspondence is not upper and lower semicontinuous it is not continuous. 2

Frederick Robinson


3. This is a closed correspondence since given arbitrary θ we have D(θ) closed. Moreover, since given arbitrary θ we have D(θ) bounded we may surmise that this is a compact correspondence. It is easy to see that it is upper semicontinuous at each point of the region [0, 1/2) ∪ (1/2, 1] since on these regions it is compact, linear. It therefore remains only to check the point { 1/2}. The function is upper semicontinuous since there exists δ > 0 such that θ ∈ Bδ 1/2 ∩ Θ ⇒ Φ(θ) ⊂ B Φ(1/2). In particular θ = 1/2 ± δ ⇒ Φ(1/2 ± δ ) ⊂ B3 δ Φ(1/2). That is, given δ the condition is satisfied by taking  = 3δ . This is easy to see since Φ(1/2) = {0} so B3δ 0 in particular includes all points in the image of B δ 1/2 under Φ. ·

However it is not lower semicontinuity since for  > 0 there does not exists δ > 0 such that θ ∈ Bδ 1/2 ∩ Θ ⇒ Φ(1/2) ⊂ B Φ(θ). That is; the continuity condition fails at θ = 1/2 Since the correspondence is not upper and lower semicontinuous it is not continuous. 4. This is a closed correspondence since given arbitrary θ we have D(θ) closed. Moreover, since given arbitrary θ we have D(θ) bounded we may surmise that this is a compact correspondence. As this set is the union of finitely many (two) continuous functions: namely f (θ) = 0 and f (θ) = θ it is continuous and therefore upper and lower continuous.

2 2.1


Let Θ = [0, 1] = S , and h : S × Θ → R be defined by f (x, θ) = 3 + 2x − 3θ − 5xθ. Here, D(θ) = [0, 1] for all θ. Find f (θ) and D (θ) for each value of θ. Using the f and D you have found, discuss why f (θ) is a continuous function and D (θ) is a usc correspondence. ∗

∗

∗

∗

∗

∗

3

Frederick Robinson

2.2


Answer

The relevant definitions are f (θ) = sup{f (x, θ) : x ∈ D(θ)} ∈ R ∗

D (θ) = arg max{f (x, θ) : x ∈ D(θ)} ⊂ D(θ) ⊂ Rn ∗

So in this case the members of f (θ) are those x which maximize the given function on the unit interval, given θ. In particular we have ∗

f (x, θ0 ) = 3 + 2x − 3θ0 − 5xθ0 Since this function is linear in x it is maximized either at 0 or 1. For θ0 < 2/5 it is maximized at 1 since the function is increasing in x. For θ0 > 2/5 it is maximized at 0 and for θ 0 = 2/5 it is maximized everywhere. Therefore we have

 1 D (θ) =  [0,0 1]

θ < 2/5 θ = 2/5 θ > 2/5

 5 − 8θ f (θ) =  9/5 3 − 3θ

θ < 2/5 θ = 2/5 θ > 2/5

∗

and correspondingly we see by substituting these values in f from above ∗

Having computed these explicitly it is easy to see that f is continuous and D is usc as desired. In particular f is a piecewise defined, linear. Thus, since it agrees at the boundaries of each piece [check: 5-8(2/5)=9/5=3-3(2/5)] it is continuous. Finally, to confirm that D is usc as claimed we need only verify at θ = 2/5 since the rest is linear. However, since at this point the entire codomain is the result of taking Φ(2/5) we see that for any  > 0 there exists δ > 0 such that θ ∈ Bδ θ0 ∩ Θ ⇒ Φ(θ) ⊂ BΦ(θ0 ). In particular taking any δ > 0 fulfills this definition since, as every point of the codomain is in Φ(θ0 ) clearly every point is in B  Φ(θ0 ) for any  > 0. ∗

∗

∗

4

Frederick Robinson

3


Problem 9.3 0.5

0.0

.5 1.

0.5

0.0

0. 1.0

0.5

1.

0.0

0.5

1.0

1.0

1.0

0.5

0.0

0.5

1.0 0.4

0.2

0.0

0.2

0.4

f is compact and D is continuous, and compact valued. Thus, the maximum theorem applies as desired. 2. Given 1/2 > θ > 0 the function f has maximum at x = 1 − θ since for a fixed θ the maximum must occur at an endpoint of one of the piecewise defined regions by linearity. Then, by plugging in we find 5

Frederick Robinson


that the above is the correct (maximal) such endpoint. On this set the function takes value θ. With θ = 0 the function is just constant valued (with value 0) so any value in D is a maximizing value. Proceeding similarly for 0 > θ > −1/2, we check each endpoint since linearity guarantees that one such point gives the maximum of the function for fixed θ. In this manner we see that the maximum occurs at |θ| − 1 = x and the value of the function here is just | θ|. Now, sincethe maximizing points from above are not necessarily contained in D we must find which points of D maximize the function in these cases. However, a quick check reveals that they are all contained. (See first figure below) To summarize we have

 1 − θ D (θ) =  [−−θ1,−1]1 ∗

θ < 0 θ = 0 θ > 0

and correspondingly we see by substituting these values in f from above

 −θ f (θ) =  0θ ∗

6

θ < 0 θ = 0 θ > 0

Frederick Robinson


1.0

0.5

0.0

0.5

1.0 0.4 0.0

0.2  0.5

0.0

0.2

0.4

.

0.5 1.0 1.0

0.5

0.0

0.5 0.5

0.0

0.5

1.0

3. The answer to both questions is yes. f is continuous since it is just the absolute value function, which we know to be continuous. Moreover D is nonempty for every θ (it is defined above). Finally D is both lsc and usc on the relevant interval since as in the previous problem we know that at the ends (θ > 0 or θ < 0) it is continuous and we can verify θ = 0 by definition exactly as we did in the previous problem. ∗

∗

∗

7

Frederick Robinson


Homework Frederick Robinson 8 March 2010

Contents 1 Problem 11.3

1


6

2.1 Question . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 1.1

1 1 6 6


Redo the consumers multiperiod utility maximization problem of Section 11.6 with u(c) = c . (cf. 11: 1a) 1 3

1.2

Answer

As in Section 11.6 we have a consumer faced with a T -period planning horizon. where T is a finite positive integer. He has initial wealth w R+ . We say that he begins period-t with a wealth wt , and consumes ct (wt ct 0) and begins the the next period with ( wt ct )(1 + r ), where r 0 is the interest rate. In unlike in the example however we have that the consumption of c in any period give a utility of u(c) = c .

−

1 3

1

∈ ≥ ≥ ≥

Frederick Robinson


For the purposes of this problem the state space S will be the set of possible wealth levels ( R+ ) and the action space A will be the set of consumption levels (again R+ ). The reward function is just rt (w, c) = u (c) = c . The transition function f t is f t (w, c) = (w c)(1 + r ). Finally we have the feasible action correspondence Φt (w) = [0, w] for all w. We shall let k = (1 + r ) as in the book for ease of notation. In the last period, given a wealth level w the consumer solves: 1 3

−

max u(c).

c∈[0,w]

But, clearly since u(c) is strictly increasing the the unique solution for any given w is to consume everything. hence we conclude that the (unique) optimal strategy for one time period is just g T (w) = w for all w S and the one period value function is given by

∈

1

V T (w) = w ,

w

3

∈ S.

Now we consider the case of two periods. In this case we need to maximize the reward function by solving 1

1

max c + (k (w c)) . c∈[0,w] Since this optimization problem is strictly convex we can just employ first order conditions. Thus we see we must have ∂ 1/3 k 1 =0 c + (k ( c + w))1/3 = 2/3 ∂c 3c 3(k( c + w))2/3

{



−

3



− √ − w − kw ⇒ c = −1 + k

3

}

−

or c =

− √

−w + kw −1 + k

however this first solution may be discarded since for w, k in the appropriate ranges, this solution corresponds to negative c. So, the corresponding value function is just V (w) =

 √   √  1/3

w

1+

k = w

1/3

 √  1+

k

2/3

1+ k Now we can compute again for the three-period problem. In this case we must solve the following optimization problem. max

c∈[0,w]

    −    √   √       2

k (w

c)

1/3

1/3

c

+

1+

Period 3

k

1+

Periods 1 & 2

2

k

.

Frederick Robinson


again applying first order conditions we see that ∂ ∂c

   √   √   − − 1/3

c

1+

k + k 2 (w

1+

k

k2

1

=

c√ 1+ k

3

− c)

2/3

3 (k 2 (

 1/3

=0

2/3

c + w))

√ kw ⇒ c = 1 + √ k + k w +

and V (w) =

=

=

=

 √   √ √ − √  √   √ √ − √   √     √  √   √    √ √ w +

k 2 (w

1+

k + k

w +

k 2 (w

k + k

+

k

1+

k + k

)

  √  √ √  1/3

k + k )(1 +

1+

k)

(w + kw )(1 + k )2 (1 + k + k )

+

w

1/3

+

k +

k )1/3

1/3

k3 w

1+

k w1/3 + kw 1/3

(1 +

(1 +

kw

1/3

1+

1+

)

w +

1/3

kw

1+ 3

1/3

kw

k + k

= 1+

k + k

2/3

w1/3

So, it seems likely that in general we have the relations



V T (w) = 1 +

and

gT (w) = w



√

k +

··· + k

T /2

√ · ·· √ · ··



2/3

w1/3

1 + k + + k (T −1)/2 1 + k + + k T /2



We will use induction to prove this relation. If we assume that the above is true for T = n we can compute the case corresponding to T = n + 1 fairly by the following optimization: max c∈[0,w]



(w

− c)k

 

n+1 1/3

+ 1+ 3

√

k +

··· + k

n/2

  2/3

c1/3

k

Frederick Robinson


again, as the above is a convex optimization we can just apply first order conditions. ∂ ∂c

=

−



(w

− c)k

  √ · ··  √ ···  √ ···  √ ···

n+1 1/3

k 1+n

2/3

3 (k1+n (

−c + w))

⇒−

k 1+n

+ 1+

+ 1+

(k 1+n ( c + w))2/3

−

⇒−

k

(w

1 3

(1+n)

2/3

− c)

+

k +

1+

k + c

1

k (1+n) 1+ = w c 2

√ k + ·· · + k

c1/3

1

2/3

3c2/3

=0

2/3

=0

n/2

⇒ − c ⇒ w −c c = 1 + √ kk + ··· + k √ ··· + k ⇒ wc = 1 + √ kk + · ·· + k + 11 ++ √ kk + + ··· + k √ ⇒ wc = 1 + 1k+ +√ ···k ++··k · ++k k √ 1 + k + · ·· + k ⇒ c = w 1 + √ k + ··· + k + k 1 2

=0

2/3

+ kn/2

+ k n/2

k + c

2/3

n/2

+ k n/2

k +

1+

+

+k

    

(1+n)

n/2

1 2

(1+n)

n/2

n/2

n/2 n/2

(n+1)/2

n/2



n/2

n/2

(n+1)/2



This agrees with what we expect for gn+1 (w). Finally we substitute to check V n+1 (w)

 −   √ ···    √ ···   √ − ···   √ ···   √ √ ··· c)k n+1

(w

=

1+

w

k +

w

1/3

+ 1+

k +

+ kn/2

1 + k + + k n/2 1 + k + + kn/2 + k(n+1)/2

+ k n/2

2/3

w1/3

2/3

c1/3

1/3

k

n+1

+

1 + k + + k n/2 1 + k + + k n/2 + k (n+1)/2

···

4

1/3

Frederick Robinson


√ 1 + k + ·· · + k √ k = w 1− + 1 + k + ··· + k + k √ 1 + k + ··· + k w √ 1 + k + ··· + k + k √ √ 1 + k + ··· + k + k 1 + k + ·· · + k √ w − 1 + √ k + ··· + k + k 1 + k + ··· + k + k √ 1 + k + ··· + k w √ 1 + k + ··· + k + k √ k 1 + k + ··· + k √ +w w √ 1 + k + ··· + k + k 1 + k + ·· · + k + k √ 1 + k + ··· + k + k = w √ 1 + k + ··· + k + k



n/2

=

n/2



n/2

n/2

(n+1)/2

n/2

(n+1)/2

(n+1)/2

=

1/3

n/2

(n+1)/2

 

1/3

k

n+1

n/2



n/2

3(n+1)/2

n/2

1/3



n/2

1/3



n+1

(n+1)/2

1/3



 

1/3

n/2

(n+1)/2

   



(n+1)/2

1/3

√

k +

1/3

n/2

1/3



n/2

n/2

= w 1/3 1 +



··· + k

   

(n+1)/2

(n+1)/2

n/2

(n+1)/2

n/2

+ k (n+1)/2

1/3

2/3

which agrees with our prediction for V n+1 (w). Thus, we are done with the inductive step and we have confirmed that



V T (w) = 1 +

and

gT (w) = w



√

k +

··· + k

T /2



2/3

√ 1 + k + ··· + k − √ 1 + k + ··· + k

w1/3

(T 1)/2 T /2

5



.



1/3

+

Frederick Robinson

2 2.1



Determine the longest path from A to H in the network in the following figure. The arrows indicate which direction an edge can be traversed.   B 

8 

4 

  F          6            10                  6 A  7    H                     9    4   6             E  G C  5  

D 

6

2.2

4

Answer

This problem is small enough that it can be solved by brute force. Towards this we shall list all possible such paths and their lengths. Path A A A A A A A A A

→ B → D → F → H → B → D → G → F → → B → D → G → H → B → E → F → H → B → E → G → F → → B → E → G → H → C → E → F → H → C → E → G → F → → C → E → G → H

Length 5+8+4+6 H 5 + 8 + 10 + 6 + 6 5 + 8 + 10 + 4 5+7+9+6 H 5 + 7 + 4 + 6 + 6 5+7+4+4 6+6+9+6 H 6 + 6 + 4 + 6 + 6 6+6+4+4

6

Sum 23 35 27 27 28 20 27 28 20

Solution 1 A First Course in Optimization Theory. Rangarajan K. SundaramCambridge University Press (1996).pdf

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