TOPIC VI UNCONSTRAINED OPTIMIZATION I

Transcription

1 [1] Motivation TOPIC VI UNCONSTRAINED OPTIMIZATION I y Consider Dom (f) = {0 7}. Global ma: y = 8 at = 7 gma Global min: y = 3 at = 5 gmin Local ma: y = 6 at = 3 lma Local min: y = 3 at = 5 lmin We wish to find global ma or min points. When global ma or min points happen to be also local ma or min, we can find the global points using calculus. VI-1

2 [] Properties of Local Minima and Maima (1) Stationary points 1) Local minimum points f( ) = 0, and in a neighborhood of, f() as. 0 o ) Local maimum points. f( ) = 0, and in a neighborhood of, f() as. 0 o VI-

3 3) Inflection points f( ) = 0, f() as 0 4) Summary: If f( ) = 0, (, f( )) is called stationary point Three possibilities for a stationary point. local ma. local min. inflection point. VI-3

4 [] Conditions for Local Ma. or Min. (1) First-Order (necessary) Condition: Find such that f( ) = 0. o 0 Note: For ma at, in the neighborhood of, f() as 0 o For min at, in the neighborhood of, f() as 0 o For inflection point, in the neighborhood of, f() as o () Second-order (sufficient) condition: d Let f() = d f (). For such that f( ) = 0, o 0 f( ) < 0 local ma. 0 f( ) > 0 local min. 0 f( ) = 0 local min or local ma or inflection point 0 Note If is inflection point, then f( ) = However, f( ) = 0 does not mean that is an inflection point. 0 0 EX) VI-4

5 f( ) = 0, but local ma. 0 EX 1) (profit) = TR - TC = R(Q) - C(Q). FOC for ma. : d dq dr dq dc dq 0 MR - MC = 0 get Q* SOC for ma. : EX ) d dq d R dq d C dq <0 * dmr at Q dq dmc dq <0 * at Q. Demand: Q = P 3 Total Cost: TC = Q - 7Q + 111Q + 50 VI-5

6 P = Q TR = P Q = (100 - Q)Q = 100Q - Q 3 = TR - TC = (100Q - Q ) - Q + 7Q - 111Q = -Q + 6Q - 11Q FOC for profit maimization d = -Q + 1Q - 11 = 0 dq * Q - 1Q + 11 = 0 (Q - 11)(Q - 1) = 0 Q = 11 or 1. SOC: d = -Q + 1. dq d At Q = 1, = 10 > 0 dq d At Q = 11, = -10 < 0 dq * Thus, local profit ma. Q = 11 * * * = TR(Q ) - C(Q ) = 111. EX 3) y = a + b + c, a 0 VI-6

7 Find local ma. or min. FOC: y = a + b Set y = 0 * = b. a SOC: y = a If a > 0 y > 0, y* = a b + b b + c is local min. a a If a < 0 y < 0, y* is local ma. EX 4) 3 y = f() = FOC: Set f() = 3-6 = 0 SOC: - = 0 ( - ) = 0 1 = 0, = f() = 6-6 f(0) = -6 < 0 ; f() = 1-6 = 6 > 0 f(0) = is local ma.; f() = = - is local min. VI-7

8 EX 5) About cost function C C(Q) Q 3 C(Q) = aq + bq + CQ + d What restrictions should be imposed on this cubic total cost function? C(0) = d fied cost d 0 Marginal cost 0 for all Q. C MC = = 3aQ + bq + C 0 Q It must be that MC 0 min Find minimum cost. MC FOC: = 6aQ + b = 0 Q VI-8

9 Q* = b. 3a SOC: MC Q = 6a For Q* to be the cost-minimizing output, a should be positive. * [If a < 0, Q is a locally cost-maimizing output level.] MC MC(Q) for some Q, MC < 0 Q If a > 0, MC b b min = 3a + b + c = 3a 3a 3ac b 3a MC min 0 3ac - b 0. In short, 3 restrictions are required. d 0; a > 0; 3ac b VI-9

10 [3] Conditions for Global Ma. and Min. Definition: If f() > 0 for all, f() is called conve function. If f() < 0 for all, f() is called concave function. Theorem: For a function y = f(), suppose that there eists * such that f(*) = 0. If f() > 0 for all, then f(*) is local min. and global min. If f() < 0 for all, then f(*) is local ma. and global ma. EX 1) y = f() = - + * FOC: Set f() = - = 0-1 = 0 = 1. SOC: f() = > 0 = 1 is the global minimum point. y = 1 is the global minimum of y. EX ) y = * FOC: Set f() = - + = = 0 = 1 SOC: f() = - < 0. = 1 is the global maimum point. y = 3 is the global maimum of y. VI-10

11 [4] Taylor s Series Short Digression n! = n (n-1) (n-) 1. EX: 8! = ! = = 4. 0! = 1. End of Digression Taylor Epansion: y = f(). Choose a point. Then, 0 f() = f( ) + 0 f ( 0 ) ( 1! 0 ) f ( 0 )! ( 0 ) f (n) ( 0 ) + + ( n! 0 ) n + R n (Remainder) Generally, R 0 as n. n EX: f() = a + b + c Choose = 0 0 f() = a + b f(0) = b VI-11

12 f() = a f(0) = a (4) f() = f () = = 0 f(0) = c. f ( 0 ) ( 1! 0 ) f ( 0 ) (! 0 ) a f( 0) + = c + b + = c + b + a = f(). Choose = : 0 f() = a + b f() = 4a + b f() = a f() = a a f() = (4a + b + c) + (4a + b)( - ) + ( - ) = a + b + c.! Theorem: (Lagrangean form of Remainder) f (n1) (p) R (n1)! ( n = 0 )n1, where p is between and 0. Implication: f (n) ( 0 ) f() = f( ) + f( )( - ) + + ( n! 0 ) n R n. Question: Choose n = 0; then f() = f( ) + f(p)( - ). Does such p eist? 0 0 Mean-Value Theorem: VI-1

13 Suppose that a function y = f() is differentiable. Then, there eists a value f() f( 0 ) p (between 0 and ) such that f(p) =. 0 <Intuitive Proof> f() y = f() f()-f( ) 0 f( ) 0 - X 0 Corollary 1: R = f(p)(- ), where p is between and <Proof> f() f( 0 ) f(p) = f() - f( 0) = f(p)( - 0) 0 f() = f( ) + f(p)( - ) R = f(p)(- ) Corollary : VI-13

14 f (n1) (p) R (n 1)! ( n = 0 )n1. Theorem: Consider a function y = f(). Suppose that (n) f( 0) = 0; f( 0) = 0;... ; f ( 0) 0. Then, the following is true: (n) f( 0) is local ma. if n is even and f ( 0) < 0. (n) f( 0) is local min. if n is even and f ( 0) > 0. (n) f( 0) is inflection point if n is odd and f ( 0) 0. <Sketch of the proof> Taylor s epansion around : 0 f() f( ) + f (n) ( 0 ) f (n) ( ( n! 0 ) n 0 ) f()-f( ) ( n! 0 ) n 0 0 (n) If n is even: f() - f( 0) 0 when f ( 0) > 0 0 is a local min. point. (n) f() - f( 0) 0 when f ( 0) < 0 0 is a local ma. point. If n is odd, either f() - f( ) 0 or f() - f( ) > EX1) y = f() = (7 - ) 3 f() = 3(7 - ) (-1) = -3(7 - ) Set f() = 0 = 7. 0 f() = -3 (7 - )(-1) = 6(7 - ) f( ) = 0 0 f() = -6 < 0. VI-14

15 = 7 is an inflection point EX) y = f() = (7 - ) 4 f() = -4(7 - ) 3 Set f() = 0. = 7. 0 f() = 1(7 - ) f(7) = 0 f() = -4(7 - ) f(7) = 0 (4) f () = 4 > 0 = 7 is the local min. point. The local minimum value of y = 0. VI-15

16 [5] Eponential Function (1) Eponential function: y = b, b > 1, R 0 When = 0, y = b = 1 As, y ; and as -, y 0 A popular choice of b = e (natural #) = y y=e Fact: e = lim 1 1 = <Proof> By Taylor epansion using the fact that de /d = e. See book (pp ). VI-16

17 Economic meaning of e: Suppose that a Bank pays annual interest rate = i. You deposit M o. Suppose the bank pays interests only once a year: M 1 = M 0 + im 0 = M 0(1 + i) Suppose the bank pays interests every 6 months. After 6 months, you have M 0 1 i. After 1 year, M 1 M 0 1 i 1 i M 0 1 i. Suppose the bank pays interests every 1 month, M 1 M 0 1 i 1 1. Suppose everyday, M 1 M 0 1 i If the interests can be continuously compounded, M 1 lim M 0 1 i. VI-17

18 Implication: If i = 1 (100% interest), M 1 = Moe. Since e =.718, 100% compounded interest rate = 171.8% one-shot interest rate. Theorem: 1 i i lim = e. <Proof> Let y = /i. Then, = yi. Note that as y,. lim 1 i lim 1 1 = y y iy lim y 1 1 y y i e i Implication: i M 1 = M0e Question: When interests are continuously compounded, how much money will you have t years later? Answer: i i i i M = M1e = (M0e )e = M0e 3i M 3 = M0e VI-18

19 it M t= M0e Future and present value of an asset: Let V t be the future value of your asset at time t. Let A be the current (present) value of your asset. it V t = Ae it -it -it Ae e = Ve t -it A = V t e Notation: e 3 f() e = ep(f()) = ep( + + 3) () Logarithm y = b, b > 1, R, yr+ monotonically increasing function can define inverse f n y = b = logby, y > 0 or y = b y = logb Usually, we use log ln(). e Graphical Relation between eponential and logarithmic functions VI-19

20 y y=b 45 1 y=log b 1 For y = log, < 1 y < 0 b = 1 y = 0 > 1 y > 0 Rules for logarithmic functions: ln(b) ln(b) Rule I: b = e. [b = e.] Rule II: logbb = 1. n Rule III: log = n log Rule IV: Rule V: b log (uv) = log u + log v b b b log b(u/v) = logbu - logbv log b ln() ln(b) b EX 1) ln() = e VI-0

21 EX ) log 17 = EX 3) log = log1010 = 3log1010 = 3 3 EX 4) log 10(3 10 ) = log log1010 = log EX 5) log 10 e log e e log e 10 1 ln10 EX 6) log 10 1 ln1 ln10 EX 7) Solve ab - c = 0 ab = c ln(ab ) = ln c ln a + ln b = ln c t rt EX 8) V = Ae (r = i) lnv lna r lnc lna lnb ln V = ln A + rt ln e = ln A + rt Conversions of base: 1) a = e. [Why? a = = e a =[e ] = e.] ln(a) e log e a ln(a) ln(a) ln(a) t ln t t ln EX 1) y = = (e ) = e t ln 3 t t ln 3 EX ) y = 3 = (e ) = e ) log b ln. lnb VI-1

22 EX1) y log t lnt. ln 7 lnt ln10 7 ln10 lnt 7 EX) y = 7log t = = ln10 (lnlnt) 10. Derivatives: 1) d dt (e t ) e t ) d dt (lnt) 1 t d dt (b t ) b t t tln(b) t tln(b) t 3) lnb [Why?: b = e d(b )/dt = e ln(b) = b ln(b).] 1 d dt (log b t) 1 t 4) [Why?: logbt = ln(t)/ln(b) d(logbt)/dt =.] tlnb ln(b) y 5) e, where y = f(t): d dt (e y ) d dy (e y ) dy dt y f(t) = e f(t) = e f(t). 6) ln(y), where y = f(t): d dt (lny) d dy (lny) dy dt dy 1 y dy dt =. dt y VI-

23 dy dt is called instantaneous growth rate. y EX 1) y = 1 1-t (1-t)ln(1) y = e. Let z = (1 - t)ln(1) z Then, y = e, z = (1 - t)ln(1). dy dy dz z (1-t)ln(1) (1-t) = e [-ln(1)] = -e ln 1 = -1 [ln(1)]. dt dz dt EX 3) y = log 17-1 y = [ln(17)] ln() dy d 1 [ln(7)]1 (1/). ln17 EX 4) y = ln(6 + 3) z = y = ln(z), z = dy d dy dz 1 dz d z (1 3) 1 3 = = (6 3) rt EX 5) V = Ae. Derive the instantaneous growth rate of V? ln V = ln A + rt dlnv growth rate = r. dt VI-3

24 EX 6) y = a e k-c. dy d a a1 e kc a e kc k a e kc a = k. dlny About : dln Let w = ln(y) and z = ln() dw 1 dz 1 dw = dy d ln y = dy; dz = d d ln = d. dy y d 1 dlny dln y dy dy 1 d y d y,. EX) Demand: Q = 1 P dq dp 100 P QP P Q dq dp P P = -1 P dlnq ln Q = ln ln p = dlnp 1 QP (A lot easier to compute). VI-4