This document was generated at 7:34 PM, 07/27/09 Copyright 2009 Richard T. Woodward

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1 his documen was generaed a 7:34 PM, 07/27/09 Copyrigh 2009 Richard. Woodward 15. Bang-bang and mos rapid approach problems AGEC Summer 2009 here are some problems for which he opimal pah does no involve a smooh approach o he seady sae or gradual changes over ime. wo imporan classes of such problems are known as "bang-bang" problems and mos rapid approach problems. In such problems he consrains play a cenral role in he soluion. I. Bang-bang OC problems A. Bang-bang example #1: A sae variable consrain Consider he following problem in which we seek o maximize discouned linear uiliy obained from a nonrenewable sock (someimes referred o as a cake-eaing problem): max x z 0 = z e 0 ( 0) = x0 r z d s.. x Wha does inuiion sugges abou he soluion o he problem? Will we wan o consume he resource sock x gradually? Why or why no? Le's check our inuiion. Following he framework laid ou in he previous lecure, we se up he Lagrangian by adding he consrain on he sae variable o he Hamilonian, i.e., L=H+φ(consrain). Using he curren-value specificaion, his give us L = z z + φ x he FOCs for he problem are: L = 0 : 1 = 0 z L = r ɺ : φ = r ɺ x he firs of hese implies ha =1. Since his holds no maer he value of, we know ha ɺ = 0 for all. he wo condiions boil down o =1 and φ =r. he second of hese is mos ineresing. I shows us ha φ, he Lagrange muliplier, is always posiive. Wha does his ell us abou he consrain on x and, herefore abou he answer o he problem? Yes, i implies ha he consrain is always binding, x =0 always. Bu hold on! We know his isn' acually rue a =0. However, a =0, x is no variable i is parameric o our problem, so ha poin in ime doesn coun. Bu a every insan excep he immediae saring value, x =0.

2 15-2 So how big is z a zero? he firs hough is ha i mus equal x 0 bu his isn' quie righ. o see his, suppose ha we found ha he consrain sared o bind, no immediaely, bu afer 10 seconds. o ge he x o zero in 10 seconds, z per second would have o equal x 0 /10. Now ake he limi of his a he denominaor goes o zero z goes o infiniy. Hence, wha happens is ha for one insan here is a spike of z of infinie heigh and zero lengh ha pushes x exacly o zero. his ype of soluion is known as a bang-bang problem because he sae variable jumps disconinuously a a single poin BANG- BANG! Since, in he real world i's prey difficul o push anyhing o infiniy, we would ypically inerpre his soluion as "consume i as fas as you can." his is formalized in he framework of mos-rapid-approach pah problems below. B. Bang-Bang Example #2 (based on Kamien and Schwarz p. 205) A conrol variable consrain Le x be a producive asse ha generaes oupu a he rae rx. his oupu can eiher be sold or reinvesed. he porion ha is reinvesed will be called z so [1-z ] is he porion ha is sold. We assume ha he ineres can be consumed, bu he principal canno be ouched. 1 Our quesion is, Wha porion of he ineres should be invesed and wha porion should be consumed over he inerval [0,]? Formally, he problem is: max 0 z x z [ 1 z ] 0 = z rx 1 ( 0) = x0 rx d s.. his ime we have wo consrains: z 1 and z 0. Hence, our Lagrangian is L = 1 z rx + λz rx + φ 1 z + φ z [ ] 1 2 So ha he necessary condiions are L = 0 rx + λrx φ1 + φ2 = 0 z L = ɺ λ ɺ λ = [ 1 z ] r + λzr x he ransversaliy condiion in his problem is λ =0 since x is unconsrained wih he Kuhn-ucker condiions, K 1 : φ 1 0 & φ 1 (1 z )=0, and K 2 : φ 2 0 & φ 2 z=0. From he K 1, we know ha if φ 1 >0, hen he firs consrain binds and z =1. Similarly, from K 2, if φ 2 >0, hen he second consrain binds and z=0. i.e. 1 his problem is very similar o one looked a in Lecure 3. Comparing he wo you ll see one key difference is ha here uiliy is linear, while in lecure 3 uiliy was logarihmic.

3 15-3 φ 1 >0 z = 1 φ 2 >0 z = 0. φ 1 =0 z < 1 φ 2 =0 z > 0. Clearly, i is no possible for boh φ 1 and φ 2 o be posiive a he same ime. he firs FOC can be rewrien ( λ 1) rx φ1 + φ2 = 0. We know ha rx will always be posiive since consumpion of he capial sock is no allowed. Hence, we can see ha hree cases are possible: 1) if λ=1 φ 1 =0 φ 2 =0 no consrain binds 2) if λ>1 φ 1 >0 φ 2 =0 z =1 3) if λ<1 φ 1 =0 φ 2 >0 z =0. From he second FOC, ɺ λ = z r + λ z r. {[ 1 ] } Since everyhing in he brackes is posiive, he RHS of he equaion is negaive λ is always falling. By he ransversaliy condiion we know ha evenually λ mus hi λ =0. Hence, evenually we'll reach case 3 where, λ <1 and z =0 and we sell all of our oupu. he quesion is, When do we sar selling righ away or afer x has grown for a while? We know from equaion 2 ha a λ =1 neiher consrain binds. Suppose ha a =n λ =1. For <n λ >1 and z=1. For >n λ <1 and z=0. An imporan quesion hen is when is n? We can figure his ou by working backwards from λ =0. From he second FOC, we know ha in he final period, (when λ <1) z=0, in which case ɺ λ = r. Solving his differenial equaion yields λ = r + A. Using he ransversaliy condiion, λ = r + A = 0 A = r λ = r + r = r Hence, λ n =1 if r n = 1 = ( 1) n r r

4 15-4 Hence, we find ha he opimal sraegy is o inves everyhing from =0 unil = n = r 1 r. Afer =n consume all of he ineres. If ( r 1 ) r < 0 hen i would be opimal o sell everyhing from he very ouse. For ( r 1 ) r > 0, we can graph he soluion: n Z X λ Wha would be he soluion as? Does his make inuiive sense? Wha is i abou he specificaion of he problem ha makes i inconsisen wih our economic inuiion? II. Mos Rapid Approach Pah problems Anoher ype of problem commonly found in economics is mos-rapid-approach pah problems (MRAP). 2 Here, he soluion o he problem is o ge as quickly as possible o he opimal seady sae where benefis are maximized. Consider he firs example, wouldn a MRAP ype soluion be more inuiively appealing han he bang-bang soluion? A. MRAP example (Kamien & Schwarz p. 211) A very simple firm generaes oupu from is capial sock wih he funcion f(x ) wih he lim f ' x =. he profi rae, herefore, is propery ha x 0 π = p f x c z where x is he firm's capial sock and z is invesmen, p and c are exogenously evolving uni price and uni cos respecively. he capial sock ha sars wih x(0)=x 0, depreciaes a he rae b so ha = z bx. he firm's problem, herefore, is o maximize he presen value of is profis, 0 e r [ p f ( x ) c z ] = z bx, d subjec o wih he addiional consrains: i) x() 0 ii) z 0 2 Someimes he erm bang-bang is also used o describe MRAP problems.

5 15-5 iii) p f ( x ) c z 0 Le's use economic inuiion o help us decide if we need o explicily include all he consrains in solving he problem? he consrain on x almos cerainly does no need o be imposed because as long as f' ges big as x 0, he opimal soluion will always avoid zero. he consrains on z, on he oher hand migh be relevan. Bu, we'll sar by assuming ha neiher consrain binds, and hen see if we can figure ou acual he soluion based on he assumed inerior soluion or, if no, we'll need o use he Kuhn- ucker specificaion. Noe ha if here does exis a seady sae in x, hen, as long as b>0, z mus be greaer han zero. Hence, we anicipae ha much migh be learned from he inerior soluion. Similarly, he profi consrain migh also bind, bu we would expec ha in he long run, profis would be posiive. So again, we sar by solving for an inerior soluion, π = p f x c z. assuming π>0 where B. he inerior soluion he curren value Hamilonian of he problem (assuming an inerior soluion w.r.. z and x wih π>0) is H c = p f ( x ) c z + ( z bx ) he necessary condiions for an inerior are: H c = 0 c + = 0 z ( x ) H f c = r ɺ p b = r ɺ x x Over any range where he consrains on z do no bind, herefore, we have c = and, herefore, i mus also hold ha c ɺ = ɺ. Since changes in c over ime are exogenous, his means ha changes in are similarly exogenously deermined. Subsiuing c for and rearranging, he second FOC becomes f ( x ) 1. p = ( r + b) c cɺ x over any inerval where z>0. (remember, p, c and cɺ are exogenous). We see, herefore, ha he opimum condiions ell us abou he opimal level of x, say x *. We can hen use he sae equaion o find he value of z ha mainains his relaion.

6 15-6 o see his more clearly, consider he case where c and p are consan. In his case he pf '( x) capial sock will be held a a consan level and 1 reduces o r + b = c, which is an known as he modified golden rule. Le's hink abou his condiion for a momen. In a saic economy, he opimal choice would be o choose x where he marginal produc of increasing x is equal o he marginal cos, i.e., where pf ' = c. In an infinie-horizon economy, if we could increase x a all poins in ime his would pf ' have a discouned presen value of. However, since he capial sock depreciaes r over ime, his depreciaion rae diminishes he presen value of he gains ha can be obained from an increase in x oday, hence he presen value of he benefi of a pf ' marginal increase in x is. r + b If p and c are no consan bu grow in a deerminisic way (e.g., consan and equal inflaion) hen we could de-rend he values and find a real seady sae. If p and c boh grow a a consan rae, say w, hen here will be a unique and seady opimal value of x for all z>0. C. Corner soluions All of he discussion above assumed ha we are a an inerior soluion, where 0 < z < p f ( x ) c. Bu, we ended up finding ha he inerior soluion only holds when he sae variable x is a he poin defined by equaion 1. Hence, if we're no a x * a =0, hen i mus be ha we're a a corner soluion, eiher z =0 or p f ( x ) c z = 0. If x 0 >x * hen i will follow ha z will equal zero unil x depreciaes o x *. If x 0 < x * hen z p will be as large as possible f ( x ) = z unil x * is reached. c Hence, economic inuiion and a good undersanding of he seady sae can ell us where we wan o ge and how we're going o ge here in he mos rapid approach possible. D. Some heory and generaliies regarding MRAP problems he general principles of MRAP problems are discussed by Wilen (1985, p. 64) Spence and Sarre show ha for any problem whose augmened inegrand (derived by subsiuing he dynamic consrain ino he original inegrand) can be wrien as A ( K Kɺ ) = M ( K ) + N( K )Kɺ Π, he opimal soluion reduces o one of simply reaching a seady sae level K=K * as quickly as possible.

7 15-7 Where K is he sae variable and by "inegrand" hey mean he objecive funcion, profis in he case considered here. How does his rule apply here? he inegrand is f ( x ) c z bx +ɺ x = z, he inegrand can be wrien ( x ) c ( bx + ) = p f ( x ) cbx c p f. Convering his o he noaion used by Wilen, M K = p f x c bx and N K K ɺ = c x ɺ. Hence his problem fis ino he general class of MRAP problems. III. References p. Using he sae equaion Kamien, Moron I. and Schwarz, Nancy Lou Dynamic Opimizaion : he Calculus of Variaions and Opimal Conrol in Economics and Managemen. New York, N.Y. : Elsevier. Wilen, James E Bioeconomics of Renewable Resource Use, In A.V. Kneese and J.L. Sweeney (eds.) Handbook of Naural Resource and Energy Economics, vol. I. New York: Elsevier Science Publishers B.V. Spence, Michael and David Sarre Mos Rapid Approach Pahs in Accumulaion Problems. Inernaional Economic Review 16(2):