3.2 Non-renewable esources A. Are socks of non-renewable resources fixed? eserves measures have an economic componen eg. wha could be exraced a curren prices? - Locaion and quaniies of reserves of resources like oil and gas are no known and can be expanded wih increased exploraion effor. - Base resource: he mass ha is hough o exis in he earh s crus - esource poenial: Tha which could be exraced given curren and expeced echnologies - World reserve base: esimaes of he upper bounds of resource socks (including reserves no ye discovered) ha are economically recoverable under reasonable expecaions of fuure prices, coss, and echnological possibiliies - eserves: quaniies ha are economically recoverable given presen coss and prices See Perman, p 59. Also he BP Saisical eview of World Energy, 24, has useful informaion on reserves versus producion for various non-renewable energy sources. B. A wo period model Two periods: period and period 1 A fixed sock of known size of a non-renewable resource Iniial sock a period is S. () is exraced in period. Inverse demand funcion: P = a b Demand goes o zero a price a. esource is eiher non-essenial or i possesses a subsiue. Ec 655, 3.2 Nonrenewable resources 1
P a-b Toal benefi o consumers from consuming : B( ) = ( a b) d = a b 2 2 (2.1) Exracions coss: C = c (2.2) Toal ne social benefi: NSB = B C (2.3) b NSB( ) = a c (2.4) 2 2 Assume he social uiliy in each period is equal o he ne social benefis in each period (a srong assumpion). Opimizaion problem: Ec 655, 3.2 Nonrenewable resources 2
MaxW 1 = + (2.5) 1 ρ, + 1 NSB NSB subjec o: + 1 = S (2.6) Se up he Lagrangian and opimize wih respec o and 1. Try his yourselves. We ge a wo period version of Hoelling s rule: ρ = ( P1 c) ( P c) ( P c) (2.7) An efficien exracion profile requires he ne price of he resource o grow a he social discoun rae. The opimal exracion program also requires ha he wo gross prices, P and P 1, saisfy he following: P = a b P = a b 1 1 + = S 1 ( P c)(1 ρ) = ( P c) 1 (2.8) These condiions will uniquely define he wo prices and quaniies of resources o be exraced for welfare maximizaion. Try problem 1 in Perman, p. 533. Ec 655, 3.2 Nonrenewable resources 3
C. A non-renewable muli-period resource model Noe on ransversaliy condiions wih a non-negaiviy consrain on he sock: T free; x( T) x min (i) x*( T) > xmin hen x(t) is free and λ(t)= or λ(t)=f (x(t)) (F is scrap value). O (ii) x*(t)=x min, hen x(t) is fixed and λ(t) is free. In summary: λ( T ), xt ( ) xmin, λ( T)[ x( T) x ] = min Also noe ha for infinie horizon problems for he ransversaliy condiions, we ake he limi as T. Ec 655, 3.2 Nonrenewable resources 4
Assume: - a finie sock of he resource - a social planner - i coss nohing o exrac he resource - demand funcions: () = D( P()) (2.1) 1 ( ) = ( ( )) inverse demand funcion P D Flow of social welfare a a poin in ime: Objecive: Subjec o: s= () W () = Psds ( ) (2.11) T s= ρ max J= e W( ) d T () ρ = e p() s ds d S = () S() = S S (), () Find he curren valued Hamilonian. We use λ o denoe he curren valued cosae variable in his secion. (2.12) (2.13) FOC s: () (2.14) H = P() s ds+ λ( ()) Ec 655, 3.2 Nonrenewable resources 5
H = P ( ( )) λ( ) = H λ λρ = = S λ = λρ (2.15) (2.16) Solving Equaion (2.16): λ() λ() e ρ = (2.17) From Equaion (2.15), This is Hoelling s rule again. Transversaliy condiions: P() = P() e ρ (2.18) (2.19) P() = ρp() λ ( T) S( T) = (2.2) HT ( ) = (2.21) We can conclude ha S*(T)=. I is opimal o exhaus he resource. Proof by conradicion: Suppose S*(T)>. Then holding T consan, we could have increased consumpion *(T), and his would make social welfare higher since here are no cocs o consumpion. Thus i canno be opimal o leave S*(T)>. Hence S*(T)=. To find he opimal T, we wan he Hamilonian a T o be exacly zero: Ec 655, 3.2 Nonrenewable resources 6
HT ( ) = T ( ) Psds () λ( TT ) ( ) = (2.22) Using he FOC s we can wrie: T ( ) P( sds ) = PT ( ) T ( ) (2.23) Wha happens o P as he resource is exhaused? Two possibiliies: P Case I P() Ec 655, 3.2 Nonrenewable resources 7
P Case II P() lim P( ) Case I: = ; *(T)=. If we are o finish wih a zero rae of exracion, price mus rise infiniely high. This will ake an infinie amoun of ime. Our opimal ime horizon is T. lim P( ) K Case II: = where K is a choke price. (T)=, P*(T)=K. There will be a finie erminal ime. A backsop echnology creaes a ceiling on he price of he non-renewable resource. Case I: T* P( T) = P() e P() = P() e S*( T) = ρt ρ (2.24) () = S (2.25) Ec 655, 3.2 Nonrenewable resources 8
Equaion (2.25) follows because oal consumpion mus equal he sock of he resource. This allows us o solve for a paricular P()*. How? We know ()=D(P(T))=D(P()e ρ ). Subsiue ino Equaion (2.25) gives: ρ DP ( () e ) d= S (2.26) This is one equaion in one unknown. I may be difficul o solve could be highly nonlinear. Case II: P( T) = K P() = P() e S*( T) = ρ (2.27) We don know T or p(). We can use he firs wo equaions of Equaion (2.27) o solve for P(). Then we solve for T* using: P() Ke ρt = (2.28) T () d= S (2.29) () = D( P()) P () = Ke ρ ( T ) ρ ( T ) () DKe ( ) = (2.3) We can subsiue for () in Equaion (2.29) and solve for T*. This gives us he ime o swich o he backsop echnology. Ec 655, 3.2 Nonrenewable resources 9
Example Suppose he inverse demand curve is: P Ke a = (2.31) Noe ha K is he choke price. P K U() = shaded area Ke -a Quaniy of resource exraced, Figure 15.2 A resource demand curve, and he oal uiliy from consuming a paricular quaniy of he resource. Solve for P*(), *(), T*, *(). Ec 655, 3.2 Nonrenewable resources 1
Figure 15.3 Ne price P P T =K Demand P P 45 T Time Area = S = oal resource sock T Time Ec 655, 3.2 Nonrenewable resources 11