Key Terms and Components of Dynamic Systems: Dynamic Systems: Systems that contain time as a parameter; suchsystems "evolve" over time.State Variable: A state variable describes the status, or "state ofbeing," of one of the variables in the system.Initial Conditions: Values that the state variables take on at thebeginning of the time period of interest.Control Variable: A control variable is a variable that is under thecontrol of some individual or group.Random variables (noise variables): Uncontrolled variables whichcan assume several values with certain probabilities.Constraints: Equations (or inequalities) which limit the values thatstate variables or control variables can take on.Equation of Motion: The equation of motion describes how a variablechanges over time.Solution of a System: The solution of a dynamic system is a set ofequations, where the equations are in terms of the system parameters,including time, such that all of the original equations in the system aresatisfied. Thus, a dynamic system may have many solutions, dependingon the specific initial conditions of the resource.Objective Function: An objective function is an equation thatmeasures how well the system is attaining some goal or objective,usually expressed in terms of the state variables, control variables,and parameters of the system.Example: Set-up for Natural Resource Dynamic SystemState variables:(St): Denotes the level of a stock at time t; (e.g., the quantity of waterstored in the reservoir behind the dam at time t).(Ut): Uncontrolled inputs, (e.g., rain, snow).(Yt): Outputs; outcome of systems at time t; (e.g., crops produced)Control variables:(Xt): Inputs whose magnitudes we can choose in our attempt to reachour objectives. (e.g., the amt. of water used for irrigation).Parameters: (P): Items that can be taken as constant with respect to the problem athand. (e.g., the production elasticity of irrigated water)Equation of motion:Next period water stock = This period water stock + rainfall -irrigation water:St+1 = St + Ut - XtObjective Function:max[ ( )] ( )( )Xt t t tttTtNPVB Y P C Xr=−+∑=10Dynamic Models of Nonrenewable ResourcesNonrenewable resources are resources that have a finite stock andthat do not grow naturally.Key Issues:• Determining optimal resource allocation and pricing.• Sources of market failure and policies to correct market failure. t = time (the initial period: t=0; the future period: t=1)r = interest rateS0 = initial stock of nonrenewable resourceXt =control variable, the amt. of the resource consumed in period tB(Xt) = benefit of consuming Xt Economics of scarcity:• Scarcity: Imposes an opportunity cost on using resources today. Ina natural resource system, we refer to dynamic opportunity cost as auser cost.• User Cost: The Present Value of foregone opportunity. (e.g., if youuse a unit of a natural resource today, you forego the opportunity touse it tomorrow)Nonrenewable Resources (cont.)The User Cost decreases as r increases:• The higher the interest rate, the less valuable tomorrow’s benefits andthe smaller the opportunity cost of using more of the resource today.• at r = infinity, resources left for tomorrow are worth nothing and usercost = 0.• Similarly, when there is enough of the resource to go around, so thatscarcity is not an issue, the user cost = 0. The dynamic model yieldsthe same outcome as two separate static models.Discounting: The use of discounting is important in determining theoptimal extraction rate of a nonrenewable resource, because therevenue a resource owner receives in period 1 is not worth as much asthe revenue received in period 0.• the NPV of benefits in period 1 in terms of the current period 0: NPV = 111+↵√rB X( )Dynamic Efficiency: An allocation of resources is said to bedynamically efficient when it maximizes the NPV of benefits.Max. L = B(X) - C(X),• B(X) is now a stream of benefits through time, B X BrBrBrBNN( ) ...= ++ ++ + ++0 122111111• C(X) is now a stream of costs through timeDynamic Efficiency: The Two Period Caseassume zero costs are associated with consuming the resource.Objective function: MaxX0,X1NPV = B X0( )+11+ rB X1( ).Equation of motion (constraint): S0= X0 + X1.Note: by assuming X0 + X1 exactly equals S0 (resource stock is usedup), we are implicitly assuming unsatiated demand.the optimization problem is:MaxX0,X1NPV = B X0( )+11+ rB X1( )subject to: S0= X0 + X1..The Lagrangian expression is:L = B X0( )+11 + rB X1( )+ λ ⋅ S0− X1− X0( ).To maximize the Lagrangian expression we find the F.O.C.'s:(1)dLdX0= BxX0( )− λ = 0(2)dLdX1= BxX1( )11+ r      −λ = 0(3)dLdλ= S0− X1− X0= 0Two-period Dynamic Efficency (cont.)The system can be solved for X0, X1 and λ in terms of the parametersof the system. An often useful step in this process is to set FOC (1) =FOC (2) and eliminate λ to obtain:(4)BxX0( )=11+ rBxX1( )• then use (3) and (4) to solve for X0 and X1, and• substitute X0 into (1) to find λ.We can find P0 and P1 by recalling that:(5) Bx(Xt) = MB of X at time t = Price at time t = PtRearranging (4), we get: (1+ r)⋅Bx(X0) = Bx(X1)Substituting P0 for Bx(X0) and P1 for Bx(X1), we find: P PPr1 00−=Two-period Dynamic Efficency (cont.)Conclusions:• when dynamic efficiency is met, the price increases at the rate ofinterest. • the shadow price of S0, λ , is equal to P0. the shadow value is alsoequal to the present value of P1. In other words, λ = P0 =P1/(1+r). Thus, the solution to the nonrenewable resource problemequates the NPV of benefits across all time periods in the horizon• If P0 > P1/(1+r), the owner should extract more today; invest themoney at r.• If P0 < P1/(1+r), the owner should leave more in the ground toextract tomorrow• the rate of return of holding resource stock in the ground is: IRR > r.• Therefore, in equilibrium, it must be the case that P0 = P1/(1+r).-Produce today until MB0 = PV(MB1)Note: The intuition for λ is that, λ = the user cost of the resource!The solution to the dynamic problem equates the user cost of extractingthe resource across all time periods.A Numerical example:Suppose B(X) = a Xthen Bx(X)=a2 x.noting that X1 = S0 - X0 from (3),X0can be found by using Bx(X) with eqn's