Two-Period Renewable Resources ModelWith Non-Zero Interest Rate• suppose we want to maximize net present value (NPV)• suppose that there is not open access to the resource• do not assume, prior to solving the model, that we are in steady-state.-- it may or may not be the solutionIn general, we would solve the problem over the "planning horizon" ofour firm or agency.A planning horizon is the number of time periods into the future forwhich the firm or agency makes plans.To keep the mathematics manageable, we will focus on a two timeperiod planning horizon.If demand is very low or harvesting costs are very high, then stock mayremain at the end of the planning horizon.A salvage value function gives the value of any remaining stock atthe end of the planning horizon.Note: salvage value is a function of stock, not harvest. In the case of afishery, what would be sold at the end of the planning horizon is notcaught fish (i.e., not "X"), but rather the right to catch the remaining fishin the sea, (i.e., the resource asset "S").Two-period Model (cont.)initial stock stock at beginning final stock(stock at beginning of period 1 (stock at endof period 0) of period 1)------+---------------------------------+------------------------+----> time So S1 S2 | | | X0 X1Definitions:• t: time period t = 0,1,2• St: resource stock at time t• g(St): the growth function of the resource stock• Xt: harvest at time t• B(Xt): total benefits from harvest at time t• C(Xt, St): total costs of harvest at time t• F(S2): salvage value function• r: interest rateThe objective is to maximize the net present value of harvest in period0, harvest in period 1 and salvage value in period 2. The choicevariables are X0, X1, S0 and S1.maxX0,X1,S1,S2NPV = B(X0) −C(X0,S0) +B(X1)−C(X1,S1)1+ r+F(S2)(1+ r)2net benefit discounted net benefit discounted "salvageperiod 0 period 1 value" of final stocksubject to:(1) g(S0)= S1− S0+ X0, equation of motion between periods 0 and 1(2) g(S1) = S2− S1+ X1, equation of motion between periods 1 and 2Also, So is the given, initial stock.Two-period Model (cont.)The Lagrangian expression for this problem is:maxX0,X1S1,S2λ0,λ1L = B(X0) −C X0,S0( )+B X1( )−C X1,S1( )1+r+F(S2)(1+ r)2+λ0g(S0)− S1+S0− X0( )+λ11 + rg(S1)− S2+S1−X1( )where λ0 and λ1 are Lagrange multipliers.FOC's(1)dLdX0= BX0(X0) −CX0X0,S0( )−λ0= 0-- MB(0) - MC(0) - user cost(0) = 0(2)dLdX1=BX1(X1)1+ r−CX1X1,S1( )1 + r−λ11+ r= 0-- NPV(MB(1)) - NPV(MC(1)) - NPV(user cost(1)) = 0(3)dLdS1=−CS1X1,S1( )1 + r− λ0+λ11 + rdgdS1+1      = 0rearranging : λ0=−CS1X1,S1( )1+ r+λ11+ r+λ11+ rdgdS1      -- user cost = NPV(increase in + NPV(user cost + NPV (user cost in period 0 harvesting cost) associated with of having lesshaving one less growth inunit of stock period 1Two-period Model (cont.)FOCs: (4)dLdS2=FS2(S2)(1+ r)2−λ11+r= 0-- NPV (marginal salvage value) - NPV (user cost(1)) = 0(5)dLdλ0= g S0( )− S1+ S0− X0= 0-- equation of motion between periods 0 and 1 must be satisfied.(6)dLdλ1= g S1( )−S2+ S1− X1= 0-- equation of motion between periods 1 and 2 must be satisfied.• keeping a unit of stock unharvested has three effects:(1) Loss of interest from not harvesting the stock today.(2) Savings in extraction cost, because Cs(Xt,St) < 0.(3) Additional growth of the resource stock. The present value ofthe additional growth in stock is λ11 + rdgdS1      .--Note: dg/dS can be positive or negative depending on whether S is < or > maximum sustainable stockTwo-period Model (cont.)In steady state, the shadow value of the resource remains constant:λt+1 − λt = 0and MC of delayed harvest (lost interest) = MB of delayed harvest(growth + cost savings): rλt = Xtcs(St) + gs(St) λtThe marginal benefit of delaying the harvest is the sum of reducedharvesting cost (Xt cs(St)) plus the value of the added growth of thestock, gs (St) λt. Thus, at steady state,r − gs(St)[ ]λt= Xtcs(St).(7)Let Pt be optimal price, i.e., Pt = BX(Xt):At an optimal resource allocation, from FOC (2):B (X ) P c(S )X t t t ttuser costextraction cos= = +λ(8)Since at steady state,St+1 − St = 0 and λt+1 − λt = 0, then Pt+1 − Pt = 0.Furthermore, with optimal prices, the steady state equation forλt+1 − λt = 0 can be expressed as a function of S and X by re-arranging equations (7) and (8).[r −gs(S)] [B(X) −c(S)] + Xcs(S) = 0.Phase Plane AnalysisPhase plane analysis is a graphical method of analyzing the dynamicbehavior of a bioeconomic system. It is useful for determining whethera bioeconomic system will be in steady-state, will drive the resource toextinction, or will cycle.The curve below gives all the points at which "the biology is in steadystate," i.e., where the stock will not rise or fall. This curve is simply thegrowth function g(S) that we have seen before. On the phase plane it islabeled "dS/dt = 0", i.e., the change in stock over time is zero.The arrows in the figure show that, if harvest is above the dS/dt = 0 line,i.e., if harvest is above steady-state harvest, then stock will fall, and ifharvest is below the dS/dt = 0 line, stock will rise. The line, dS/dt = 0,is commonly referred to as an iso-cline.Phase Plane Analysis (cont.)The second curve gives all the points at which "the economics is insteady state," i.e., where the economic agent has no incentive to eitherincrease or decrease harvest X. The second curve is derived from thefirst order conditions when assuming the system is in economic steadystate.On the phase plane the second curve is either "dX/dt = 0," i.e., thechange in harvest over time is zero, or "dλ/dt = 0," the change in(undiscounted) user cost over time is zero.The arrows in the figure show that, if stock is to the right of the dX/dt =0 line, i.e., if stock is high, fish are easy to find in the ocean andharvesting costs are therefore low, then harvest will rise, and if stock isto the left of the dX/dt = 0 line, harvest will fall in order to equate MB =MC of fishing.Phase Plane Analysis (cont.)Putting the two steady-state curves together, we get:Phase Plane (cont.)The curve, X = g(s), denotes all X, S combinations that lead