Optimal rotation age (ORA) • Dynamic optimization problem • Long discussed in economic literature • Shorter rotation – benefit arrives earlier – earlier replanting opportunity – planting more frequent – timber yield is lower Forest scientist's ORA • doesn't like cutting down trees … • Maximizes sustained gross yield Max[f (T) / T] T • Solution f (Tg ) f '(Tg ) Tg "Economic" forest scientist's ORA • takes into account planting cost • maximizes sustained net yield Max[f (T) / T (W / P) L / T] T • Solution f (Tn ) (W / P) L f '(Tn ) Tn Forest economists' ORA • Maximize profit • Economic Literature: – Maximizing present discounted value over one cycle (Von Thünen, Irving Fisher) – Maximizing internal rate of return (Boulding) – Maximizing present discounted value over infinite cycles Max NPV over 1 Period Max[e r T T f (T) (W / P) L] FOC (r) e r T1 f (T1 ) e r f '(T1 ) / f (T1 ) r T1 f '(T1 ) 0 Max Internal Rate of Return ri r e rT f (T) (W / P)L 0 ri T ln f (T) P /(W L) 1 Max[T ln f (T) P /(W L) ] 1 T FOC f '(Ti ) / f (Ti ) T ln f (Ti ) P /(W L) 1 i Max NPV over all Periods Max[ P f (T) e rT W L 1 e Max[ P f (T) e rT W L / 1 e T T rT rT e ] FOC f '(T ) r f (T ) r P f (T ) e rT W L / 1 e rT rT 2 ... ] Optimal Rotation Age Ti < T < T1 < Tg < Tn ORA - Assumptions • future prices, wages, interest rates are known • future technologies (yields, input requirements) are known • growth rate initially increasing later decreasing (I.e. cubic growth function) ORA - Example f (t) = b*t^2 + a*t^3 a = -1/800 b = 0.2 W[age] = 16 L[abor] = 25 P[rice] = 20 r = 0.06 = 6% Growth Function Timber f(t) 700 600 500 400 300 f (t) = 0.2 t2 - 1/800 t3 200 100 20 40 60 80 100 Time (t) 120 140 ORA Example, Results (see Mathematika output) Maximization Criterion Optimal Rotation Age Gross sustainable yield 80.00 Net sustainable yield 81.21 Net present value of one period Net present value of all periods Internal rate of returns 29.56 28.62 26.94
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