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Continuous time formalism

Continuous time formalism. Exhaustible Problem. Max  (e -rt p(t) h(t)) s.t. dx/dt = -h Hamiltonian: Objective function + multiplier * rhs of diffeq. H = e -rt p(t) h(t) – l h. Rules. Exhaustible Resource. Since x >0 it can’t be that h=inf for any measurable length of time.

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Continuous time formalism

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  1. Continuous time formalism

  2. Exhaustible Problem • Max  (e-rt p(t) h(t)) s.t. dx/dt = -h • Hamiltonian: • Objective function + multiplier * rhs of diffeq. • H = e-rt p(t) h(t) – l h

  3. Rules

  4. Exhaustible Resource

  5. Since x >0 it can’t be that h=inf for any measurable length of time. • If h=0 forever then x doesn’t go to zero, so by transverality l must = 0, and by costate it is always zero, which can’t be greater than or equal to pe-rt unless p is zero. • It must be that at least for long enough to exhaust h = any and l = pe-rt

  6. During the time h = any, l = e-rtp • By costate: -re-rtp + e-rt dp/dt= 0 • Or dp/dt = rp • Which is Hotellings rule.

  7. Plausibility for costate • Let length of time between periods be n. We will make n small and see what happens to costate equation in discrete time problem.

  8. Steady State Fish • State equation—hold constant for n seconds • xt+n = xt +f(xt)n - htn. • U(x,h) = St(1+r)-t pt ht n • here take p as constant • L(x,h,l) = St(1+r)-t pt ht n • +St lt (xt +f(xt)n - htn -xt+1*n) • Lx = lt(1+f’(xt)n ) - lt-1*n=0

  9. lt(1+f’(xt)n ) - lt-n=0 • lt - lt-n= - ltf’(xt)n • Limn->0 (lt - lt-n)/n = - ltf’(xt) • dl/dt=-lf’ • Hamiltonian: H = pe-rth + l(f(x)-h) • dH/dx = - lf’ = dl/dt Costate equation!

  10. Finish Fish • H = pe-rth + l(f(x)-h) • restrict h to [hmin, hmax] • maxh H implies • pe-rt – l = 0 and h = any • pe-rt – l < 0 and h = hmin • pe-rt – l > 0 and h = hmax

  11. Exceptional Control • Suppose pe-rt – l = 0 and h = any • hence -r pe-rt – dl/dt = 0 • hence -r pe-rt + lf’ = 0 • hence -r l+ lf’ • hence r= f’(x*) • So the only way to have an interval with h not at hmin or hmax is for x to be x*.

  12. Suppose that x(0) is > x*. • Set h = hmax until x = x* is an optimal control • x above x* means that f’(x) < f’(x*) so • dl/dt > - lr • when x = x*, pe-rt – l • so l falls slower than pe-rt it must have started below it, which is what is needed for h=hmax

  13. Optional: At home finish this up—show what happens if x(0) < x*, a moratorium. Also show that once one gets to x* it is indeed optimal to stay there. (h = f(x*) so dx/dt = 0.

  14. Note that the “moratorium” is a function of the linearity of the problem in h. • Alternative-- monopoly: max  (p(h)*h e-rt) • Try with p= h-a and compare competition and monopoly • Competition for renewable resource. Berck Jeem 1981

  15. Competition Sketch • max int( p(t)h(t)e-rt s.t. dx/dt = f(x) – h • H = p(t)h(t)e-rt + l(f(x) – h) • Costate:

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