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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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exhaustible problem
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
slide5
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
slide6
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.
plausibility for costate
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.
steady state fish
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
slide9
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!
finish fish
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
exceptional control
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*.
slide12
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
slide13
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.
slide14
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
competition sketch
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: