Natural Resources

Future value diminshes so discount rate (= to delta or d) is the opportunity cost of waiting (less now, more later)

FV/PV ^ 1/n(T)

Also called a perpetuity. A[(1+d)/d] where d = discount rate

[(1+d)/d]A(1-[1/((1+d)^T+1)]

sum of (1/1+d)*Nt

F'(X) + deriv(profit)/derivX / deriv(profit)/derivY = discount rate

Basically: Market return = Nature return. F'(X) is the marginal net growth rate. deriv(profit)/derivX / deriv(profit)/derivY is the marginal stock effect and measures marginal value of the stock relative to marginal value of harvest. These sum to d, which is the resources' rate of return for maintaining the stock at X*.

Harvest = Net Growth

Xt+1 - Xt = F(Xt)-Yt

Xss = k(r - a)

max profit = PH(E) - CE, where P = price, H(E) is harvest as function of effort, C = cost, E = effort

max profit = P ei/E * H(E) - Cei, where ei is individual effort

Profit is zero!

F(X) = rX(1-X/K), H(E,X) = qEX. F(X) is growth, r is intrinsic rate of growth, X is stock size, K is carrying capacity.

E = Einfinity, cost of effort is being covered but there is nothing left to pay the other important factor of production: stock. Access / rent = free, fish stock is reduced until worthless.

Maximum sustainable yield: traditional method that maximizes revenue. Max F(X) s.t. F(X) = H(E,X)

Maximum economic yield, maximizes profit. Max profit = PH(E,X) - C(E) s.t. F(X) = H(E,X). Remember H = qEX

Open access. Max F(X) s.t. F(X) = H(E,X) and profit = zero. TC = TR.

Less than or equal to that of effort for MSY

Greater than or equal to stock size MSY and OA

Greater than effort for MEY and MSY

Greater, equal, or less than MEY but less than MSY

1. Intrinsic growth rate r > alpha so Xss > 0, and 2. Convergence

Maximum annual increment. Max Q(t)/t, get Q'(t) = Q(t)/t or marginal volume over time equal to average annual volume per acre

Additional benefit of reduced cost of fishing / effort from more stock

Max profit(t) = PQ(t)e ^ -dt (discount rate * time). Result is PQ'(t) = dPQ(T).

PQ'(t)

dPQ(t)

Optimal rotation model. Max profit(t) = [PQ(t) - C]/ [(e^-dt)-1] and get PQ'(t) = d(PQ(t)-C) + d(profit)

Opportunity cost of delaying harvest (faustmann rotation). You forego interest payments on net revenue & interest of reseeded land.

P or d increase: rotation time decreases and supply increases. C increase: rotation time increases and supply decreases.

Short run only looks at change in Q(T)/T in response to T changes. When T decreases, rotation is shorter and (now) overmature timber gets cut so supply increases. When T increases, rotation is longer which delays cutting so supply decreases.

P or C increase: ambiguous. d increase: supply decreases.

Long run looks at change in Q(T)/T AND land use in response to T changes. Same short run scenario + land considerations with d(profit). When P increases, supply decreases and then increases if more land is turned to forestry = ambiguous. When C increases, supply increases and then decreases with d(profit) = ambiguous. Only for sure is d(profit).

1. any resource left behind is worthless
2. stock independent extraction cost
3. finite time T
4. extraction in time T qt
5. Initial stock Ro

Rt+1 - Rt = -qt

Max utility [sum T, t=0 (1/1+d)t*Ut*(qt)] s.t. Initial reserves Ro - [sum T, t=0 qt] = 0. Lagrangian lambda = (1/1+d)t*U'*(qt) which is shadow price. FOC, U'(qt +1) = (1+d)U'(qt). In perfect competition, U' = P, so we can say that Pt+1 = (1+d)Pt. Optimal extraction rate = a[1-(1-d)^t-Tc

Logical consequence of optimizing utility. In perfect competition, Pt+1 - Pt / Pt = d or rate of price change equals the discount rate.

Profit = PtQt because monopolist what to maximize present value of revenue. Profit = (a-bqt)qt because monopolist sets price based on demand.
Max profit [sum T, t=0 (1/1+d)*(a-qt - bqt^2) s.t. Initial reserves - [sum T, t=0 qt] = 0. Optimal extraction rate qt = a/2b[1-(1-d)^t-Tm

Monopolist can increase profit by restricting production, thereby causing the market price to rise. However, both PC and M have to exhaust resource. If we assume stock independent and same starting point for both, then Tm > Tc.

Now Ct = C(qt, Rt) -- Rt = 0 may or may not be optimal, and last few units will be above market price. Max [sum T, t-0 (1/1+d)*(Ptqt - C(qt, Rt)) s.t. Rt+1 - Rt = -qt.
Essentially the stock dependent extraction cost rate of price increase is less than d.

1. two time periods and can only extract in t=1
2. E = probability of fail, 1-E is probability of success.
3. S is state of the world in t=1. No discovery, S = 0 and yes discovery, S = 1.
4. Discovery function = dSeo