For logistic growth, we have

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1 Dr. Lozada Econ For logistic growth, we have. Harvest H depends on effort E and on X. With harvesting, Ẋ = F (X) H. Suppose that for a fixed level of effort E, H depends linearly on X: If E increases, this line moves up: E If H = F (X) then X will not change, so there will be a steady state. So let s get H and F (X) on one graph, so we can make them equal. (In the graph below, E 1 < E 2 < E 3.). K = sustainable yield It follows that if E = E 1, steady-state harvest is H 1 ; if E = E 2, steady-state harvest is H 2 ; and if E = E 3, steady-state harvest is H 3. 1

2 Graphically: Clearly more effort does not always yield more fish. Algebraic Example. Suppose F (X) = X (1 X) and H = XE 1/2. Find the steady-state relationship between H and E. Answer: In the steady state, F (X ) = H, so and therefore H = X E 1/2 X (1 X ) = X E 1/2 1 X = E 1/2 X = 1 E 1/2 = (1 E1/2 )E1/2 = E1/2 E. In any case, we have in general something like. Total revenue is price times quantity produced, namely P H. We d like to graph TR versus E assuming a competitive industry (that is, an industry whose firms all take price P as given). If P = 1, then TR = (1) H = H, so TR vs. E looks just like H vs. E :. If P = 1/10, then TR = (1/10) H, so the graph would look like. If P = 5, 2

3 then TR = 5H, so the graph would look like. As for total cost, suppose. Put the TC graph together with the three TR graphs (for low, medium, and high prices): 1. Competitive, Open-Access Fishery First consider the top graph. Suppose no one owns the fish an open access fishery and there is no government regulation. Then TR = TC, because if TR < TC, firms would leave the industry, and if TR > TC, firms would enter the industry. I ve marked the TR = TC places in the top graph with. For low, medium, and high prices, the equilibrium effort levels are E l, E m, and E h. These imply through the bottom graph which is just from page 2 flipped upside down harvest levels of H l, H m, and H h. Since from the graph H l < H h < H m, we have: 3

4 Thus we have a backward-bending supply curve. Suppose demand for this fish rises from D 1 to D 2 to D 3 to D 4 : From D 1 to D 2, P and H. From D 2 to D 3, P and H : as more people like this fish, the steady-state harvest of it falls! If demand further rises to D 4, supply can never equal demand, and there is no steady-state equilibrium. The concludes our analysis of open-access equilibrium, except for one point: since this occurs at in, where profit equals zero, the industry is unable to take advantage of the opportunity to produce at, where TR TC, so profit is positive. The point is better for the firms and better for the fish (less E more fish), but the open access externality namely that the harvest of Firm 1 affects X, so it increases the cost of Firm 2 s harvesting leads to the worse outcome. 2. Competitive, Private-Property, Net-Present-Value- Maximizing Fishery One might think that the competitive, private property solution could be obtained where short-run profit is maximized, for example, approximately 4

5 the points marked X in this diagram: X X X However, a proper analysis requires acknowledging the intertemporal aspects of the problem, even if we choose to concentrate on the steady state. The profit of each firm is Π(H t, X t ) = TR t (H t ) T C(E(H t, X t )) (o1) where H is the harvest, X is the stock size, TR is the total revenue, T C is the total cost, and E is fishing effort, all at time t. The objective of the firm is to Π t max (1 + δ) t s.t. (o2) t=0 X t+1 X t = F (X t ) H t (o3) where F is the natural excess of births over deaths. (o3) represents an infinite number of constraints on (o2). Using k 1, k 2,..., to denote the Lagrange multipliers, the Lagrangian is 5

6 L = Π Π 6(H 6, X 6 ) (1 + δ) 6 + Π 7(H 7, X 7 ) (1 + δ) 7 + Π 8(H 8, X 8 ) (1 + δ) 8 (1) + Π 9(H 9, X 9 ) (1 + δ) 9 + Π 10(H 10, X 10 ) (1 + δ) k 1 (X 1 X 0 F (X 0 ) + H 0 ) + + k 6 (X 6 X 5 F (X 5 ) + H 5 ) + k 7 (X 7 X 6 F (X 6 ) + H 6 ) + k 8 (X 8 X 7 F (X 7 ) + H 7 ) + k 9 (X 9 X 8 F (X 8 ) + H 8 ) + k 10 (X 10 X 9 F (X 9 ) + H 9 ) +. We wish to maximize this with respect to X t and H t for all t. For example, 0 = L X 8 = Π 8/ X 8 (1 + δ) 8 + k 8 + k 9 ( 1 F (X 8 )) (2) 0 = L H 8 = Π 8/ H 8 (1 + δ) 8 + k 9 (3) 0 = L H 7 = Π 7/ H 7 (1 + δ) 7 + k 8. (4) (3) and (4) can easily be solved for k 9 and k 8. Substituting these values into (2) yields so 0 = Π 8/ X 8 (1 + δ) 8 Π 7/ H 7 (1 + δ) 7 + Π 8/ H 8 (1 + δ) 8 [ 1 + F (X 8 ) ], (5) 0 = Π 8 X 8 (1 + δ) Π 7 H 7 + [ 1 + F (X 8 ) ] Π 8 H 8. (6) From (o1), Π 8 = TR 8 (H 8 ) T C(H 8, X 8 ), so Π 8 X = T C 8 X ; call this C 8 X8 for short. By definition, Π 7 H = MΠ 7 and Π 8 7 H = MΠ 8. Also, let F (X 8 ) be 8 abbreviated by F 8. Then substituting these results into (6) yields which can be rewritten as or as 0 = C X8 (1 + δ)mπ 7 + [1 + F 8] MΠ 8 (6.5) (1 + δ)mπ 7 = [1 + F 8] MΠ 8 C X8 (7) (1 + δ)mπ 7 = [1 + F 8] MΠ 8 + Π 8 X 8. (7.5) If, in (7.5), there is a steady state, then this equation becomes (1 + δ)mπ = [1 + F ] MΠ + Π X, (20) 6

7 which simplifies to δ MΠ = F MΠ + Π (20.5) X or δ = F + 1 Π MΠ X. (20.6) Finally, to show that this is consistent with what your textbook has, recall that by definition, C X = T C/ X. Your book, in (16.13), assumes that T C = c(x)h. (Your book uses C instead of c, but I think c is less confusing.) Maintaining this assumption, C X = ( c(x)h ) / X = c (X)H. In a steady state, X t+1 = X t, so from (o3), in a steady state, F (X) = H. Making this substitution results in C X = c (X)F (x). (21) In addition, in your book, equation (16.13) has π = P H c(x)h, so MΠ = π H = P c(x). (22) Substitute (21) and (22) into (20.6), remembering that Π/ X = C X : δ = F + c (X)F (X) P c(x). (23) This is (16.16) of your textbook. Steady state with TC/ X = 0 ( schooling ): (20.6) δ = F (X). If δ is too large (if δ > r), X = 0 (extinction). Steady state with TC/ X < 0 ( search ): (20.6) F = δ+(c X /MΠ). 7

8 Comparative Statics: P MΠ δ + C X /MΠ increases and moves closer to δ x, H probably (a completely backward-bending supply curve) but H could at first. TC/ H ( MC ) MΠ, the opposite results from P. TC/ X more negative f X. = δ + C X /MΠ moves further below δ Note: if C X = 0, none of these hold since all that matters is δ = F (X) (unless this means π < 0). 8

Chapter 4. Systems of Linear Equations; Matrices. Opening Example. Section 1 Review: Systems of Linear Equations in Two Variables

Chapter 4. Systems of Linear Equations; Matrices. Opening Example. Section 1 Review: Systems of Linear Equations in Two Variables Chapter 4 Systems of Linear Equations; Matrices Section 1 Review: Systems of Linear Equations in Two Variables Opening Example A restaurant serves two types of fish dinners- small for $5.99 and large for