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NR Unit 2.1

renewable resources/fisheries

QuestionAnswer
Renewable resources and examples regenerated naturally forests, fisheries, grassland, water
differences between flow and stock resources flow resources such as solar, wave wind are non depletable, stock resources such as living organisms, inanimate systems are capable of being fully exhausted
issues with renewable resources - private property rights do not exist or are poorly defined, resource stocks tend to be overexploited as Subject to open access
Without human Predation, what is the rate of change of the population over time ds/dt= gS, where g is the growth rate and S is is population stock
Integratiion of rate of change St=S0e^(gt) Population grows exponentially over time at rate of g, if g is greater than zero
what is THE ACTUAL POPULATION GROWTH, denoted by x, depend on and hence what is the rate of change of stock over time depends on stock size S dot = x(S)S
What is density-dependent growth we find growth rate by dividing S dot by S, so density dependent growth is given by S . /S = x(S)
Carrying Capacity and implications carrying capacity of a biological species is the maximum population size of the species that the environment can sustain indefinitley given food, habitat etc Limit on population growth
Notation for carrying capacity S max
Logistic net growth of population i.e what is x(s) equal to using carrying capacity x(S) =g(1-s/Smax)
Rewriting S dot as a result S dot/G(s) = g(1 - S/smax)*S Gives net effect of natural changes and human predation
implications of This formula Fish population growth rate slows when there are very high levels of St (approaching Smax) and when there are only a few fish Fish population grows in size slowly when there are only a few fish
graphical illustration of logistic biological growth S on x-axis, G(S) i.e population growth rate on y Parabola, maxima, starts at zero goes up to maximum at S/2(SMSY), then falls back to zero to Smax
MSY Maximum Sustainable Yield
Actual rate of change of the stock s for in terms of growth and harvesting S dot = G- H Where G is rate at which population regenerates itself, and H is the rate of harvest
where does steady state harvest occur when s dot =0 (the resource stock size remains constant over time) , and hence G=H
Steady state harvest graph S on x-axis, and G.H on y At peak, we have Smax/2 at GMSY=HMSY on y-axis We have a horizontal line running from below smax/2, creating a rectangl, corresponds to G1=H1on y-axis, which relate to S1L*unsustainable* (on left hand side) and S1U on RHSm *sus
What do these new points mean Quantity of net natural growth is at a maximum (GMSY) when the stock size is at the maximum sustainable yield , steady state here NOTE: Any stock between 0 and Smax can support steady state harvest, long as harvesting is at corresponding level
What does S1L represent Low stock equilibrium : unstable, if growth falls below Harvesting rate, stock collapses, left side of equilibrium If stock rises slightly, and growth exceeds harvesting, move away from this
what does S1U represent high stock equilibrium : stable, if growth goes slightly above/below harvesting, will return back and correct itself, rightside of hump
where is overfishing shown on the graph, and what is an overfished equlibrium At SMSY, but above hump, so actual harvest is above peak, so stock will fall (overfished) Overfished equilibrium is below SMSY, but where H=G, but at a lower level
What is open access fishery rival in consumption but NOT EXCLUDABLE
types of externalities in fisheries Contemporaneous - “if i don’t fish, someone else will”, leads to overfishing, lower rate of return for everyone, current net benefit decreases, inefficient for fishermen Intergenerational externality: overfishing reduces available future stock , future
cont future net benefit decreases, dynamically inefficient
what do open access fisheries imply no property rights To the in situ fishery resources, included fish in water, so free for all
two components of open-access model 1. A biological sub-model (natural growth process of the fishery) 2, an economic sub model (behaviour of fishermen)
What occurs in our steady state (bioeconomic equilibrium) the resource stock SIZE is unchanging over time (b) The fishing fleet is constant with no net inflow or outflow of vessels (e)
Setting up the model (1) - defining resource stock level G(S) = g(1 -s/Smax)S
setting up the model (2) - defining model for HARVESTING H(S, E) = eES Harvesting depends on harvest effort (E), and the prevailing stock (S). Small e is a productivity parameter
Setting up the model (3) fish stock growth function w/harvesting S dot = G(S) -H
Setting up the model (4) harvesting costs C = wE linear function of effort, where w is cost per unit of effort
Setting up the model (5) revenue from harvest B=PH Price times quantity harvested
setting up the model (6) fishing profits NB = B-C
Showing whether or not we have every or exit from the fishery industry (7) if profits are positive -> entry Profits are negative —> exit Differentiate effort with respect to time to get &*NB, where & is a positive parameter
Where does biological equilibrium occur (8) When G=H, steady state
where does economic equilibrium occur (9) Where rents/profits driven to zero i.e NB = B-C=0, which implies PH=wE
Using G=H to get S in terms of Smax, e,g, and E (biological substitution) Pre-cursor for optimality conditions for S (10) biological equilibrium G=H We know G=g(1-s/Smax)*S And H=eES So put them together and solve for S=Smax(1-(e/g)E)
Using PH=wE for economic substitution To find S* (11) In open access equilibrium profits is zero, P(eES)= wE Solving for S (this is S*) , S*=w/Pe
Using our result for S to find the yield-effort curve (the upside down U curve coming up). Use H =eES H=eESmax(1-(e/g)E)
Finding E* set two answers for S together i.e 10 and 11 and solve for E to get E*=g/e(1-w/PeSmax)
finding H* sub in E* and S*into H=eES and simplify to get H*=gw/Pe(1-w/(peSmax))
Steady state Equilibrium yield-effort relationship graph (effort on x-axis) Effort E on x-axis, harvest on Y-axis U shaped curve (yield-effort curve) given by H=eESmax(1-e/g(E) running from 0 to g/e Straight line H=(w/P)E, upward sloping, POI at EOA and HOA, derived for zero economic profit
Steady state Equilibrium yield-effort relationship graph (stock on x-axis) Stock on x-axis , harvest levels on y U shaped yield effort curve (Intersection with HMSY=eEMSYS upward sloping line) at SMSY. When E increases, shift of line to left, steeper gradient, greater level of stock and harvest at equi, lower effort
cont shift of H line to right higher level of stock
Comparative Static results, s.s equilibrium, S*=w/Pe Determining whether relationship is positive or negative by differentiating wrt P, w, e negative, positive, negative
Comparative Static results E*=g/e(1-w/PeSmax) Determining whether relationship is positive or negative by differentiating wrt P, w, e, g, & positive, negative, ?, positive, zero
Comparative Static results H*=gw/Pe(1-w/PeSmax) Determining whether relationship is positive or negative by differentiating wrt P, w, e, g, & ?, ?, ?, positive, 0
cont BUT on stock harvest graph, as H0A shifts to the left as E INCREASES, steeper gradient of H=Ees, lower stock and harvest
cont now move to graph with harvest on y-axis to determine effect on harvest level, with stock on x-axis. Increase in Eoa results in the harvest line shifting right, and H0a decreasing
Doing comparative analysis with the effect of price on harvest level - case 2, E0Ay to the left of MSY, then increase in price £ on y axis, effort on x Yield-effort curve total revenue curve, harvest line TC Increase in price shifts TR upwards, EOA is higher, same as case 1 here
cont BUT on stock harvest graph, as H0A shifts leftwards closer to SMSY, harvest level increases
Created by: aliyah s
 

 



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