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NR Unit 2.1
renewable resources/fisheries
| Question | Answer |
|---|---|
| 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 |