Chapter 7 Project Comments Emily, Conor, and Logan Very good work! A. 4/4 B. 21/22 -- This is mostly very good, but you missed that there are actually two different equilibrium values when 0 < h < 14 or so. There's the stable one that your solutions tend toward, but there's a second, unstable, one too. If P(0) is below the unstable equilibrium, that's when you get solutions tending to 0 (population crashes). See the solutions posted on the course homepage. We'll also discuss this in class on Monday, 11/27. C. 20/20 -- P = 0 is also an (unstable) equilibrium in the proportional harvesting case. For part (5), you can also think about what happens over a long time. If your solution tends to the stable equilibrium, then the average amount taken per year will tend to (.3) times that stable equilibrium too. D. 2/2 -- Yes, proportional harvesting is the "way to go" See solutions for more on this. Total: 47/48 Christian, Genevieve, and Andrew A. 4/4 B. 18/22 -- The situation is a bit more intricate than you realized here. When 0 < h < 14 or so, there are two different equilibrium solutions, one stable (the one your solutions tended toward, if P(0) was big enough), and also a second equilibrium that is unstable. You don't "see" that one in the same way, because unless P(0) = that value exactly, then the solution tends away from that value as n increases. To determine it, though, you can proceed as in the solutions posted on the course homepage and as the problem was suggesting in part (3) of B. What you said in (5) is the right idea, but you can actually take h even slightly bigger than 14 and still get a stable equilbrium. If P(0) is large enough, then you can take that much per year by the constant harvesting. We'll also discuss this in class on Monday, 11/27. C. 20/20 -- P = 0 is also an (unstable) equilibrium in the proportional harvesting case. For part (5), you can also think about what happens over a long time. If your solution tends to the stable equilibrium, then the average amount taken per year will tend to (.3) times that stable equilibrium too. D. 2/2 -- Yes, proportional harvesting is the "way to go" See the solutions for more on this. Total: 44/48 Lily, Joe, Kyle, and Madison A. 2/4 -- You did not answer the second part here. If you take out 77 - 1 = 76 of the fish biomass, it takes about 11 years to regrow to 77 from the 1 that is left. So that means you are getting 76/11 = approx. 6.9 x 10^6 kg per year on average. B. 17/22 -- This is mostly good, but there are a couple of problems with what you said here. When 0 < h < 14 or so, there are always two different equilibrium solutions (not three or more), one stable (the one your solutions tended toward, if P(0) was big enough), and also a second equilibrium that is unstable. You don't "see" that one in the same way, because unless P(0) = that value exactly, the solution tends away from that value as n increases. To determine it, though, you can proceed as in the solutions posted on the course homepage and as the problem was suggesting in part (3) of B. What you said in part (5) is not correct. The maximum harvesting level for which a stable equilibrium exists is actually even a bit larger than 14. As long as P(0) is large enough you can take that much out per year with constant harvesting. So you don't want to divide by the number of years since you're taking that much every year. We'll also discuss this in class on Monday, 11/27. C. 19/20 -- Same comment as on B (2) -- There are actually only two equilibrium solutions for each value of the harvesting proportion p. (I think you are getting confused by the difference between an equilibrium and a solution tending toward an equilibrium. An equilibrium is a constant population level that corresponds to a solution of the difference equation that does not change with n. See the solutions.) What you said here in (5) is not correct either because you don't want to divide by the 11 years here either, though. The idea here is that the amount you take out this way will be roughly p x the equilibrium level if you do the fishing over a long period of years. D. 2/2 -- Yes, proportional harvesting is the "way to go" See the solutions for more on this. Total: 40/48 Elizabeth, Angel, and Matthew A. 4/4 B. 20/22 -- This is mostly very good, but when you say ``there are 73 equilibrium values'' in case h = 5, that's not correct. You mean to say the ``equilibrium value is (about) 73.'' It's also true that there are actually two different equilibrium values when 0 < h < 14 or so. There's the stable one that your solutions tend toward, but there's a second, unstable, one too. If P(0) is below the unstable equilibrium, that's when you get solutions tending to 0 (population crashes). See the solutions posted on the course homepage for the algebraic way to determine the equilibrium solutions. We'll also discuss this in class on Monday, 11/27. C. 18/20 -- Your description for C (2) really goes with B (2). There is not a threshold value for the proportional harvesting the same way there is one for constant harvesting. (That is, there are no positive levels for P(0) where the population crashes -- at least as long as p < .71.) Doing the proportional harvesting lowers the stable equilibrium but it leaves the unstable one at P = 0. D. 2/2 -- Yes, proportional harvesting is the "way to go" See the solutions for more on this. Total: 44/48 Ryan, Grace, Najee, and Eve A. 4/4 B. 21/22 -- This is mostly very good, but you missed that there are actually two different equilibrium values when 0 < h < 14 or so. There's the stable one that your solutions tend toward, but there's a second, unstable, one too. If P(0) is below the unstable equilibrium, that's when you get solutions tending to 0 (population crashes). See the solutions posted on the course homepage. We'll also discuss this in class on Monday, 11/27. C. 19/20 -- There is an (unstable) equilibrium at P = 0 too in this case. Also, in part (5), you need to think about what happens over a long time. If your solution tends to the stable equilibrium, then the average amount taken per year will tend to that .3 x stable equilibrium, in the long run. The .3 x 46 is closer to 14 than it is to 16. D. 1/2 -- A is not really a viable strategy in the real world. Who would want to eat halibut that has been sitting around in a freezer for up to 11 years(?!) That's what it would take to use that approach. The proportional harvesting (C) is actually better from a number of perspectives. You can also get a larger average harvest per year (see comment on C (5) above) than you can with either (A -- ``greedy'') or (B -- ``constant'') harvesting. Total -- 45/48