MATH 136 -- Calculus 2  

Lab 3 -- Logistic Models with Harvesting (Environmental Modeling) 

November 14 and 16, 2016 

 

Question A:  The logistic equation and the solution with  Q(0) = 1.3 Mg 

 

logistic := diff(Q(t), t) = `+`(`*`(.1, `*`(Q(t), `*`(`+`(1, `-`(`*`(`/`(1, 160), `*`(Q(t))))))))); 1 

diff(Q(t), t) = `+`(`*`(.1, `*`(Q(t), `*`(`+`(1, `-`(`*`(`/`(1, 160), `*`(Q(t))))))))) (1)
 

with(DEtools); -1 

DEplot(logistic, Q(t), t = 0 .. 130, [[Q(0) = 1.3]], Q = 0 .. 170, linecolor = black); 1 

Plot_2d
 

A 2)  The formula for the solution of the logistic equation is:
solQ := proc (t) options operator, arrow; `+`(`/`(`*`(160), `*`(`+`(1, `*`(a, `*`(exp(`*`(-1, `*`(.1, `*`(t)))))))))) end proc
 

proc (t) options operator, arrow; `+`(`/`(`*`(160), `*`(`+`(1, `*`(a, `*`(exp(`*`(-1, `*`(.1, `*`(t)))))))))) end proc (2)
 

solve(solQ(0) = 1.3, a); 1 

122.0769231 (3)
 

solQQ := proc (t) options operator, arrow; `+`(`/`(`*`(160), `*`(`+`(1, `*`(122.08, `*`(exp(`*`(-1, `*`(.1, `*`(t)))))))))) end proc; 1 

proc (t) options operator, arrow; `+`(`/`(`*`(160), `*`(`+`(1, `*`(122.08, `*`(exp(`*`(-1, `*`(.1, `*`(t)))))))))) end proc (4)
 

These are the biomass values at the times t = 0, 10, 20, ... , 150 years:

seq(solQQ(`+`(`*`(10, `*`(t)))), t = 0 .. 15); 1
 

1.299967501, 3.485024683, 9.131517693, 22.60523877, 49.44416728, 87.78819224, 122.8306886, 143.9726027, 153.7052655, 157.6252388, 159.1181002, 159.6744328, 159.8800764, 159.9558616, 159.9837596, 159.9...
1.299967501, 3.485024683, 9.131517693, 22.60523877, 49.44416728, 87.78819224, 122.8306886, 143.9726027, 153.7052655, 157.6252388, 159.1181002, 159.6744328, 159.8800764, 159.9558616, 159.9837596, 159.9...
(5)
 

A 3)  The recovery time is the time it takes the forest to regrow to 99% of the carrying capacity, or (.99)(160) = 158.4: 

T := solve(solQQ(t) = `*`(.99, 160), t); 1 

93.99796418 (6)
 


Thus the recovery time is about 94 years.
 

 

B)  
 

1)  The "virgin state" would be represented by the initial value Q(0) = 160.

2 and 3)  With 3.2 Mg of biomass harvested each year, the differential equation becomes:

harvest := diff(Q(t), t) = `+`(`*`(.1, `*`(Q(t), `*`(`+`(1, `-`(`*`(`/`(1, 160), `*`(Q(t)))))))), `-`(3.2)); 1
 

diff(Q(t), t) = `+`(`*`(.1, `*`(Q(t), `*`(`+`(1, `-`(`*`(`/`(1, 160), `*`(Q(t)))))))), `-`(3.2)) (7)
 

DEplot(harvest, Q(t), t = 0 .. 130, [[Q(0) = 160]], Q = 0 .. 170, linecolor = black); 1 

Plot_2d
 

4) Note that the solution seems to be tending to a limiting value of about Q = 116 as  
5)  With the harvesting level at 6.4 Mg/year, the differential equation becomes
 

 

harvestmore := diff(Q(t), t) = `+`(`*`(.1, `*`(Q(t), `*`(`+`(1, `-`(`*`(`/`(1, 160), `*`(Q(t)))))))), `-`(6.4)); 1 

diff(Q(t), t) = `+`(`*`(.1, `*`(Q(t), `*`(`+`(1, `-`(`*`(`/`(1, 160), `*`(Q(t)))))))), `-`(6.4)) (8)
 


DEplot(harvestmore, Q(t), t = 0 .. 130, [[Q(0) = 158.4]], Q = 0 .. 170, linecolor = black); 1
 

Plot_2d
 

The solution reaches the value Q = 0 at about t = 47 years.  At that time, all the biomass in the forest 

has been removed.  

C)  To find the equilibria of these differential equations with the harvesting terms, we need to
 

set the right-hand sides equal to zero and solve for Q.  This can be done in Maple as follows. 

For the harvesting level of 3.2 Mg/year, we use: 

 

> solve(`+`(`*`(.1, `*`(Q, `*`(`+`(1, `-`(`*`(`/`(1, 160), `*`(Q))))))), `-`(3.2)), Q); 1
 

44.22291236, 115.7770876 (9)
 

 

`*`(These, `*`(equilibrium, `*`(levels, `*`(are, `*`(visible)))))in the plot of the solutions for the equation with the harvesting 

level of 3.2 Mg/year above.   The solutions tend toward the 115.78 level, they tend away from the 44.22 

level.  These values can also be derived by using the quadratic formula to solve this quadratic equation 

for Q:   `+`(`-`(`*`(`/`(1, 1600), `*`(`^`(Q, 2)))), `*`(.1, `*`(Q)), `-`(3.2)) = 0 

 

 

> (-.1 + sqrt((.1)^2 - 4*(-1/1600)*(-3.2)))/(2*(-1/1600));
 

44.22291236 (10)
 

> `/`(`*`(`+`(`-`(.1), `-`(sqrt(`+`(`^`(.1, 2), `*`(3.2, `*`(`*`(4, -`/`(1, 1600))))))))), `*`(`*`(2, -`/`(1, 1600)))); 1
 

115.7770876 (11)
 

On the other hand, with the harvesting term 6.4 Mg/year, the quadratic equation `+`(`-`(`*`(`/`(1, 1600), `*`(`^`(Q, 2)))), `*`(.1, `*`(Q)), `-`(6.4)) = 0 

`+`(`-`(`*`(`/`(1, 1600), `*`(`^`(Q, 2)))), `*`(.1, `*`(Q)), `-`(6.4)) = 0 (12)
 

has no real roots, so there are no equilibria. 

 

D)  With selective harvesting strategy 1, the biomass level is tending to 115.78 Mg in the long run, 

      so 3.2 Mg can be harvested every year, and the average yield per year is 3.2 Mg/year. 

 

     With the clear-cutting strategy (after the first cycle), we would be taking 158.4 - 1.3 Mg 

     then letting the forest regrow for 94 years, so the average yield per year is  

     `+`(`*`(`/`(1, 94), `*`(`+`(158.4, -1.3)))) = 1.67Mg/year.

     With the selective harvesting strategy 2, there are 6.4 Mg harvested in the first 47 or so years,
 

     then the forest is allowed to regrow for 94 years with no harvesting, then the cycle repeats.   

     This gives  

     an average yield per year of  `/`(`*`(`+`(`*`(6.4, 47), `*`(0, 94))), `*`(`+`(47, 94))) = 2.13Mg/year.

E)  
     1)  The maximum sustainable harvesting level is exactly 4.0 Mg/year (per hectare).  This can be seen
 

          algebraically, since the largest value of   h  for which the quadratic equation  

     `+`(`-`(`*`(`/`(1, 1600), `*`(`^`(Q, 2)))), `*`(.1, `*`(Q)), `-`(h)) = 0has real roots is when  so h = 4.  The equilibrium level here is Q = 80.  If Q(0) > 80, the solution 

          curve will tend to 80 as t goes to infinity.
 

      2) The question is asking if it is possible to tell a forest that is undergoing a  

          sustainable harvesting strategy apart from a "virgin" forest.  It's true that  

          there are logging operations going on, so the forest is not undisturbed, but  

          the deeper question is:  Can we tell something measurable that is different?   

          The best answer here is that the sustainable equilibrium level of the biomass  

          decreases even with sustainable harvesting as in Selective Harvesting Strategy 

          1, so the forest is qualitatively different (it would be  less "thick" or "dense"  

          forest).  This can be seen by comparing the sustainable equilibrium of 115.78 Mg  

          per hectare from Strategy 1 with the "virgin" level of  160 Mg per hectare.  It's  

          not the same forest with harvesting! 

 

    3)  What this question was getting at was to compare Strategies 1 and 2.  From  

         question D, the average yield per hectare per year with Strategy 1 was larger   

         than with Strategy 2 even though the harvesting level was smaller with Strategy  

         1 as compared with Strategy 2.  The loggers would have something to do every  

         year in Strategy 1, while they would not have any employment at all during the  

         94-year recovery period that is part of Strategy 2.  In other words, lower harvesting  

         levels do not necessarily mean fewer jobs in the long run.