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Class | Value in Dollars | Height | |

1 | |||

2 | 12 | ||

3 | 18 | ||

4 | 25 | ||

5 | 35 |

We will use a vector to represent the state of the farm
at a particular point in time as follows. The component of is the number
of trees in the class. For example, at a particular
instant in time *x _{2}* would be the
number of trees that are at least 4 feet, but less than 6 feet tall.

To keep things simple, we suppose that the farmer only measures the trees once a year. That is, every year he measures his trees and records the results. This produces a sequence of vectors , for , where would represent the first year of the farm, when all the trees were seedlings. To model the growth of the trees, we make the following assumptions.

- 1.
- A fixed fraction of the trees in the class grow into the class during each year. This fraction depends on the class, but does not vary over time.
- 2.
- No trees die.
- 3.
- No tree can move up more than one class in a single year. For example, a tree beginning the year in class 2 could grow enough to be in class 3 at the end of the year, but it cannot grow enough to make it to class 4 or 5 in a single year.

Given a starting vector , this model behaves very much like a Markov Chain model, as you will see in the exercises.

- 1.
- No trees from class 1 (seedlings) are ever harvested.
- 2.
- For every tree harvested, a seedling is planted. This means that the total number of trees is constant, as you will be asked to show in the exercises.
- 3.
- The vector now represents the number of trees in each class at the end of the year, but with that year's harvesting and replacement included.
- 4.
- We define
*y*_{i}, to be the number of trees in class*i*that are harvested at the end of each year. These numbers are assumed to not change from year to year, and are to be non-negative. - 5.
- We define
*y*to be the number of seedlings planted each year to replace the harvested trees. This means that_{1}*y*=_{1}*y*+_{2}*y*+_{3}*y*+_{4}*y*_{5}

Introducing harvesting makes the computations more difficult, as shown below. If we start the first year with the state vector , then the state vectors in subsequent years would be given by the following.

and, in general,
To help make this clear, consider a simplified model with
only two classes and assume that half of the trees grow from class 1
to class 2 in a year. Also assume that the total number of trees is
fixed at 200. Then the matrix *G* would be given by

More generally, the equation for fixed points in our model with 5 classes would look like

(1) |

(2) |

This seems like a very difficult task, but there turns out to be a
general result from a branch of mathematics called linear programming
that simplifies the task tremendously. This result says the
following. *The optimum sustainable yield is achieved by
harvesting all of the trees from one particular height class and none
of the trees from any other height class.*

To get a feel for what this means, let us return to the simple example
of only two classes at the end of the previous section. The linear
programming result says that the optimal result is to harvest all of
the class 2 trees. This is the only possible choice, since we can't
harvest the class 1 trees. This means that we should have *x _{2}* = 0,

More generally what this result says is that there are a finite number
of candidates for
optimal sustainable harvest. The harvest and replacement vectors all
have a special form with only two nonzero components. The value of
*y _{1}*, the number of seedlings being planted, is given by

It turns out that using harvest and replacement vectors of this particular form greatly simplifies the calculation of sustainable harvest policies. For example, suppose that trees of class 3 were to be completely harvested. Then it is not too hard to show that the equation for fixed points in reduced form (equation 2) is

which can easily be solved to give- 1.
- Consider the matrix
*G*given by and suppose that the tree farm is started from scratch with 2400 seedlings. If no harvesting or replacement is allowed, compute the populations of the five classes after 2 years, 5 years, and 10 years. What happens if you let the number of years get large? Does this make sense? - 2.
- Show that if is a five dimensional vector
satisfying the condition
*x*+_{1}*x*+_{2}*x*+_{3}*x*+_{4}*x*=_{5}*N*, then the components of the vector , where*G*is a growth matrix, add up to the same sum of*N*. How do you interpret this result as it applies to our model of tree growth without harvesting or replacement? - 3.
- Now consider the model with growth and replacement, using the
matrix
*G*given in the first exercise. If the components of the starting vector are (600, 1400, 400, 0, 0) and the harvest and replacement vector has components (100,0,0,-100,0), compute the populations after harvest at the end of 2 years, 5 years, 10 years, and 20 years. What happens to the populations? - 4.
- Repeat the previous exercise, but use the harvest and replacement vector . What happens to the populations?
- 5.
- Explain the steps needed to go from equation 1 to equation 2.
- 6.
- Show that if is a five dimensional vector
satisfying the condition
*x*+_{1}*x*+_{2}*x*+_{3}*x*+_{4}*x*=_{5}*N*, and is a harvest and replacement vector satisfying*y*=_{1}*y*+_{2}*y*+_{3}*y*+_{4}*y*, then the components of the vector , where_{5}*G*is a growth matrix, add up to the same sum of*N*. How do you interpret this result as it applies to our model of tree growth with harvesting and replacement? - 7.
- Consider the matrix
*G*from the first exercise and let*N*=2400. Show that if all of the trees of class 3 are to be harvested and no other classes of trees are to be harvested, the fixed point has components (960, 1440, 0, 0, 0) and the harvest and replacement vector has components (480,0, -480, 0, 0). - 8.
- Again using the matrix
*G*from the first exercise and letting*N*=3150, find the optimum sustainable yield. That is, find the values of the steady state and corresponding harvest and replacement vector that maximize the dollar value of the trees harvested.

9/17/1998