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Subsections


Forest Management

Introduction

This project deals with a linear model of forest management, describing growth and harvesting of a single species of tree. Most of the background material you need is in this document, but sections 8.3 on Markov Chains and 8.6 on Linear Economic Models might be helpful. (This project is based on an example taken from Applications of Linear Algebra by Rorres and Anton.)

Background

Tree classification and growth

Consider a Christmas Tree farm, consisting of a single species of tree. We classify the trees by height into five different groups by height in feet, as shown in the following table. The table also shows the dollar values of the trees in each class.

Class Value in Dollars Height  
1   $0 \leq h < 4$  
2 12 $4 \leq h < 6$  
3 18 $6 \leq h < 7$  
4 25 $7 \leq h < 9$  
5 35 $9 \leq h $  

We will use a vector $\mathbf{x}$ to represent the state of the farm at a particular point in time as follows. The $i\mbox{th}$component of $\mathbf{x}$ is the number of trees in the $i\mbox{th}$ class. For example, at a particular instant in time x2 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 $\mathbf{x}^{(k)}$, for $k=0,1,2,\ldots$, where $\mathbf{x}^{(0)}$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 $i\mbox{th}$ class grow into the $(i+1)\mbox{st}$ 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.
Under these assumptions, we can define a growth matrix G that models changes in the vector $\mathbf{x}$ from year to year. That is, the vector $\mathbf{x}^{(k+1)}$ would be calculated from the vector $\mathbf{x}^{(k)}$ from the following equation.

\begin{displaymath}
\mathbf{x}^{(k+1)} = G \mathbf{x}^{(k)} \end{displaymath}

The matrix G would have the following structure,

\begin{displaymath}
G = \left[ \begin{array}
{ccccc}
1-g_1 & 0 & 0 & 0 & 0 \\ g_...
 ... 0 & g_3 & 1-g_4 & 0 \\ 0 & 0 & 0 & g_4 & 1 \end{array} \right]\end{displaymath}

where gi is the fraction of the trees in the $i\mbox{th}$ class that grow into the $(i+1)\mbox{st}$ class in a year.

Given a starting vector $\mathbf{x}^{(0)}$, this model behaves very much like a Markov Chain model, as you will see in the exercises.

Harvesting

In this section we consider the behavior of our model if harvesting is introduced. To keep things as simple as possible, we make the following assumptions.
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 $\mathbf{x}^{(k)}$ 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 yi, $i=2 \ldots 5$ 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 y1 to be the number of seedlings planted each year to replace the harvested trees. This means that

y1 = y2+y3+y4+y5

Under these assumptions, our model becomes

\begin{displaymath}
\mathbf{x}^{(k+1)} = G \mathbf{x}^{(k)} + \mathbf{Y} \end{displaymath}

where the vector $\mathbf{Y}$ includes harvesting and replacement, and is given by

\begin{displaymath}
\mathbf{Y} = \left[ \begin{array}
{c}
 y_1 \\  -y_2 \\  -y_3 \\  -y_4 \\  -y_5 
 \end{array} \right]\end{displaymath}

Note that the negative signs are necessary, because harvesting must reduce the populations, and the variables yi were defined to be non-negative.

Introducing harvesting makes the computations more difficult, as shown below. If we start the first year with the state vector $\mathbf{x}^{(0)}$, then the state vectors in subsequent years would be given by the following.

\begin{displaymath}
\mathbf{x}^{(1)} = G \mathbf{x}^{(0)} + \mathbf{Y} \end{displaymath}

\begin{displaymath}
\mathbf{x}^{(2)} = G \mathbf{x}^{(1)} + \mathbf{Y} = G^2
\mathbf{x}^{(0)} + G \mathbf{Y} + \mathbf{Y} \end{displaymath}

\begin{displaymath}
\mathbf{x}^{(3)} = G \mathbf{x}^{(2)} + \mathbf{Y} = G^3
\mathbf{x}^{(0)} + G^2 \mathbf{Y} + G \mathbf{Y} + \mathbf{Y} \end{displaymath}

and, in general,

\begin{displaymath}
\mathbf{x}^{(k)} = G^k \mathbf{x}^{(0)} + G^{k-1} \mathbf{Y} +
G^{k-2} \mathbf{Y} + \ldots + \mathbf{Y}\end{displaymath}

Sustainable Harvest Policy

In this section we consider the question of whether our model can have steady states or not. That is, we consider if there can be values of the state vector $\mathbf{x}$ and the harvest and replacement vector $\mathbf{Y}$ that satisfy the equation

\begin{displaymath}
\mathbf{x} = G \mathbf{x} + \mathbf{Y} \end{displaymath}

which would correspond to harvesting being exactly balanced by replacement and growth. We can write this as a linear system as follows.

\begin{displaymath}
(I - G) \mathbf{x} = \mathbf{Y} \end{displaymath}

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

\begin{displaymath}
G = \left[ \begin{array}
{cc}
 1/2 & 0 \\  1/2 & 1
 \end{array} \right]\end{displaymath}

and the harvest and replacement vector $\mathbf{Y}$ can be written as

\begin{displaymath}
\mathbf{Y} = \left[ \begin{array}
{c}
 a \\  -a
 \end{array} \right]\end{displaymath}

where a is the number of trees from class 2 to be harvested. So the linear system for the fixed point is

\begin{displaymath}
\left[ \begin{array}
{cc} 
 1/2 & 0 \\  -1/2 & 0 
 \end{arra...
 ...ght]
= \left[ \begin{array}
{c}
 a \\  -a 
 \end{array} \right]\end{displaymath}

The solution is easily obtained as x1 = 2 a. The value of x2 seems to be arbitrary, but recall that the total number of trees is fixed at 200 so we have x2 = 200 - x1. The value of y1 = a is arbitrary, but note that it cannot be greater than 100.

More generally, the equation for fixed points in our model with 5 classes would look like  
 \begin{displaymath}
\left[ \begin{array}
{ccccc}
g_1 & 0 & 0 & 0 & 0 \\ -g_1 & g...
 ...+ y_5 \\  -y_2 \\  -y_3 \\  -y_4 \\  -y_5 
 \end{array} \right]\end{displaymath} (1)
where we have used our model assumption that y1= y2 + y3 +y4 + y5. This linear system can easily be put into the following form that is close to reduced row echelon form.  
 \begin{displaymath}
\left[ \begin{array}
{ccccc}
g_1 & 0 & 0 & 0 & 0 \\ 0 & g_2 ...
 ..._3 +y_4 + y_5 \\  y_4 + y_5 \\  y_5 \\  0 
 \end{array} \right]\end{displaymath} (2)
This form shows that the set of equations is consistent, and that an infinite number of solutions exist. In fact, there seems to be too much freedom.

Optimal Sustainable Yield

By Optimal Sustainable Yield, we mean to maximize the dollar value of the trees harvested. That is, we want a sustainable harvest (that is, a fixed point) that produces as much income as possible. Income is calculated based on the value of the harvested trees. For example, using the dollar values in the table at the beginning of this writeup, we would calculate the income V as V = 12y2 + 18 y3 +25 y4 + 35 y5, where the yi are the components of $\mathbf{Y}$.

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 x2 = 0, x1 = 200 and a=100 for the fixed point. If you think about it for a while, this result does make sense. This provides the largest possible harvest, since a had to be less than or equal to 100.

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 y1, the number of seedlings being planted, is given by y1=a and the component -yk of $\mathbf{Y}$ corresponding to the class being completely harvested is given by -yk = -a.

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

\begin{displaymath}
\left[ \begin{array}
{ccccc}
g_1 & 0 & 0 & 0 & 0 \\ 0 & g_2 ...
 ...egin{array}
{c}
 a \\  a \\  0 \\  0\\  0 
 \end{array} \right]\end{displaymath}

which can easily be solved to give x3=x4=x5 = 0 and the two simple equations g1 x1 = a and g2 x2 = a. Solving these for x1 and x2 and using the condition that the sum of the populations has to add up to the fixed total tree population N gives the equation

\begin{displaymath}
\frac{a}{g_1} + \frac{a}{g_2} = N\end{displaymath}

which can be solved to give a in terms of N.

Exercises

1.
Consider the matrix G given by

\begin{displaymath}
G = \left[ \begin{array}
{ccccc}
1/2 & 0 & 0 & 0 & 0 \\ 1/2 ...
 ... & 0 & 1/4 & 4/5 & 0 \\ 0 & 0 & 0 & 1/5 & 1 \end{array} \right]\end{displaymath}

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 $\mathbf{x}$ is a five dimensional vector satisfying the condition x1+x2+x3+x4+x5 = N, then the components of the vector $G \mathbf{x}$, 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 $\mathbf{x}^{(0)}$ are (600, 1400, 400, 0, 0) and the harvest and replacement vector $\mathbf{Y}$ 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 $\mathbf{Y} = (400,0,-400,0,0)$. What happens to the populations?

5.
Explain the steps needed to go from equation 1 to equation 2.

6.
Show that if $\mathbf{x}$ is a five dimensional vector satisfying the condition x1+x2+x3+x4+x5 = N, and $\mathbf{Y}$is a harvest and replacement vector satisfying y1 = y2+y3+y4+y5, then the components of the vector $G \mathbf{x} + \mathbf{Y}$, 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 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 $\mathbf{x}$ has components (960, 1440, 0, 0, 0) and the harvest and replacement vector $\mathbf{Y}$ 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 $\mathbf{x}$ and corresponding harvest and replacement vector $\mathbf{Y}$ that maximize the dollar value of the trees harvested.

next up previous
Next: About this document ... Up: No Title Previous: No Title

William W. Farr
9/17/1998