# media/2/257/files/Linear Algebra_Application based Jason Rypkema

Michelle Johnson

Andrew Eben

Linear Algebra

Application Based Paper

Forest Management

Chapter 10.9

The problem we investigated is finding the optimal sustainable yield when harvesting a forest. This means that we seek the perfect time to harvest each class of trees in order to earn the most money while at the same time preserving the forest enough so that the next year there will be other trees left to produce another good yield. The taller the tree is when we cut it down, the more money we earn. Every year we will only cut down trees of a certain standard height. The height interval that achieves the most value will be based on the table shown below.

Class Value (dollars) Height Interval

1 (seedlings)

2

3

⋮ n-1

N

None 𝑝

2 𝑝

3

⋮ 𝑝 𝑛−1 𝑝 𝑛

[0, ℎ

1

)

[ ℎ

1

, ℎ

2

)

[ℎ

2

, ℎ

3

)

[ ℎ 𝑛−2

, ℎ 𝑛−1

)

[ ℎ 𝑛−1

, )

In order to maintain a sustainable harvest, the forest will need to be returned to its 𝑥

1 previous state each year. We define the non-harvest vector to be the vector x =

[ 𝑥

2

]

where x i 𝑥 𝑛

is the number of trees in the ith class that remain after each harvest. Because the number of the trees in the forest never changes we can use the equation 𝑥

1

+ 𝑥

2

+

+ 𝑥 𝑛

= 𝑠

, where s is the fixed amount of trees determined according to how much land and space are available. We assume that all of the trees grow and none of the seeds die. Also, we assume that each tree can only move at most one height class in any given year.

To show the cycle of one growth period of trees in the forest, we first need to establish a matrix equation for:

(Configuration at the end of growth period) – (harvest) + (new seedling replacement) = (Configuration at the beginning of growth period) (Equation 1).

First we define the growth matrix G as:

G =

[

1 − 𝑔

1 𝑔

0

0

0

1

0

1 − 𝑔

2 𝑔

2

0

0

0

0

1 − 𝑔

3

0

0

1 − 𝑔 𝑛−1 𝑔 𝑛−1

0

0

0

0

1]

Where g i

equals the fraction of trees in the ith class that grows into the (i+1) st class during a year.

We then multiply the growth matrix by the non-harvest vector(x) in order to get all of the trees in the n classes after the growth period.

Therefore Gx is equal to:

(1 − 𝑔

1

)𝑥

1

Gx = 𝑔 𝑛−2 𝑔

1 𝑥

1 𝑔

2 𝑥

2 𝑥 𝑛−2

[ 𝑔

+ (1 − 𝑔

2

+ (1 − 𝑔

3

+ 𝑥

)𝑥 𝑛

2

]

)𝑥

3

+ (1 − 𝑔 𝑛−1

)𝑥 𝑛−1 𝑛−1 𝑥 𝑛−1

We let y be the harvest vector, which is the number of trees removed at each harvest. It is 𝑦

1 represented by the vector: y=

[ 𝑦

2

] 𝑦 𝑛

We subtract y from Gx to get the number of trees left. Since we want the same number of trees that we started with, we will add the replacement matrix times the harvest vector. The replacement matrix introduces trees in the lowest height category (as seedlings), making the top row equal to one. R= [

1

0

0

1

0

0

1

0

]

0

Therefore through matrix multiplication Ry=

[ 𝑦

1

+𝑦

2

+ ⋯ +𝑦 𝑛

0

0

0

]

The matrix Ry gives the total number of new seedlings that need to be planted. We are now in a position to model equation (1) by: Gx-y + Ry = x.

Using these results we can calculate that the yield of the harvest is given by:

𝑌𝑙𝑑 = 𝑝

2 𝑦

2

+ 𝑝

3 𝑦

3

+ ⋯ + 𝑝 𝑛 𝑦 𝑛

, and from this result using linear programming we can find the optimal sustainable yield for the forest. In Theorem 10.9.1 it says that the optimal sustainable yield is achieved by harvesting all the trees from one particular height class and none of the trees from any other height class. Although to find the optimal sustainable yield we use Theorem 10.9.2 which states that the optimal sustainable yield is the largest value of

1 𝑔

1

+

1 𝑔

2 𝑝 𝑘 𝑠

+ ⋯ +

1 𝑔 𝑘−1 for k = 2, 3, …, n. The corresponding value of k is the number of the class that is completely harvested.

In conclusion, we have learned that through matrix operations we are able to find the yield of a forest’s harvest without depleting the forest but optimizing the harvest results. From this result we know how to get the most profitable yield every year for the forest. We have also found that even simple matrices and vectors can solve many real life problems.

Appendix:

Here we show how to perform matrix addition and multiplication, the two operations utilized in this paper.

11

[ 𝑎

21 𝑎 𝑛1 𝑎

12 𝑎

22

⋮ 𝑎 𝑛2

⋯ 𝑎

1𝑛 𝑎

2𝑛

⋮ 𝑎 𝑛𝑛 𝑏

] + [ 𝑏

21

⋮ 𝑏

11 𝑛1 𝑏

12 𝑏

22

⋮ 𝑏 𝑛2

⋯ 𝑏

1𝑛 𝑎

11 𝑏

2𝑛

] = [ 𝑎

21 𝑏 𝑛𝑛 𝑎 𝑛1

+ 𝑏

11

+ 𝑏

21

+ 𝑏 𝑛1 𝑎

12 𝑎

22 𝑎 𝑛2

+ 𝑏

12

+ 𝑏

22

+ 𝑏 𝑛2

⋯ 𝑎

1𝑛 𝑎

2𝑛 𝑎 𝑛𝑛

+ 𝑏

1𝑛

+ 𝑏

2𝑛

]

+ 𝑏 𝑛𝑛

Matrix Multiplication: 𝑎

11

[ 𝑎

21 𝑎 𝑛1 𝑎

12 𝑎

22

⋮ 𝑎 𝑛2

⋯ 𝑎

1𝑛 𝑎

2𝑛

⋮ 𝑏

11

] ∗ [ 𝑏

21 𝑎 𝑛𝑛 𝑏 𝑛1 𝑏

12 𝑏

22

⋮ 𝑏 𝑛2

⋯ 𝑏

1𝑛 𝑏

2𝑛

] 𝑏 𝑛𝑛

= [ 𝑎

11 𝑏

11 𝑎

21 𝑏

11 𝑎 𝑛1 𝑏

11

+ 𝑎

12 𝑏

21

+ 𝑎

22 𝑏

21

+ 𝑎 𝑛2 𝑏

21

+ ⋯ + 𝑎

1𝑛 𝑏 𝑛1

+ ⋯ + 𝑎

2𝑛 𝑏 𝑛1

+ ⋯ + 𝑎 𝑛𝑛 𝑏 𝑛2 𝑎

11 𝑏

12

+ 𝑎

12 𝑏

22

+ ⋯ + 𝑎

1𝑛 𝑏 𝑛2 𝑎 𝑛1 𝑏

12

+ 𝑎 𝑛2 𝑏

22

+ ⋯ + 𝑎 𝑛𝑛 𝑏 𝑛2

⋯ 𝑎

1𝑛 𝑏

1𝑛 𝑎 𝑛1 𝑏

1𝑛

+ 𝑎

12 𝑏

2𝑛

+ ⋯ + 𝑎

1𝑛 𝑏 𝑛𝑛

]

+ 𝑎 𝑛2 𝑏

2𝑛

+ ⋯ + 𝑎 𝑛𝑛 𝑏 𝑛𝑛

Works Cited

Anton, Howard, and Chris Rorres.

Elementary Linear Algebra: Applications Version

. 10th ed. Hoboken,

NJ: Wiley, 2010. Print.