Optimal fishing strategy

To protect the natural environment on which humans depend, development of renewable resources (such as fishery and forestry resources) must be moderate. A reasonable and simplified Strategy is to achieve the maximum output or the best benefit while achieving sustainable gains.

Consider the optimal fishing strategy for catfish: Suppose this fish is divided into four age groups, known as 1-age fish, 2-age fish, 3-age fish, 4-age fish. The average quality of each fish in each age group is 5.07, 11.55, 17.86, 22.99 (G), and the natural mortality rate of each age group is 0.8 (1/year ), this type of fish breeds eggs in a seasonal manner. The average egg production of each 4-year-old fish is 1.109 × 1011 (pieces). The egg production of three-year-old fish is half of the total, 2-year-old and 1-year-old fish do not lay eggs. The period of spawning and hatching is the last 4 months of each year, survival rate (the ratio of the number of 1-age fish lines to the number of eggs produced n) was 1.22 × 1011/(1.22 × 1011 + n ).

The fisheries authority stipulates that fishing can be performed only eight months before the spawning period. If the annual fishing capacity (such as the number of fishing vessels and number of times of downloading the network) remains unchanged, the catch per unit time will be proportional to the number of fish lines in each age group. The ratio coefficient may be set to the fishing intensity coefficient. A 13mm mesh pull network is usually used. This network can only catch 3-and 4-age fish, and the ratio of the two fishing coefficients is 0.42: 1. Fisheries refer to this method as a fixed amount of effort.

① Establish a mathematical model to analyze how to achieve sustainable fishing (that is, when fishing starts every year, the number of fish in each age group remains unchanged), and obtain the highest annual harvest (the total quality of fishing ).

② A Fishery Company has contracted the fishing business for this kind of fish for five years. The contract requires that the production capacity of fish in five years cannot be greatly damaged. It is known that the number of fish in each age group is 122, 29.7, 10.1, 3.29 (× 109). If a fixed amount of effort is still used for fishing, what strategy should the company adopt to maximize the total harvest.

(1) problem analysis and Model Establishment

Question hypothesis

① Although the increase in the total number of fish is discrete, for large-scale fish, the change in the total number of fish can be consecutive over time.

② According to the conditions given by the question, it is possible to set up a school of fish to lay eggs instantly at the end of August each year, and all eggs are incubated at the end of December.

③ I + 1-year-old fish grow into I + 1-year-old fish, I =, 3 respectively.

④ At the end of the year, the number of 4-year-old fish remained at a very small proportion of the total number. It can be assumed that all fish died.

⑤ Continuous capturing: the number of fish in each age group changes cyclically. The cycle is one year. You can only consider the number of fish in one year.

Problem Analysis

① Symbol description.

XI (t) -- number of I-aged fish lines at t time, I = 1, 2, 3, 4;

K -- fishing strength coefficient of 4-age fishes;

N-Annual egg volume;

Ai-Number of I-aged fish in the early stage of each year, I = 1, 2, 3, 4.

② Understanding of mortality. The natural mortality rate of the fish given in the question is 0.8 (1/year). We understand it as the average mortality rate, which is the ratio of the number of fish deaths per unit time to the number of existing fish. We can see from the assumption that, it is a constant irrelevant to other factors such as the environment. The number of fish is continuously changing, and the number of 1-year-old fish, 2-year-old fish and 3-year-old fish in the next 4 months is only related to mortality, the number of fish of all ages is sufficient.

, T, [] (I =)

, T' [2/3, 1] (I =)

③ Understanding of fishing intensity coefficient. The catch of 4-year-old fish is proportional to the total number of 4-year-old fish. The ratio coefficient is K. It is certain and only within the catch period (that is, the first 8 months of each year) 3-and 4-age fish are caught. Therefore, on the one hand, the fishing intensity coefficient determines the quantity of three-and four-age fish during the fishing period.

, T, [/3]

, T, [/3]

On the other hand, it is also determined that the number of three-and four-age fish caught at t time is 0.42kx3 (T) and kx4 (t), respectively ).

④ Understanding of survival rate. Because only three-and four-age fish lay eggs at the end of August each year, the annual egg N can be expressed

 

The survival rate is shown in the question, so the number of 1-year-old fish at the beginning of each year is

 

(2) matlab simulation program and results

The model is as follows:

 

S. t

, T, [0, 1]

, T, [0, 1]

, T, [/3]

, T' [2/3, 1]

, T, [/3]

, T' [2/3, 1]

Computer program simulation experiment program: first create two M-files.

% [Buyu. m] optimal fishing strategy Problem

Function Y = buyu (X)

Global A10 A20 A30 A40 total K;

Syms K A10;

X1 = dsolve ('dx1 =-0.8 * x1', 'x1 (0) = a10 ');

T = 1; A20 = Subs (X1 );

X2 = dsolve ('dx2 =-0.8 * X2 ', 'x2 (0) = a20 ');

T = 1; A30 = Subs (X2 );

X31 = dsolve ('dx31 =-(0.8 + 0.42 * k) * x31', 'x31 (0) = a30 ');

T = 2/3; A31 = Subs (x31 );

X32 = dsolve ('dx32 =-0.8 * x32', 'x32 (2/3) = a31 ');

T = 1; A40 = Subs (x32 );

X41 = dsolve ('dx41 =-(0.8 + k) * x41', 'x41 (0) = a40 ');

T = 2/3; a41 = Subs (x41 );

X42 = dsolve ('dx42 =-0.8 * x42 ', 'x42 (2/3) = a41 ');

Nn = 1.109*10 ^ 5*(0.5 * A31 + a41 );

EQ1 = a10-nn * 1.22*10 ^ 11/(1.22*10 ^ 11 + nn );

S = solve (EQ1, A10); A10 = S (2 );

Syms T;

T3 = Subs (subs (INT (0.42 * K * x31, T,/3 )));

T4 = Subs (subs (INT (K * a41, T, 0, 2/3 )));

Total = 17.86 * T3 + 22.99 * T4;

K = X;

Y = Subs (-total );

 

% [Buyu1.m] optimal fishing strategy Problem

Global A10 A20 A30 A40 total;

[K, Mtotal] = fminbnd ('buyu ', 16,18 );

Ezplot (total, 0, 25)

Xlabel ('fishing intensity coefficient ')

Ylabel ('total harvest (g )')

Title ('fishing intensity-total yield curve graph ')

Format long;

K

Total =-Mtotal

A10 = eval (A10)

A20 = eval (A20)

A30 = eval (A30)

A40 = eval (A40)

Format short

Clear

 

In Matlab, enter the command: buyu1 and press Enter:

K = 17.86984648968516, total = 3.869962667986214e + 011,

A10 = 1.192332637390619e + 011, A20 = 5.357495888418817e + 010

A30 = 2.407278077805501e + 010, A40 = 7.262288087797806e + 007

At the same time, we also get the image-figure 1.

When K = 17.86984648968516, the highest annual harvest is totalt = 3.869962667986214e + 011 (g). At this time, the number of fish of four ages at the beginning of each year is:

1.192332637390619e + 011

5.357495888418817e + 010

2.407278077805501e + 010

7.262288087797806e + 007

The solution to the second problem is omitted here.

 

References

[1] Li Zhilin, ou yigui, mathematical modeling and typical case analysis, Beijing: Chemical Industry Press, 2006.12

 

Reference page number and request number

[1] page22; o141.4/L. Z. L

 

Important Note: Use matlab programming to solve mathematical problems.

Optimal fishing strategy