# Discrete Optimal Control Method Based on the Optimal Strategy of Fishing

Authors: Lichun Zhang; Qingdao Huang
Journal: Discrete Dynamics in Nature and Society
 (2015)

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## Abstract

Consideration was given to the discrete optimal control method for the optimal fishing strategy. Our method is new and efficient for discrete optimal control problem, which is different from the other optimal methods such as the traditional variational method, the Pontryagin principle of maximum, and the dynamic programming. The basic construction of the model is the traditional logistic function relating to the growth of fry. The discrete optimal control method for optimal fishing strategy was used to construct the optimal rate of each fishing strategy; the main focus of our work is on the rigorous mathematical analysis of the optimal control problem. The analysis allows one to obtain the optimal initial investment amount of the fry and the optimal size of the total catch. Furthermore, when the initial investment amount of the fry is below or above the optimal value and the intrinsic growth rate of fishR is too small, we derive that fishing operations should not be started in the last few years to make the overall fishing amount optimal. At last, several typical examples are given to illustrate the obtained results.


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## Body

## 1. Introduction

Optimal control problem has been studied for many years, and in recent years, the optimal harvesting strategy of some species has been studied extensively by many authors [1–3]. An optimal harvesting policy is given using Pontryagin’s maximum principle by many authors [1, 2]. They consider the following fishery model:
(1)x˙=Gx-qEx,
where x is biomass of fish population, G(x) is the biological net growth rate, q is catch ability coefficient, and E is fishing effort. This model was originally developed by Schaefer [4] as a management tool for the Eastern Tropical Pacific Tuna Fishery. In the original Schaefer model, G(x) was specified in “logistic” form
(2)Gx=rx1-xK,
where r and K are positive parameters called the “intrinsic growth rate” and the “carrying capacity,” respectively. The term h=qEx represents the rate of mortality imposed by the fishery, that is, the rate of catch, corresponding to a given input of fishing “effort” E. Their objective is to determine
(3)max⁡⁡h=qEx.
The study of predator-prey models with harvesting has also attracted the attention of researchers; see, for example, Brauer and Soudack [5–7], Beddington and Cooke [8], Dai and Tang [9], Hogarth et al. [10], Myerscough et al. [11], and Goh [12].Extensive and unregulated harvesting of fishes can even lead to the dropping of the overall number of fishing. Some associated documents have shown that the optimal initial investment amount of the fryN/2 is the optimal value to guarantee the optimal overall fishing amount [2], but they did not give the optimal rate of each fishing strategy and more important, however, is that when the initial investment amount of the fry is not N/2, how can we decide the optimal fishing strategy ?In this paper, we will give the discrete optimal control method of the optimal fishing strategy. This method is new and efficient for discrete optimal control problem, which is different from the other optimal methods such as the traditional variational method, the Pontryagin principle of maximum, and the discrete dynamic programming. The basic construction of the model is the traditional logistic function relating to the growth of fry. The discrete optimal control method for optimal fishing strategy was used to construct the optimal rate of each fishing strategy; the main focus of our work is on the rigorous mathematical analysis of the optimal control problem. The analysis allows one to obtain the optimal initial investment amount of the fry and the overall optimal fishing amount. Furthermore, when the initial investment amount of the fry is below or above the optimal value and the intrinsic growth rate of fishR is too small, we derive that fishing operations should not be started in the last few years to make the overall fishing amount optimal. However, the other optimal methods such as the traditional variational method require the controlled variable un not to be restricted; hence the variational method is not available. The discrete dynamic programming method is not convenient in the calculation for the logistic model is complex and not linear. The maximum principle in the calculation of the discrete control optimization is not convenient too. Hence, we pointed out that our method is new and efficient for discrete optimal control problem. Finally, several typical examples are given to illustrate the obtained results.The aim of this paper is to undertake the mathematical analysis of the optimal control problem introduced above, namely, to maximize (7) subject to (4)–(6). We also provide some numerical results that illustrate the theoretical results by several typical examples. We remark that there are few optimal control studies in the literature for discrete optimal control problem.An outline of the paper is as follows. In Section2, discrete optimal control method of the optimal fishing strategy is given. This method is used to construct the optimal rate of each fishing strategy and then obtain the optimal initial investment amount of the fry and the overall optimal fishing amount. In Section 3, we discuss that when the initial investment amount of the fry is not the optimal value and the intrinsic growth rate of fish R is too small, fishing operations should not be started in the last few years to make the overall fishing amount optimal, and then, several typical examples are given to illustrate the obtained results. In Section 4, we give a detailed table to show some figures to illustrate the obtained results. Finally, in Section 5 we draw a conclusion.


## 2. Discrete Optimal Control Method to Construct the Optimal Fishing Rate

Letxn denote the original amount of fry, which is the density of fish in the nth years. Let
(4)Fxn=R1-xnNxn
be the natural growth rate of fish, where R>0 is the intrinsic growth rate of fish, N>0 is the maximum carrying capacity of environment, which is assumed to be fixed within three years, and n=1,2,3 is the year under consideration. Let
(5)xn+1=1-unFxn
be the remaining amount of fish after n years, where un∈[0,1] is the degree of each fishing strategy in the nth years; choose the control set U⊂Rn, defined as
(6)U=u1,u2,u3∣0≤un≤1,n=1,2,3.We define the objective functional(7)Jun=∑n=13‍unFxn,
where unF(xn) represents the amount of fishing in the nth years. Our goal is to maximize J(un), that is, to find a un such that
(8)Jun*=max⁡uJun,
where un satisfies conditions (4)–(6).Theorem 1.
Assume thatn=3,R≥2, and conditions (4)–(6) hold; then there exists the optimal initial investment amount of the fry x1*=N/2, and the optimal rate of each fishing strategy is u1*=u2*=1-2/R, u3*=1 making the overall amount of fishing the optimal value J*=3RN/4-N.Proof.
The way we consider this problem is similar to the dynamic programming; to make the overall amount of fishing the optimal value, we letu3*=1, which means that we fish all in the last year:
(9)J3=u3Fx3=Fx3=R1-x3Nx3.
It is easy to see that J3 is a quadratic function about x3, when x3*=N/2, J3 to be the maximum J3*=RN/4, we have completed the third fishing progress.
From condition (5), we have
(10)N2=1-u2Fx2
and then we can derive that
(11)J2=u2Fx2=u2N/21-u2.
Derive J2, J2′=(N/2)/(1-u2)2>0. Obviously, J2 is an increasing function on u2. Since
(12)Fx2=R1-x2Nx2≤RN4
and when x2=N/2, the inequality could be equality. Adjoining this inequality with (10) gives
(13)N2=1-u2Fx2≤1-u2RN4
and this relation yields u2≤1-2/R. When u2*=1-2/R, then
(14)J2*=u2Fx2=R-2RRN4=R-2N4,
where x2*=N/2; we have completed the second fishing progress.
Similarly, from condition (5), we have
(15)N2=1-u1Fx1.
We can derive that
(16)J1=u1Fx1=u1N/21-u1.
Derive J1, J1′=(N/2)/(1-u1)2>0, where J1 is an increasing function on u1. Since
(17)Fx1=R1-x1Nx1≤RN4,
then
(18)N2=1-u1Fx1≤1-u1RN4.
We are able to obtain that u1≤1-2/R. When u1*=1-2/R, then
(19)J1*=u1Fx1=R-2RRN4=R-2N4,
where x1*=N/2; we have completed the first fishing progress.
To sum up, we can determine the optimal rate of each fishing strategy as(20)u1*=u2*=1-2R,u3*=1
and the overall amount of fishing to be the optimal value as
(21)J*=J1*+J2*+J3*=R-2N4+R-2N4+RN4=3RN4-N.
The optimal initial investment amount of the fry is x1*=N/2.Corollary 2.
Assume thatn=k,R≥2, and conditions (4)–(6) hold; then there exists the optimal initial investment amount of the fry x1*=N/2, and the optimal degree of each fishing strategy is
(22)u1*=u2*=⋯=uk-1*=1-2R,uk*=1
making the overall amount of fishing the optimal value
(23)J*=J1*+J2*+⋯+Jk-1*+Jk*=k-1R-2N4+RN4.Note 1.
In Theorem1 we take N to be fixed, since the capacity of environment cannot change greatly in three years. When time is more than three years or when it has been a long time, we can consider that N is possible to be changed, and recounting is necessary.Note 2.
In front of the proof, we say that the way we consider this problem is similar to the discrete dynamic programming. Actually, this problem cannot be solved by other optimal methods such as the traditional variational method, the Pontryagin principle of maximum, and the dynamic programming. This is because we need a hypothesis in discrete optimal control problem:δx(k) and δu(k) are linearly independent, which can be assumed for discrete optimal control problem of free terminal, but for discrete optimal control problem of fixed terminal, this assumption is not established for the state equation x(k+1)=f(x(k),u(k),k).From Theorem1 we can derive that the optimal initial investment amount of the fry is x1*=N/2, and the optimal rate of each fishing strategy is u1*=u2*=1-2/R, u3*=1, and then the overall amount of fishing is the optimal value J*=3RN/4-N. However, in realistic application the initial investment amount of the fry may be above or below the optimal value x1*=N/2 and R≥2 may not hold; then how can we decide the optimal degree of each fishing strategy un to make the overall amount of fishing the optimal value? These questions are investigated below.


## 3. Optimal Fishing Strategy When the Initial Investment Amount of the Fry Is Not the Optimal ValueN/2

Theorem 3.
Suppose thatn=3, x1=cN, c∈(0,1), 0<R1<1/2a, and a=c(1-c) where
(24)R1=13a27a-42-42-27a-441/3+13a27a-42-42+27a-441/3
and conditions (4)–(7) hold. Then one has the following optimal fishing strategy.(1)
WhenR∈(0,R1), one does not fish in the two previous kinds of fishing progress.(2)
WhenR∈[R1,1/2a)(R1<1/2a), one does not fish in the first fishing progress.(3)
WhenR∈[1/2a,+∞), one fishes in each kind of progress.
In every kind of fishing progress, one can determine the optimal fishing degreeun and obtain the overall optimal amount of fishing J.Proof.
Assumex1=cN, where c∈(0,1) is the density coefficient of the initial fry. From condition (5),
(25)Fx1=R1-x1Nx1=R1-ccN=RaN,
where a=c(1-c)∈(0,1/4). From condition (5),
(26)x2=1-u1Fx1.
Suppose that u1=0; then x2=F(x1)=RaN; when x2=RaN<N/2 (i.e., R<1/2a), u1 should be 0, or x2<F(x1)=RaN<N/2 does not reach the optimal investment value in the next fishing process.Step 1.
In the first fishing process, we divided the situation into two circumstances.(1)
Whenx2=RaN<N/2 (i.e., R<1/2a), we have u1*=0, J1*=u1*F(x1)=0(R<1/2a).(2)
Whenx2=RaN≥N/2 (i.e., R≥1/2a), we have u1*≠0; to decide u1* we set x2=x2*=N/2, from (5), where
(27)N2=1-u1RaN.We get
(28)u1*=1-12Ra
and then
(29)J1*=u1*Fx1=1-12RaRaN=N22Ra-1R≥12a.
It is easy to see that whenx2=RaN≥N/2 (i.e., R≥1/2a), our fishing amount J1*=(N/2)(2Ra-1)=RaN-N/2 is exactly equal to x2-x2*=RaN-N/2, which is just the exceeded amount.Step 2.
In the second fishing process, we divided the situation into two circumstances too.(1)
WhenR<1/2a, x2=RaN<N/2, from condition (4), we obtain
(30)Fx2=R1-x2Nx2=R1-RaRaN=R2aN1-Ra.From condition (5),
(31)x3=1-u2Fx2.
Suppose that u2=0, and then x3=F(x2)=R2aN(1-Ra); when x3=R2aN(1-Ra)<N/2 (i.e., R<R1<1/2a), u2 should be 0; otherwise x3<F(x2)<N/2 does not attain the optimal investment value in the next fishing process.(a)
Whenx3=R2aN(1-Ra)<N/2  (R<R1<1/2a), we have
(32)u2*=0,J2*=u2*Fx2=0R<R1<12a,where
(33)R1=13a27a-42-42-27a-441/3+13a27a-42-42+27a-441/3
which is the solution of x3=R2aN(1-Ra)=N/2 (i.e., a2R3-aR2+1/2=0).
To sum up, whenR∈(0,R1)(R1<1/2a), we have
(34)u1*=u2*=0,J1*=J2*=0x1=cN,x2=RaN<N2,x3=R2aN1-Ra<N2.(b)
Whenx3=R2aN(1-Ra)≥N/2  (R1≤R≤1/2a), we have u2*≠0; similar to Step 1 we set x3=x3*=N/2; from (5),
(35)N2=1-u2R2aN1-Ra.We get
(36)u2*=1-12R2a1-Ra
and then
(37)J2*=u2*Fx2=1-12R2a1-RaR2aN1-Ra=N22R2a1-Ra-1R1≤R≤12a.
Obviously, whenx3=R2aN(1-Ra)≥N/2  (R1≤R≤1/2a), we can verify that J2*=x3-x3*=R2aN(1-Ra)-N/2, which is just the exceeded amount.
To sum up, whenR∈[R1,1/2a)(R1<1/2a), we have
(38)u1*=0,J1*=0;u2*=1-12R2a1-Ra,J2*=N22R2a1-Ra-1x1=cN,x2=RaN<N2,x3*=N2.(2)
WhenR≥1/2a, x2=N/2 along with condition (4), we obtain
(39)Fx2=R1-x2Nx2=RN4.
From condition (5),
(40)x3=1-u2Fx2.
Suppose that u2=0, and then x3=F(x2)=RN/4; when x3=RN/4<N/2 (i.e., R<2), u2 should be 0; otherwise x3<F(x2)<N/2 does not attain the optimal investment value in the next fishing process.(a)
Whenx3=RN/4<N/2  (R<2), we have
(41)u2*=0,J2*=u2*Fx2=0R<2.Since c∈(0,1), then a=c(1-c)∈(0,1/4); we get 1/2a≥2, and therefore, R<2 and R≥1/2a have no intersection set.(b)
Whenx3=RN/4≥N/2  (R≥2), we have u2*≠0; we set x3=x3*=N/2; from (5),
(42)N2=1-u2RN4.We get
(43)u2*=1-2R
and then
(44)J2*=u2*Fx2=1-2RRN4=R-24NR≥2.
Since 1/2a≥2, therefore, R≥2 and R≥1/2a have the intersection set R≥1/2a.
Obviously, whenx3=RN/4≥N/2  (R≥2), we can verify that J2*=x3-x3*=RN/4-N/2, which is just the exceeded amount.
To sum up, whenR∈[1/2a,+∞), we have
(45)u1*=1-12Ra,J1*=N22Ra-1;u2*=1-2R,J2*=R-24Nx1=cN,x2*=N2,x3*=N2.Step 3.
In the third fishing process, we divided the situation into two circumstances.(1)
WhenR∈0,R1(R1<1/2a), we have x3=R2aN(1-Ra)<N/2 along with condition (4); we obtain
(46)Fx3=R1-x3Nx3=R1-R2a1-RaR2a1-RaN=R3a1-RaN1-R2a1-Ra.In the last fishing process, we let u3*=1, and then
(47)J3*=u3Fx3=R3a1-RaN1-R2a1-Ra.(2)
WhenR∈[R1,1/2a) or R∈[1/2a,+∞)(R1<1/2a), x3*=N/2 along with condition (4); we obtain
(48)Fx3=R1-x3Nx3=RN4.In the last fishing process, we let u3*=1, and then
(49)J3*=u3Fx3=RN4.
In summary, we obtain the following conclusion.(1)
WhenR∈(0,R1)(R1<1/2a), we have
(50)u1*=u2*=0,u3*=1;J1*=J2*=0,J3*=u3F(x3)=R3a1-RaN1-R2a1-RaJ*=J1*+J2*+J3*=0+0+u3Fx3=R3a1-RaN1-R2a1-Rax1=cN,x2=RaN<N2,x3=R2aN1-Ra<N2.It means that when R∈(0,R1)(R1<1/2a), we do not fish in the two previous kinds of fishing progress and fish all in the last fishing progress. This has no sense from a practical point of view. Therefore, we consider that not going in for fish farming is the better strategy.(2)
WhenR∈[R1,1/2a)(R1<1/2a), we have
(51)u1*=0,u2*=1-12R2a1-Ra,u3*=1;J1*=0,J2*=N22R2a1-Ra-1,J3*=u3Fx3=RN4J*=J1*+J2*+J3*=0+N22R2a1-Ra-1+RN4.x1=cN,x2=RaN<N2,x3*=N2.It means that when R∈[R1,1/2a)(R1<1/2a), we do not fish in the first fishing progress, among which it is not difficult to verify J2*=x3-x3* for x3≥x3*.(3)
WhenR∈[1/2a,+∞), we have
(52)u1*=1-12Ra,u2*=1-2R,u3*=1;J1*=N22Ra-1,J2*=R-24N,J3*=RN4J*=J1*+J2*+J3*=N22Ra-1+R-24N+RN4.x1=cN,x2*=N2,x3*=N2.It means that when R∈[1/2a,+∞), we fish in each kind of progress. In every kind of fishing progress, we can determine the optimal fishing degree un and obtain the overall optimal amount of fishing J, among which it is not difficult to verify J1*=x2-x2*, J2*=x3-x3* for x2≥x2* and x3≥x3*, respectively.Example 4.
Consider Theorem3  c=1/2 and x1=cN; determine the optimal strategy of fishing using the discrete optimal control method.
Sincec=1/2, then a=c(1-c)=1/4, 1/2a=2; we obtain R1=2 by calculating the function f(R)=R3-4R2+8=0, which come from the equality x3=(R2N/4)(1-R/4)=N/2. A more detailed procedure is described below.
Sincex1=cN=N/2,
(53)Fx1=R1-x1Nx1=R1-ccN=RaN=RN4
and then
(54)x2=1-u1Fx1.
Suppose that u1=0; then x2=F(x1)=RaN=RN/4; when x2=RaN=RN/4<N/2 (i.e., R<2), u1 should be 0.Step 1.
In the first fishing process, we divided the situation into two circumstances.(1)
WhenR<2, we have u1*=0, J1*=0.(2)
WhenR≥2, we have u1*≠0; we set x2=x2*=N/2; from (5),
(55)N2=1-u1RN4.We get
(56)u1*=1-2R
and then
(57)J1*=u1*Fx1=1-2RRN4=R-24N.Step 2.
In the second fishing process, we divided the situation into two circumstances too.(1)
WhenR<2, x2=RN/4<N/2; from condition (4), we obtain
(58)Fx2=R1-x2Nx2=R2N41-R4.From condition (5),
(59)x3=1-u2Fx2.
Suppose that u2=0, and then x3=F(x2)=(R2N/4)(1-R/4); when x3=(R2N/4)(1-R/4)<N/2  (i.e., R<2), u2 should be 0.
Note. To solve x3=(R2N/4)(1-R/4)<N/2, we only find the solution of the following problem:
(60)fR=R3-4R2+8>0.
In the derivatives of R,
(61)f′R=3R2-8R=R3R-8
and when R<0 or R>8/3, then f(R) is an increasing function; it is easy to see that f(2)=0, when 0<R<2, and the above inequality is right. It means that R1=1/2a=2.
To sum up, whenR∈(0,2), we have
(62)u1*=u2*=0,J1*=J2*=0x1=12N,x2=RN4<N2,x3=R2N41-R4<N2.(2)
WhenR≥2, we have u2*≠0, and x2*=N/2 along with condition (4); we obtain
(63)Fx2=R1-x2Nx2=RN4and we set x3=x3*=N/2; from condition (5),
(64)N2=1-u2RN4.
We get
(65)u2*=1-2R
and then
(66)J2*=u2*Fx2=R-2N4.
We have
(67)u1*=1-2R,J1*=R-2N4;u2*=1-2R,J2*=R-2N4x1*=N2,x2*=N2,x3*=N2.Step 3.
In the third fishing process, we divided the situation into two circumstances.(1)
WhenR∈(0,2), we have x3=(R2N/4)(1-R/4)<N/2 along with condition (4); we obtain
(68)Fx3=R1-x3Nx3=R31-R/4N41-R21-R/44.In the last fishing process, we let u3*=1, and then
(69)J3*=u3Fx3=R31-R/4N41-R21-R/44.(2)
WhenR∈[2,+∞), x3=N/2 along with condition (4); we obtain
(70)Fx3=RN4.In the last fishing process, we let u3*=1, and then
(71)J3*=u3Fx3=RN4.
In summary, we obtain the following conclusion.(1)
WhenR∈(0,2), we have
(72)u1*=u2*=0,u3*=1;J1*=J2*=0,J3*=R31-R/4N41-R21-R/44.J*=J1*+J2*+J3*=0+0+u3Fx3=R31-R/4N41-R21-R/44.x1=12N,x2=RN4<N2,x3=R2N41-R4<N2.It means that when R∈(0,2), we do not fish in the two previous kinds of fishing progress. It is easy to see that
(73)J*=J1*+J2*+J3*=0+0+u3Fx3≤RN4<N2=x1*.
It means that the maximum harvesting amount J* is less than the initial optimal investment amount x1*=N/2 within three years; this has no sense from a practical point of view, let alone x1≠N/2. Therefore, we consider that when the intrinsic growth rate of fish R<2, not going in for fish farming is the better strategy. On the contrary, when R≥2, we consider that setting x1*=N/2 is the best policy decision and then determine un and Jn.(2)
WhenR∈[2,+∞), we have
(74)u1*=1-2R,u2*=1-2R,u3*=1;J1*=R-2N4,J2*=R-2N4,J3*=RN4J*=J1*+J2*+J3*=R-2N4+R-2N4+RN4=3RN4-N.x1*=N2,x2*=N2,x3*=N2.It means that when R∈[2,+∞), we fish in each kind of progress. In every kind of fishing progress, we can determine the optimal fishing degree un and obtain the overall optimal amount of fishing Jn.
Note. When R∈[2,+∞), c=1/2, the result of Example 4 is just the very thing of Theorem 1. It means that Theorem 1 is a special case of Theorem 3.Example 5.
Consider Theorem3  x1=cN; we set c=1/3 and determine the optimal strategy of fishing using the discrete optimal control method.
For convenience, we only give the final result. Sincec=1/3, then a=c(1-c)=2/9, 1/2a=9/4=2.25; we obtain R1≈2.02 by calculating the function f(R)=2R3-9R2+20.25=0, which come from the equality x3=(2R2(9-2R)/81)N=N/2.(1)
WhenR∈(0,2.02), we have
(75)u1*=u2*=0,u3*=1;J1*=J2*=0,J3*=2R39-2RN81-2R29-2R812J*=J1*+J2*+J3*=2R39-2RN81-2R29-2R812x1=N3,x2=2RN9<N2,x3=2R2N9-2R81<N2.It means that when R∈(0,2.02), we do not fish in the two previous kinds of fishing progress. It is easy to see that
(76)J*=J1*+J2*+J3*=0+0+u3Fx3≤RN4,
where J* may be greater than N/2=x1 only when R∈[2,2.02). It means that the maximum harvesting amount J1*+J2* is less than the initial optimal investment amount x1=N/2 within two years, and the maximum harvesting amount J* is less likely to be more than the initial optimal investment amount x1=N/2 within three years; this has no sense from a practical point of view. Therefore, we consider that when the intrinsic growth rate of fish R<2.02, not going in for fish farming is the better strategy.(2)
WhenR∈[2.02,2.25), we have
(77)u1*=0,u2*=1-814R29-2R,u3*=1;J1*=0,J2*=4R29-2R-8181×2N,J3*=RN4J*=J1*+J2*+J3*=0+4R29-2R-8181×2N+RN4.x1=N3,x2=2RN9<N2,x3=N2.It means that when R∈[2.02,2.25), we do not fish in the first fishing progress. It is easy to see that the maximum harvesting amount J1* is less than the initial optimal investment amount x1=N/2 within one year; this has no sense from a practical point of view.(3)
WhenR∈[2.25,+∞), we have
(78)u1*=1-94R,u2*=1-2R,u3=1;J1*=4R-918N,J2*=R-24N,J3*=RN4J*=J1*+J2*+J3*=4R-918N+R-24N+RN4.x1=N3,x2*=N2,x3*=N2.It means that when R∈[2.25,+∞), we fish in each kind of progress. In every kind of fishing progress, we can determine the optimal fishing degree un and obtain the overall optimal amount of fishing Jn.
Note. When Example 5  x1=cN, c=1-1/3=2/3, we can obtain the same optimal strategy of fishing as Example 5. This is because no matter whether c=1/3 or 2/3, a=c(1-c)=(1/3)(2/3)=2/9 and F(x1)=R(1-x1/N)x1=2N/9 is unchanging.Corollary 6.
No matter whether Theorem3  x1=c1N or x1=c2N, one can obtain the same optimal strategy of fishing on condition that c1=1-c2.Corollary 7.
Some people mistakenly think that “more is better” when they give the initial investment amount, but in fact this is not the case. Only whenx1=N/2 can one obtain the optimal overall fishing amount.


## 4. Discussion

From Sections2 and 3 and Examples 4 and 5, we set n=3 and obtain the following conclusions.When the initial investment amount of fryx1=N/2, R≥2, we can determine un(n=1,2,3) to make the overall fishing amount optimal. It is easy to see that when R<2, we cannot guarantee RN/4≥N/2. It means that the maximum harvesting amount unF(xn)=RN/4 is less than the initial optimal investment amount N/2, which has no sense from a practical point of view, let alone x1≠N/2. Therefore, we consider that when the intrinsic growth rate of fish R<2, not going in for fish farming is the better strategy. On the contrary, when R≥2, setting x1*=N/2 is the best policy decision. Assume that x1≠N/2; if R is large enough, we only have J1<J1*, and the other Jn is possible to reach Jn*(n=2,3); we do not cause great losses except for the first year.In Section3, we set n=3 and obtain a series of conclusions. If n>3, we can also gain some similar conclusions. When the initial investment amount of the fry is below or above the optimal value and the intrinsic growth rate of fish R is too small, we derive that fishing operations should not be started in the last few years to make the overall fishing amount optimal. To illustrate our conclusions clearly, we list the optimal total fishing amount J as follows.Assume thatn=3, x1=cN, and a=c(1-c), when c=1/2,1/3 or 2/3 and N=1000, detailed in Table 1.Table 1
x
1
1
/
2
a
R
1
R
(
0
<
R
1
≤
1
/
2
a
)
J
1
J
2
J
3
J
N
/
2
2
2
R
=
1
∈
(
0,2
)
0
0
0.15
N
0.15
N
=
150
R
=
2
∈
[
2
,
+
∞
)
0
0
N
/
2
0.5
N
=
500
R
=
2.20
∈
[
2
,
+
∞
)
0.05
N
0.05
N
0.55
N
0.65
N
=
650
R
=
4
∈
[
2
,
+
∞
)
N
/
2
N
/
2
N
2
N
=
2000
N
/
3 or 2N/3
2.25
2.02
R
=
1
∈
(
0,2.02
)
0
0
0.14
N
0.14
N
=
140
R
=
2
∈
(
0,2.02
)
0
0
0.49
N
0.49
N
=
499
R
=
2.20
∈
[
2.02,2.25
)
0
0.05
N
0.55
N
0.6
N
=
600
R
=
4
∈
[
2.25
,
+
∞
)
7
N
/
18
N
/
2
N
1.89
N
=
1890From Table1 we see that the overall optimal amount J of fishing is different when the initial investment amount of fry x1 is different; only when initial investment amount of fry is x1=N/2, is the overall fishing amount optimal. On the other hand, in every kind of fishing progress, we can determine the optimal fishing degree un and obtain the overall optimal amount of fishing J. It is easy to see that when R<R1, we do not fish in the two previous kinds of fishing progress; when R∈[R1,1/2a)(R1<1/2a), we do not fish in the first fishing progress; when R∈[1/2a,+∞), we fish in each kind of progress.


## 5. Conclusion

From the rigorous mathematical analysis of the optimal control problem, we obtain the optimal initial investment amount of the fry and the optimal size of the total catch. When the initial investment amount of fry attains the optimal valuex1=N/2, we can determine un(n=1,2,3) to make the overall fishing amount optimal. Furthermore, when the initial investment amount of the fry is below or above the optimal value x1=N/2 and the intrinsic growth rate of fish R is too small, we derive that fishing operations should not be started in the last few years to make the overall fishing amount optimal.



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*Source: 727531-2015-01-11.xml*