DESIMAL: JURNAL MATEMATIKA

Rice is one of the most widely consumed food sources in Indonesia. The success of the rice harvest can be influenced by several factors, one of which is plant pests. This can threaten national food security. Predator-prey mathematical model can be constructed to explain the relationship between rice plants with plant pests. The predator prey model consist of ODE system describing 2 predators level one (rats and sparrows), 1 predator level two (snake) and prey (rice plants). In this model, nine equilibrium points are obtained with analysis of the behavior of the model at each of its equilibrium points. We successfully simulated the model using hypothetical parameters and the results are agrees with the analysis behavior. Several factors will make plant pest populations decrease even lost from the population is the natural death and interaction of rice with plant pests.


INTRODUCTION
The Indonesia is a agrarian country where most of the people work as a farmer (Anitasari, 2019;Mulyadi et al, 2020).According to the Central Bureau of Statistics (BPS) consumption of rice every year always increasing in Indonesia.However, data from BPS also shows that production rice fell in 2019-2021.This matter will be a problem if inadequate rice availability to meet the demand for rice.The decline in the level of rice production can caused by crop failure as a result of plant pests.Manueke et al (2017) in his research proves that sparrow (Passer sp.) and mice (Rattus argentiventer) is a pest main rice crop.In its habitat at fields, rats and birds are prey for snakes.(Sabrina, et al. 2015).

atter will be
a problem if inadequate rice availability to meet the demand for rice.The decline in the level of rice production can caused by crop failure as a result of plant pests.Manueke et al (2017) in his research proves that sparrow (Passer sp.) and mice (Rattus argentiventer) is a pest main rice crop.In its habitat at fields, rats and birds are prey for snakes.(Sabrina, et al. 2015).

The relationship between rice plants with plant pests as well as predators level two can be constructed into a mathematical model that is a predatorprey model (Ardestani, 2020;Padilah et al, 2018;Sun et al, 2016).The predator are organisms that eat prey.The assumptions used in this model is prey growth satisfying logistic growt The relationship between rice plants with plant pests as well as predators level two can be constructed into a mathematical model that is a predatorprey model (Ardestani, 2020;Padilah et al, 2018;Sun et al, 2016).The predator are organisms that eat prey.The assumptions used in this model is prey growth satisfying logistic growth model (Streipert et al, 2014).The predator-prey model in this study discusses the 1 prey model (rice plants), 2 predators of level one that is the pests of rice plants (rats and sparrow) and 1 predator level two (snake).In this research, a predator-prey model will be constructed in the form of a system of linear differential equations.We will determine the equilibrium point and carry out a stability analysis at the equilibrium point to determine the behavior of the system.Based on the behavior of the system, several influencing factors will be investigated so that the number of pests can decrease from the population.We perform numerical simulations using the MATLAB software to support our analysis.
model (Streipert et al, 2014).The predator-prey model in this study discusses the 1 prey model (rice plants), 2 predators of level one that is the pests of rice plants (rats and sparrow) and 1 predator level two (snake).In this research, a predator-prey model will be constructed in the form of a system of linear differential equations.We will determine the equilibrium point and carry out a stability analysis at the equilibrium point to determine the behavior of the system.Based on the behavior of the system, several influencing factors will be investigated so that the number of pests can decrease from the population.We perform numerical simulations using the MATLAB software to support our analysis.

Therefore, the purpose of this research is to analyze the behavior of the mathematical predator-prey model of rice plants, sparrows, rats and snake.The organization of this paper are as follows.Method are described in Sec 2.Here we provide the research step.In Sec. 3 we construct a system of linear ordinary differential equation of pr Therefore, the purpose of this research is to analyze the behavior of the mathematical predator-prey model of rice plants, sparrows, rats and snake.The organization of this paper are as follows.Method are described in Sec 2.Here we provide the research step.In Sec. 3 we construct a system of linear ordinary differential equation of predator-prey model.In Sec. 4 we conduct a stability analysis whereas the simulation is conducted in Sec. 5 Conclusions and remarks will be given in the last section.
dator-prey model.In Sec. 4 we conduct a stability analysis whereas the simulation is conducted in Sec. 5 Conclusions and remarks will be given in the last section.


METHOD


Predator-Prey Model

Organisms in this predator-prey model were divided into: rice (P), sparrow (B), rat (T), and snake (U).The assumptions used in this model are as follows, (1) prey and predator populations are closed so there is no migration; (2) There is no competition between level 1 predators; (3) no disease in both prey an

Predator-Prey Model
Organisms in this predator-prey model were divided into: rice (P), sparrow (B), rat (T), and snake (U).The assumptions used in this model are as follows, (1) prey and predator populations are closed so there is no migration; (2) There is no competition between level 1 predators; (3) no disease in both prey and predators; (4) The parameters used are positive while the variables are nonnegative; (5) Prey growth follows logistic growth.

predato
s; (4) The parameters used are positive while the variables are nonnegative; (5) Prey growth follows logistic growth.


Changes in Population of Rice (Prey) with respect to time

Changes in the rice population P were influenced by interactions with 2 first-level predators, namely rats and birds.If there is no interaction, the rice will grow by following the logistic growth model (assuming a maximum rice growth of 100%).The interaction between rats and rice with an interaction level of  and the interaction between sparrows and rice with an interaction level of  causes a reduction in the amount of rice.Therefore, the model of population change over time is given as follows:
𝑑𝑃 𝑑𝑡 = 𝑝𝑃(1 −

Changes in Population of Rice (Prey) with respect to time
Changes in the rice population P were influenced by interactions with 2 first-level predators, namely rats and birds.If there is no interaction, the rice will grow by following the logistic growth model (assuming a maximum rice growth of 100%).The interaction between rats and rice with an interaction level of  and the interaction between sparrows and rice with an interaction level of  causes a reduction in the amount of rice.Therefore, the model of population change over time is given as follows:

Changes in Population of Rats (Predator) with respect to time
nges in Population of Rats (Predator) with respect to time

The growth of the rat population is due to the large number of interactions between rats and rice for .Meanwhile, the reduction in the number of rat populations was affected by the number of rats interacting with level 2 predators, i.e. snakes with the amount of δ and natural deaths experienced by rats for .Based on this description, the model of rats population change over ti The growth of the rat population is due to the large number of interactions between rats and rice for .Meanwhile, the reduction in the number of rat populations was affected by the number of rats interacting with level 2 predators, i.e. snakes with the amount of δ and natural deaths experienced by rats for .Based on this description, the model of rats population change over time is given as follows:

is given as follows:

(2)


Changes in Population of Spar
(2)

Changes in Population of Sparrows (Predator) with respect to time
ows (Predator) with respect to time

The growth of the sparrow population is due to the large number of interactions between the sparrow and rice by .Meanwhile, the reduction in the number of sparrow populations was affected by the large number of sparrows interacting with level 2 predators, i.e. snakes for  and natural deaths experienced by sparrows for .Based on this description, the model of sparrows population change over time is given as follows:
𝑑𝐵 𝑑𝑡 = 𝑎𝑃𝐵 − 𝜔𝑈𝐵 − 𝑏𝐵 (3)

Changes in Populati The growth of the sparrow population is due to the large number of interactions between the sparrow and rice by .Meanwhile, the reduction in the number of sparrow populations was affected by the large number of sparrows interacting with level 2 predators, i.e. snakes for  and natural deaths experienced by sparrows for .Based on this description, the model of sparrows population change over time is given as follows:

ator level 2) with respect to time

The growth of the snake po
The growth of the snake population is due to the large number of interactions between snakes and sparrows of  and interactions between snakes and rats of .Meanwhile, the reduction in the number of snake populations is influenced by the number of snakes that experience natural death of .Based on this description, a model of changes in the rat population over time can be made as follows: ulation is due to the large number of interactions between snakes and sparrows of  and interactions between snakes and rats of .Meanwhile, the reduction in the number of snake populations is influenced by the number of snakes that experience natural death of .Based on this description, a model of changes in the rat population over time can be made as follows:
𝑑𝑈 𝑑𝑡 = 𝑤𝑈𝐵 − 𝑑𝑈𝑇 − 𝑢𝑈(4)
From ( 1)-( 4), w From ( 1)-( 4), we can construct mathematical model of predator-prey between rice plants, sparrows, rats and snakes as follows:

Stability Analysis of Equilibrium Points
Stability analysis was carried out at nine equilibrium points by looking at the eigenvalues at each point to determine the behavioral of the model.Due to limitations and capabilities, there are several equilibrium points where it cannot be known whether the eigenvalues of that point are negative or positive.In addition, there are several equilibrium points that have negative values, so further analysis is not carried out, because population values in real life will not be negative.a) Stability analysis at  1 If the equilibrium point  1 is substituted into the Jacobian matrix, we get

lues at each point to determine the behavioral of the model.Due to limit
tions and capabilities, there are several equilibrium points where it cannot be known whether the eigenvalues of that point are negative or positive.In addition, there are several equilibrium points that have negative values, so further analysis is not carried out, because population values in real life will not be negative.a) Stability analysis at  1 If the equilibrium point  1 is subst to the Jacobian matrix, we get
𝐽 1 ≔ [ 𝑝 0 0 0 0 −𝑏 0 0 0 0 −𝑡 0 0 0 0 −𝑢 ]
with eigen values: , −, −, −.Because the values of with eigen values: , −, −, −.Because the values of all the parameters used are positive, the first eigenvalue will always be positive, so that the stability at point  1 becomes unstable.b) Stability analysis at  2  2 = ( * ,  * ,  * ,  * ) = (0,   , 0, −   ) ve, the first eigenvalue will always be positive, so that the s point  1 becomes unstable.b) Stability analysis at  2  2 = ( * , Because the equilibrium point value of  * is negative due to the assumption that all parameters are positive, for stability at point  2 no further analysis is carried out.c) Stability analysis at  3  3 = ( * ,  * ,  * ,  * ) = (0,0,   , −   ) um point value of  * is ne

positive, for stability at point  2 no f
rther analysis is carried out.c) Stability analysis at  3  3 = ( * ,  * ,  * ,  * ) = (0,0,   , −   )

Because the equilibrium point value of  * is negative due to the assumption that all parameters are positive, for stability at point  3 no further analysis is carried out.d) Stability analysis at  4

If the equilibrium point  4 is substituted into the Jacobian matrix, we get   with eigen values:
𝑎𝑡 − 𝑏𝑦 𝑦 , − 𝑑𝑝𝑡 − 𝑑𝑝𝑦 + 𝛾𝑢𝑦 𝛾𝑦 , 1 2 −𝑡𝑝 + √𝑝 2 𝑟 2 + 4𝑝𝑡 2𝑦 − 4𝑝𝑡𝑦 2 𝑦 , − 1 2 𝑡𝑝 + √𝑝 2 𝑟 2 + 4𝑝𝑡 2𝑦 − 4𝑝𝑡𝑦 2 𝑦 Because the equilibrium point value of  * is negative due to the assumption that all parameters are positive, for stability at point  3 no further analysis is carried out.d) Stability analysis at  4 If the equilibrium point  4 is substituted into the Jacobian matrix, we get   with eigen values: Taking into account that all parameter values are positive, so that the first eigenvalue is negative if  <   , the second eigenvalue will be negative if  < − (−+)  , the third eigenvalue will be negative if  > 0, and the fourth eigenvalue will be negative if  >  The stability of point  5 depends on the four eigenvalues, so that the stability of point  5 is local asymptotically stable if  < / ,  < −(− + ) /,  > 0, and  > .Because the equilibrium point value of  * is negative due to the assumption that all parameters are positive, for stability at point  6 no further analysis is carried out.
aking into account that all parameter values are positive, so that the first eigenvalue is negative if  <   , the second eigenvalue will be negative if  < − (−+)  , the third eigenvalue will be negative if  > 0, and the fourth eigenvalue will be negative if  >  The stability of point  5 depends on the four eigenvalues, so that the stability of point  5 is local asymptotically stable if  < / ,  < −(− + ) /,  > 0, and  > .Because the equilibrium point value of  * is negative due to the assumption that all parameters are positive, for stability at point  6 no further analysis is carried

t.

g) Stability analysis a
g) Stability analysis at  7 If the equilibrium point  7 is substituted into the Jacobian matrix, we get 7 If the equilibrium point  7 is substituted into the Jacobian matrix, we 𝛼 0 0 − 𝑝(𝑎 − 𝑏)𝜔 𝑎𝛼 0 0 𝑏𝑦 𝑎 − 𝑡 0 0 0 0 𝑝(𝑎 − 𝑏)𝑤 𝑎𝛼 − 𝑢 ]
with eigen values:
− 𝑎𝑡 − 𝑏𝑦 𝑎 , − 𝑎𝛼𝑢 − 𝑎𝑝𝑤 + 𝑏𝑝𝑤 𝑎𝛼 , 1 2 −𝑏𝑝 + √−4𝑎 2 𝑏𝑝 + 4𝑎𝑏 2 𝑝 + 𝑏 2 𝑝 2 𝑎 , − 1 2 𝑏𝑝 + √−4𝑎 2 𝑏𝑝 + 4𝑎𝑏 2 𝑝 + 𝑏 2 𝑝 2 𝑎
Taking into account that all parameter values are positive, so that the first eigenvalue is negative if  < / , the second eigenvalue will be negative if  < −( − )/, the third eigenvalue will be negative if  > 0, and the fourth eigenvalue will be negative if  > with eigen values: Taking into account that all parameter values are positive, so that the first eigenvalue is negative if  < / , the second eigenvalue will be negative if  < −( − )/, the third eigenvalue will be negative if  > 0, and the fourth eigenvalue will be negative if  >  The stability of point  5 depends on the four eigenvalues, so that the stability of point  5 is locally asymptotically stable if  < / ,  < −( − )/ ,  > 0, and  > .h) Stability analysis at  8 If the equilibrium point  8 is substituted into the Jacobian matrix, we get y of point  5 depends on the four eigenvalues, so that the stability of poi , and  > .h) Stability analysis at  8 If the equilibrium point  8 is substituted into the Jacobian matrix, we get
𝐽 8 ≔ [ 𝛼𝑢 − 𝑝𝑤 𝑤 𝛼(𝛼𝑢 − 𝑝𝑤) 𝑝𝑤 𝛾(𝛼𝑢 − 𝑝𝑤) 𝑝𝑤 0 𝑎𝑢 𝑤 0 0 − 𝑢𝜔 𝑤 0 0 𝑎𝛼𝛿𝑢 − 𝑎𝛿𝑝𝑤 − 𝛼𝜔𝑢𝑦 + 𝑏𝛿𝑝𝑤 − 𝜔𝑝𝑡𝑤 + 𝜔𝑝𝑤𝑦 𝜔𝑝𝑤 0 0 − 𝑎𝛼𝑢 − 𝑎𝑝𝑤 + 𝑏𝑝𝑤 𝜔𝑝 − 𝑑(𝑎𝛼𝑢 − 𝑎𝑝𝑤 + 𝑏𝑝𝑤) 𝜔𝑝𝑤 0 ]
The characteristic equations and eige The characteristic equations and eigenvalues are quite long and tedious so they will not be analyzed and presented here.For this reason, the stability analysis at  8 will be investigated with numerical assistance.i) Stability analysis at  9 If the equilibrium point  9 is substituted into the Jacobian matrix, we get The characteristic equations and eigenvalues are quite long and tedious so they will not be analyzed and presented here.For this reason, the stability analysis at  9 will be investigated with numerical assistance.
ed into the Jacobian matrix, we get The characteristic equations and eigenvalues are quite long and tedious so they will not be analyzed and presented here.For this reason, the stability analysis at  9 will be investigated with numerical assistance.


RESULTS AND DISCUSSION

Here we conduct a numerical simulation using hypothetical parameter to support and to see the suitability of our previous analysis.The simulati

RESULTS AND DISCUSSION
Here we conduct a numerical simulation using hypothetical parameter to support and to see the suitability of our previous analysis.The simulation for  1 will not be carried out because at point  1 stability is not achieved, while at points  2 ,  3 , and  6 the simulation will not be carried out because one of the values at the equilibrium point is negative.

for  1 will not be carried out be
ause at point  1 stability is not achieved, while at points  2 ,  3 , and  6 the simulation will not be carried out because one of the values at the equilibrium point is negative.


Simulation in Equilibrium Point 𝐸 4

In this model simulation, the following parameters are used: Parameter values in Table 1 give simulation results as plotted in Figure 1 Figure 1., , , and  plots against time ,

Simulation in Equilibrium Point 𝐸 4
In this model simulation, the following parameters are used: Parameter values in Table 1 give simulation results as plotted in Figure 1 Figure 1., , , and  plots against time , , , and  will converge a value: 0.9999; 5.9672e-06; 9.5439e-47; and 9.7566e-08.This is in accordance with the search for the equilibrium point  4 , where the rice population will live and the others will become extinct.

, and  will converge a value: 0.99
9; 5.9672e-06; 9.5439e-47; and 9.7566e-08.This is in accordance with the search for the equilibrium point  4 , where the rice population will live and the others will become extinct.


Simulation in Equilibrium Point 𝐸 5

In this model simulation, the following parameters are used:  , , , and  will converge a value: 0.1211; 0.8826; 4.9041e-08; and 0.2913.This is in accordance with the search for

Simulation in Equilibrium Point 𝐸 5
In this model simulation, the following parameters are used:  , , , and  will converge a value: 0.1211; 0.8826; 4.9041e-08; and 0.2913.This is in accordance with the search for the equilibrium point  8 , where the rice, sparrows dan snakes populations will live and rats will become extinct.Although the stability analysis cannot be determined, numerical simulations can determine that the equilibrium point is towards local asymptotic stability.

e equilibrium point  8 , where the
rice, sparrows dan snakes populations will live and rats will become extinct.Although the stability analysis cannot be determined, numerical simulations can determine that the equilibrium point is towards local asymptotic stability.


Simulation in Equilibrium Point 𝐸 9

In this model simulation, the following parameters are used: Parameter values in Table 5 give simulation results as plo

Simulation in Equilibrium Point 𝐸 9
In this model simulation, the following parameters are used: Parameter values in Table 5 give simulation results as plotted in Figure 5 Figure 5. , , , and  plots against time , , , and  will converge a value: 0.7835; 0.6192; 1.08614; and 0.7459.This is in accordance with the search for the equilibrium point  9 , where all of the populations will live.Although the stability analysis cannot be determined, numerical simulations can determine that the equilibrium point is towards local asymptotic stability.

ed in Figure 5 Figure 5. , , , and
plots against time , , , and  will converge a value: 0.7835; 0.6192; 1.08614; and 0.7459.This is in accordance with the search for the equilibrium point  9 , where all of the populations will live.Although the stability analysis cannot be determined, numerical simulations can determine that the equilibrium point is towards local asymptotic stability.


CONCLUSIONS AND SUGGESTIONS

Based on the description of the results and discussion, it can be concluded: (1) the predator-prey model that has been constructed is a system of linear differential equations

CONCLUSIONS AND SUGGESTIONS
Based on the description of the results and discussion, it can be concluded: (1) the predator-prey model that has been constructed is a system of linear differential equations with four variables; (2) 9 equilibrium points are obtained where the behavior of the model at point  1 is unstable,  2 ,  3 ,  6 is not analyzed further because one of the equilibrium point values is negative and  4 ,  5 ,  7 ,  8 ,  9 is locally asymptotically stable, well proven through eigenvalues and numerical simulations; (3) The results of the model simulation using MATLAB are in accordance with the analysis.

ith four variables; (2) 9 equilibr
um points are obtained where the behavior of the model at point  1 is unstable,  2 ,  3 ,  6 is not analyzed further because one of the equilibrium point values is negative and  4 ,  5 ,  7 ,  8 ,  9 is locally asymptotically stable, well proven through eigenvalues and numerical simulations; (3) The results of the model simulation using MATLAB are in accordance with the analysis.

We can conclude the influence factors are the natural death rate and the interaction level of rice with both plant pests and the level of interaction between pests and second-level predators.Our results can be improved and elaborated in future resea We can conclude the influence factors are the natural death rate and the interaction level of rice with both plant pests and the level of interaction between pests and second-level predators.Our results can be improved and elaborated in future research to simulate real world data in any province.

h to simulate real world dat
in any province.



The research steps in this paper are as follows: (1) Determine model assumptions.Assumptions are used to form the limits of modelling; (2) Determine the variables and parameters of the model; (3) Construct a model by presenting the relationship between prey and predator into a system of linear differential equations; (4) Determine the equilibrium point with the help of Maple software; (5) Analyze the behavior of the model by looking at the stability of the eigenvalues of the characteristic equation; (6) Simulation are conducted using MATLAB with hypothetical The research steps in this paper are as follows: (1) Determine model assumptions.Assumptions are used to form the limits of modelling; (2) Determine the variables and parameters of the model; (3) Construct a model by presenting the relationship between prey and predator into a system of linear differential equations; (4) Determine the equilibrium point with the help of Maple software; (5) Analyze the behavior of the model by looking at the stability of the eigenvalues of the characteristic equation; (6) Simulation are conducted using MATLAB with hypothetical parameter to support and to see the suitability between the analysis and simulation; (7) Interpretation and discussion are given.

Table 1 .
Parameter values at point  4

Table 2 .
Parameter values at point  5

Table 5 .
Parameter values at point  9