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Talla de Madurez

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Talla de Madurez

Benjamin Suarez

1) Determinar talla media de primera madurez ($ L_{50} $) mediante el calculo de la ojiva de madurez sexual para Jurel en distintas zonas aplicando verosimilitud.

Para calcular la proporcion estimada de ejemplares maduros se utilizo la siguiente formula:

\[ P_{L} = \frac{1}{(1 + exp^{-(B_{0} + B_{1} \ast L_{t})})} \]

La funcion de verosimilitud se define como:

\[ LogLike = M \ast log(P_{L}) + (I \ast log(1 - P_{L})) \]

donde $ M $ es la cantidad de ejemplares maduros, $ I $ la cantidad de ejemplares inmaduros y $ P_{L} $ la proporcion estimada de ejemplares maduros.

2) Determinar por que la talla media de primera madurez es igual al cuociente negativo de los parametros de la ojiva de madurez.

\[ L_{50} = \frac{-B_{0}}{B_{1}} \]

Esta ecuacion resulta del despeje realizado en la ecuacion mediante la que se obtiene la proporcion estimada de ejemplares maduros. Se ha establecido que la talla media de primera madurez sexual corresponde al momento en que el 50% de los ejemplares muestreados se encuentran maduros (Cerna & Oyarzun, 1998; Leal & Oyarzun 2003; Leal et al. 2011), por lo que el despeje resulta de la siguiente manera:

\[ P_{L} = \frac{1}{(1 + exp^{-(B_{0} + B_{1} \ast L_{t})})} \]

\[ 0.5 = \frac{1}{(1 + exp^{-(B_{0} + B_{1} \ast L_{t})})} \]

\[ 0.5 \ast (1 + exp^{-(B_{0} + B_{1} \ast L_{t})}) = 1 \]

\[ 0.5 + 0.5 \ast exp^{-(B_{0} + B_{1} \ast L_{t})} = 1 \]

\[ 0.5 \ast exp^{-(B_{0} + B_{1} \ast L_{t})} = 1 - 0.5 \]

\[ 0.5 \ast exp^{-(B_{0} + B_{1} \ast L_{t})} = 0.5 \]

\[ exp^{-(B_{0} + B_{1} \ast L_{t})} = \frac{0.5}{0.5} \]

\[ exp^{-(B_{0} + B_{1} \ast L_{t})} = 1 \]

\[ - B_{0} + B_{1} \ast L_{t} = Ln(1) \]

\[ - B_{0} + B_{1} \ast L_{t} = 0 \]

\[ B_{1} \ast L_{t} = B_{0} \]

\[ L_{t} = \frac{B_{0}}{B_{1}} \]

A continuacion se procede a desarrollar el calculo de la $ L_{50} $:

Incidencia de hembras de jurel maduras e inmaduras microscopicamente por estrato de talla, colectadas en la zona norte y sur de Chile durante 1999.

Lt <- c(20:35) frec <- c(2, 2, 52, 102, 151, 216, 250, 208, 83, 24, 32, 21, 17, 9, 7, 3) m <- c(0, 0, 17, 39, 70, 144, 207, 185, 80, 23, 32, 21, 17, 9, 7, 3) I <- rep(NA, 16) mad.obs <- rep(NA, 16) for (i in 1:16) { I[i] <- frec[i] - m[i] mad.obs[i] <- m[i]/frec[i] } data.frame(Lt, frec, m, I, mad.obs)

## Lt frec m I mad.obs ## 1 20 2 0 2 0.0000 ## 2 21 2 0 2 0.0000 ## 3 22 52 17 35 0.3269 ## 4 23 102 39 63 0.3824 ## 5 24 151 70 81 0.4636 ## 6 25 216 144 72 0.6667 ## 7 26 250 207 43 0.8280 ## 8 27 208 185 23 0.8894 ## 9 28 83 80 3 0.9639 ## 10 29 24 23 1 0.9583 ## 11 30 32 32 0 1.0000 ## 12 31 21 21 0 1.0000 ## 13 32 17 17 0 1.0000 ## 14 33 9 9 0 1.0000 ## 15 34 7 7 0 1.0000 ## 16 35 3 3 0 1.0000

plot(Lt, mad.obs, xlim = c(10, 35), ylim = c(0, 1), xlab = "Longitud total (cm)", ylab = "Proporcion Madura") # Modelo no lineal con verosimilitud negloglike <- function(parms) { a <- parms[1] b <- parms[2] pl <- 1/(1 + exp(-(a + b * Lt))) -sum(m * log(pl) + (I * log(1 - pl))) } m1 <- optim(par = c(22, -2), fn = negloglike, method = "BFGS", hessian = TRUE) LM1 <- -m1$par[1]/m1$par[2] lt <- seq(10, 35, 0.1) mad.est <- 1/(1 + exp(-(m1$par[1] + m1$par[2] * lt))) lines(lt, mad.est) text(16, 0.8, expression(P(L) == frac(1, 1 + exp(-(a + b * L))))) text(16, 0.6, expression(LM1 == 23.86))

Incidencia de hembras de jurel maduras e inmaduras microscopicamente por estrato de talla, colestadas en la zona norte durante 1999.

Lt <- c(20:35) frec <- c(2, 2, 46, 42, 51, 78, 115, 103, 60, 13, 16, 17, 11, 7, 5, 3) m <- c(0, 0, 17, 25, 40, 65, 111, 103, 60, 13, 16, 17, 11, 7, 5, 3) I <- rep(NA, 16) mad.obs <- rep(NA, 16) for (i in 1:16) { I[i] <- frec[i] - m[i] mad.obs[i] <- m[i]/frec[i] } data.frame(Lt, frec, m, I, mad.obs)

## Lt frec m I mad.obs ## 1 20 2 0 2 0.0000 ## 2 21 2 0 2 0.0000 ## 3 22 46 17 29 0.3696 ## 4 23 42 25 17 0.5952 ## 5 24 51 40 11 0.7843 ## 6 25 78 65 13 0.8333 ## 7 26 115 111 4 0.9652 ## 8 27 103 103 0 1.0000 ## 9 28 60 60 0 1.0000 ## 10 29 13 13 0 1.0000 ## 11 30 16 16 0 1.0000 ## 12 31 17 17 0 1.0000 ## 13 32 11 11 0 1.0000 ## 14 33 7 7 0 1.0000 ## 15 34 5 5 0 1.0000 ## 16 35 3 3 0 1.0000

plot(Lt, mad.obs, xlim = c(10, 35), ylim = c(0, 1), xlab = "Longitud total (cm)", ylab = "Proporcion Madura") # Modelo no lineal con verosimilitud negloglike <- function(parms) { a <- parms[1] b <- parms[2] pl <- 1/(1 + exp(-(a + b * Lt))) -sum(m * log(pl) + (I * log(1 - pl))) } m2 <- optim(par = c(22, -2), fn = negloglike, method = "BFGS", hessian = TRUE) LM2 <- -m2$par[1]/m2$par[2] lt <- seq(10, 35, 0.1) mad.est <- 1/(1 + exp(-(m2$par[1] + m2$par[2] * lt))) lines(lt, mad.est) text(16, 0.8, expression(P(L) == frac(1, 1 + exp(-(a + b * L))))) text(16, 0.6, expression(LM2 == 22.71))

Incidencia de hembras de jurel maduras e inmaduras microscopicamente por estrato de talla, colestadas en la zona sur durante 1999.

Lt <- c(22:34) frec <- c(6, 60, 100, 138, 135, 105, 23, 11, 16, 4, 6, 2, 2) m <- c(0, 14, 30, 79, 96, 82, 20, 10, 16, 4, 6, 2, 2) I <- rep(NA, 13) mad.obs <- rep(NA, 13) for (i in 1:13) { I[i] <- frec[i] - m[i] mad.obs[i] <- m[i]/frec[i] } data.frame(Lt, frec, m, I, mad.obs)

## Lt frec m I mad.obs ## 1 22 6 0 6 0.0000 ## 2 23 60 14 46 0.2333 ## 3 24 100 30 70 0.3000 ## 4 25 138 79 59 0.5725 ## 5 26 135 96 39 0.7111 ## 6 27 105 82 23 0.7810 ## 7 28 23 20 3 0.8696 ## 8 29 11 10 1 0.9091 ## 9 30 16 16 0 1.0000 ## 10 31 4 4 0 1.0000 ## 11 32 6 6 0 1.0000 ## 12 33 2 2 0 1.0000 ## 13 34 2 2 0 1.0000

plot(Lt, mad.obs, xlim = c(10, 35), ylim = c(0, 1), xlab = "Longitud total (cm)", ylab = "Proporcion Madura") # Modelo no lineal con verosimilitud negloglike <- function(parms) { a <- parms[1] b <- parms[2] pl <- 1/(1 + exp(-(a + b * Lt))) -sum(m * log(pl) + (I * log(1 - pl))) } m3 <- optim(par = c(22, -2), fn = negloglike, method = "BFGS", hessian = TRUE) LM3 <- -m3$par[1]/m3$par[2] lt <- seq(10, 35, 0.1) mad.est <- 1/(1 + exp(-(m3$par[1] + m3$par[2] * lt))) lines(lt, mad.est) text(16, 0.8, expression(P(L) == frac(1, 1 + exp(-(a + b * L))))) text(16, 0.6, expression(LM3 == 24.86))

Zona <- c("Ambas", "Norte", "Sur") B0 <- c(m1$par[1], m2$par[1], m3$par[1]) B1 <- c(m1$par[2], m2$par[2], m3$par[2]) LMad <- c(LM1, LM2, LM3) data.frame(Zona, B0, B1, LMad)

## Zona B0 B1 LMad ## 1 Ambas -16.20 0.6789 23.86 ## 2 Norte -21.87 0.9632 22.71 ## 3 Sur -17.34 0.6974 24.86

Referencias:

Cerna J. & C. Oyarzun. 1998. Talla de primera madurez sexual y fecundidad parcial de la merluza común (Merluccius gayi, Guichenot 1848) del área de la pesquería industrial de la zona de Talcahuano, Chile. Investigaciones Marinas, Valparaíso 26: 31-40.

Leal E. & C. Oyarzun. 2003. Talla de madurez y época de desove de la reineta (Brama australis Valenciennes, 1836) en la costa central de Chile. Investigaciones Marinas, Valparaíso 31(2): 17-24.

Leal E., Canales M., Aranis A. & M. Gonzales. 2011. Actividad reproductiva y longitud de madurez de sardina austral Sprattus fuegensis en el mar interior de Chiloé, Chile. Revista de Biología Marina y Oceanografía 46(1): 43-51.

#maturity

Biología de Poblaciones: La matriz de Leslie

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La dinámica estructurada

La dinámica de una población estructurada puede ser representada por las siguientes matrices:

\[ \mathbf{N_{t+1}} = \mathbf{P}\times \mathbf{N_t} \]

donde $ \mathbf{N_{t}} $ y $ \mathbf{N_{t+1}} $ es la matriz de abundancia en el tiempo $ t $ y $ t+1 $, respectivamente; y $ \mathbf{P} $ es la matriz de Leslie. Esta última se representa la fertilidad ($ F $) en la primera fila de la matriz, mientras que en la diagonal que comienza a partir de la segunda fila hasta la última fila, queda representada la sobrevivencia ($ S $). Por ejemplo, en una población con tres estados dada por la abundancia de los estados $ X_t $, $ Y_t $ y $ Z_t $, i.e.,

\[ \mathbf{N_t} = \begin{bmatrix} X_t \\ Y_t \\ Z_t \end{bmatrix} \]

con la siguiente matriz de Leslie:

\[ \mathbf{P} = \begin{bmatrix} F_x & F_y & Fz\\ S_x & 0 & 0 \\ 0 & S_y & 0 \end{bmatrix} \]

Al multiplicar $ \mathbf{P} $ por $ \mathbf{N_t} $, se tendría que la siguiente dinámica de los estados:

\[ X_{t+1} = F_{x} \times X_{t} + F_{y} \times Y_{t} + F_{z}\times Z_{t} \] \[ Y_{t+1} = S_{x} \times X_{t} \] \[ Z_{t+1} = S_{y} \times Y_{t} \]

Por ejemplo, en R considere la siguiente matriz de Leslie:

(P <- matrix(c(0.4271, 0.8498, 0.1273, 0.9924, 0, 0, 0, 0.9826, 0), nrow = 3, ncol = 3, byrow = TRUE))

## [,1] [,2] [,3] ## [1,] 0.4271 0.8498 0.1273 ## [2,] 0.9924 0.0000 0.0000 ## [3,] 0.0000 0.9826 0.0000

La proyeccion de la abudancia a 10 años puede ser calculada de la siguiente manera, considerando que en el primer año la abundancia está representada por $ X_1 = 1, Y_t = 0, y Z_t = 0 $, i.e.,

years <- 10 Nt <- matrix(0, nrow(P), ncol = years) Nt[, 1] <- matrix(c(1, 0, 0), ncol = 1) # Proyeccion for (i in 1:(years - 1)) { Nt[, i + 1] <- P %*% Nt[, i] }

Una vez realizada la proyección, se puede graficar la matriz de abundancia (\mathbf{N_t}) en el tiempo con 'matplot', i.e.,

matplot(1:years, t(Nt), type = "l", lty = c(1, 2, 3), xlab = "Años", ylab = "Abundancia") legend(2, 2.5, legend = c("X", "Y", "Z"), lty = c(1, 2, 3), col = c(1, 2, 3))

Parámetros poblacionales

Los cambios de abundancia no son los importantes desde el punto de vista del análisis de la matriz de Leslie para la dinámica de la población. En efecto, una vez construida la matriz de Leslie, el análisis se debe enfocar en obtener por ejemplo: la tasa de crecimiento poblacional ($ \lambda $), la estructura estable de la población ($ \mathbf{w} $), y el valor reproductivo ($ \mathbf{v} $). Esto se logra con un análisis 'eigen' en R, tal que se puedan extraer los valores propios (eigenvalues) y los vectores propios (eigenvectors) usando algebra lineal.

Por ejemplo, para calcular la tasa intrinseca de crecimiento poblacional, la estructura estable poblaciona ($ \mathbf{w} $), se debe considerar la siguiente igualdad:

\[ \mathbf{P}\times \mathbf{w} = \lambda \times \mathbf{w} \]

La igualdad se resuelve mediante el comando 'eigen' a la matriz de Leslie, i.e,

# Eigenanalysis eigsP <- eigen(P)

El modo dominante, o del valor máximo, de los valores propios representará la tasa de crecimiento poblacional, que generalmente es un valor positivo y en la primera posición, i.e.,

# Posición pos.do <- which.max(eigsP$values) # Tasa de crecimiento poblacional (Lambda <- eigsP$values[pos.do])

## [1] 1.209

# Tasa intrisica de crecimiento poblacional (r <- log(Lambda))

## [1] 0.1901

La estructura estable de la población, que resulta de la dinámica de la matriz de Leslie, viene dada por la parte real de los vectores propio (la parte imaginaria no se usa). Para ello, se utiliza la función 'Re' para obtener los números reales del vector propio asociado al modo dominante, i.e.,

(w <- Re(eigsP$vectors)[, pos.do])

## [1] -0.6871 -0.5639 -0.4582

# La estructura estable se puede expresar en porcentaje,i.e. (eee <- w/sum(w) * 100)

## [1] 40.20 32.99 26.81

Para obtener el valor reproductivo ($ \mathbf{v} $), se utiliza la siguiente igualdad, y considerando la transpuesta de la matriz de Leslie, i.e.,

\[ \mathbf{v}\times (\mathbf{P}^T) = \lambda \times \mathbf{v} \]

Para ello se aplica 'eigen' a la traspuesta de $ \mathbf{P} $, se extrae el vector propio asociado con el modo dominante, y el valor reproductivo se expresa como porcentaje del primer estado, i.e.,

M <- eigen(t(P)) v <- Re(M$vectors[, which.max(M$values)]) # El valor reproductivo se obtiene re-escalando por el primer valor (VRep <- v/v[1] * 100)

## [1] 100.00 78.82 10.53

Sensibilidad y elasticidad

El análisis de sensibilidad y elasticidad permiten evaluar la importancia relativa de cada transición (de cada elemento de la matriz) en la tasa de crecimiento poblacional ($ \lambda $). Para ello, se debe combinar la estructura estable $ \mathbf{w} $ y el valor reproductivo $ \mathbf{v} $. Se trata de evaluar pequeños cambios en cada elemento de la matriz, i.e.,

\[ \frac {dL}{da} = \frac{{v_{i,j} \times w_{i,j}}}{\mathbf{v} \cdot \mathbf{w}} \]

El producto pareado de cada combinación pareada de los elementos de los vectores propios dominantes izquierdo y derecho, $ \mathbf{v} $ y $ \mathbf{w} $. El producto punto, es la suma de productos pareados de cada uno de los elementos del vector.

El numerador se calcula de la siguiente manera, i.e.,

(vw <- v %*% t(w))

## [,1] [,2] [,3] ## [1,] 0.53781 0.44133 0.35859 ## [2,] 0.42392 0.34787 0.28265 ## [3,] 0.05661 0.04646 0.03775

Luego la suma del denominador, para después dividir el numerador para obtener la sensibilidad (Se), i.e.,

(Se <- vw/as.numeric(v %*% w))

## [,1] [,2] [,3] ## [1,] 0.58241 0.47793 0.38832 ## [2,] 0.45907 0.37672 0.30609 ## [3,] 0.06131 0.05031 0.04088

Finalmente, se calcula la elasticidad

Elas <- (P/Lambda) * Se round(Elas, 3)

## [,1] [,2] [,3] ## [1,] 0.206 0.336 0.041 ## [2,] 0.377 0.000 0.000 ## [3,] 0.000 0.041 0.000

Curso Herramientas Cuantitativas en Ecología

Luis A. Cubillos

Biólogo Pesquero

blog: HCE

blog: code4fisheries

twitter: @LuisCubillos

Análisis de sistemas dinámicos

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Integración de ecuaciones diferenciales ordinarias acompladas

En el siguiente código se utiliza el paquete deSolve para resolver un sistema dinámico, representado por ecuaciones diferenciales ordinarias (ODE) acompladas.

En este ejemplo, se tiene un sistema dinámico entre una presa ($ u $) y un depredador ($ v $), del tipo Lotka-Volterra, con una interacción tipo II de Holling, i.e,

\[ \alpha uv/(u+d) \]

donde $ \alpha $ es la tasa de ataque máxima por unidad de presa del depredador, y $ d $ es la abundancia de presa en la cual la tasa de ataque es la mitad del nivel de saciedad. Esta versión del tipo II de Holling es la versión tipo Michaelis-Menten.

El siguiente sistema de EDO acopladas es el siguiente:

\[ du/dt = u(1-u) - \alpha uv/(u+d) \] \[ dv/dt = bv(1-v/u) \]

Las condiciones iniciales al tiempo 0 son $ u(0) = 1.2 $ y $ v(0)= 0.2 $, y los parámetros son $ b=0.2 $, $ \alpha = 1 $, y $ d=0.1 $.

El código es el siguiente. Primero Se debe llamar al programa de Solve, i.e.

library(deSolve)

Luego se debe escribir la función para la relación de Holling tipo II, i.e.

f <- function(x, y, d) { x * y/(x + d) } # Holling tipo II

Ahora el sistema de EDO:

model <- function(t, xx, parms) { u <- xx[1] v <- xx[2] with(as.list(parms), { du <- u * (1 - u) - alpha * f(u, v, d) dv <- b * v * (1 - v/u) list(c(du, dv)) }) } # La secuencia temporal que se integrará (100 años): times <- seq(0, 100, 0.5) # Los parámetros del sistema: parms <- c(b = 0.2, alpha = 1, d = 0.2) # El estado inicial del sistema: xstart <- c(u = 1.2, v = 0.2) # La solución del sistema dinámico se realiza con ode y Runge-Kutta out <- as.data.frame(ode(xstart, times, model, parms, method = "rk4"))

El comando 'ode' permite resolver el sistema de ecuaciones diferenciales, en este caso se usó el método Rung-Kutta.

El sistema en el tiempo muestra el típico comportamiento depredador-presa, i.e.,

El comportamiento dinámico, también se puede observar al graficar la abundancia de la presa en función del depredador, i.e.,

Curso Herramientas Cuantitativas en Ecología

Luis A. Cubillos

Biólogo Pesquero

blog: HCE

blog: code4fisheries

twitter: @LuisCubillos

#Integración #deSolve #sistema dinámico #ODE #rstats

Tarea 01: Derivadas y optimización

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Tarea 01: Derivadas y optimización

Solución de derivadas con R

Una de las ventajas de R es encontrar una solución analítica para derivadas comunes mediante lenguaje simbólico. La siguiente es una función que debe ser corrida antes:

mi.derivada <- function(funcion.matematica, var) { temp <- substitute(funcion.matematica) name <- deparse(substitute(var)) D(temp, name) }

Por ejemplo, la solución para la función:

\[ f(X) = aX^b \]

es la siguiente:

\[ f(X)' = abX^{b-1} \]

Para comprobar se debe ejecutar la siguiente instrucción

mi.derivada(a * X^b, X)

## a * (X^(b - 1) * b)

A usted se le pide obtener la derivada de las siguientes funciones: \[ f(x)=(x^2+5x)^3 \] \[ f(x)=exp(x^2) \] \[ f(x)=k*(1-x^q) \] \[ f(x)=log(1-x)^2 \] \[ f(x)=k*exp(-k*x) \] \[ f(x)=(1-exp(-k*x)) \] \[ f(x)=a*x*exp(-b*x) \] \[ f(x)=a*(1-exp(-(b*(x-c))))^3 \]

Newton-Raphson

Uno de los métodos más utilizados para buscar la solución numérica de derivativas es la formula de Newton-Raphson, que está representado por: \[ X_{i+1}=X_{i} - f(X)'/f(X)'' \] donde $ X_{i+1} $ es el valor de X actualizado, el subíndice $ i $ denota iteración, $ f(X)' $ es la primera derivada de la función y $ f(X)'' $ es la segunda derivada de la función.

Ejemplo

Se pide buscar el valor de X que minimiza la siguiente función $ f(x)=exp(-x)+x^4 $.

# Se escribe la función f <- function(x) { exp(-x) + x^4 } # Esta función tiene la siguiente forma: curve(f, from = -1, to = 3)

Sabemos que la primera derivada de la función es $ f(x)= -exp(-x)+4*x^3 $, y que la segunda derivada es $ f(x)''=exp(-x)+12*x^2 $.

Para buscar la solución, aplicamos Newton-Raphson en R:

# Primera y segunda derivada de la función: fprima <- function(x) -exp(-x) + 4 * x^3 fprima2 <- function(x) exp(-x) + 12 * x^2 # Para la solución con 6 iteraciones se necesita evaluar la funcion en # cada valor de x (se comienza con x=1), i.e., x <- c(1, rep(NA, 6)) feval <- rep(NA, 7) fprimaeval <- rep(NA, 7) fprimaeval2 <- rep(NA, 7) # Se realizan 6 iteraciones: for (i in 1:6) { feval[i] <- f(x[i]) fprimaeval[i] <- fprima(x[i]) fprimaeval2[i] <- fprima2(x[i]) x[i + 1] <- x[i] - fprimaeval[i]/fprimaeval2[i] } data.frame(x, feval, fprimaeval, fprimaeval2)

## x feval fprimaeval fprimaeval2 ## 1 1.0000 1.3679 3.632e+00 12.368 ## 2 0.7063 0.7424 9.161e-01 6.480 ## 3 0.5650 0.6703 1.529e-01 4.399 ## 4 0.5302 0.6675 7.678e-03 3.962 ## 5 0.5283 0.6675 2.276e-05 3.938 ## 6 0.5283 0.6675 2.019e-10 3.938 ## 7 0.5283 NA NA NA

Se pude observar que la solución es $ x=0.5283 $.

A usted se le pide encontrar el máximo rendimiento sostenido para una pesquería, sabiendo que la función de rendimiento es:

\[ f(x)=100*x-10*x^2 \]

Es una función cuadrática, i.e,

f <- function(x) { 100 * x - 10 * x^2 } curve(f, from = 0, to = 10)

A usted se le pide que busque la primera y segunda derivada, y encuentre el valor de x que maximiza la función con el método de Newton-Raphson.

Instrucciones

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blog: HCE

blog: code4fisheries

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#Newton-Raphson #optimización #derivadas #R #RStudio

Estimating VBG parameters

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Fitting von Bertalanffy Growth Function

The von Bertalanffy Growth Function describes the body growth in length of most fish, i.e.

\[ L_t = L_\inf (1 - \exp(-K(t-t_0))) \]

where $ L_t $ is the body length at age $ t $, $ L_\inf $ is the asymptotic length, $ K $ is the growth rate coefficient, and $ t_0 $ is the age at $ L=0 $. All you need to estimate the three parameters of the model are observed data-pairs of length and age, i.e.,

DATA_SECTION

age <- 1:12 lt <- c(10, 19, 24, 28, 32, 34, 38, 40, 41, 42, 42, 43) dat <- data.frame(age, lt) dat

## age lt ## 1 1 10 ## 2 2 19 ## 3 3 24 ## 4 4 28 ## 5 5 32 ## 6 6 34 ## 7 7 38 ## 8 8 40 ## 9 9 41 ## 10 10 42 ## 11 11 42 ## 12 12 43

EXPLORATORY_DATA_SECTION

To explore the data, you need the ggplot2 library

library(ggplot2)

A plot of the length against age is describing the observed growth for a fish population, i.e.

qplot(age, lt, data = dat, xlab = "Age (yrs)", ylab = "Length (cm)", xlim = c(0, 13), ylim = c(0, 50))

PARAMETERS_SECTION

You need to guess initial parameters, which can be in a vector called $ theta $, i.e.,

theta <- c(45, 0.2, -0.5)

PROCEDURE_SECTION

You need a function to estimate the parameters with $ optim $, i.e.,

SSQ <- function(theta, x) { Linf <- theta[1] K <- theta[2] t0 <- theta[3] epsilon <- rep(0, length(age)) lpred <- rep(0, length(age)) for (i in 1:length(age)) { lpred[i] <- Linf * (1 - exp(-K * (age[i] - t0))) epsilon[i] <- (lt[i] - lpred[i])^2 } ssq <- sum(epsilon) return(ssq) }

Once ran the previous function, then you can use $ optim $ to solve for the parameters, i.e.,

out <- optim(theta, fn = SSQ, method = "BFGS", x = age, hessian = TRUE) out$V <- solve(out$hessian) #solve the hessian out$S <- sqrt(diag(out$V)) #Standard Error out$R <- out$V/(out$S %o% out$S) #Correlation

REPORT_SECTION

To known the parameters than have been estimated, you can list the following

out$par

## [1] 45.78855 0.23959 -0.08253

out$S

## [1] 0.82927 0.01577 0.11253

Finally, you can add the following to see the fitted curve to observed data:

lp <- out$par[1] * (1 - exp(-out$par[2] * (age - out$par[3]))) plot(age, lt, xlab = "Age (yrs)", ylab = "Length (cm)", xlim = c(0, 13), ylim = c(0, 50)) lines(sort(age), sort(lp), col = 3, lwd = 2)

SUGGESTIONS

The example here is based on least square.

You can obtain the same results by fitting with $ nls $, but the advantage of using a function is the possibility of changing it to consider maximum likelihood estimates, or considering a multiplicative error structure for observed length at age.

Who Am I?

Luis A. Cubillos

Fisheries Biologist

blog: EPOMAR

blog: Cambio Climático Pesca y Acuicultura en Chile

github: LuisACubillos

twitter: @LuisCubillos

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