1 Comparaison en termes d’accroissement

par(mfrow = c(1,1))
K <- 8000
N0 <- 100
r0 <- 0.85
#exponentiel
curve(r0*x, ylim = c(0, 10000),from = 0, to = 10000, las = 1, main = "Comparaison des vitesses des 4 modèles", xlab = "Effectif de la population N(t)", ylab = "Vitesse d'accroissement de la population")
#logistique
curve(r0*x*(1-x/K),from = 0, to = 10000,lty = 2, add = T)
#Gompertz
curve(r0*x*log(K/x), lty = 3, add = T)
#von Bertalanffy
curve(r0*(K-x), lty = 4, add = T)
legend(3000, 10000, leg = c("Exponentiel", "Logistique", "Gompertz", "von Bertalanffy"), lty = 1:4)
abline(h = 0)

2 Comparaison des courbes représentatives

N0 <- 100
K <- 8000
r0 <- 0.85
curve(K/(1+(K/N0-1)*exp(-r0*x)), from = 0, to = 15, las = 1, col = "black", lwd = 2, xlab = "Temps (h)", ylab = "Population bactérienne (UFC/ml)")
abline(h = K/2, lwd = 2, lty = 2)
curve(K*exp(log(N0/K)*exp(-r0*x)), from = 0, to = 15, lwd = 2, col = "blue", add = TRUE)
abline(h = K/exp(1), lwd = 2, lty = 2, col = "blue")
curve(K-(K-N0)*exp(-r0*x), from = 0, to = 15, lwd = 2, col = "green", add = TRUE)
legend(10,2000, legend = c("Verhulst","Gompertz","von Bertalanffy"), lty = 1, col = c("black","blue","green"), bty = "n", text.col = c("black", "blue", "green"), lwd = 2)

On peut mettre les deux graphes ci-dessus côte à côte avec R. Il suffit de les précéder de l’instruction par(mfrow = c(1,2)).