Table 1.

Models, parameter values ($\mathit{\theta },\mathrm{\Omega },\mathit{\sigma }$) and effect sizes (${\mathit{\theta }}_{\text{H}}^{\ast }$) used for the evaluation of the algorithms

Structural model			Parameter model	Parameter values
				${\mathit{\theta }}_{\text{H}}^{\ast }$	$\mathit{\theta }$	$\mathrm{\Omega }$	$\mathrm{\Sigma }$
1. Binary
$P(y_{ij}|{\mathit{\eta }}_i)=\left\{\begin{array}{cc}p(t_j)\hfill & \text{if}{y}_{ij}=1\hfill \\ 1-p(t_j)\hfill & \text{if}{y}_{ij}=0\hfill \end{array}\right.$		$p(t)=\text{expit}(b+at)$	$\begin{array}{cc}\hfill b& ={\mathit{\theta }}_1+{\mathit{\eta }}_{i1}\hfill \\ \hfill a& =({\mathit{\theta }}_2+{\mathit{\eta }}_{i2})(1-{\mathit{\theta }}_{\text{H}}{z}_i)\hfill \end{array}$	0.3	$\left(\begin{array}{c}-1\\ 4\end{array}\right)$	$\left(\begin{array}{cc}0.4& 0\\ 0& 4\end{array}\right)$	–
2. Time-to-event (TTE)
${\displaystyle P(y_i)=\left\{\begin{array}{cc}h(y_i)S(y_i)\hfill & \text{if}{y}_i<T\hfill \\ S(T)\hfill & \text{if}{y}_i=T\hfill \end{array}\right.}$		$\begin{array}{cc}\hfill lclh(t)& =\mathit{\lambda }\mathit{\gamma }{(\mathit{\lambda }·t)}^{\mathit{\gamma }-1}\hfill \\ \hfill S(t)& =exp(-{\int }_0^{t}h(x)dx)\hfill \end{array}$	$\begin{array}{cc}\hfill \mathit{\lambda }& ={\mathit{\theta }}_{1}exp({\mathit{\theta }}_{\text{H}}{z}_i)\hfill \\ \hfill \mathit{\gamma }& ={\mathit{\theta }}_2\hfill \end{array}$	0.4	$\left(\begin{array}{c}0.2\hfill \\ 2\hfill \end{array}\right)$	–	–
3. Count
${\displaystyle P(y_{ij}|{\mathit{\eta }}_i)=\frac{\mathit{\lambda }{(t_j)}^{y_{ij}}}{y_{ij}!}{e}^{-\mathit{\lambda }(t_j)}}$		$\mathit{\lambda }(t)=b+A(1-exp(-k·t))$	$\begin{array}{cc}\hfill b_i& ={\mathit{\theta }}_{1}exp({\mathit{\eta }}_{i1})\hfill \\ \hfill A_i& ={\mathit{\theta }}_{2}exp({\mathit{\eta }}_{i2})\hfill \\ \hfill k_i& ={\mathit{\theta }}_{3}exp({\mathit{\eta }}_{i3}+{\mathit{\theta }}_{\text{H}}{\tilde{z}}_i)\hfill \\ \hfill \end{array}$	0.3	$\left(\begin{array}{c}1\hfill \\ 4\hfill \\ 2\hfill \end{array}\right)$	$\left(\begin{array}{ccc}0.09\hfill & 0\hfill & 0\hfill \\ 0\hfill & 0.09\hfill & 0\hfill \\ 0\hfill & 0\hfill & 0.09\hfill \end{array}\right)$	–
4. PK [12]
${\displaystyle y_{ij}=\frac{A_1(t_j)}{V}(1+{\mathit{\epsilon }}_{ij})}$		$\begin{array}{c}\frac{d{A}_1}{dt}=\frac{Q}{V_p}{A}_2-\left(\frac{Cl}{V}+\frac{Q}{V}\right)A_1\hfill \\ \frac{d{A}_2}{dt}=\frac{Q}{V}{A}_1-\frac{Q}{V_p}{A}_2\hfill \end{array}$	$\begin{array}{cc}\hfill Cl& ={\mathit{\theta }}_{1}exp({\mathit{\eta }}_{i1})(1+{\mathit{\theta }}_{\text{H}}{\tilde{z}}_i)\hfill \\ \hfill V& ={\mathit{\theta }}_{2}exp({\mathit{\eta }}_{i2})\hfill \\ \hfill Q& ={\mathit{\theta }}_{3}exp({\mathit{\eta }}_{i3})\hfill \\ \hfill V_p& ={\mathit{\theta }}_4\hfill \end{array}$	0.2	$\left(\begin{array}{c}0.04\hfill \\ 0.14\hfill \\ 3.62\hfill \\ 2.9\hfill \end{array}\right)$	$\left(\begin{array}{ccc}0.09\hfill & 0\hfill & 0.16\hfill \\ 0\hfill & 0.04\hfill & 0\hfill \\ 0.16\hfill & 0\hfill & 1.23\hfill \end{array}\right)$	0.06
5. PKPD [8]
$\begin{array}{cc}\hfill y_{ij}^{\prime }& =\frac{A_2(t_j)}{V}(1+{\mathit{\epsilon }}_{1,ij})\hfill \\ \hfill y_{ij}^{″}& =R(t_j)+{\mathit{\epsilon }}_{2,ij}\hfill \end{array}$		$\begin{array}{c}\frac{d{A}_1}{dt}=-k_a{A}_1\hfill \\ \frac{d{A}_2}{dt}=k_a{A}_1-\frac{Cl}{V}{A}_2\hfill \\ \frac{dR}{dt}=k_o\frac{A_2}{V}/(C_{50}+\frac{A_2}{V})-k_{o}R\hfill \end{array}$	$\begin{array}{cc}\hfill Cl& ={\mathit{\theta }}_{1}exp({\mathit{\eta }}_{i1})\hfill \\ \hfill V& ={\mathit{\theta }}_{2}exp({\mathit{\eta }}_{i2})\hfill \\ \hfill k_a& ={\mathit{\theta }}_3\hfill \\ \hfill k_o& ={\mathit{\theta }}_{4}exp({\mathit{\eta }}_{i3})\hfill \\ \hfill C_{50}& ={\mathit{\theta }}_{5}exp({\mathit{\eta }}_{i4})(1+{\mathit{\theta }}_{\text{H}}{z}_i)\hfill \end{array}$	0.3	$\left(\begin{array}{c}10\hfill \\ 100\hfill \\ 2\hfill \\ 0.2\hfill \\ 0.3\hfill \end{array}\right)$	$\left(\begin{array}{cccc}0.49\hfill & 0\hfill & 0\hfill & 0\hfill \\ 0\hfill & 0.49\hfill & 0\hfill & 0\hfill \\ 0\hfill & 0\hfill & 0.49\hfill & 0\hfill \\ 0\hfill & 0\hfill & 0\hfill & 0.49\hfill \end{array}\right)$	$\left(\begin{array}{c}0.04\hfill \\ 0.04\hfill \end{array}\right)$