over the unoccupied lattice site in front of the door occur. At time $t$, with probability $1-\mu$ one of these particles is able to move. In the next time step $t+1$ this particle will escape through the door with probability $1$. Neglecting conflicts elsewhere in the system, because of the big jam in front of the door the configuration shown in fig. \[plot\_8\] will be restored at time $t+2$. Therefore, repeating the above sequence a typical representation of the time evolution of $N(t)$ is shown in fig. \[plot\_9\].

![An example for the time evolution of the number of evacuated persons.[]{data-label="plot_9"}](\DIR/fig16.eps){width=".95\columnwidth"}

Using simple combinatorics, the expectation value of $N(t)$ can be written as $$\langle N(t) \rangle=\frac{1}{C}\sum_{n=0}^{[t/2]}n \left(\begin{array}{c}
t-n \\
n
\end{array}\right)
 (1-\mu)^n \mu^{t-2n},
\label{mean}$$ where $C$ is the normalization factor defined by $$\begin{aligned}
 C(t)&=&\sum_{n=0}^{[t/2]} \left(\begin{array}{c}
t-n \\
n
\end{array}\right)
 (1-\mu)^n \mu^{t-2n}
\nonumber\\
&=&\mu^t \sum_{n=0}^{[t/2]} \left(\begin{array}{c}
t-n \\
n
\end{array}\right)
 \left(\frac{1-\mu}{\mu^2}\right)^{n}\nonumber\\
&=&\frac{1-(\mu-1)^{t+1}}{2-\mu}.\label{equat5}\end{aligned}$$ In the last step we have used the identity (\[for\]) which is derived in Appendix \[proof\]. Due to (\[mean\]) and (\[equat5\]), $\langle N(t)\rangle$ is related to $C$ by the equation $$\frac{d}{d\mu}C=\frac{t}{\mu}~C+\frac{\mu-2}{\mu(1-\mu)}~\langle 
N(t)\rangle~C$$ or $$\langle N(t)\rangle=\frac{\mu(1-\mu)}{\mu-2}\left(
\frac{d}{d\mu}\log C -\frac{t}{\mu}\right).$$ Finally an analytical expression for $\langle N(t)\rangle$ is obtained as $$\begin{aligned}
&&\langle 