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Set up all initial variables for all elements:

Global variables $$a=0.266ppl/m^2$$
Staircase $$ \begin{aligned} L_{stairs} &= \sqrt{(10\times7)^2 +(10\times 11)^2} \\ &=3.3m \\ \\ k&=1.08m/s \\ \\ f_{max}&=1.01 ppl/s/m \\ \\ W_{eff}&=1.8-0.3 \\ &=1.5m \\ \\ F_{staircase, \;max}&=1.515 ppl/s \end{aligned}$$
Corridor $$ \begin{aligned} k &= 1.4 m/s \\ f_{max}&=1.3 ppl/s/m \\ W_{eff}&=1.8-0.4 \\ &=1.4m \\ F_{corridor, \;max}&=1.82 ppl/s \end{aligned}$$
Door $$ \begin{aligned} f_{max}&=1.3 ppl/s/m\\ W_{eff} &=1.3-0.3 \\ &=1m \\ F_{door, \;max}&=1.3 ppl/s \end{aligned}$$


Step 1: Find the time for the first individual to descend stairs $$\begin{aligned} T_{stairs}&=\frac{L_{stairs}}{S_{stairs}} \\ &=\frac{L_{stairs}}{(1-aD)k} \\ &=\frac{3.3}{0.64908} \\ &=5.1s \end{aligned}$$ Step 2: Calculate the flow rate for the group through the stairs $$ \begin {aligned} F_{stairs}&=f_{stairs} W_{eff} \\ &=(1-aD)kDW_{eff} \\ &=1.46ppl/s \end {aligned}$$ The calculated flow is less than $F_{max}$, so there is no restriction to flow. Also, we should check the delay for the last person to start moving down the stairs ($T_{delay}$). $$\begin{aligned} T_{delay}&=\frac{N}{F_{stairs}} \\ T_{delay}&=\frac{50}{1.46} \\ T_{delay}&=34.26s \end{aligned}$$

Step 3: Find the density of the group in the corridor $$ \begin{aligned} F_{in}&=F_{out} \\ F_{in}&=(1-aD)kDW_{eff} \end{aligned}$$ Solving this for D yields: $$D=1.03ppl/m^2, \;or\;2.72ppl/m^2$$ We take the lower value going forward.


Step 4: Find the speed (and therefore traversal time) for an individual moving in the corridor $$\begin{aligned} T_{corridor}&=\frac{L_{corridor}}{S_{corridor}} \\ &=\frac{L_{corridor}}{(1-aD)k} \\ &=\frac{10}{1.016428}\\ &=9.84s \end{aligned}$$ Step 5: Find the flow rate through door for group $$\begin{aligned} F_{in}&=F_{out} \end{aligned}$$ Note here that $F_{in}=1.46ppl/s$, which is bigger than $F_{door, \;max}$, so a queue occurs, and the time to get the group through is: $$\begin{aligned} T_{door}&=\frac{N}{F_{door,\;max}} \\ &=\frac{50}{1.3}\\ &=38.5s \end{aligned}$$ As the duration of the queue ($T_{door}$) is longer than the delay for the last person to start moving down the stairs ($T_{delay}$), it is the limiting factor.

Step 6: Gather all information together

Since the queue at the door takes longer than the delay to start moving down the stairs, it is the limiting factor. So, we need to find the time it takes for the first person to get to the door, then add on the time it takes for the whole group to get through the door. $$\begin{aligned} T_{total}&=T_{stairs}+T_{corridor} +T_{queue}\\ T_{total}&=5.1s+9.84s+38.5s \\ T_{total}&=53.4s \end{aligned}$$