Derivation of developer productivity parameters

At $t=0$ $P_{0}$ is defined to be developer productivity without legacy obligations (1) $P(0)=P_{0}$

At time $t$ the developer productivity is given by the following equation. $r$ denotes the rate in which productivity declines over time (see Derivation of developer productivity decline function) (2) $P(t)=P_{0}\cdot e^{-rt}dt$

We know that after 1 year a developer has produced 10KLoc: (3a) $\int _{0}^{1}P_{0}\cdot e^{-rt}dt$ = ${ProductionPerYear}=10$

We know that over the full life time a developer can maximally maintain 50KLoc: (4a) $\int _{0}^{\infty }P_{0}\cdot e^{-rt}dt$ = ${MaxMaintenance}=50$

Using the Substitution rule we can see that: (5a) $u=-rt$ (5b) $t=0=>u=0$ (5c) $t=\infty =>u=-r\cdot \infty$ (5d) $t=1=>u=-r$ (5e) $du=-rdt=>dt=$ $-{\frac {1}{r}}$$du$

So that: (3b) $-P_{0}{\frac {1}{r}}\int _{0}^{-r}e^{u}du$ = ${ProductionPerYear}$ (3c) $-P_{0}{\frac {1}{r}}(e^{-r}-e^{0})$ = ${ProductionPerYear}$ (3d) $P_{0}{\frac {1-e^{r}}{r}}$ = ${ProductionPerYear}$ (3e) $P_{0}=$Failed to parse (syntax error): {\displaystyle \frac} {{ProductionPerYear} \cdot r}{1-e^{-r}}

(4b) $-P_{0}{\frac {1}{r}}\int _{0}^{-\infty }\cdot e^{u}du$ = ${MaxMaintenance}$ (4c) $-P_{0}{\frac {1}{r}}(e^{-\infty }-e^{0})$ = ${MaxMaintenance}$ (4d) $-P_{0}{\frac {1}{r}}(0-1)$ = ${MaxMaintenance}$ (4e) ${\frac {P_{0}}{r}}$ = ${MaxMaintenance}$ (4f) $P_{0}$ = ${MaxMaintenance}\cdot r$

Combining (3e) and (4f) we get: (6a) Failed to parse (syntax error): {\displaystyle {MaxMaintenance} \cdot r = \frac} {{ProductionPerYear} \cdot r}{1-e^{-r}}$(6b)$\large {1-e^{-r}} = \frac{{ProductionPerYear} \cdot r}{{MaxMaintenance} \cdot r}$(6c)$\large -e^{-r} = \fracTemplate:ProductionPerYearTemplate:MaxMaintenance -1$(6d)$\large e^{-r} = 1 - \fracTemplate:ProductionPerYearTemplate:MaxMaintenance $(6e)$\large -r = ln(1- \frac{ProductionPerYear}Template:MaxMaintenance)$(6f)$\large r = -ln(1- \frac{ProductionPerYear}Template:MaxMaintenance)

Which results in the following values for $r$ and $P_{0}$ (6f) $r=-ln(1-10/50)=0.22314355$ (4g) $P_{0}=50*0.22314355=11.1571775657$