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| − | At <math>t = 0</math> <math>P_0</math> is defined to be developer productivity without legacy obligations
| + | Derivation of developer productivity parameters |
| − | (1) <math>P(0) = P_0</math>
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| − | At time <math>t</math> the developer productivity is given by the following equation. <math>r</math> denotes the rate in which productivity declines over time (see [[Derivation of developer productivity decline function|Derivation of developer productivity decline function]])
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| − | (2) <math>P(t) = P_0 \cdot e^{-r t} dt</math>
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| − | We know that after 1 year a developer has produced 10KLoc:
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| − | (3a) <math>\int^1_0 P_0 \cdot e^{-r t} dt</math> = <math>{ProductionPerYear} = 10</math>
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| − | We know that over the full life time a developer can maximally maintain 50KLoc:
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| − | (4a) <math>\int^\infty_0 P_0 \cdot e^{-r t} dt</math> = <math>{MaxMaintenance} = 50</math>
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| − | Using the [[Substitution rule|Substitution rule]] we can see that:
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| − | (5a) <math>u = -rt</math>
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| − | (5b) <math>t = 0 => u = 0</math>
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| − | (5c) <math>t = \infty => u = - r \cdot \infty</math>
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| − | (5d) <math>t = 1 => u = - r</math>
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| − | (5e) <math>du = -rdt => dt = </math> <math>-\frac{1}{r}</math><math>du</math>
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| − | So that:
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| − | (3b) <math>-P_0 \frac {1}{r} \int^{-r}_0 e^u du</math> = <math>{ProductionPerYear}</math>
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| − | (3c) <math>-P_0 \frac {1}{r} (e^{-r} - e^0)</math> = <math>{ProductionPerYear}</math>
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| − | (3d) <math>P_0 \frac {1-e^r}{r}</math> = <math>{ProductionPerYear}</math>
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| − | (3e) <math>P_0 = </math><math>\frac</math>{{ProductionPerYear} \cdot r}{1-e^{-r}} <math></math>
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| − | (4b) <math>-P_0 \frac {1}{r} \int^{-\infty}_0 \cdot e^u du</math> = <math>{MaxMaintenance}</math>
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| − | (4c) <math>-P_0 \frac {1}{r} (e^{-\infty} - e^0)</math> = <math>{MaxMaintenance}</math>
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| − | (4d) <math>-P_0 \frac {1}{r} (0 -1)</math> = <math>{MaxMaintenance}</math>
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| − | (4e) <math>\frac {P_0}{r}</math> = <math>{MaxMaintenance}</math>
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| − | (4f) <math>P_0</math> = <math>{MaxMaintenance} \cdot r </math>
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| − | Combining (3e) and (4f) we get:
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| − | (6a) <math>{MaxMaintenance} \cdot r = \frac</math>{{ProductionPerYear} \cdot r}{1-e^{-r}}<math>
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| − | (6b) </math>\large {1-e^{-r}} = \frac{{ProductionPerYear} \cdot r}{{MaxMaintenance} \cdot r}<math>
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| − | (6c) </math>\large -e^{-r} = \frac{{ProductionPerYear}}{{MaxMaintenance}} -1<math>
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| − | (6d) </math>\large e^{-r} = 1 - \frac{{ProductionPerYear}}{{MaxMaintenance}} <math>
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| − | (6e) </math>\large -r = ln(1- \frac{ProductionPerYear}{{MaxMaintenance}})<math>
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| − | (6f) </math>\large r = -ln(1- \frac{ProductionPerYear}{{MaxMaintenance}})<math></math>
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| − | Which results in the following values for <math>r</math> and <math>P_0</math>
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| − | (6f) <math>r = -ln(1-10/50) = 0.22314355</math>
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| − | (4g) <math>P_0 = 50 * 0.22314355 = 11.1571775657 </math>
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| | [[Category:math]] | | [[Category:math]] |
