Derivation of developer productivity parameters - Revision history

Dirkjan at 19:51, 27 August 2020

2020-08-27T19:51:38Z

Dirkjan: Replaced content with "Derivation of developer productivity parameters Category:math"

2020-08-27T19:16:55Z

Replaced content with "Derivation of developer productivity parameters Category:math"

Dirkjan: Created page with "At $t = 0$ $P_0$ is defined to be developer productivity without legacy obligations (1) $P(0) = P_0$ At time $t$ the developer pro..."

2020-08-25T07:55:34Z

Created page with "At <math>t = 0</math> <math>P_0</math> is defined to be developer productivity without legacy obligations (1) <math>P(0) = P_0</math> At time <math>t</math> the developer pro..."

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At <math>t = 0</math> <math>P_0</math> is defined to be developer productivity without legacy obligations
(1) <math>P(0) = P_0</math>

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]])
(2) <math>P(t) = P_0 \cdot e^{-r t} dt</math>

We know that after 1 year a developer has produced 10KLoc:
(3a) <math>\int^1_0 P_0 \cdot e^{-r t} dt</math> = <math>{ProductionPerYear} = 10</math>

We know that over the full life time a developer can maximally maintain 50KLoc:
(4a) <math>\int^\infty_0 P_0 \cdot e^{-r t} dt</math> = <math>{MaxMaintenance} = 50</math>

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

So that:
(3b) <math>-P_0 \frac {1}{r} \int^{-r}_0 e^u du</math> = <math>{ProductionPerYear}</math>
(3c) <math>-P_0 \frac {1}{r} (e^{-r} - e^0)</math> = <math>{ProductionPerYear}</math>
(3d) <math>P_0 \frac {1-e^r}{r}</math> = <math>{ProductionPerYear}</math>
(3e) <math>P_0 = </math><math>\frac</math>{{ProductionPerYear} \cdot r}{1-e^{-r}} <math></math>

(4b) <math>-P_0 \frac {1}{r} \int^{-\infty}_0 \cdot e^u du</math> = <math>{MaxMaintenance}</math>
(4c) <math>-P_0 \frac {1}{r} (e^{-\infty} - e^0)</math> = <math>{MaxMaintenance}</math>
(4d) <math>-P_0 \frac {1}{r} (0 -1)</math> = <math>{MaxMaintenance}</math>
(4e) <math>\frac {P_0}{r}</math> = <math>{MaxMaintenance}</math>
(4f) <math>P_0</math> = <math>{MaxMaintenance} \cdot r </math>

Combining (3e) and (4f) we get:
(6a) <math>{MaxMaintenance} \cdot r = \frac</math>{{ProductionPerYear} \cdot r}{1-e^{-r}}<math>
(6b) </math>\large {1-e^{-r}} = \frac{{ProductionPerYear} \cdot r}{{MaxMaintenance} \cdot r}<math>
(6c) </math>\large -e^{-r} = \frac{{ProductionPerYear}}{{MaxMaintenance}} -1<math>
(6d) </math>\large e^{-r} = 1 - \frac{{ProductionPerYear}}{{MaxMaintenance}} <math>
(6e) </math>\large -r = ln(1- \frac{ProductionPerYear}{{MaxMaintenance}})<math>
(6f) </math>\large r = -ln(1- \frac{ProductionPerYear}{{MaxMaintenance}})<math></math>

Which results in the following values for <math>r</math> and <math>P_0</math>
(6f) <math>r = -ln(1-10/50) = 0.22314355</math>
(4g) <math>P_0 = 50 * 0.22314355 = 11.1571775657 </math>
[[Category:math]]

← Older revision		Revision as of 19:16, 27 August 2020
Line 1:		Line 1:
−	~~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>~~
−
−	~~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]])~~
−	~~(2) <math>P(t) = P_0 \cdot e^{-r t} dt</math>~~
−
−	~~We know that after 1 year a developer has produced 10KLoc:~~
−	~~(3a) <math>\int^1_0 P_0 \cdot e^{-r t} dt</math> = <math>{ProductionPerYear} = 10</math>~~
−
−	~~We know that over the full life time a developer can maximally maintain 50KLoc:~~
−	~~(4a) <math>\int^\infty_0 P_0 \cdot e^{-r t} dt</math> = <math>{MaxMaintenance} = 50</math>~~
−
−	~~Using the [[Substitution rule\|Substitution rule]] we can see that:~~
−	~~(5a) <math>u = -rt</math>~~
−	~~(5b) <math>t = 0 => u = 0</math>~~
−	~~(5c) <math>t = \infty => u = - r \cdot \infty</math>~~
−	~~(5d) <math>t = 1 => u = - r</math>~~
−	~~(5e) <math>du = -rdt => dt = </math> <math>-\frac{1}{r}</math><math>du</math>~~
−
−	~~So that:~~
−	~~(3b) <math>-P_0 \frac {1}{r} \int^{-r}_0 e^u du</math> = <math>{ProductionPerYear}</math>~~
−	~~(3c) <math>-P_0 \frac {1}{r} (e^{-r} - e^0)</math> = <math>{ProductionPerYear}</math>~~
−	~~(3d) <math>P_0 \frac {1-e^r}{r}</math> = <math>{ProductionPerYear}</math>~~
−	~~(3e) <math>P_0 = </math><math>\frac</math>{{ProductionPerYear} \cdot r}{1-e^{-r}} <math></math>~~
−
−	~~(4b) <math>-P_0 \frac {1}{r} \int^{-\infty}_0 \cdot e^u du</math> = <math>{MaxMaintenance}</math>~~
−	~~(4c) <math>-P_0 \frac {1}{r} (e^{-\infty} - e^0)</math> = <math>{MaxMaintenance}</math>~~
−	~~(4d) <math>-P_0 \frac {1}{r} (0 -1)</math> = <math>{MaxMaintenance}</math>~~
−	~~(4e) <math>\frac {P_0}{r}</math> = <math>{MaxMaintenance}</math>~~
−	~~(4f) <math>P_0</math> = <math>{MaxMaintenance} \cdot r </math>~~
−
−	~~Combining (3e) and (4f) we get:~~
−	~~(6a) <math>{MaxMaintenance} \cdot r = \frac</math>{{ProductionPerYear} \cdot r}{1-e^{-r}}<math>~~
−	~~(6b) </math>\large {1-e^{-r}} = \frac{{ProductionPerYear} \cdot r}{{MaxMaintenance} \cdot r}<math>~~
−	~~(6c) </math>\large -e^{-r} = \frac{{ProductionPerYear}}{{MaxMaintenance}} -1<math>~~
−	~~(6d) </math>\large e^{-r} = 1 - \frac{{ProductionPerYear}}{{MaxMaintenance}} <math>~~
−	~~(6e) </math>\large -r = ln(1- \frac{ProductionPerYear}{{MaxMaintenance}})<math>~~
−	~~(6f) </math>\large r = -ln(1- \frac{ProductionPerYear}{{MaxMaintenance}})<math></math>~~
−
−	~~Which results in the following values for <math>r</math> and <math>P_0</math>~~
−	~~(6f) <math>r = -ln(1-10/50) = 0.22314355</math>~~
−	~~(4g) <math>P_0 = 50 * 0.22314355 = 11.1571775657 </math>~~
	[[Category:math]]		[[Category:math]]

← Older revision		Revision as of 19:51, 27 August 2020
Line 1:		Line 1:
−	Derivation of developer productivity ~~parameters~~	+	At <math>t = 0</math> <math>P_0</math> is defined to be developer productivity without legacy obligations
		+	(1) <math>P(0) = P_0</math>
		+
		+	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]])
		+	(2) <math>P(t) = P_0 \cdot e^{-r t} dt</math>
		+
		+	We know that after 1 year a developer has produced 10KLoc:
		+	(3a) <math>\int^1_0 P_0 \cdot e^{-r t} dt</math> = <math>{ProductionPerYear} = 10</math>
		+
		+	We know that over the full life time a developer can maximally maintain 50KLoc:
		+	(4a) <math>\int^\infty_0 P_0 \cdot e^{-r t} dt</math> = <math>{MaxMaintenance} = 50</math>
		+
		+	Using the [[Substitution rule\|Substitution rule]] we can see that:
		+	(5a) <math>u = -rt</math>
		+	(5b) <math>t = 0 => u = 0</math>
		+	(5c) <math>t = \infty => u = - r \cdot \infty</math>
		+	(5d) <math>t = 1 => u = - r</math>
		+	(5e) <math>du = -rdt => dt = </math> <math>-\frac{1}{r}</math><math>du</math>
		+
		+	So that:
		+	(3b) <math>-P_0 \frac {1}{r} \int^{-r}_0 e^u du</math> = <math>{ProductionPerYear}</math>
		+	(3c) <math>-P_0 \frac {1}{r} (e^{-r} - e^0)</math> = <math>{ProductionPerYear}</math>
		+	(3d) <math>P_0 \frac {1-e^r}{r}</math> = <math>{ProductionPerYear}</math>
		+	(3e) <math>P_0 = </math><math>\frac</math>{{ProductionPerYear} \cdot r}{1-e^{-r}} <math></math>
		+
		+	(4b) <math>-P_0 \frac {1}{r} \int^{-\infty}_0 \cdot e^u du</math> = <math>{MaxMaintenance}</math>
		+	(4c) <math>-P_0 \frac {1}{r} (e^{-\infty} - e^0)</math> = <math>{MaxMaintenance}</math>
		+	(4d) <math>-P_0 \frac {1}{r} (0 -1)</math> = <math>{MaxMaintenance}</math>
		+	(4e) <math>\frac {P_0}{r}</math> = <math>{MaxMaintenance}</math>
		+	(4f) <math>P_0</math> = <math>{MaxMaintenance} \cdot r </math>
		+
		+	Combining (3e) and (4f) we get:
		+	(6a) <math>{MaxMaintenance} \cdot r = \frac</math>{{ProductionPerYear} \cdot r}{1-e^{-r}}<math>
		+	(6b) </math>\large {1-e^{-r}} = \frac{{ProductionPerYear} \cdot r}{{MaxMaintenance} \cdot r}<math>
		+	(6c) </math>\large -e^{-r} = \frac{{ProductionPerYear}}{{MaxMaintenance}} -1<math>
		+	(6d) </math>\large e^{-r} = 1 - \frac{{ProductionPerYear}}{{MaxMaintenance}} <math>
		+	(6e) </math>\large -r = ln(1- \frac{ProductionPerYear}{{MaxMaintenance}})<math>
		+	(6f) </math>\large r = -ln(1- \frac{ProductionPerYear}{{MaxMaintenance}})<math></math>
		+
		+	Which results in the following values for <math>r</math> and <math>P_0</math>
		+	(6f) <math>r = -ln(1-10/50) = 0.22314355</math>
		+	(4g) <math>P_0 = 50 * 0.22314355 = 11.1571775657 </math>
	[[Category:math]]		[[Category:math]]

Derivation of developer productivity parameters - Revision history

Dirkjan at 19:51, 27 August 2020

Dirkjan: Replaced content with "Derivation of developer productivity parameters Category:math"

Dirkjan: Created page with "At t = 0 P_0 is defined to be developer productivity without legacy obligations (1) P(0) = P_0 At time t the developer pro..."

Dirkjan: Created page with "At $t = 0$ $P_0$ is defined to be developer productivity without legacy obligations (1) $P(0) = P_0$ At time $t$ the developer pro..."