Derivation of developer productivity decline function - Revision history

Dirkjan at 19:51, 27 August 2020

2020-08-27T19:51:31Z

2020-08-27T19:16:55Z

Replaced content with "Derivation of developer productivity decline function Category:math"

← Older revision		Revision as of 19:16, 27 August 2020
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−	~~The first assumption is that~~ developer productivity ~~declines with a constant rate <math>r</math>:~~	+	Derivation of developer productivity decline function
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−	~~(1) <math>\frac{dP(t)}{dt}</math> = <math>-r \cdot P(t)</math>~~
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−	~~Divide by <math>P(t)</math>:~~
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−	~~(2) <math>\frac{1}{P(t)}</math> <math>\cdot</math> <math>\frac{dP(t)}{dt}</math> = <math>-r</math>~~
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−	~~(3) <math>\frac{dP(t)}{P(t)}</math> = <math>-r dt</math>~~
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−	~~(4) <math>\int \frac{dP(t)}{P(t)}</math> = <math>-r \int t dt</math>~~
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−	~~(5) <math>ln(P(t))= -rt +C </math>~~
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−	~~Exponantiate both sides:~~
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−	~~(6) <math>P(t) = e^{-rt}e^C </math>~~
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−	~~(7) <math>P(t) = e^C \cdot e^{-rt} =e^C \cdot 1</math>~~
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−	~~As P(0) = P_0 we get:~~
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−	~~(8) <math>P(0) = e^C \cdot e^{-r \cdot 0} = e^C \cdot 1</math>~~
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−	~~So <math>P(0) = e^C</math> giving:~~
−	~~(9) <math>P(t) = P_0 \cdot e^{-rt}</math>~~
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−	~~See: http://en.wikipedia.org/wiki/Exponential_decay#Solution_of_the_differential_equation~~
	[[Category:math]]		[[Category:math]]

← Older revision		Revision as of 19:51, 27 August 2020
Line 1:		Line 1:
−	~~Derivation of~~ developer productivity ~~decline function~~	+	The first assumption is that developer productivity declines with a constant rate <math>r</math>:
		+
		+	(1) <math>\frac{dP(t)}{dt}</math> = <math>-r \cdot P(t)</math>
		+
		+	Divide by <math>P(t)</math>:
		+
		+	(2) <math>\frac{1}{P(t)}</math> <math>\cdot</math> <math>\frac{dP(t)}{dt}</math> = <math>-r</math>
		+
		+	(3) <math>\frac{dP(t)}{P(t)}</math> = <math>-r dt</math>
		+
		+	(4) <math>\int \frac{dP(t)}{P(t)}</math> = <math>-r \int t dt</math>
		+
		+	(5) <math>ln(P(t))= -rt +C </math>
		+
		+	Exponantiate both sides:
		+
		+	(6) <math>P(t) = e^{-rt}e^C </math>
		+
		+	(7) <math>P(t) = e^C \cdot e^{-rt} =e^C \cdot 1</math>
		+
		+	As P(0) = P_0 we get:
		+
		+	(8) <math>P(0) = e^C \cdot e^{-r \cdot 0} = e^C \cdot 1</math>
		+
		+	So <math>P(0) = e^C</math> giving:
		+	(9) <math>P(t) = P_0 \cdot e^{-rt}</math>
		+
		+	See: http://en.wikipedia.org/wiki/Exponential_decay#Solution_of_the_differential_equation
	[[Category:math]]		[[Category:math]]