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I want to calculate the rate at which work gets done on a project.

On each week some work is done and some new work is generated. The new work generated is directly proportional to the amount of work remaining in the project.

Work done on project

$dW = (cW - R) dt$

W = work remaining R = rate of work c = constant representing churn rate 

How do I calculate when the project ends?

EDIT

Edited to remove original sign error where I had: $dW = (R - cW) dt$

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    Sorry, there was an error in the calculation; I've corrected the answer.2012-01-06

1 Answers 1

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The general solution of this ordinary linear differential equation is

$W=\frac Rc+a\mathrm e^{ct}\;.$

Thus, if the initial workload is above $R/c$, work will pile up exponentially ($a\gt0$), whereas if the initial workload is below $R/c$, it will decrease exponentially ($a\lt0$), but the rate won't exceed $R$ before the work becomes $0$. The time $T$ to finish as a function of the initial workload is given by

$W_0=W(t=0)=\frac Rc+a\;,$

$a=W_0-\frac Rc\;,$

$0=W(t=T)=\frac Rc+a\mathrm e^{cT}=\frac Rc+\left(W_0-\frac Rc\right)\mathrm e^{cT}\;,$

$\left(\frac Rc-W_0\right)\mathrm e^{cT}=\frac Rc\;,$

$\mathrm e^{cT}=\frac{\frac Rc}{\frac Rc-W_0}=\frac{1}{1-\frac{cW_0}R}\;,$

$cT=\log\frac{1}{1-\frac{cW_0}R}=-\log\left(1-\frac{cW_0}R\right)\;,$

$T=-\frac{\log\left(1-\frac{cW_0}R\right)}c\;,$

where $\log$ denotes the natural logarithm.

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    @Peter: You're very welcome. My degree was in physics, too :-)2012-01-06