Is is possible to condense a formula to calculate the number of failures from a failure rate per year?

Question

I have been asked to help simplify a formula for a report that helps estimate the number of failures of an install base per year.

The information I was given was:

First order failures: IB * failure rate (how many units there are) * (how often they fail)

Second order failures (repeat failures of first order failures) IB * failure rate * failure rate * ½ (first two terms are same as above, providing the new “IB” that has already failed once) * (failure rate for these items) * (1/2, since the average opportunity for a second level failure would be ½ of the time interval if the failures are occurring randomly through the time interval)

Third order failures (repeat failures of second order failures) IB * failure rate * failure rate * ½ * failure rate * ¼ (first four terms are IB from above) * (failure rate again ) * (1/4, average opportunity here would be ¼ of the time period)

If: x = Install Base y = failure rate

xy + (xyy0.5) + (xyy0.5y0.25) + (xyy0.5y0.25y0.125)...

Example: Install Base of 1000 machines and failure rate of 0.2

The first year, 200 fail (1000 * 0.2) The second year, of the 200 were replaced, 20 failed (1000 * 0.2 * 0.2 * .5)

Grand total of 220 failures. (1000 * 0.2) + (1000 * 0.2 * 0.2 * .5)

The number of failures is usually insignificant after the 3rd year is calculated, but I was asked if the formula could be cleaned up and shortened to work with an unknown number of years.

This is well beyond my ability to write an equation that might help. I am an art major working in an engineering group to help perform tests on equipment while they analyze the data. I enjoy math, but my head feels like it will explode after looking at this for the past few days.

Answer 1

Starting from @Brian Moehring's answer $$S(y)=\sum_{k=1}^\infty y^k\,\,2^{\frac{k-k^2}{2}} $$ can be bounded by two integrals $$f(y)=\int_1^\infty y^k\,\,2^{\frac{k-k^2}{2}}\,dk < S(y) <\int_0^\infty y^k\,\,2^{\frac{k-k^2}{2}}\,dk=g(y)$$

$$f(y)=\sqrt{\frac{\pi y}{2^{3/4} \log (2)} }\,\,\exp\left(\frac{\log ^2(y)}{2\log (2)} \right)\,\,\left(1+\text{erf}\left(\frac{2 \log (y)-\log (2)}{2 \sqrt{2 \log (2)}}\right)\right)$$

$$g(y)=\sqrt{\frac{\pi y}{2^{3/4} \log (2)} }\,\,\exp\left(\frac{\log ^2(y)}{2\log (2)} \right)\,\,\left(1+\text{erf}\left(\frac{2 \log (y)+\log (2)}{2 \sqrt{2 \log (2)}}\right)\right)$$

Some numbers

$$\left( \begin{array}{cccc} y & f(y) & S(y) & g(y) \\ 1 & 1.11173 & 1.64163 & 2.17154 \\ 2 & 4.34308 & 5.28327 & 5.87138 \\ 3 & 11.0943 & 12.4012 & 13.0215 \\ 4 & 23.4855 & 25.1331 & 25.7750 \\ 5 & 44.4880 & 46.4580 & 47.1159 \\ 6 & 78.1290 & 80.4075 & 81.0780 \\ 7 & 129.733 & 132.309 & 132.990 \\ 8 & 206.200 & 209.064 & 209.754 \\ 9 & 316.325 & 319.471 & 320.168 \\ 10 & 471.159 & 474.580 & 475.283 \\ \end{array} \right)$$

It seems that $$S(y) \sim \sqrt{\frac{\pi y}{2^{3/4} \log (2)} }\,\,\exp\left(\frac{\log ^2(y)}{2\log (2)} \right)\,\,\left(1+\text{erf}\left(\frac{ \log (y)}{\sqrt{2 \log (2)}}\right)\right)=h(y)$$ would not be a bad idea. In fact $$h(y)=\int_{\frac 12}^\infty y^k\,\,2^{\frac{k-k^2}{2}}\,dk $$ $$\left( \begin{array}{ccc} y & h(y) & S(y) \\ 1 & 1.64163 & 1.64163 \\ 2 & 5.23649 & 5.28327 \\ 3 & 12.3126 & 12.4012 \\ 4 & 25.0068 & 25.1331 \\ 5 & 46.2974 & 46.4580 \\ 6 & 80.2154 & 80.4075 \\ 7 & 132.087 & 132.309 \\ 8 & 208.816 & 209.064 \\ 9 & 319.196 & 319.471 \\ 10 & 474.280 & 474.580 \\ \end{array} \right)$$

Edit

If you want a series solution, let $$y=e^{z \sqrt{2 \log (2)}} \quad \implies \quad S(z)=\sqrt{\frac{\pi }{2 \log (2)}}\,\exp\Bigg[\left((z+\frac{1}{2} \sqrt{\frac{\log (2)}{2}}\right)^2 \Bigg]\,(1+\text{erf}(z))$$ and write $$\log(h(z)=\frac{1}{8} \log \left(\frac{\pi ^4}{8 \log ^4(2)}\right)+\sum_{n=1}^\infty a_n\, z^n$$ the first coefficients being $$\left( \begin{array}{cc} n & a_n \\ 1 & \frac{2}{\sqrt{\pi }}+\sqrt{\frac{\log (2)}{2}} \\ 2 & \frac{\pi -2}{\pi } \\ 3 & -\frac{2 (\pi -4)}{3 \pi ^{3/2}} \\ 4 & \frac{4 (\pi -3)}{3 \pi ^2} \\ 5 & \frac{96+\pi (3 \pi -40)}{15 \pi ^{5/2}} \\ 6 &-\frac{4 (120+\pi (7 \pi -60))}{45 \pi ^3} \\ 7 &\frac{5760+\pi ((532-15 \pi ) \pi -3360)}{315 \pi ^{7/2}} \\ 8 & \frac{8 (\pi (280+\pi (3 \pi -56))-420)}{105 \pi ^4} \\ 9 & \frac{645120+\pi (\pi (116928+\pi (105 \pi -9328))-483840)}{11340 \pi ^{9/2}}\\ 10 &-\frac{4 (120960+\pi (\pi (28560+\pi (83 \pi -3040))-100800))}{4725 \pi ^5} \\ \end{array} \right)$$

Making them rational, they are (starting with the constant term)

$$\left\{\frac{863}{1741},\frac{1578}{919},\frac{129}{355},\frac {63}{613},\frac{18}{941},-\frac{1}{4780},-\frac{2}{1179},- \frac{1}{1694},-\frac{1}{38656},\frac{1}{15499},\frac{1}{335 81}\right\} $$ To give you an idea, using only the terms given above $$\Phi=\int_0^2 \big[\log(h(z))-\text{series}_{10}\big]^2\, dz=1.4844\times 10^{-5}$$ Notice that to $z=2$ corresponds $y=10.5362$.

Answer 2

Answering the question you've written, the series would be written $$\sum_{k=1}^\infty xy^k2^{(-k^2+k)/2}$$ As far as I'm aware, there is no closed form for this sum or its partial sums. You could probably find asymptotics if you wished, but I don't know how to do that.

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Answering the question as I think it should have been, which amounts to changing the factors corresponding to how long the $k$th replacement has to break again, the series would be written $$\sum_{k=1}^\infty \frac{xy^k}{k!} = x(e^y-1)$$ though the partial sums may still be messy.

Answer 3

In my previous answer, by lack of attention and/or ignorance, I did not take into account the very restricted range $0\leq y \leq 1$ reported by @Brian Moehring in comments.

Asuming that $$S(y) \sim \int_a^\infty y^k\,\,2^{\frac{k-k^2}{2}}\,dk$$ $$S(y)\sim\sqrt{\frac{\pi y}{2^{3/4} \log (2)} }\,\exp\left(\frac{\log ^2(y)}{2 \log (2)} \right)\,\left(1+ \text{erf}\left(\frac{(1-2 a) \log (2)+2 \log (y)}{2 \sqrt{2 \log (2)}}\right)\right)$$ Computing the values for $y=1$ $$\frac{\vartheta _2\left(0,\frac{1}{\sqrt{2}}\right)}{2^{7/8}}=\sqrt{\frac{\pi }{2^{3/4} \log (2)} }\,\left(1+\text{erf}\left(\frac{1-2 a}{2} \sqrt{\frac{\log (2)}{2}}\right)\right)$$ gives $$a=\frac 12+1.2911\times 10^{-12}$$ confirms the previous result.

A few numbers for $a=\frac 12$ which gives $$S(y)\sim\sqrt{\frac{\pi y}{2^{3/4} \log (2)} }\,\exp\left(\frac{\log ^2(y)}{2 \log (2)} \right)\,\left(1+ \text{erf}\left(\frac{ \log (y)}{ \sqrt{2 \log(2)}}\right)\right)=h(y)$$

$$\left( \begin{array}{ccc} y & h(y) & S(y) \\ 0.1 & 0.13509 & 0.10513 \\ 0.2 & 0.25313 & 0.22103 \\ 0.3 & 0.37899 & 0.34850 \\ 0.4 & 0.51574 & 0.48841 \\ 0.5 & 0.66502 & 0.64163 \\ 0.6 & 0.82812 & 0.80910 \\ 0.7 & 1.00620 & 0.99179 \\ 0.8 & 1.20039 & 1.19073 \\ 0.9 & 1.41181 & 1.40697 \\ 1.0 & 1.64163 & 1.64163 \\ \end{array} \right)$$ which is not fantastic.

Using asymptotics $$S(y)=y+\frac 12 y^2+\frac 18y^3+\frac 1{60}y^4 +O(y^5)$$

$$\left( \begin{array}{ccc} y & \text{approximation} & S(y) \\ 0.1 & 0.10513 & 0.10513 \\ 0.2 & 0.22103 & 0.22103 \\ 0.3 & 0.34851 & 0.34850 \\ 0.4 & 0.48843 & 0.48841 \\ 0.5 & 0.64167 & 0.64163 \\ 0.6 & 0.80916 & 0.80910 \\ 0.7 & 0.99188 & 0.99179 \\ 0.8 & 1.19083 & 1.19073 \\ 0.9 & 1.40706 & 1.40697 \\ 1.0 & 1.64167 & 1.64163 \\ \end{array} \right)$$