How to calculate the ideal maintaining time of a machine? I have a question regarding a cost calculation problem:
I have the following curve, which shows, that with increasing number of tasks without maintaining a machine, the error probability for the produced parts increases.

The x-Axis represents the counter of tasks without maintaining the machine. This means the point (10;0.2) says: When the machine wasn’t maintained the last 10 cycles, the error rate is 20%.
Maintaining the machine costs, let’s say 1000 € and one error 15 €. Now I need to find out the perfect maintaining counter, when to maintain the machine.
I thought about multiplying the curve with 15€ and then integrating the function, to take into account the already increased probability of the previous cycles at a specific count. However, I am not sure about this integration step.
Maybe anyone else has any idea how to handle this problem.
Suppose the probability of an error in the $n$th cycle since the most recent maintenance is
$a+bn$, and let the number of tasks (cycles) between maintenance be $x$. I'm assuming here that the tasks can be treated as discrete.
The way I would approach the optimization is to calculate both the total maintenance cost and the expected total error cost over some fairly large number of cycles - say 1,000.  What number is chosen here is arbitrary, but I find it more intuitive to think of total costs in that context than averages per cycle, although each should lead to the same result.
Total maintenance costs are:
$$\frac{1000}{x}\times 1000=\frac{1000000}{x}$$
Expected error costs between two successive maintenance events, using the triangle number formula to sum the $b$'s, are:
$$\sum_{n=1}^x 15(a+bn)=15\Big(ax+b\Big(\frac{x(x+1)}{2}\Big)\Big)$$
Hence total expected error costs are:
$$\frac{1000}{x}\times 15\Big(ax+b\Big(\frac{x(x+1)}{2}\Big)\Big)=15000\Big(a+\Big(b\frac{x+1}{2}\Big)\Big)$$
Total costs $TC$ (including both maintenance and expected error costs) are therefore:
$$TC = \frac{1000000}{x} + 15000\Big(a+\Big(b\frac{x+1}{2}\Big)\Big)$$
Setting the first derivative equal to zero to find the minimum:
$$\frac{dTC}{dx}=\frac{-1000000}{x^2}+15000\Big(\frac{b}{2}\Big)=0$$
$$-2000000+15000bx^2=0$$
$$-400+3bx^2=0$$
$$x=\sqrt{\frac{400}{3b}}$$
To confirm this is a minimum:
$$\frac{d^2TC}{dx^2}=\frac{2000000}{x^3}&gt;0$$
Putting (as approximately suggested by the chart) $a=0.18, b=0.001$, this yields:
$$x=\sqrt{\frac{400}{0.003}}\approx365$$
Note that in this case $a+bx=0.18+(0.001\times365)=0.545 &lt; 1$.  The formula would need modifying should the values imply a probability of error exceeding 1 before reaching the next maintenance event.
Addendum 3/11/2020
In response to your comment re integration, if the tasks are taken to be discrete then integration does not give an exactly correct formula for the total expected error costs.  The integral needed is:
$$\int_0^x 15(a+bn)dn=15\Big(ax+\frac{bx^2}{2}\Big)$$
This has $x^2$ where the calculation above has $x(x+1)$.  However, this difference vanishes on differentiating TC since:
$$\frac{d(bx)}{dx}=\frac{d(b(x+1))}{dx}=b$$
Thus this approach leads to the same optimum value for $x$.