A component has a mean to to failure $U$ and a mean time to repair $D$. Two of these components are combined into a parallel system. In other words, we need just one of the components to be working for the system to be working. A natural question is what the mean time to failure and mean time to repair for the system become. On page 614 of "Introduction to probability models" by Sheldon Ross (9th ed), this exact scenario is described (as a consequence of equations 9.32 and 9.33 we get the MTBF of the system ($\bar{U}$)), which is also calculated on that page for a parallel system:
$$\bar{U} = \frac{1-\left(\frac{D}{U+D}\right)^2}{\left(\frac{D}{U+D}\right)^2} \frac{D}{2}$$
Now, what happens when the mean time to repair of the components, $D$ becomes very large? When I calculate the limit of the above expression as $D \to \infty$, I surprisingly get $\bar{U} = U$.
This doesn't make sense since if we consider the time to repair of the components as exponential, we should see one failure of the system at the max of the times to repair of the two components. And the average of this time should be double the average of the time to failure of the components, not the same. What am I missing?
It should not be surprising because most of the time the system is running one of the components is broken. In that case the MTBF of the system is the same as the MTBF of one component. In fact, most of the time the system is down completely. You get one component repaired and it runs until it fails. Then both are broken and the system is down.
As an example, imagine MTBF of a component is 1 hour and MTTR is 1000 hours. You start with both working. After about a half hour one of the two components fails. After about another hour the other fails and the system is down. You are down for about 1000 hours until one component is repaired. Unless the other gets repaired in the next hour, you will only run for about an hour before the newly repaired one fails. On average you are up for 2 hours out of every 1000, about 1 hour at a time.
Checking limiting cases like this is a good check of formulas. If you made an error in the derivation, it often shows up in one of the limiting cases. You can often find answers to the limiting cases without using your derivation, as I did here.