Let arrivals in a $M/G/1$ queue be $\lambda$, and service rate exponential with rate $\mu$ but may be interrupted by an independent crisis event with independent duration uniformly distributed between $1$ and $2$, this is a renewal process. The time between the end of a crisis and the start of the next crisis is exponential with rate $\theta$. And service is not done during a crisis event. Find the expected service time $E[S]$.
Letting $ \displaystyle S = \sum_{i=1}^n S_i$ be the sum of the service time split up by the crisis event, $m(t)$ be the duration of the renewal process, I have tried setting up the renewal process as follows:
$$ \displaystyle E[S] = E[\sum_{i=1}^n S_i] + m(t) $$
Since the crisis event has duration $E[U(1,2)] = 1.5$ and begins again with rate $\theta$ the renewal process should have duration:
$$ m(t) = 1.5 + \frac{1}{\theta} $$
And I think if we split up the service rate into $n$ parts summing them together we will return to the sum of them being $\mu$, so
$$ E[S] = \frac{1}{mu} + 1.5 + \frac{1}{\theta} $$
I don't think this is right though, as I next need to calculate $E[S^2]$ which I don't see how to apply from this computation. would it just be changing $\frac{1}{\mu}$ to $\frac{2}{\mu^2}$
I assume that the service time $S$ includes the time spent in crisis. Denote by $C$ the time in a single crisis and let $E_x$ be an exponentially distributed random variable with rate $x$.
Assume we are in a non-crisis situation. Then, we start a 'race' between two exponential random variables: either we end the service, or we go into a crisis, after which we can start anew. The time spent in this 'race' is $E_{\mu + \theta}$ (the minimum of two exponential random variables), after which we end the service with probability $\mu/(\mu + \theta)$ or we go into crisis and have to start over with probability $\theta/(\mu + \theta)$. So, in conclusion
\begin{equation} S = \begin{cases} E_{\mu + \theta}, & \text{with probability } \frac{\mu}{\mu + \theta}, \\ E_{\mu + \theta} + C + S, & \text{with probability } \frac{\theta}{\mu + \theta}. \end{cases} \end{equation}
Can you take it from here?