Expected time to failure 
A machine needs two types of components in order to function. We have a
  stockpile of $n$ type-$1$ components and $m$ type-$2$ components. Type-$1$
  components last for an exponential time with rate $\mu_i$ before
  failing. Compute the mean length of time the machine is operative if a
  failed components is replaced by one the same type from the stockpile;
  that is, compute $$\mathbb E\left[\min\left\{\sum_{t=1}^nX_t,\;\sum_{t=1}^mY_t\right\}\right]$$ where the
  $X_t, Y_t$ have exponential distribution with rates $\mu_1, \mu_2$ respectively.

I was looking again the book and I see it
Suppose we start out with a sequence {$X_n,n\geq 1$} of independent identically distributed exponential random variables each having mean $\frac{1}{\lambda}$. Now let us define a counting process by saying that the nth event of this process occurs at time $S_n$ where 
$$S_n=X_1+,\dots,+X_n$$
The resultant counting process {$N(t),t\geq 0$} will be Poisson with rate $\lambda$ 
So what I tried was
Let $$S_n=X_1+\cdots+X_n$$
$$S_m=Y_1+\cdots+Y_m$$
So I will have two Poisson process, where $\{N_1(t) : t\geq 0\}$ is Poisson with rate $\mu_1$ and $\{N_2(t) : t\geq 0\}$ is Poisson with rate $\mu_2$, and I have that 
$$P\left(\{N_1(t)=n\}\cup \{N_2(t)=m\}\right)=P(N_1(t)=n)+P(N_2(t)=m)$$
But I do not know how to calculate the expectation, because it may be that $n\neq m$, and I would have trouble with the summation index in this case.
 A: Extended Comment:  
The first sum $T_1$ is distributed as $Gamma(n_1, rate=\mu_1)$
and the second sum $T_2$ as $Gamma(n_2, rate=\mu_2)$. Because the shape
parameters are integers, these are also Erlang distributions.
The problem is to find the mean of the minimum $U$ of the two Erlang
distributions. The reliability function of $U$ is
$$1 - F_U(u) = R_U(u) = R_{T_1}(u) \times R_{T_2}(u) 
= P(T_1 > u)P(T_2 > u).$$
In connection with your speculation about Poisson random variables, it may be helpful to look at Wikipedia on 'Erlang' for the form
of the the two factors, but I'm not sure if that leads to a solution.
Simulation in R is trivial. It may help to have pretty good
approximate answers against which to check analytic results.
  mu1 = 1/2;  n1 = 20;  mu2 = 1/5;  n2 = 10;  B = 10^6
  mn = pmin(rgamma(B, n1, mu1), rgamma(B, n2, mu2))
  mean(mn); sd(mn)
  ## 36.94078
  ## 8.125646

  mu1 = 1/2;  n1 = 20;  mu2 = 1/2;  n2 = 20;  B = 10^6
  mn = pmin(rgamma(B, n1, mu1), rgamma(B, n2, mu2))
  mean(mn); sd(mn)
  ## 34.99401
  ## 6.691445

