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Am assuming a machine needs 2 types of parts to work which are type A and type B. The machine has one part of each type to begin, and there are also 2 spare A parts and 1 spare B part. I assume that, when a part fails, it is replaced by a spare part of the same type, if available, instantaneously. I'm supposing that the lifetimes (time in service) of all parts are independent whereby parts of type A are Exp(λ) distributed while parts of B are Exp(μ)

distributed.

I now want to determine the expected time until the machine fails for lack of needed parts.

In my opinion, I will proceed as follows:

Let the event that machine fails because of lack of part A be X while Y because it fails because of lack of part B Thus, I need to compute E(X+Y) = $E(X)$ + $E(Y)$. Since $X$ and $Y$ follow the exponential distribution with mean $1/λ$ and $1/μ$ respectively, we compute $2(1/λ)$ + (1/μ).

Is this correct?

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    You do *not* want to compute $\mathsf E(X+Y)$. If the machine fails for lack of either part, then you will want to compute $\mathsf E(\min\{X,Y\})$. If the machine fails only if it lacks both parts, then you will want $\mathsf E(\max\{X,Y\})$. You would only require the expectation of the sum if parts A and B are different versions of the same component, and parts B are only installed after the supply of parts A is exhausted. – 2017-02-10
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    Thank You Graham For that insight. I think I was wrong from the word go. How do I go about to compute the E(min{X,Y})? – 2017-02-10
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    You seek $\mathsf E(\min\{X,Y\})$, where [$X\sim\mathcal{Erlang}(3,\lambda)$](https://en.wikipedia.org/wiki/Erlang_distribution) and $Y\sim\mathcal{Erlang}(2,\mu)$, independently. $$\mathsf E(\min\{X,Y\})=\int_0^\infty \mathsf P(\min\{X,Y\}> s)\operatorname d s =\int_0^\infty (1-F_X(s))(1-F_Y(s))\operatorname d s $$ Which apparently someone else answered while I typed this so... – 2017-02-10
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    Thank you so much. I get it now – 2017-02-10

1 Answers 1

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Let $A_t$ denote the number of defective part A's at time $t$ and $B_t$ denote the number of defective part B's at time $t$. Set $T_A^3=\inf\{t:A_t=3\}$ and $T_B^2=\inf\{t:B_t=2\}$. We are interested in $E(\min\{T_A^3,T_B^2\})$. Fortunately the distributions of $T_A^3$ and $T_B^2$ are known explicitly since $A_t$ and $B_t$ can be regarded as independent Poisson processes with rates $\lambda$ and $\mu$ respectively.

Note that

$ \begin{align*} P(T_A^3\leq t)&=P(A_t\geq3) \\ &=\sum_{n=3}^{\infty}\frac{(\lambda t)^n}{n!}e^{-\lambda t} \end{align*} $

Similarly $P(T_B^2\leq t)=\sum_{m=2}^{\infty}\frac{(\mu t)^m}{m!}e^{-\mu t}$.

By independence we get $P(\min\{T_A^3,T_B^2\}>t)=P(T_A^3>t)P(T_B^2>t)$. Hence

$ \begin{align*} E(\min\{T_A^3,T_B^2\})&=\int_0^{\infty}P(\min\{T_A^3,T_B^2\}>t)\,dt \\ &=\int_0^{\infty}(1-\sum_{n=3}^{\infty}\frac{(\lambda t)^n}{n!}e^{-\lambda t})(1-\sum_{m=2}^{\infty}\frac{(\mu t)^m}{m!}e^{-\mu t})\,dt \\ &=\int_0^{\infty}(1+\lambda t+\frac{(\lambda t)^2}{2})(1+\mu t)e^{-(\lambda+\mu)t}\,dt \\ &=\int_0^{\infty}(1+(\lambda+\mu)t+(\lambda\mu+\frac{\lambda^2}{2})t^2+\frac{\lambda^2\mu}{2}t^3)e^{-(\lambda+\mu)t}\,dt \\ &=\frac{2}{\lambda+\mu}+\frac{\lambda(\lambda+2\mu)}{(\lambda+\mu)^3}+\frac{3\lambda^2\mu}{(\lambda+\mu)^4}. \end{align*} $