Poisson Process with Expected Time (Stochastic Processes)

Question

Excited by the recent warm weather Jill and Kelly are doing spring cleaning at their apartment. Jill takes an exponentially distributed amount of time with mean 30 minutes to clean the kitchen. Kelly takes an exponentially distributed amount of time with mean 40 minutes to clean the bath room. The first one to complete their task will go outside and start raking leaves, a task that takes an exponentially distributed amount of time with a mean of one hour. When the second person is done inside, they will help the other and raking will be done at rate 2. (Of course the other person may already be done raking in which case the chores are done.) What is the expected time until the chores are all done?

My current though process is to define the Kitchen ~ exp(1/30), Bathroom ~ exp(1/40), and Rake1 ~ exp(1/60) / Rake 2 ~ exp(1/30). From here, we can see that if cleaning the bathroom or kitchen takes longer than cleaning the other room + raking then obviously the expected time is just the expected time to clean that room. Outside of that, we can calculate the E(max(Kitchen,Bathroom)) = 30 + 40 - 1/(30 + 40). But after this I am lost as to how to get the final result.

Answer 1

Hint: Let's call $K$ time spent to clean kitchen, $B$ time spent to clean bathroom, $R$ time spent raking.

You've got two posibilities:

1. One of them finishes his individual task + rakes and still finishes earlier.
2. One of them finishes his individual tasks, starts raking and the other arrives at some point while he is raking.

In scenario 1, the time taken will be the time it took for the slowest one to complete his individual task. Hence $t = \max \{ B,K \}$.

In scenario 2, the time taken will be $t = \max \{ B,K \} + \frac{1}{2}\left[ R - \left( \max \{ B,K \} - \min \{ B,K \} \right) \right]$ So, $$t = \frac{1}{2}\left( R + \max \{ B,K \} + \min \{ B,K \} \right)$$

Combining scenario $1$ and $2$ the total time taken will be:

$$t = \max \left\{ \max \{ B,K \}, \frac{1}{2}\left( R + \max \{ B,K \} + \min \{ B,K \} \right) \right\}$$

Answer 2

Let $K\sim\text{Exp}(1/30)$, $B\sim\text{Exp}(1/40)$, and $R\sim\text{Exp}(1/60)$ be the times to clean the kitchen, bathroom, and rake the leaves. Let $F\equiv K\wedge B\equiv\min\{K,B\}$ and $S\equiv K\vee B\equiv\max\{K,B\}$.

If one person is able to finish their first chore and raking the leaves before the other person is able to finish their first chore, the total time at which all chores are done is $S$. Otherwise, we can assume the total time is $(S+F+R)/2$, corresponding to the midpoint between the time at which the first person would be finished raking without help and the other person is finished their first chore.

Summarizing, $$ T=\begin{cases} S & \text{if }S\geq F+R\\ \frac{1}{2}\left(S+F+R\right) & \text{if }S<F+R. \end{cases} $$ We can write the above in terms of $K$ and $B$ by noting that $$ S+F=K+B\qquad\text{and}\qquad S-F=\left|K-B\right|. $$

Then $\mathbb{E}T=I_{1}+I_{2}$ where \begin{multline*} I_{1}=\int\left(K\vee B\right)\boldsymbol{1}_{\{|K-B|\geq R\}}d\mathbb{P}=\int K\boldsymbol{1}_{\{K-B\geq R\}}+B\boldsymbol{1}_{\{B-K\geq R\}}d\mathbb{P}\\ =\frac{1}{72000}\left(\int_{0}^{\infty}\int_{0}^{k}\int_{0}^{k-b}k\exp\left(-\frac{k}{30}-\frac{b}{40}-\frac{r}{60}\right)drdbdk\right)\\ +\frac{1}{72000}\left(\int_{0}^{\infty}\int_{0}^{b}\int_{0}^{b-k}b\exp\left(-\frac{k}{30}-\frac{b}{40}-\frac{r}{60}\right)drdkdb\right) \end{multline*} and \begin{multline*} I_{2}=\frac{1}{2}\int\left(K+B+R\right)\boldsymbol{1}_{\{|K-B|<R\}}d\mathbb{P}\\ =\frac{1}{2 \cdot 72000}\left[\int_{0}^{\infty}\int_{r}^{\infty}\int_{k-r}^{k+r}+\int_{0}^{\infty}\int_{0}^{r}\int_{0}^{k+r}\right]\left(k+b+r\right)\exp\left(-\frac{k}{30}-\frac{b}{40}-\frac{r}{60}\right)dbdkdr. \end{multline*} It should all be mechanical from here.