I have these two problems regarding the reneval processes.
Excercise 1:
Assume that bike is being replaced if it's broken or has an age of $5$. New bike costs $4$ and the old one is repaired for $1$. If bike still works after $5$ years is gets sold for $2$. Let's denote the "time of life" of the bike by random variable $X\sim \mathcal{U}[2,7]$. Calculate the avarage cost of ownership of the bike in a long period of time.
Excercise 2:
In some warehouse, the stock is replenished according to the scheme $(s,s+S)$ and demands of the clients (independent of time period between their arrivals) are random variable with exponential distribution with parameter $\alpha$. Assume that probability of the stock level in the warehouse being being at least $y$ is less of greater than $1/3$. Estimate the value of $y$.
Here is my solution for the first one:
We have the $X\sim\mathcal{U}[2,7]$. Denote the time between the replacements of bikes by $Y$ and the cost incurred in that time by $R$. Then: $$ Y= \begin{cases} X,\ X<5\\ 5,\ X\geq 5. \end{cases} $$ and $$ R= \begin{cases} 4+1=5,\ X<5\\ 4-2=2,\ X\geq 5. \end{cases} $$
From that we have: $$\mu=\mathbb{E}[Y]=\mathbb{E}[\min(X,5)]=\int_0^5(1-F_X(s)ds=5-\int_0^5\frac{s-2}{5}=5-\frac{25}{10}-2=\frac{1}{2}$$ $$\rho=\mathbb{E}[R]=5\cdot P(X<5)+2\cdot P(X\geq 3)=5\cdot\frac{5-2}{7-2}+2\cdot\left( 1-\frac{3-2}{7-2}\right)=3+2-\frac{2}{5}=\frac{21}{5}$$
The limit theorem says that if $\rho,\mu<\infty$ then $$\frac{R(t)}{t}\rightarrow\frac{\rho}{\mu},$$ when $t\rightarrow\infty$ with the probability $1$. From that limit theorem we get: $$\frac{\rho}{\mu}=\frac{21}{5}\cdot 2=\frac{42}{5}$$
For the second one I don't really have any clever idea. All I have found in my notes was that for given $y,S,s,\mu$ if $n_i\sim Exp(\mu)$, the probability of stock level being on the level equal or greter than y is given by formula:
$$P_y=\frac{\mu(S+s-y)+1}{\mu S+1}$$
Applying the condision given in the problem we get the inequality:
$$\frac{1}{3}\leq \frac{\alpha(S+s-y)+1}{\alpha S+1}\leq 1,$$ but from that we get the following: $$\frac{1}{3}\leq 1+\frac{\alpha(s-y)}{\alpha S+1}\leq 1,$$ $$-\frac{2}{3}\leq \frac{\alpha(s-y)}{\alpha S+1}\leq 0,$$ $$-\frac{2}{3}\cdot\frac{\alpha S+1}{\alpha}\leq s-y\leq 0,$$ $$-\frac{2}{3}\cdot\frac{\alpha S+1}{\alpha}-s\leq -y\leq -s,$$ $$\frac{2}{3}\cdot\frac{\alpha S+1}{\alpha}+s\geq y\geq s,$$ which gives us some kind of estimated value of $y$. I cannot find any justification fo the fromula though.
Can someone advise on my solutions? Is the first problem correctly calculated? What is the source of the formula in the second excercise? And if it's not the correct way to do is - what would be other way to approuch this? Thank you in advance.