Using Markov Process for this question assume that the space is finite.

$\displaystyle X_t\epsilon\begin{Bmatrix} 1,2,3,4,...,m \end{Bmatrix}$

The stochastic process which determines the future of the process from a probabilistic standpoint is defined as follows.

Let 

$\displaystyle T_i=\text{time spent in state }i$

$\displaystyle Pr{\begin{Bmatrix} T_i>t+s|T_i>s \end{Bmatrix}}=Pr\begin{Bmatrix} T_i>t \end{Bmatrix}$

Recall that the exponential distribution, and thus the $\displaystyle T_i$ could have the density function,

$\displaystyle F_i(t)=\lambda_ie^{-\lambda_it}$

Now formulate the Markov Process specifically for this problem.

$\displaystyle X_t\epsilon\begin{Bmatrix} 0,1,2,..., 29\end{Bmatrix}$

The transitions of computers in and out of the shop are,

$\displaystyle X_t=i$ to $\displaystyle X_t=i+1$ or $\displaystyle X_t=i-1$

Now calculate the rates up and down.

To calculate $\displaystyle \lambda$ take the total number of computers fixed last year and divide it by the total months in the year. To calculate $\displaystyle \mu$ assume there are 22 working days in a month and it takes on average, two days to fix a computer.

$\displaystyle \\\lambda=5.6 \\\mu=11$

Its key to know that at zero we cannot move down a state and like wise when at 29, we cannot move up a state.

$\displaystyle [\text{Rate In}]=[\text{Rate Out}]$

$\displaystyle \\\lambda P_0=\mu P_1 \\(\mu +\lambda)P_1=\lambda P_0+\mu P_1 \\(\mu +\lambda)P_2=\lambda P_2+\mu P_3\\ \begin{matrix} .\\ .\\ . \end{matrix} \\\mu P_{29}=\lambda P_{28}$

Solving with 

$\displaystyle \sum P_i=1$

$\displaystyle EX_t=\sum iP_i$

Continuing in this fashion results in the following.

$\displaystyle \\P_n=(\lambda / \mu)P_{n-1} \\P_n=(\lambda / \mu)^nP_0 \\\\\sum_{n=0}^{29}P_n=P_0\sum_{n=0}^{29}\left(\frac{\lambda}{\mu} \right )^n=1 \\\\P_0=\frac{1-\rho }{1-\rho^{30}}$

Now, writing the sum of a finite geometric series is

$\displaystyle P_n=\rho^nP_0=\frac{\rho^n(1-\rho)}{1-\rho^{30}}$

Now 

$\displaystyle \rho=\frac{\lambda}{\mu}=\frac{5.6}{11}=0.509$

therefore,

$\displaystyle 1-\rho^{30}=1-0.509^{30}\approx 0.9999999984$

Next, 

$\displaystyle Pr\begin{Bmatrix} X_t>0 \end{Bmatrix}=1-P_0=\rho\approx 0.509$

$\displaystyle \\EX_t=\sum_{n=0}^{29}nP_n \\EX_t=\sum_{n=0}^{29}n\rho^n(1-\rho) \\EX_t\approx 1.037$

Using Markov Process for this question assume that the space is finite.

$\displaystyle X_t\epsilon\begin{Bmatrix} 1,2,3,4,...,m \end{Bmatrix}$

The stochastic process which determines the future of the process from a probabilistic standpoint is defined as follows.

Let 

$\displaystyle T_i=\text{time spent in state }i$

$\displaystyle Pr{\begin{Bmatrix} T_i>t+s|T_i>s \end{Bmatrix}}=Pr\begin{Bmatrix} T_i>t \end{Bmatrix}$

Recall that the exponential distribution, and thus the $\displaystyle T_i$ could have the density function,

$\displaystyle F_i(t)=\lambda_ie^{-\lambda_it}$

Now formulate the Markov Process specifically for this problem.

$\displaystyle X_t\epsilon\begin{Bmatrix} 0,1,2,..., 32\end{Bmatrix}$

The transitions of computers in and out of the shop are,

$\displaystyle X_t=i$ to $\displaystyle X_t=i+1$ or $\displaystyle X_t=i-1$

Now calculate the rates up and down.

To calculate $\displaystyle \lambda$ take the total number of computers fixed last year and divide it by the total months in the year. To calculate $\displaystyle \mu$ assume there are 22 working days in a month and it takes on average, two days to fix a computer.

$\displaystyle \\\lambda=5.9 \\\mu=7.3$

Its key to know that at zero we cannot move down a state and like wise when at 29, we cannot move up a state.

$\displaystyle [\text{Rate In}]=[\text{Rate Out}]$

$\displaystyle \\\lambda P_0=\mu P_1 \\(\mu +\lambda)P_1=\lambda P_0+\mu P_1 \\(\mu +\lambda)P_2=\lambda P_2+\mu P_3\\ \begin{matrix} .\\ .\\ . \end{matrix} \\\mu P_{32}=\lambda P_{31}$

Solving with 

$\displaystyle \sum P_i=1$

$\displaystyle EX_t=\sum iP_i$

Continuing in this fashion results in the following.

$\displaystyle \\P_n=(\lambda / \mu)P_{n-1} \\P_n=(\lambda / \mu)^nP_0 \\\\\sum_{n=0}^{32}P_n=P_0\sum_{n=0}^{32}\left(\frac{\lambda}{\mu} \right )^n=1 \\\\P_0=\frac{1-\rho }{1-\rho^{33}}$

Now, writing the sum of a finite geometric series is

$\displaystyle P_n=\rho^nP_0=\frac{\rho^n(1-\rho)}{1-\rho^{30}}$

Now 

$\displaystyle \rho=\frac{\lambda}{\mu}=\frac{5.9}{7.3}=0.808$

therefore,

$\displaystyle 1-\rho^{33}=1-0.808^{33}\approx 0.9991198131$

Next, 

$\displaystyle Pr\begin{Bmatrix} X_t>0 \end{Bmatrix}=1-P_0=\rho\approx 0.808$

$\displaystyle \\EX_t=\sum_{n=0}^{32}nP_n \\EX_t=\sum_{n=0}^{32}n\rho^n(1-\rho) \\EX_t\approx 4.176$

Using Markov Process for this question assume that the space is finite.

$\displaystyle X_t\epsilon\begin{Bmatrix} 1,2,3,4,...,m \end{Bmatrix}$

The stochastic process which determines the future of the process from a probabilistic standpoint is defined as follows.

Let 

$\displaystyle T_i=\text{time spent in state }i$

$\displaystyle Pr{\begin{Bmatrix} T_i>t+s|T_i>s \end{Bmatrix}}=Pr\begin{Bmatrix} T_i>t \end{Bmatrix}$

Recall that the exponential distribution, and thus the $\displaystyle T_i$ could have the density function,

$\displaystyle F_i(t)=\lambda_ie^{-\lambda_it}$

Now formulate the Markov Process specifically for this problem.

$\displaystyle X_t\epsilon\begin{Bmatrix} 0,1,2,..., 23\end{Bmatrix}$

The transitions of computers in and out of the shop are,

$\displaystyle X_t=i$ to $\displaystyle X_t=i+1$ or $\displaystyle X_t=i-1$

Now calculate the rates up and down.

To calculate $\displaystyle \lambda$ take the total number of computers fixed last year and divide it by the total months in the year. To calculate $\displaystyle \mu$ assume there are 22 working days in a month and it takes on average, two days to fix a computer.

$\displaystyle \\\lambda=2.2 \\\mu=4.4$

Its key to know that at zero we cannot move down a state and like wise when at 29, we cannot move up a state.

$\displaystyle [\text{Rate In}]=[\text{Rate Out}]$

$\displaystyle \\\lambda P_0=\mu P_1 \\(\mu +\lambda)P_1=\lambda P_0+\mu P_1 \\(\mu +\lambda)P_2=\lambda P_2+\mu P_3\\ \begin{matrix} .\\ .\\ . \end{matrix} \\\mu P_{23}=\lambda P_{22}$

Solving with 

$\displaystyle \sum P_i=1$

$\displaystyle EX_t=\sum iP_i$

Continuing in this fashion results in the following.

$\displaystyle \\P_n=(\lambda / \mu)P_{n-1} \\P_n=(\lambda / \mu)^nP_0 \\\\\sum_{n=0}^{23}P_n=P_0\sum_{n=0}^{23}\left(\frac{\lambda}{\mu} \right )^n=1 \\\\P_0=\frac{1-\rho }{1-\rho^{24}}$

Now, writing the sum of a finite geometric series is

$\displaystyle P_n=\rho^nP_0=\frac{\rho^n(1-\rho)}{1-\rho^{30}}$

Now 

$\displaystyle \rho=\frac{\lambda}{\mu}=\frac{2.2}{4.4}=0.5$

therefore,

$\displaystyle 1-\rho^{24}=1-0.5^{24}\approx 0.9999999404$

Next, 

$\displaystyle Pr\begin{Bmatrix} X_t>0 \end{Bmatrix}=1-P_0=\rho\approx 0.5$

$\displaystyle \\EX_t=\sum_{n=0}^{23}nP_n \\EX_t=\sum_{n=0}^{23}n\rho^n(1-\rho) \\EX_t\approx 0.999$