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Flashcards: Stochastic Models

A computer company has one service repair man and has space for 29 computers in the shop at one time. Last year the shop repaired 67 computers with an average repair time of 2 days per computer. Formulate a Markov process model for X_t $\displaystyle X_t$ which represents the number of computers in the shop for repair at time $\displaystyle t$ months and calculate the EX_t $\displaystyle EX_t$.

$EX_t\approx 1.307$ $\displaystyle EX_t\approx 1.307$

$EX_t\approx 1.037$ $\displaystyle EX_t\approx 1.037$

$EX_t\approx 1.067$ $\displaystyle EX_t\approx 1.067$

$EX_t\approx 1.133$ $\displaystyle EX_t\approx 1.133$

$EX_t\approx 1.036$ $\displaystyle EX_t\approx 1.036$

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Flashcards: Stochastic Models

$EX_t\approx 1.307$ $\displaystyle EX_t\approx 1.307$

$EX_t\approx 1.037$ $\displaystyle EX_t\approx 1.037$

$EX_t\approx 1.067$ $\displaystyle EX_t\approx 1.067$

$EX_t\approx 1.133$ $\displaystyle EX_t\approx 1.133$

$EX_t\approx 1.036$ $\displaystyle EX_t\approx 1.036$

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Flashcards: Stochastic Models

Your answer:

$EX_t\approx 1.307$ $\displaystyle EX_t\approx 1.307$

Your answer:

$EX_t\approx 1.067$ $\displaystyle EX_t\approx 1.067$

Your answer:

$EX_t\approx 1.133$ $\displaystyle EX_t\approx 1.133$

Your answer:

$EX_t\approx 1.036$ $\displaystyle EX_t\approx 1.036$

Correct answer:

$EX_t\approx 1.037$ $\displaystyle EX_t\approx 1.037$

Explanation:

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

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

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

Let

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

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

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

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

Now formulate the Markov Process specifically for this problem.

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

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

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

Now calculate the rates up and down.

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

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

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.

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

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

$\sum P_i=1$ $\displaystyle \sum P_i=1$

$EX_t=\sum iP_i$ $\displaystyle EX_t=\sum iP_i$

Continuing in this fashion results in the following.

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

Now, writing the sum of a finite geometric series is

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

Now

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

therefore,

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

Next,

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

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

Flashcards: Stochastic Models

A computer company has one service repair man and has space for 32 computers in the shop at one time. Last year the shop repaired 71 computers with an average repair time of 3 days per computer. Formulate a Markov process model for X_t $\displaystyle X_t$ which represents the number of computers in the shop for repair at time $\displaystyle t$ months and calculate the EX_t $\displaystyle EX_t$.

$EX_t\approx 4.716$ $\displaystyle EX_t\approx 4.716$

$EX_t\approx 4.236$ $\displaystyle EX_t\approx 4.236$

$EX_t\approx 4.761$ $\displaystyle EX_t\approx 4.761$

$EX_t\approx 4.176$ $\displaystyle EX_t\approx 4.176$

$EX_t\approx 4.426$ $\displaystyle EX_t\approx 4.426$

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Flashcards: Stochastic Models

Your answer:

$EX_t\approx 4.716$ $\displaystyle EX_t\approx 4.716$

Your answer:

$EX_t\approx 4.236$ $\displaystyle EX_t\approx 4.236$

Your answer:

$EX_t\approx 4.761$ $\displaystyle EX_t\approx 4.761$

Your answer:

$EX_t\approx 4.426$ $\displaystyle EX_t\approx 4.426$

Correct answer:

$EX_t\approx 4.176$ $\displaystyle EX_t\approx 4.176$

Explanation:

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

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

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

Let

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

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

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

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

Now formulate the Markov Process specifically for this problem.

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

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

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

Now calculate the rates up and down.

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

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.

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

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

$\sum P_i=1$ $\displaystyle \sum P_i=1$

$EX_t=\sum iP_i$ $\displaystyle EX_t=\sum iP_i$

Continuing in this fashion results in the following.

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

Now, writing the sum of a finite geometric series is

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

Now

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

therefore,

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

Next,

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

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

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Flashcards: Stochastic Models

All Math Modeling Resources

Flashcards: Stochastic Models

Flashcards: Stochastic Models

Flashcards: Stochastic Models

Flashcards: Stochastic Models

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Flashcards: Stochastic Models

All Math Modeling Resources

Flashcards: Stochastic Models

Flashcards: Stochastic Models

Flashcards: Stochastic Models

Flashcards: Stochastic Models

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