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Math Modeling : Markov Chains & Processes

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Example Question #3 : Probability 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 which represents the number of computers in the shop for repair at time months and calculate the EX_t .

Possible Answers:

$EX_t\approx 1.036$

$EX_t\approx 1.307$

$EX_t\approx 1.037$

$EX_t\approx 1.067$

$EX_t\approx 1.133$

Correct answer:

$EX_t\approx 1.037$

Explanation:

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

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

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

Let

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

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

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

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

Now formulate the Markov Process specifically for this problem.

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

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

X_t=i to X_t=i+1 or X_t=i-1

Now calculate the rates up and down.

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

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

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.

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

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

Solving with

$\sum P_i=1$

$EX_t=\sum iP_i$

Continuing in this fashion results in the following.

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

Now, writing the sum of a finite geometric series is

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

Now

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

therefore,

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

Next,

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

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

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Example Question #4 : Probability 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 which represents the number of computers in the shop for repair at time months and calculate the EX_t .

Possible Answers:

$EX_t\approx 4.716$

$EX_t\approx 4.426$

$EX_t\approx 4.236$

$EX_t\approx 4.176$

$EX_t\approx 4.761$

Correct answer:

$EX_t\approx 4.176$

Explanation:

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

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

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

Let

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

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

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

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

Now formulate the Markov Process specifically for this problem.

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

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

X_t=i to X_t=i+1 or X_t=i-1

Now calculate the rates up and down.

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

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.

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

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

Solving with

$\sum P_i=1$

$EX_t=\sum iP_i$

Continuing in this fashion results in the following.

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

Now, writing the sum of a finite geometric series is

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

Now

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

therefore,

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

Next,

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

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

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Example Question #5 : Probability Models

A computer company has one service repair man and has space for 23 computers in the shop at one time. Last year the shop repaired 51 computers with an average repair time of 5 days per computer. Formulate a Markov process model for X_t which represents the number of computers in the shop for repair at time months and calculate the EX_t .

Possible Answers:

$EX_t\approx 0.999$

$EX_t\approx 1.999$

$EX_t\approx 0.989$

$EX_t\approx 1.989$

$EX_t\approx 0.997$

Correct answer:

$EX_t\approx 0.999$

Explanation:

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

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

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

Let

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

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

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

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

Now formulate the Markov Process specifically for this problem.

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

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

X_t=i to X_t=i+1 or X_t=i-1

Now calculate the rates up and down.

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

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.

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

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

Solving with

$\sum P_i=1$

$EX_t=\sum iP_i$

Continuing in this fashion results in the following.

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

Now, writing the sum of a finite geometric series is

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

Now

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

therefore,

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

Next,

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

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

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Math Modeling : Markov Chains & Processes

Study concepts, example questions & explanations for Math Modeling

All Math Modeling Resources

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Example Question #3 : Probability Models

Example Question #4 : Probability Models

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