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Math Modeling : Probability Models

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Example Question #1 : Probability Models

A beauty supply company manufactures a variety of brushes. Quality control engineers work to ensure that the defected brushes in the factory will be detected prior to shipping them. It is estimated that approximately 0.2% of the brushes made will be defected. Tests can be done individually on the brushes or on batches of the brushes. If the test on a batch of brushes fails, it means that one or more of the brushes in that particular batch are defected. The estimated cost is 4 cents for a single brush, and 3+n cents for a group of n>1 brushes. If a batch fails then all brushes in that batch must be tested individually. Find the most cost-effected quality control procedure for detecting defected brushes.

Possible Answers:

By testing brush batches in groups of 12 will reduce testing costs without sacrificing the quality.

By testing brush batches in groups of 10 will reduce testing costs without sacrificing the quality.

By testing brush batches in groups of 20 will reduce testing costs without sacrificing the quality.

By testing brush batches in groups of 9 will reduce testing costs without sacrificing the quality.

By testing brush batches in groups of 15 will reduce testing costs without sacrificing the quality.

Correct answer:

By testing brush batches in groups of 15 will reduce testing costs without sacrificing the quality.

Explanation:

First identify the known variables and assumptions.

$\\n=\text{number of brushes per test batch} \\C=\text{testing cost for one group(cent)} \\A=\text{average testing cost (cents/brush)}$

$n=1\rightarrow A=4$

If a batch of brushes is tested and if the test shows that all the brushes are good then,

$n>1\rightarrow C=3+n$

If the batch test shows that there is a defected brush in the batch then,

$A=\text{(Average value of C)}/n$

Use a discrete probability model to find the most cost-effected quality control procedure for detecting defected brushes.

Consider the random variable

$X\epsilon\begin{Bmatrix} x_1, x_2, x_3,... \end{Bmatrix}$

that has a probability of

$\text{Pr}\begin{Bmatrix} X=x_i \end{Bmatrix}=p_i$

$\\\Sigma p_i=1 \\EX=\Sigma x_ip_i$

If the probability of a brush being good is then the probability of a brush being defected is 1-p . Then the average expected value of is as follows:

EC=(3+n)p+[(3+n)+4n](1-p)

Now there are brushes and the probability that one brush is defected is 0.002 thus assuming independence, the probability of all brushes in one test group are good is p=0.998^n .

Therefore the expected value of the random variable is,

$\\EC=(3+n)0.998^n+[(3+n)+4n](1-0.998^n) \\EC=(3+n)+4n(1-0.998^n) \\EC=3+5n-4n(0.998^n)$

Therefore the average testing cost is,

$A=\frac{3}{n}+5-4(0.998)^n$

Using the law of large numbers minimizing results in

A=1.31, n=15

Now answer the question.

By testing brush batches in groups of 15 will reduce testing costs without sacrificing the quality.

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Example Question #1 : Probability Models

An oil spill occurs at a factory 12 kilometers up stream from a town. One hour after the spill happens it reaches the stream and a 1600 meters long patch of oil begins to flow towards the town at a rate of 2 kilometers per hour. The maximum concentration of the oil in the water is 15 times the acceptable level. What is the maximum concentration expected in town and when will it arrive?

Possible Answers:

P=6.12, t=5

P=6.12, t=6

P=6.15, t=6.7

P=6.14, t=6.3

P=6.10, t=5.4

Correct answer:

P=6.12, t=6

Explanation:

Identify what is known and the assumptions.

This is a diffusion problem and thus a diffusion equation will be used with a term of relative concentration.

The relative concentration is denoted as

$\\C(x,t) \\x=\text{location of oil} \\t=\text{time}$

This function has been normalized resulting in the following.

$\int C(x,t)dx=1$

The law of conservation of mass will also be useful is solving this problem.

$\frac{\partial C}{\partial t}=-\frac{\partial q}{\partial x}$

$\\t=\text{Time since release of oil} \\\mu=\text{Distance travelled by the patches center} \\x=\text{Distance between patches center and town} \\s=\text{Patches spread at time t} \\\\ \mu=2t \\x=12-\mu \\P=15, t=1 \\s=1600, t=1$

The goal is to calculate the maximum pollution level in town.

The diffusion equation is,

$\frac{\partial C}{\partial t}=\frac{D}{2}\frac{\partial ^2C}{\partial x^2}$

Using the Fourier transforms to solve the diffusion equation is as follows.

$\hat C(k,t)=\int_{-\infty}^{\infty} e^{-ikx}C(x,t)dx$

$\frac{d \hat C}{dt}=\frac{D}{2}(ik)^2 \hat C=-\frac{D}{2} k^2\hat C$

$C(x,t)=\frac{1}{\sqrt{2\pi Dt}} e^{-\frac{x^2}{2Dt}}$

For this particular function

$\mu =2, \sigma=0.500$

so the interval will be

$\mu + 2\sigma\rightarrow s=4\sigma$

$D=\sigma^2=0.25$ $C(x,t)=\frac{1}{\sqrt{0.5 \pi t}}e^{-\frac{x^2}{0.5t}}$

$P=15=P_0\frac{1}{\sqrt{0.5\pi }}e^{-\frac{0}{0.5}}$

Calculate given P_0

$\\P=P_0C \\ P_0=15\sqrt{0.5\pi} \\\\P=\frac{15}{\sqrt{t}}e^{-\frac{(12-2t)^2}{(0.5t)}}$

Now find where the maximum occurs in time.

P=6.12, t=6

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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 : Probability Models

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