First identify the known variables and assumptions.

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

If 

\(\displaystyle n=1\rightarrow A=4\)

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

\(\displaystyle n>1\rightarrow C=3+n\)

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

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

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

Consider the random variable

\(\displaystyle X\epsilon\begin{Bmatrix} x_1, x_2, x_3,... \end{Bmatrix}\)

that has a probability of

\(\displaystyle \text{Pr}\begin{Bmatrix} X=x_i \end{Bmatrix}=p_i\)

\(\displaystyle \\\Sigma p_i=1 \\EX=\Sigma x_ip_i\)

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

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

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

Therefore the expected value of the random variable \(\displaystyle C\) is,

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

Therefore the average testing cost is,

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

Using the law of large numbers minimizing \(\displaystyle A\) results in

\(\displaystyle A=1.31, n=15\)

Now answer the question.

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

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 

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

This function has been normalized resulting in the following.

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

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

\(\displaystyle \frac{\partial C}{\partial t}=-\frac{\partial q}{\partial x}\)

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

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

The diffusion equation is,

\(\displaystyle \frac{\partial C}{\partial t}=\frac{D}{2}\frac{\partial ^2C}{\partial x^2}\)

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

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

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

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

For this particular function

\(\displaystyle \mu =2, \sigma=0.500\)

so the interval will be

\(\displaystyle \mu + 2\sigma\rightarrow s=4\sigma\)

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

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

Calculate \(\displaystyle P\) given \(\displaystyle P_0\)

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

Now find where the maximum occurs in time.

\(\displaystyle P=6.12, t=6\)

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, 

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

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