Reliability of a manufacturing network. A team of industrial management university professors investigated the reliability of a manufacturing system that involves multiple production lines (Journal of Systems Sciences & Systems Engineering, March 2013). An example of such a network is a system for producing integrated circuit (IC) cards with two production lines set up in sequence. Items (IC cards) first pass through Line 1, then are processed by Line 2. The probability distribution of the maximum capacity level (x) of each line is shown below. Assume the lines operate independently.

a. Verify that the properties of discrete probability distributions are satisfied for each line in the system.

b. Find the probability that the maximum capacity level for Line 1 will exceed 30 items.

c. Repeat part b for Line 2.

d. Now consider the network of two production lines. What is the probability that a maximum capacity level exceeding 30 items is maintained throughout the network? [Hint: Apply the multiplicative law of probability for independent events.]

e. Find the mean maximum capacity for each line. Interpret the results practically.

f. Find the standard deviation of the maximum capacity for each line. Then, give an interval for each line that will contain the maximum capacity with probability of at least .75.

a. The probabilities observed in Line 2 are positive which one of the two properties is fulfilled and the summation of the probabilities is shown below:

$\begin{array}{rcl}\mathbf{\text{Summation}}& \mathbf{=}& \mathbf{0}\mathbf{.}\mathbf{01}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{02}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{02}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{95}\\ & \mathbf{=}& \mathbf{1}\\ & & \end{array}$

The summation is found to be 1 in the case of Line 1 which means the second property is also fulfilled to regard Line 1 to be a discrete probability distribution.

The probabilities observed in Line 2 are positive which one of the two properties is fulfilled and the summation of the probabilities is shown below:

$\begin{array}{rcl}\mathbf{\text{Summation}}& \mathbf{=}& \mathbf{0}\mathbf{.}\mathbf{002}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{002}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{996}\\ & \mathbf{=}& \mathbf{1}\\ & & \end{array}$

The summation is also found to be 1 in the case of Line 2.In short, the second property is also fulfilled for this line.

In this context, probability refers to the possibility of the maximum capacity level reaching a certain value. For example, the possibility of reaching 12 is only 0.02 out of all other possible values.

The maximum capacity level above 30 is found to be 36 in the case of Line 1 1.Therefore, as the associated probability of 36 is 0.95, it can be said that the probability that it will reach above 30 has to be 0.95.

Probability, in this case, refers to the actual chance of the maximum capacity level reaching one of the mentioned values. For example, the chance of reaching 35 is only 0.002 out of all other possible values.

c. The associated probability of 35 is 0.002 and for 70 is 0.996.Therefore, the summation is 0.998, and so it can be said that the probability that it will reach above 30 has to be 0.998.

d. The multiplicative law of probability takes two events into consideration and the events can be dependent or independent. Whenever it is independent, the product of the probabilities of the two events is considered.

e. In the calculation below, $\mathbf{\text{P}}\left(\text{1} \text{and} \text{2}\right)$ is the required probability, $\mathit{P}\left(1\right)$and $\mathit{P}\left(2\right)$are the probability that maximum capacity will remain above 30 for Line 1 and 2 respectively.

$\begin{array}{rcl}\mathbf{\text{P}}\left(1 and 2\right)& \mathbf{=}& \mathit{P}\left(1\right)\mathbf{\times }\mathit{P}\left(2\right)\\ & \mathbf{=}& \mathbf{0}\mathbf{.}\mathbf{95}\mathbf{\times }\left(0.002+0.996\right)\\ & \mathbf{=}& \mathbf{0}\mathbf{.}\mathbf{95}\mathbf{\times }\mathbf{0}\mathbf{.}\mathbf{998}\\ & \mathbf{=}& \mathbf{0}\mathbf{.}\mathbf{9481}\\ & & \end{array}$

Therefore, the value of $\mathit{P}\left(1and2\right)$is found to be 0.9481.

The calculation of means of maximum capacity for Lines 1 and 2 are shown below:

$\begin{array}{rcl}\mathbf{\text{Mean maximum capacity for Line}}\mathbf{1}& \mathbf{=}& \left(0\times 0.01\right)\mathbf{+}\left(12\times 0.02\right)\mathbf{+}\left(24\times 0.02\right)\mathbf{+}\left(36\times 0.95\right)\\ & \mathbf{=}& \mathbf{0}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{24}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{48}\mathbf{+}\mathbf{34}\mathbf{.}\mathbf{2}\\ & \mathbf{=}& \mathbf{34}\mathbf{.}\mathbf{92}\\ \mathbf{\text{Mean maximum capacity for Line}}\mathbf{2}& \mathbf{=}& \left(0\times 0.002\right)\mathbf{+}\left(35\times 0.002\right)\mathbf{+}\left(70\times 0.996\right)\\ & \mathbf{=}& \mathbf{0}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{070}\mathbf{+}\mathbf{69}\mathbf{.}\mathbf{72}\\ & \mathbf{=}& \mathbf{69}\mathbf{.}\mathbf{79}\\ & & \end{array}$

Therefore, the values of means of maximum capacity for Lines 1 and 2 are 34.92 and 69.79 respectively.

Whenever the items are passed through line 1, there is a possibility of observing that the maximum capacity level will be 34.92. When the items will be further in Line 2, there is a possibility of observing that the maximum capacity level will be 69.79.

f. For line 1, the standard deviation ($\mathit{\sigma }$) and the interval are calculated as shown below where the mean is shown by $\mathit{\mu }$:

$\begin{array}{rcl}\mathit{\sigma }& \mathbf{=}& \sqrt{\left[{\left(0-34.92\right)}^{2}0.01+{\left(12-34.92\right)}^{2}0.02+{\left(24-34.92\right)}^{2}0.02+{\left(36-34.92\right)}^{2}0.95\right]}\\ & \mathbf{=}& \sqrt{\left[0+10.51+2.39+1.11\right]}\\ & \mathbf{=}& \mathbf{3}\mathbf{.}\mathbf{74}\\ \mathit{\mu }\mathbf{+}\mathbf{2}\mathit{\sigma }& \mathbf{=}& \mathbf{34}\mathbf{.}\mathbf{92}\mathbf{+}\mathbf{2}\mathbf{\times }\mathbf{3}\mathbf{.}\mathbf{74}\\ & \mathbf{=}& \mathbf{42}\mathbf{.}\mathbf{40}\\ \mathit{\mu }\mathbf{-}\mathbf{2}\mathit{\sigma }& \mathbf{=}& \mathbf{34}\mathbf{.}\mathbf{92}\mathbf{-}\mathbf{2}\mathbf{\times }\mathbf{3}\mathbf{.}\mathbf{74}\\ & \mathbf{=}& \mathbf{27}\mathbf{.}\mathbf{44}\\ & & \end{array}$

3.74 is the standard deviation and the interval ranges from 27.44 to 42.40.

For line 2, the standard deviation ($\mathit{\sigma }$) and the interval are calculated as shown below where the mean is shown by $\mathit{\mu }$:

\(\begin{aligned}{c}\sigma &= \sqrt {\left( {{{\left( {0 - 69.79} \right)}^2}0.002 + {{\left( {35 - 69.79} \right)}^2}0.002 + {{\left( {70 - 69.79} \right)}^2}0.996} \right)} \\ &= \sqrt {\left( {0 + 2.42 + 0.21} \right)} \\ &= 1.62\\\mu + 2\sigma &= 69.79 + 2 \times 1.62\\ &= 73.03\\\mu - 2\sigma &= 69.79 - 2 \times 1.62\\ &= 66.55\end{aligned}\)

1.62 is the standard deviation and the interval ranges from 66.55 to 73.03.