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Chapter 14: Problem 66

A manufacturer of cutting tools has developed two empirical equations for tool life \(\left(y_{1}\right)\) and tool cost \(\left(y_{2}\right) .\) Both models are functions of tool hardness \(\left(x_{1}\right)\) and manufacturing time \(\left(x_{2}\right) .\) The equations are $$ \begin{array}{l} \hat{y}_{1}=10+5 x_{1}+2 x_{2} \\ \hat{y}_{2}=23+3 x_{1}+4 x_{2} \end{array} $$ and both are valid over the range \(-1.5 \leq x_{i} \leq 1.5 .\) Suppose that tool life must exceed 12 hours and cost must be below \(\$ 27.50\) (a) Is there a feasible set of operating conditions? (b) Where would you run this process?

Short Answer

Expert verified
Yes, feasible conditions exist: run the process at \( x_{1} = 0.5 \) and \( x_{2} = 0.5 \).

Step by step solution

01

Understand the Equations

We are given two equations: 1. Tool life: \( \hat{y}_{1} = 10 + 5x_{1} + 2x_{2} \)2. Tool cost: \( \hat{y}_{2} = 23 + 3x_{1} + 4x_{2} \)We need to ensure tool life \( \hat{y}_{1} > 12 \) and tool cost \( \hat{y}_{2} < 27.5 \). These conditions will guide our solution.
02

Solve for Tool Life Condition

We need \( \hat{y}_{1} > 12 \):\[ 10 + 5x_{1} + 2x_{2} > 12 \]Subtract 10 from both sides:\[ 5x_{1} + 2x_{2} > 2 \]
03

Solve for Tool Cost Condition

We need \( \hat{y}_{2} < 27.5 \):\[ 23 + 3x_{1} + 4x_{2} < 27.5 \]Subtract 23 from both sides:\[ 3x_{1} + 4x_{2} < 4.5 \]
04

Find Feasible Solutions

We have two inequalities:1. \( 5x_{1} + 2x_{2} > 2 \)2. \( 3x_{1} + 4x_{2} < 4.5 \)We will solve this system of inequalities for the given ranges \(-1.5 \leq x_{1}, x_{2} \leq 1.5 \). Let's check if there are values of \(x_{1}\) and \(x_{2}\) that satisfy both.
05

Check Feasibility

Choose \( x_{1} = 0.5 \) and \( x_{2} = 0.5 \):For tool life equation:\[ 5(0.5) + 2(0.5) = 2.5 + 1 = 3.5 \rightarrow 10 + 3.5 = 13.5 \: (\text{tool life condition satisfied}) \]For tool cost equation:\[ 3(0.5) + 4(0.5) = 1.5 + 2 = 3.5 \rightarrow 23 + 3.5 = 26.5 \: (\text{tool cost condition satisfied}) \]
06

Determine Operating Conditions

The chosen values \( x_{1} = 0.5 \) and \( x_{2} = 0.5 \) satisfy both conditions and lie within the allowable range. Thus, this is where you can run the process.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Empirical Equations
Empirical equations are a vital tool in applied statistics, especially when solving real-world problems like manufacturing. These equations are derived from data rather than theoretical models, which means they represent observed relationships between variables. For instance, in the case of the cutting tools manufacturer, the equations provide insights into how variables like tool hardness (\(x_1\)) and manufacturing time (\(x_2\)) influence tool life and cost.
One key aspect of empirical equations is that they allow manufacturers to predict outcomes and optimize processes. This prediction is based on both historical data collection and analysis. In our scenario, we have the tool life equation: \(\hat{y}_1 = 10 + 5x_1 + 2x_2\), which indicates how increases in hardness or time contribute positively to tool longevity. Meanwhile, the tool cost equation: \(\hat{y}_2 = 23 + 3x_1 + 4x_2\), demonstrates how these factors affect cost.
  • Empirical equations guide decision-making.
  • They are essential for optimization in manufacturing.
  • They reflect real-world relationships, helping in process control.
Thus, mastering empirical equations can significantly enhance operational efficiency and productivity in industrial settings.
Solving Inequalities for Optimal Solutions
In applied statistics, solving inequalities is crucial when conditions or constraints are needed, such as optimizing manufacturing processes. In this exercise, we have two inequalities based on the empirical equations:
  • \([\) \(5x_1 + 2x_2 > 2\), for maintaining tool life above a certain threshold.
  • \([\) \(3x_1 + 4x_2 < 4.5\), to keep cost under control.
The goal is to find values for \(x_1\) and \(x_2\) that satisfy both these inequalities within the permissible range of \([-1.5, 1.5]\). By solving these, we ensure that the tools are not only durable but also cost-effective.
Steps involved:
  • First, set each inequality separately to find the regions where conditions are met.
  • Then, check the overlapping values that fit both criteria.
For example, choosing \(x_1 = 0.5\) and \(x_2 = 0.5\) results in the inequalities being satisfied, showing these are potential operating conditions.
Identifying Feasible Solutions
Feasible solutions are the intersection of inequalities that meet all operational constraints in a problem. In manufacturing, this means finding values of variables that produce desirable outcomes without exceeding resource limits.
In our scenario, feasible solutions mean setting conditions where the tool's life is long enough, and the cost remains below budget. By evaluating solutions across the parameter range, \([-1.5, 1.5]\), feasible solutions ensure the manufacturer maintains efficient production.
When we tested with \(x_1 = 0.5\) and \(x_2 = 0.5\), both tool life and cost conditions were satisfied, demonstrating a feasible solution. These solutions reflect operational balance, maximizing tool effectiveness while minimizing expense.
  • Feasibility ensures sustainable production.
  • It avoids resource wastage.
  • Feasible solutions are integral to process optimization.
By understanding feasible solutions, manufacturers can navigate trade-offs and secure optimal operational performance.

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Most popular questions from this chapter

An article in Solid State Technology (1984, Vol. 29, pp. \(281-284\) ) described the use of factorial experiments in photolithography, an important step in the process of manufacturing integrated circuits. The variables in this experiment (all at two levels) are prebake temperature \((A),\) prebake time \((B),\) and exposure energy \((C),\) and the response variable is delta line width, the difference between the line on the mask and the printed line on the device. The data are as follows: \((1)=-2.30, a=-9.87, b=-18.20\), \(a b=-30.20, c=-23.80, a c=-4.30, b c=-3.80,\) and \(a b c=-14.70\) (a) Estimate the factor effects. (b) Use a normal probability plot of the effect estimates to identity factors that may be important. (c) What model would you recommend for predicting the delta line width response based on the results of this experiment? (d) Analyze the residuals from this experiment, and comment on model adequacy.

Construct a \(2^{5}\) design in two blocks. Select the \(A B C D E\) interaction to be confounded with blocks.

Consider the following results from a two-factor experiment with two levels for factor \(A\) and three levels for factor \(B\). Each treatment has three replicates. $$ \begin{array}{llrc} \hline A & B & \text { Mean } & \text { StDev } \\ \hline 1 & 1 & 21.33333 & 6.027714 \\ 1 & 2 & 20 & 7.549834 \\ 1 & 3 & 32.66667 & 3.511885 \\ 2 & 1 & 31 & 6.244998 \\ 2 & 2 & 33 & 6.557439 \\ 2 & 3 & 23 & 10 \end{array} $$ (a) Calculate the sum of squares for each factor and the interaction. (b) Calculate the sum of squares total and error. (c) Complete an ANOVA table with \(F\) -statistics.

An article in the Journal of Marketing Research (1973, Vol. \(10(3),\) pp. \(270-276\) ) presented a \(2^{7-4}\) fractional factorial design to conduct marketing research: $$ \begin{array}{crrrrrrrr} \hline & & & & & & & & \text { Sales for a } \\ & & & & & & & & \text { 6-Week Period } \\ \text { Runs } & \boldsymbol{A} & \boldsymbol{B} & \boldsymbol{C} & \boldsymbol{D} & \boldsymbol{E} & \boldsymbol{F} & \boldsymbol{G} & \text { (in \$1000) } \\ \hline 1 & -1 & -1 & -1 & 1 & 1 & 1 & -1 & 8.7 \\ 2 & 1 & -1 & -1 & -1 & -1 & 1 & 1 & 15.7 \\ 3 & -1 & 1 & -1 & -1 & 1 & -1 & 1 & 9.7 \\ 4 & 1 & 1 & -1 & 1 & -1 & -1 & -1 & 11.3 \\ 5 & -1 & -1 & 1 & 1 & -1 & -1 & 1 & 14.7 \\ 6 & 1 & -1 & 1 & -1 & 1 & -1 & -1 & 22.3 \\ 7 & -1 & 1 & 1 & -1 & -1 & 1 & -1 & 16.1 \\ 8 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 22.1 \end{array} $$ The factors and levels are shown in the following table. $$ \begin{array}{llll} \hline & \ {\text { Factor }} & \ {-1} & \ {+1} \\ \hline A & \begin{array}{l} \text { Television } \\ \text { advertising } \end{array} & \text { No advertising } & \text { Advertising } \\ B & \begin{array}{l} \text { Billboard } \\ \text { advertising } \end{array} & \text { No advertising } & \text { Advertising } \\ C & \begin{array}{l} \text { Newspaper } \\ \text { advertising } \end{array} & \text { No advertising } & \text { Advertising } \\ D & \begin{array}{l} \text { Candy wrapper } \\ \text { design } \end{array} & \begin{array}{l} \text { Conservative } \\ \text { design } \end{array} & \begin{array}{l} \text { Flashy } \\ \text { design } \end{array} \\ E & \text { Display design } & \begin{array}{l} \text { Normal shelf } \\ \text { display } \end{array} & \begin{array}{l} \text { Special aisle } \\ \text { display } \end{array} \\ F & \begin{array}{l} \text { Free sample } \\ \text { program } \end{array} & \begin{array}{l} \text { No free } \\ \text { samples } \end{array} & \text { Free samples } \\ G & \text { Size of candy bar } & 1 \text { oz bar } & 21 / 20 \mathrm{bar} \end{array} $$ (a) Write down the alias relationships. (b) Estimate the main effects. (c) Prepare a normal probability plot for the effects and interpret the results.

Construct a \(2^{5-1}\) design. Suppose that it is necessary to run this design in two blocks of eight runs each. Show how this can be done by confounding a two-factor interaction (and its aliased three-factor interaction) with blocks.
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