Phillips Screw Company produces screw fittings. It is determined that T = 1 inch in diameter. Specification limits allow a variation of plus or minus 0.5 inches. Products produced at values beyond the specification limits lose $15. A sample of five units produced showed the following values: $$\begin{array} { l c }
\text { Unit Number } & \text { Actual Diameter (Y) } \
\text { Unit } 1 & 1.2 \
\text { Unit 2 } & 1.3 \
\text { Unit 3 } & 0.8 \
\text { Unit } 4 & 0.9 \
\text { Unit } 5 & 1.0
\end{array}$$ What is k, the constant, in the Taguchi loss function?

Question

Phillips Screw Company produces screw fittings. It is determined that T = 1 inch in diameter. Specification limits allow a variation of plus or minus 0.5 inches. Products produced at values beyond the specification limits lose $15. A sample of five units produced showed the following values: $$\begin{array} { l c } 
\text { Unit Number } & \text { Actual Diameter (Y) } \
\text { Unit } 1 & 1.2 \
\text { Unit 2 } & 1.3 \
\text { Unit 3 } & 0.8 \
\text { Unit } 4 & 0.9 \
\text { Unit } 5 & 1.0
\end{array}$$ What is k, the constant, in the Taguchi loss function?

Quizplus · Accepted Answer

The Taguchi loss function is given by $L(y) = k(y - T)^2$, where $L(y)$ is the loss associated with producing a unit with characteristic $y$, $T$ is the target value for the characteristic, and $k$ is a constant. The loss is incurred when the actual value $y$ deviates from the target value $T$. Given that the loss for producing units beyond the specification limits is $15 and the specification limits allow a variation of plus or minus 0.5 inches from the target $T = 1$ inch, we can calculate $k$ using the formula for the loss at the specification limit. For a deviation of 0.5 inches (either 1.5 or 0.5 inches from the target), the loss is $15. Substituting into the formula: $$15 = k(0.5)^2$$ $$15 = k(0.25)$$ Solving for $k$ gives: $$k = 15 / 0.25 = 60$$

Phillips Screw Company Produces Screw Fittings What Is K, the Constant, in the Taguchi Loss Function