A company produces screw fittings. It is determined that T = 1 inch in diameter. Specification limits allow a variation of plus or minus 0.5 cm. Products produced at values beyond the specification limits lose $10. A sample of five units produced showed the following values:
$$\begin{array} { c c }
\text { Unit Number } & \text { Actual Diameter (Y) } \
\text { Unit 1 } & 1.2 \
\text { Unit 2 } & 1.3 \
\text { Unit 3 } & 0.8 \
\text { Unit 4 } & 0.9 \
\text { Unit 5 } & 1.0
\end{array}$$
-Refer to the figure.What is k,the constant,in the Taguchi loss function?

Question

A company produces screw fittings. It is determined that T = 1 inch in diameter. Specification limits allow a variation of plus or minus 0.5 cm. Products produced at values beyond the specification limits lose $10. A sample of five units produced showed the following values:
$$\begin{array} { c c } 
\text { Unit Number } & \text { Actual Diameter (Y) } \
\text { Unit 1 } & 1.2 \
\text { Unit 2 } & 1.3 \
\text { Unit 3 } & 0.8 \
\text { Unit 4 } & 0.9 \
\text { Unit 5 } & 1.0
\end{array}$$
-Refer to the figure.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, and $k$ is a constant. The specification limits are $T \pm 0.5$ cm, or 1 inch $\pm 0.5$ cm. Since 1 inch = 2.54 cm, the target value $T$ in cm is 2.54 cm. The loss for producing a unit outside the specification limits is $10. The specification limit in cm is $2.54 \pm 0.5$ cm. Using the formula for the loss at the specification limit, $L = k(specification\ limit - T)^2$, and solving for $k$ with $L = 10$ and the deviation at the specification limit being 0.5 cm, we find $k = 40$.

A Company Produces Screw Fittings -Refer to the Figure