Question

Assignment i Help Save & Ext Sub The desired distance for Flextronchip customers is 0.6 mm between two adjacent pins. Any deviation from this value causes interference. The process of handling complaints costs the firm at approximately $40.00 per chip. The engineers of the firm expect the average customer will be likely to complain when the distance is off target by at least 0.0002 mm. At the factory, the adjustment can be made at a cost of $1.60, which includes the labor to make the adjustment and additional testing. Required: 1. What is the value of k, the cost coefficient, in the Taguchi quality loss function (OLF) based on the above data? 2. Based on a Taguchi quality loss function (QLF), what should the tolerance be before an adjustment is made at the factory? 3. Based on your answer to (requirement 2) above, what are the specification limits fupper and lower) for the production of the product in question? Complete this question by entering your answers in the tabs below Required 1 Required 2 Required 3 What is the value of k, the cost coefficient, in the Taguchi quality loss function (QUF) based on the above data? (Do not round intermediate calculation.) Value of k Required 2 > SAMSUNG

Answer 1

1) Calculation of the value of K We know what, Taguchi Quality Loss function, Total Quality cost K (Tolerance) ^2 Here, Total Quality cost $40 K Cost of coefficient And Tolerance = 0.0002mm Total Quality cost-K(Tolerance) $40-K (0.0002) $40 K= (0.0002) $40 K- 0.00000004 K-S1,000,000,000 Cost coefficient (K) $1,000,000,000

2) Calculation of tolerance before an adjustment is made at the factory By using loss function: $1.60 $1,000,000,000(X-0.6) ^2 $1.60-S1,000,000,000(X-0.6) $1.60 -(X-0.6) $1,000,000,000 $1.60 X-0.6= Vs1,000,000,000 X-0.6 10.0000000016 X-0.6 0.0000000016 X-0.0000000016 Tolerance 0.000000002 (Round off)

3) Specification limits are 0.6+ 0.00000000 16 Upper limit 0.6+0.0000000016 Upper limit 0.6000000016 And lower limit Lower limit -0.6-0.000026 Lower limit -0.599999998