Quantitative and Qualitative Forecasting
Case: Quality Management—Toyota
Quality Control Analytics at Toyota
As part of the process for improving the quality of their cars, Toyota engineers have identifi ed a potential improvement does happen to get too large, it can cause the accelerator to bind and create a potential problem for the driver. (Note: This part of the case has been fabricated for teaching purposes, and none of these data were obtained from Toyota.)
Let’s assume that, as a first step to improving the process, a sample of 40 washers coming from the machine that produces the washers was taken and the thickness measured in millimeters. The following table has the measurements from the sample:
1.9 2.0 1.9 1.8 2.2 1.7 2.0 1.9 1.7 1.8
1.8 2.2 2.1 2.2 1.9 1.8 2.1 1.6 1.8 1.6
2.1 2.4 2.2 2.1 2.1 2.0 1.8 1.7 1.9 1.9
2.1 2.0 2.4 1.7 2.2 2.0 1.6 2.0 2.1 2.2
Questions
1 If the specification is such that no washer should be greater than 2.4 millimeters, assuming that the thick-nesses are distributed normally, what fraction of the output is expected to be greater than this thickness?
The average thickness in the sample is 1.9625 and the standard deviation is .209624. The probability that the thickness is greater than 2.4 is Z = (2.4 – 1.9625)/.209624 = 2.087068 1 - NORMSDIST(2.087068) = .018441 fraction defective, so 1.8441 percent of the washers are expected to have a thickness greater than 2.4.
2 If there are an upper and lower specification, where the upper thickness limit is 2.4 and the lower thick-ness limit is 1.4, what fraction of the output is expected to be out of tolerance?
The upper limit is given in a. The lower limit is 1.4 so Z = (1.4 – 1.9625)/.209624 = -2.68337. NORMSDIST(-2.68337) = .003644 fraction defective, so .3644 percent of the washers are expected to have a thickness lower than 1.4. The total expected fraction defective would be .018441 + .003644 = .022085 or about 2.2085 percent of the washers would be expected to be out of tolerance.
3 What is the Cpk for the process?
4 What would be the Cpk for the process if it were centered between the specification limits (assume the process standard deviation is the same)?
The center of the specification limits is 1.9, which is used for X-bar in the following:
5 What percentage of output would be expected to be out of tolerance if the process were centered?
Z = (2.4 – 1.9)/.209624 = 2.385221
Fraction defective would be 2 x (1-NORMSDIST(2.385221)) = 2 x .008534 = .017069, about 1.7 percent.
6 Set up X - and range control charts for the current process. Assume the operators will take samples of 10 washers at a time.
|
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Observation |
|
|
|||||||||
|
Sample |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
X-bar |
R |
|
1 |
1.9 |
2 |
1.9 |
1.8 |
2.2 |
1.7 |
2 |
1.9 |
1.7 |
1.8 |
1.89 |
0.5 |
|
2 |
1.8 |
2.2 |
2.1 |
2.2 |
1.9 |
1.8 |
2.1 |
1.6 |
1.8 |
1.6 |
1.91 |
0.6 |
|
3 |
2.1 |
2.4 |
2.2 |
2.1 |
2.1 |
2 |
1.8 |
1.7 |
1.9 |
1.9 |
2.02 |
0.7 |
|
4 |
2.1 |
2 |
2.4 |
1.7 |
2.2 |
2 |
1.6 |
2 |
2.1 |
2.2 |
2.03 |
0.8 |
|
|
|
|
|
|
|
|
|
|
|
Mean: |
1.9625 |
0.65 |
From Exhibit 10.13, with sample size of 10, A2 = .31, D3 = .22 and D4 = 1.78
The upper control limit for the X-bar chart = 1.9625 + .31 x .65 = 2.164
The lower control limit for the X-bar chart = 1.9625 - .31 x .65 = 1.761
The upper control limit for the Range chart = 1.78 x .65 = 1.157
The lower control limit for the Range chart = .22 x .65 = .143
7 Plot the data on your control charts. Does the cur-rent process appear to be in control?
With respect to the control limits, the process appears to be in control, though it should be noted that it is difficult to have confidence in that conclusion based on just four samples. Students should also point out that there appears to be a positive trend, both in process mean and variability. Again, it would be difficult to say so with much confidence based on just four samples.
8 If the process could be improved so that the standard deviation were only about .10 millimeter, what would be the best that could be expected with the processes relative to fraction defective?
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