Training Industrial Engineering Statistical Process Control
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Statistical Process Control

24 min Industrial Engineering

Statistical Process Control

SPC uses control charts to monitor a manufacturing process and detect when it goes out of control.

Control Chart Limits

$$UCL = \bar{x} + z\frac{\sigma}{\sqrt{n}}, \qquad LCL = \bar{x} - z\frac{\sigma}{\sqrt{n}}$$

where $\bar{x}$ = process mean, $\sigma$ = standard deviation, $n$ = sample size, $z$ = number of standard deviations (usually 3).

Process Capability Index

$$C_p = \frac{USL - LSL}{6\sigma}$$

where USL and LSL are the upper and lower specification limits. $C_p \ge 1.33$ is considered "capable."

$C_{pk}$ — Centering-Adjusted Capability

$$C_{pk} = \min\left(\frac{USL - \bar{x}}{3\sigma}, \frac{\bar{x} - LSL}{3\sigma}\right)$$

Example 1

A process has $\bar{x} = 50.0$ mm, $\sigma = 0.5$ mm, sample size $n = 4$. Find the 3-sigma control limits for the $\bar{x}$-chart.

$$UCL = 50.0 + 3 \times \frac{0.5}{\sqrt{4}} = 50.0 + 0.75 = 50.75 \text{ mm}$$

$$LCL = 50.0 - 0.75 = 49.25 \text{ mm}$$

Example 2

Specification limits are 49 to 51 mm, $\sigma = 0.3$ mm. Find $C_p$.

$$C_p = \frac{51 - 49}{6(0.3)} = \frac{2}{1.8} = 1.11$$

Since $C_p < 1.33$, the process is not fully capable.

Example 3

If $\bar{x} = 50.2$ mm with the specs from Example 2, find $C_{pk}$.

$C_{pu} = (51 - 50.2)/(3 \times 0.3) = 0.8/0.9 = 0.889$

$C_{pl} = (50.2 - 49)/(0.9) = 1.333$

$$C_{pk} = \min(0.889, 1.333) = 0.889$$

The process is off-center, making capability worse.

Practice Problems

1. $\bar{x} = 100$, $\sigma = 2$, $n = 9$. Find UCL and LCL (3-sigma).
2. $USL = 105$, $LSL = 95$, $\sigma = 1.5$. Find $C_p$.
3. If $\bar{x} = 102$ for the process in #2, find $C_{pk}$.
4. A p-chart monitors defect rate. If $\bar{p} = 0.05$, $n = 100$, find UCL.
5. 10 samples of size 5 have means: 50.1, 49.8, 50.3, 49.9, 50.0, 50.2, 50.5, 49.7, 50.1, 50.0. Find $\bar{\bar{x}}$.
6. What is the probability of a point falling outside 3-sigma limits when the process is in control?
7. If $\sigma$ is reduced from 1.5 to 1.0, how does $C_p$ change in #2?
8. An R-chart has $\bar{R} = 4.5$, $D_3 = 0$, $D_4 = 2.114$ (for $n=5$). Find the control limits.
9. Define "assignable cause" vs. "common cause" variation.
10. Six Sigma aims for $C_p = 2.0$. If $USL - LSL = 12$, find the required $\sigma$.
11. A process shifts from $\bar{x} = 50$ to $\bar{x} = 50.5$. $\sigma = 0.5$, $n = 4$. Will the shift be detected by the $\bar{x}$-chart?
12. OC curve: the probability of not detecting a 1$\sigma$ shift on a single sample. If UCL is at $+3\sigma/\sqrt{n}$, find the probability for $n=4$.
Show Answer Key

1. $UCL = 100 + 3(2/3) = 102$; $LCL = 98$

2. $C_p = 10/9 = 1.11$

3. $C_{pu} = 3/4.5 = 0.667$; $C_{pl} = 7/4.5 = 1.556$; $C_{pk} = 0.667$

4. $UCL = 0.05 + 3\sqrt{0.05(0.95)/100} = 0.05 + 3(0.0218) = 0.115$

5. $\bar{\bar{x}} = (50.1+49.8+\cdots+50.0)/10 = 500.6/10 = 50.06$

6. $0.27\%$ or about 2.7 per 1,000 samples

7. $C_p = 10/6 = 1.67$ — now capable

8. $UCL = 2.114 \times 4.5 = 9.51$; $LCL = 0$

9. Assignable: identifiable, fixable causes (tool wear, operator error). Common: inherent random variation.

10. $\sigma = 12/12 = 1.0$

11. New $\bar{x} = 50.5$; UCL = 50.75. The mean is inside, but close — detection probability ≈ 16% per sample. Multiple samples needed.

12. Shift $= 1\sigma_{\bar{x}}$. New center at $+1\sigma_{\bar{x}}$. $P(\text{within limits}) \approx P(-4 < Z < 2) = 0.977$. So $\beta \approx 0.977$, detection probability ≈ 2.3%.

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