What this lesson covers
Statistical process control is one topic of Quality Assurance and Improvement, which carries 23 of the 100 scored questions on the ASQ CQI exam. Capability questions are some of the most reliable marks on the exam, because they are pure arithmetic once you know which number goes where. This free lesson works one example from raw chart values to a decision, then shows the traps that cost people the mark.
You need four things for every capability question: the upper specification limit (USL), the lower specification limit (LSL), the process mean and the process standard deviation (sigma). The question either gives sigma or gives you what you need to estimate it.
The idea in one line
Cp asks whether the spread fits; Cpk asks whether it fits where the process actually sits.
Picture the specification as a garage and the process spread (6 sigma wide) as a car. Cp compares the width of the garage with the width of the car. Cpk asks how close the car is to the nearer wall. A narrow car parked against one wall has a good Cp and a poor Cpk.
The formulas
| Index | Formula | What it tells you |
|---|---|---|
| Cp | (USL - LSL) / (6 sigma) | Potential, if the process were centered |
| Cpu | (USL - mean) / (3 sigma) | Room on the upper side |
| Cpl | (mean - LSL) / (3 sigma) | Room on the lower side |
| Cpk | the smaller of Cpu and Cpl | Actual capability, counting the mean's position |
Two facts to hold onto:
- Cpk is never larger than Cp. They are equal only when the mean sits exactly in the middle of the specification.
- Each Cpk side counts in units of 3 sigma. A Cpk of 1.00 means the nearer limit is exactly 3 sigma from the mean; 1.33 means 4 sigma; 2.00 means 6 sigma.
The worked example
The setup (invented). A shaft diameter is specified as 25.00 plus or minus 0.05 mm, so LSL = 24.95 mm and USL = 25.05 mm. The process is in statistical control on an X-bar and R chart with subgroups of 5. The chart gives X-double-bar = 25.010 mm and R-bar = 0.0349 mm. For n = 5, d2 = 2.326.
The working (5 steps)
Step 1: confirm the process is stable. Capability means something only for a process in control. The chart shows no points outside the limits and no runs or trends, so go ahead. If it were out of control, the right answer would be to find and remove the special causes first.
Step 2: estimate sigma. Within-subgroup sigma = R-bar / d2 = 0.0349 / 2.326 = 0.0150 mm. Check: 2.326 x 0.0150 = 0.0349.
Step 3: Cp. Specification width = 25.05 - 24.95 = 0.10 mm. Six sigma = 6 x 0.0150 = 0.090 mm. Cp = 0.10 / 0.090 = 1.11.
Step 4: both sides of Cpk. Three sigma = 0.045 mm.
- Cpu = (25.05 - 25.010) / 0.045 = 0.040 / 0.045 = 0.89
- Cpl = (25.010 - 24.95) / 0.045 = 0.060 / 0.045 = 1.33
Step 5: take the smaller. Cpk = 0.89, set by the upper side.
Second route to check it. The mean is 0.010 mm above the 25.00 mm midpoint. As a share of the half-width (0.05 mm) that is k = 0.010 / 0.05 = 0.20. Then Cpk = Cp x (1 - k) = 1.11 x 0.80 = 0.89. Both routes agree.
Reading the result
How much is out? The upper limit sits z = 0.040 / 0.0150 = 2.67 sigma above the mean. The area beyond z = 2.67 is about 0.0038 (from a normal table; an exam item gives it to you). So about 0.38% of shafts, or roughly 3,800 parts per million, are oversize. The lower limit is 4 sigma away (0.060 / 0.015), and the tail beyond z = 4 is tiny.
What should change? Two separate problems show up:
- The process is off center. Moving the mean down to 25.00 mm makes both sides equal, and Cpk rises to Cp: 1.11. That is often a quick adjustment.
- The spread is too wide for a 1.33 requirement. Even centered, Cp is only 1.11. Many customers ask for at least 1.33, but the number required is always the customer's. To reach Cp = 1.33 with a 0.10 mm specification, sigma must be no more than 0.10 / (6 x 1.33) = 0.0125 mm. Check: 0.10 / (6 x 0.0125) = 1.33. That is a process change (tooling, fixturing, material), not a knob on the machine.
So the answer to "what first?" is: center it (cheap, and it raises Cpk from 0.89 to 1.11), then work on variation to get from 1.11 to 1.33.
Pp and Ppk: the same formulas, a different sigma
Pp and Ppk use the overall standard deviation of all the individual readings (the ordinary sample s), not R-bar / d2. Overall sigma includes shifts between subgroups, so it is usually larger, and Pp and Ppk are usually lower.
Suppose the overall s for the shafts is 0.018 mm.
- Pp = 0.10 / (6 x 0.018) = 0.10 / 0.108 = 0.93
- Ppk = (25.05 - 25.010) / (3 x 0.018) = 0.040 / 0.054 = 0.74
- Check: Ppk / Pp = 0.74 / 0.93 = 0.80, the same (1 - k) as before, because centering does not depend on which sigma you use.
A big gap between Cpk and Ppk says the process moves between subgroups more than it varies within them.
One-sided specifications
Many features have only one limit: a maximum runout, a minimum strength. Cp needs both limits, so it cannot be calculated. Cpk is just the one side that exists.
Example (invented): runout has only a maximum of 0.50 mm, the mean is 0.20 mm and sigma is 0.05 mm. Cpk = (0.50 - 0.20) / (3 x 0.05) = 0.30 / 0.15 = 2.00. Check: 0.20 + 6 x 0.05 = 0.50, so the limit is 6 sigma above the mean.
The traps that cost the mark
- Taking the larger side. Cpk is always the smaller of Cpu and Cpl.
- Dividing by 6 sigma for Cpk. Each side is a distance over 3 sigma; only Cp uses 6 sigma.
- Using R-bar as sigma. Divide by d2 first.
- Using control limits. Capability uses the specification limits. Control limits are about stability, not about the customer.
- Skipping the stability check. An out-of-control process has no single capability to report.
- Mixing within and overall sigma. Cp and Cpk use within-subgroup sigma; Pp and Ppk use the overall one.
Try it
The practice set with this lesson has 10 questions built on these steps. Work each one by the two-sided formula, then check it by a second route: reverse the formula, or use Cpk = Cp x (1 - k).
Not affiliated with or endorsed by ASQ. ASQ is named here only to describe the exam these materials prepare you for.