4.4 Process Capability & Performance Analysis

4.4 Process Capability & Performance Analysis

Process capability analysis establishes the quantitative relationship between two independent boundaries: the Voice of the Customer (Engineering Specification Limits: Upper Specification Limit USL and Lower Specification Limit LSL) and the Voice of the Process (Statistical Control Limits: (\mu \pm 3\sigma)).

Prerequisites for Capability Analysis

Before calculating process capability indices ((C_p, C_{pk})), two mandatory statistical conditions must be satisfied:

1. Statistical Stability: The process must be in a state of statistical control (free from special causes). Computing capability indices on an unstable process yields invalid, misleading estimates.
2. Normality: Individual process observations must approximate a normal distribution.

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Process Capability ((C_p, C_{pk})) vs. Process Performance ((P_p, P_{pk}))

A fundamental ASQ CQE competency is distinguishing short-term potential capability from long-term process performance:

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Process Capability Formulas

1. Process Capability Index ((C_p))

Measures potential capability assuming the process mean is perfectly centered between specification limits: $C_p = \frac{\text{Specification Width}}{\text{Process Spread}} = \frac{USL - LSL}{6\hat{\sigma}_{within}}$

2. Off-Centering Capability Index ((C_{pk}))

Accounts for process off-centering by evaluating upper capability ((C_{pu})) and lower capability ((C_{pl})): $C_{pu} = \frac{USL - \mu}{3\hat{\sigma}_{within}}, \quad C_{pl} = \frac{\mu - LSL}{3\hat{\sigma}_{within}}$ $C_{pk} = \min(C_{pu}, \, C_{pl}) = C_p (1 - k)$ where (k = \frac{|\mu - M|}{(USL - LSL)/2}) represents the off-centering factor, and (M = \frac{USL + LSL}{2}) is the specification midpoint.

Key Index Relationships:

• When the process mean is perfectly centered ((\mu = M)), (C_{pk} = C_p).
• When the process mean drifts off-center, (C_{pk} < C_p).
• If (C_{pk} = 1.00), the 3-sigma process edge touches the specification boundary (2,700 ppm defect rate).
• If (C_{pk} < 0), the process mean lies outside specification boundaries.

3. Taguchi Capability Index ((C_{pm}))

Incorporates customer target value (T) (which may differ from midpoint (M)), penalizing process off-target deviation: $C_{pm} = \frac{USL - LSL}{6 \sqrt{\hat{\sigma}^2 + (\mu - T)^2}} = \frac{C_p}{\sqrt{1 + \left(\frac{\mu - T}{\hat{\sigma}}\right)^2}}$

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Six Sigma Quality Requirements and Benchmark Values

The 1.5-Sigma Shift Concept

In Six Sigma doctrine, long-term processes naturally experience a mean shift of approximately (1.5\sigma) over extended time horizons due to ambient temperature shifts, operator changes, and tool wear. Therefore:

• A short-term (6\sigma) capable process ((C_p = 2.00)) degrades to a long-term (C_{pk} = 1.50).
• At (C_{pk} = 1.50) ((4.5\sigma) long-term), the single-tailed normal distribution area yields exactly 3.4 Defects Per Million Opportunities (DPMO).

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Step-by-Step Worked Capability & Performance Example

A precision bushing machining line has engineering specifications of (25.000 \pm 0.150\text{ mm}) ((LSL = 24.850\text{ mm}, USL = 25.150\text{ mm}), Target (T = 25.000\text{ mm})).

A stable process study of (k = 30) subgroups ((n = 5), total (N = 150) parts) yields:

• Grand Mean (\bar{\bar{X}} = 25.030\text{ mm})
• Average Range (\bar{R} = 0.0930\text{ mm}) ((d_2 = 2.326) for (n = 5))
• Overall Sample Standard Deviation (s_{overall} = 0.0480\text{ mm})

Step 1: Calculate Within-Subgroup Standard Deviation ((\hat{\sigma}_{within}))

$\hat{\sigma}_{within} = \frac{\bar{R}}{d_2} = \frac{0.0930}{2.326} = 0.0400\text{ mm}$

Step 2: Calculate Potential Capability ((C_p))

$C_p = \frac{25.150 - 24.850}{6(0.0400)} = \frac{0.300}{0.240} = \mathbf{1.250}$

Step 3: Calculate Off-Centered Capability ((C_{pk}))

$C_{pu} = \frac{25.150 - 25.030}{3(0.0400)} = \frac{0.120}{0.120} = 1.000$ $C_{pl} = \frac{25.030 - 24.850}{3(0.0400)} = \frac{0.180}{0.120} = 1.500$ $C_{pk} = \min(1.000, \, 1.500) = \mathbf{1.000}$

Step 4: Calculate Taguchi Capability Index ((C_{pm}))

$C_{pm} = \frac{0.300}{6 \sqrt{(0.0400)^2 + (25.030 - 25.000)^2}} = \frac{0.300}{6 \sqrt{0.0016 + 0.0009}} = \frac{0.300}{6 \sqrt{0.0025}} = \frac{0.300}{6(0.050)} = \mathbf{1.000}$

Step 5: Calculate Long-Term Process Performance ((P_p, P_{pk}))

$P_p = \frac{25.150 - 24.850}{6(0.0480)} = \frac{0.300}{0.288} = \mathbf{1.042}$ $P_{pu} = \frac{25.150 - 25.030}{3(0.0480)} = \frac{0.120}{0.144} = 0.833$ $P_{pl} = \frac{25.030 - 24.850}{3(0.0480)} = \frac{0.180}{0.144} = 1.250$ $P_{pk} = \min(0.833, \, 1.250) = \mathbf{0.833}$

Interpretation of Capability Results

The process fails to meet standard manufacturing capability ((C_{pk} = 1.000 < 1.33)). The gap between (C_p = 1.250) and (C_{pk} = 1.000) proves process off-centering (mean shifted to (25.030\text{ mm})). Furthermore, (P_{pk} = 0.833 < C_{pk} = 1.000) confirms significant long-term variation drift between subgroups across production.