Question 6 of 7: Process capability against specification

Question 6: Process capability against specification (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Shewhart constants used: $d_2 = 2.326$ for $n = 5$ and $d_2 = 2.059$ for $n = 4$.

Find. For (a), whether the wrist-pin process can hold the specification and what must be assumed to say so; for (b), which of the two finishing processes is the more capable of holding the head-thickness specification.

Approach. Estimate the process standard deviation from the average range using $\hat{\sigma} = \bar{R}/d_2$, compare the natural tolerance $6\hat{\sigma}$ of the individual pieces with the specification width to obtain $C_p$, then bring in the process centring through $C_{pk}$ and convert both to a fraction non-conforming using the normal table.

1. Part (a) — estimate the process standard deviation. The range chart gives the within-subgroup spread, and for a normal population the expected range of a subgroup of $n$ is $d_2\sigma$, so $$\hat{\sigma}=\frac{\bar{R}}{d_2}=\frac{0.002}{2.326}=\boxed{0.00086\ \text{in}}$$ This is the standard deviation of individual wrist pins, which is what a specification applies to — not the standard deviation of the subgroup averages, which is smaller by $\sqrt{5}$.
2. Part (a) — compare the natural tolerance with the specification. The natural tolerance of the process is the six-sigma band that contains 99.73 per cent of individual pieces: $$6\hat{\sigma}=6(0.00086)=0.00516\ \text{in} \qquad\text{against a specification width of } 0.004\ \text{in}$$ so the capability index is $$C_p=\frac{\text{USL}-\text{LSL}}{6\hat{\sigma}}=\frac{0.004}{0.00516}=\boxed{0.78}$$ Since $C_p < 1$ the specification cannot be met: the process is inherently wider than the tolerance, and no amount of re-centring will fix it.
3. Part (a) — account for the off-centre mean. The grand average is 1.001 in, half a tolerance above the 1.000 in nominal, so the one-sided index is $$C_{pk}=\frac{\min(\text{USL}-\bar{\bar{X}},\ \bar{\bar{X}}-\text{LSL})}{3\hat{\sigma}} =\frac{1.002-1.001}{3(0.00086)}=\frac{0.001}{0.00258}=0.39$$ exactly half of $C_p$, which is the signature of a mean displaced by half the tolerance. Converting to a fraction non-conforming with $z_U=(1.002-1.001)/0.00086=1.163$ and $z_L=(1.001-0.998)/0.00086=3.489$, $$p = P(z>1.163)+P(z>3.489) = 0.1224+0.0002 = 0.1227$$ so about 12.3 per cent of pins are out of tolerance, nearly all of them oversize. Re-centring the process on 1.000 in would cut this to $2P(z>2.326) = 2.0$ per cent — a large improvement, and still an unacceptable reject rate.
4. Part (a) — state the necessary assumption and the remedy. Two assumptions are needed and the question asks for them explicitly. First and foremost, the process must be in a state of statistical control: $\bar{R}$ estimates $\sigma$ only if the within-subgroup variation is the whole of the common-cause variation, and if special causes are present between subgroups the true spread of individuals is larger than $6\hat{\sigma}$ and the capability calculation understates the problem. Second, the individual measurements must be approximately normally distributed, which is what licenses both the $6\sigma$ natural tolerance and every percentage quoted above. The practical conclusion is that centring the process is worth doing but is not sufficient; the spread itself must be reduced by at least the factor $1/0.78 = 1.29$ — through a better machine, tooling or work-holding — or the tolerance must be reviewed, or the pins must be screened 100 per cent and graded.
5. Part (b) — estimate both process standard deviations. Both studies use subgroups of four, so $d_2 = 2.059$ throughout: $$\hat{\sigma}_{\text{broach}}=\frac{0.0005}{2.059}=0.000243\ \text{in}, \qquad \hat{\sigma}_{\text{mill}}=\frac{0.001}{2.059}=0.000486\ \text{in}$$ The broach is twice as precise as the milling machine, which is what one expects of a single-pass broaching operation against a milled surface.
6. Part (b) — compare the natural tolerances. The specification width is $2(0.001) = 0.002$ in, so $$C_{p,\text{broach}}=\frac{0.002}{6(0.000243)}=\frac{0.002}{0.00146}=\boxed{1.37}, \qquad C_{p,\text{mill}}=\frac{0.002}{6(0.000486)}=\frac{0.002}{0.00291}=\boxed{0.69}$$ On inherent capability the broach comfortably clears the usual minimum of 1.33 while the milling machine is not capable at all: it would have to reduce its spread by a factor of 1.46 merely to reach $C_p = 1$.
7. Part (b) — bring in the centring. Capability and performance are not the same thing, and here they point in opposite directions. The broach is running at 4.877 in, which is 0.001 in above the upper limit of 4.876 in, so $$C_{pk,\text{broach}}=\frac{4.876-4.877}{3(0.000243)}=-1.37$$ a negative index, meaning the mean itself is outside specification and essentially the whole output — $P(z>-4.118) = 99.998$ per cent — is oversize. The milling machine is dead on nominal, so $C_{pk,\text{mill}} = C_{p,\text{mill}} = 0.69$, and with $z=0.001/0.000486=2.059$ its fraction non-conforming is $$p_{\text{mill}} = 2P(z>2.059) = 2(0.01975)=0.0395\ \Rightarrow\ 3.95\ \text{per cent}$$ split evenly between oversize and undersize.
8. Part (b) — recommend the broach, and say why. As the two processes stand the milling machine is producing the better parts, but that comparison is the wrong one: centring is a set-up adjustment and spread is a property of the process. Offsetting the broach by 0.002 in brings its mean to nominal, after which $$z = \frac{0.001}{0.000243}=4.118 \quad\Longrightarrow\quad p_{\text{broach}} = 2P(z>4.118)=3.8\times10^{-5}\ \Rightarrow\ \boxed{38\ \text{parts per million}}$$ against 3.95 per cent for the milling machine, which cannot be improved by adjustment at all. The broach is therefore the capable process and should be selected, with an immediate 0.002 in tool offset and an $\bar{X}$–R chart to hold the setting. The same assumptions as in part (a) apply: statistical control and approximate normality.