$3.$ Is the manufacturing process Cpk capable?

As the Bright Star Stamping Company of Yantai, China, is pursuing an expansion drive, it stands at a juncture where one essential quality management$-$related problem would lurk down. The main aim of this company is to expand its markets from primarily a local market it currently has into an international platform, even with all the peculiarities that come along with the Chinese business environment. It does that through its synergy with a local company engaged in trading, which is significant due to high$-$quality control $($Seidelson$,2022).$ Recently, the challenge of this task was emphasized by a massive buy order for $8$ million parts with stringent quality management standards. For this order assurance, the company has to prepare a PPAP that will imply needful awareness of measurement system analysis regarding process stability capacity and quality sampling requirements, as well as the probability of making the wrong product regarding strength competence reactions $($Seidelson$,2022).$ The PO states a Cpk of $1\mathrm{.}67$ or higher looks for the process to be controlled and stable in the Quality Plan.

Calculations

To Calculate $\backslash$sigma :

The $70$ samples are divided into $10$ groups of $7$ samples each $($from the $10$ hours of data$)$

To calculate the R$-$bar, use data from Table $2($for Operator $1)$ and Table $3($for Operator $2).$

For group $1,$ the ranges are $0\mathrm{.}6,0\mathrm{.}5,0\mathrm{.}4,0\mathrm{.}3,0\mathrm{.}7$ and $0\mathrm{.}6$

The average range for group $1$ is $(0\mathrm{.}6+0\mathrm{.}5+0\mathrm{.}4+0\mathrm{.}3+0\mathrm{.}7+0\mathrm{.}6)/6=0\mathrm{.}5$

For $10$ groups, the average ranges are

$$ Group $1$: $0\mathrm{.}5$

$$ Group $2$: $0\mathrm{.}5$

$$ Group $3$: $0\mathrm{.}4$

$$ Group $4$: $0\mathrm{.}6$

$$ Group $5$: $0\mathrm{.}7$

$$ Group $6$: $0\mathrm{.}5$

$$ Group $7$: $0\mathrm{.}4$

$$ Group $8$: $0\mathrm{.}5$

$$ Group $9$: $0\mathrm{.}5$

$$ Group $10$: $0\mathrm{.}5$

The average of these $10$ averages is R$-$bar $=(0\mathrm{.}5+0\mathrm{.}5+0\mathrm{.}4+0\mathrm{.}6+0\mathrm{.}7+0\mathrm{.}5+0\mathrm{.}4+0\mathrm{.}5+0\mathrm{.}5+0\mathrm{.}5)/10=0\mathrm{.}51$ mm

The sample size, n$,$ is $7$

From Table $5,$ the d$2$ constant for n$=7$ is $2\mathrm{.}704$ Table $2.$ Operator # $1$ Process Study $($mm$)$

Therefore, $\backslash$sigma $=$ R$-$bar$/$d$2$ Sample $1$ Sample $2$ Sample $3$ Sample $4$ Sample $5$ Sample $6$ Sample $7$

$=0\mathrm{.}51$ mm $/2\mathrm{.}704$ Hour $13030\mathrm{.}230\mathrm{.}33030\mathrm{.}230\mathrm{.}630$

$=0\mathrm{.}189$ mm Hour $230\mathrm{.}530\mathrm{.}43030\mathrm{.}33030\mathrm{.}330$

$=0\mathrm{.}19$ mm Hour $330\mathrm{.}230\mathrm{.}230\mathrm{.}43030\mathrm{.}13030$

To calculate the x$-$bar: Hour $430\mathrm{.}23030\mathrm{.}4303030\mathrm{.}630$

The sum of all $70$ measurements is: Hour $53030\mathrm{.}730\mathrm{.}63030\mathrm{.}33030$

Table $2$: $1056\mathrm{.}5$ mm Average $30\mathrm{.}1830\mathrm{.}3030\mathrm{.}3430\mathrm{.}0630\mathrm{.}1230\mathrm{.}3030\mathrm{.}00$

Table $3$: $1055\mathrm{.}6$ mm

Total $=1056\mathrm{.}5+1055\mathrm{.}6$ Table $3.$ Operator # $2$ Process Study $($mm$)$

$=2112\mathrm{.}1$ mm Sample $1$ Sample $2$ Sample $3$ Sample $4$ Sample $5$ Sample $6$ Sample $7$

The number of samples, n$,$ is $70$ Hour $130\mathrm{.}0030\mathrm{.}1030\mathrm{.}3030\mathrm{.}0030\mathrm{.}2030\mathrm{.}6030\mathrm{.}10$

Therefore, x$-$bar $=$ total$/$n $=2112\mathrm{.}1/70$ Hour $230\mathrm{.}5030\mathrm{.}4030\mathrm{.}0030\mathrm{.}4030\mathrm{.}0030\mathrm{.}3030\mathrm{.}00$

$=30\mathrm{.}17$ mm Hour $330\mathrm{.}1030\mathrm{.}0030\mathrm{.}0030\mathrm{.}0030\mathrm{.}1030\mathrm{.}1030\mathrm{.}00$

The specification limits given are: Hour $430\mathrm{.}2030\mathrm{.}0030\mathrm{.}4030\mathrm{.}0030\mathrm{.}0030\mathrm{.}2030\mathrm{.}00$

LSL $=29\mathrm{.}5$ mm Hour $530\mathrm{.}0030\mathrm{.}6030\mathrm{.}6030\mathrm{.}1030\mathrm{.}2030\mathrm{.}0030\mathrm{.}10$

USL $=30\mathrm{.}5$ mm Mean $30\mathrm{.}160030\mathrm{.}220030\mathrm{.}260030\mathrm{.}100030\mathrm{.}100030\mathrm{.}240030\mathrm{.}0400$

For Cpk calculation

Cpk $=$ min $[($USL $-$ Mean$)/(3$Sigma$),($Mean $-$ LSL$)/(3$Sigma$)]$

USL $-$ Mean $=30\mathrm{.}5-30\mathrm{.}17$ mm

$=0\mathrm{.}33$ mm

Mean $-$ LSL $=30\mathrm{.}17-29\mathrm{.}5$

$=0\mathrm{.}67$ mm

$3*$Sigma $=3*0\mathrm{.}165$ Taking average SD $(0\mathrm{.}16+0\mathrm{.}17)/2$

$=0\mathrm{.}495$ mm

Therefore:

$($USL $-$ Mean$)/(3*$Sigma$)$

$=0\mathrm{.}33/0\mathrm{.}495$

$=0\mathrm{.}667$

$($Mean $-$ LSL$)/(3*$Sigma$)$

$=0\mathrm{.}67/0\mathrm{.}495$

$=1\mathrm{.}3535$

Taking the minimum, Cpk $=0\mathrm{.}667$

Explanation

The manufacturing process is not Cpk capable, as indicated by the fact that the process can produce parts with a thickness within the specified tolerance range. The manufacturing process for the stamped parts is marginally capable of meeting specifications, with a Cpk of $0\mathrm{.}57.$ This is below the target of $1\mathrm{.}67$ specified in the purchase order. The Process mean of $30\mathrm{.}14$ mm is well centered between the LSL of $29\mathrm{.}5$ mm and USL of $30\mathrm{.}5$ mm$.$ This indicates the process is well controlled and centered. However, the process variation, estimated from the R$-$bar calculation as $0\mathrm{.}19$ mm$,$ is too high relative to the tolerance of $1$ mm $($Seidelson$,2022).$ With $3$ sigma equal to $0\mathrm{.}57$ mm and the tolerance only $1$ mm$,$ over $99\%$ of the parts will meet specifications $($as $6$ sigma covers $99\mathrm{.}73\%$ of a norm