Process Capability indices

The basic Process Capability indices commonly used are Cp, Cpk, Cpm

Cp = VOC/ VOP = Tolerance/Natural Process variation = USL - LSL/ 6 std. dev.

• Cp values greater than or equal to 1 indicates the process is technically capable.
• Cp value equal to 2 is said to represent 6 sigma performance.

However, Cp index is limited in its use since it does not address centering of a process relative to the specification limits. For that reason Cpk was developed.

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Cpk = minimum (Cp lower , Cp upper)

• Cp lower = mean - LSL / 3 std dev.
• Cp upper = USL - mean / 3 std dev.
• Cpk determines the proximity of process mean to the nearest Specification limit.
• Also, at-least one specification limit must be stated in order to compute Cpk value.
• Cpk values of 1.33 or 1.67 are commonly set as goals since they provide some room for the process to drift in relation to the nominal setting

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Cpm, also known as process capability index of mean.

It is an index that accounts for the location of the process average relative to target value and is defined as.

where, T = target value (typically center of Tolerance)

• When the process average and target values are equal, the Cpm = Cpk.
• When the process average drifts from the target value, the Cpm is less than Cpk.

Calculating DPMO in Minitab

DPMO Example 1

Calculate the DPMO for the Valve manufacturing process. Values are given in Table 1.1 below. 

Given,

Lower specification limit (LSL) = 5.95

Upper specification limit (USL) = 6.05

figure 1.4

Observe image 1.4, the area on the left side of 6.05 is 0.995599. Thus, the area on the right side of 6.05 is 0.004401 (i.e., 1 – 0.995599). 

Hence, the proportion of defectives

= Area on the left side of 5.95 + Area on the right side of 6.05

= 0.0078001 + 0.004401

= 0.012201

Hence, DPMO = 0.012201 * 1,000,000 = 12,201. or 3.4 Sigma

Valve Diameters - Table 1.1

5.99	6.04	6.01	6	6.03
5.99	6	6.03	5.97	5.98
5.99	6	6.02	6.01	6.02
6.04	5.99	5.99	5.97	6.01
6.02	6.01	5.99	5.98	6
6.01	5.99	6.01	6.05	6.03
6	5.98	5.97	6	6
5.99	5.99	5.98	5.99	6.03
6.01	5.99	6.03	6.01	5.99
5.99	5.98	6.01	5.99	6.01
6	6.01	6	6.03	6.03
6	5.99	6	6.02	6.01
5.98	6.01	6	5.99	5.99
6.01	6.02	5.98	5.98	5.99
5.98	5.96	5.95	6.01	6.03
6.01	6	5.97	6.02	6.05
5.97	5.95	5.98	6.01	5.99
6.02	5.99	6	6.01	6
6	5.99	6.03	6.03	5.98
5.97	6	5.97	5.99	5.97
6.01	6	5.99	6.02	6
6.01	6.01	5.99	5.97	6
5.99	5.96	5.99	6.05	6.01
6	5.98	6	5.97	6.01
5.99	5.98	5.98	5.98	6
6.01	6.03	6	6	6.01
5.95	6.01	6	5.97	6.02
6	5.97	5.99	5.99	6
5.98	5.98	6	6	6
5.95	6	6	6.01	6

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DPMO Example 2

Roller engineering. manufactures SS ball bearings 25 mm in diameter. Allowable tolerance as per customer is plus-minus 6mm from 25 mm. Find (DPMO).

Assume data is normally distributed and Process mean = 28 mm, Process standard deviation = 2 mm 

Solution

Process target = 25 mm (given)

Process tolerance = ± 6 mm (given)

Hence, LSL = 25 – 6 = 19 mm, and USL = 25 + 6 = 31 mm

Hence, the proportion of defectives

= Area on the left side of 19 + Area on the right side of 31

= 0.0000034 + 0.066807

= 0.0668104

Hence, DPMO = 0.0668104* 1,000,000 = 66810

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DPMO Example 3

Suppose that the Pizza King restaurant delivers pizza in 1,750 seconds on average, with a standard deviation of 90 seconds. The promised delivery time is 30 minutes or less. else you get free pizza.

 Calculate the DPMO for the pizza delivery process. Also, find the sigma level of the process? Assume delivery time is normally distributed.

Solution:

LSL = None

USL = Target = 30 min. = 1,800 seconds

Process mean = 1,750 seconds (given)

Process standard deviation = 90 seconds

Notice in image 1.9 that the area on the left side of 1,800 is 0.710743. Thus,  the area on the right side of 1,800 is 1 - 0.710743 = 0.289257

Hence, the proportion of defectives

= Area on the right side of 1,800

= 0.289257

Hence, DPMO = 0.289257 * 1,000,000 = 289,257.

DPMO value of 289,257 equates to the sigma level between 2 and 3.

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Hypothesis Testing for Quality Control at a Manufacturing Company: Case Study

Case Study 1:

DEFINE

The process engineer has the following hypothesis tests to perform:

Test 1 is to check whether the population proportion of defectives is less than the target of 5%. 

In other words,

Null Hypothesis: Population proportion of defectives = 0.05.

Alternative Hypothesis: Population proportion of defectives < 0.05.

Measure and Analyze Phases

A sample of 200 products is taken, and 3 products have been identified as defective. 

Figure 4.1 shows how to select “1 Proportion” in Minitab® to test the population proportion of defectives. 

Doing so will open the dialog box shown in Figure 4.2. Enter “3” for “Number of events” and “200” for “Number of trials”. Also, check the box for “Perform hypothesis test” and enter “0.05” for “Hypothesized proportion”. Click on “Options” and the dialog box shown in Figure 4.3 opens. Select “less than” from the drop-down menu for “Alternative” and click on “OK”. This takes you back to the dialog box shown in Figure 4.2. 

Click on “OK” and the output shown in Figure 4.4 is the result. Because P-value (0.012) is less than 0.05, reject the null hypothesis. In other words, the population proportion of defectives is lower than 5%, which is good news.