6
S U P P L E M E N T
Statistical Process Control
DISCUSSION Q UESTIONS 1. Shewhart’ Shewhart’ss two types types of variation variation:: common common and special causes are also called natural and assignable variation. 2. A process is said to be operating in statistical control when the only source of variation is natural or common causes. 3. The x-bar chart indicates whether changes have occurred in the central tendency of a process; the R-chart indicates whether a gain or a loss in uniformity has occurred. 4. A proces processs can be out out of contro controll becaus becausee of assign assignabl ablee variation variation,, which which can be traced traced to specific specific causes. Examples Examples include such factors as:
Tool wear A change in raw materials A chang changee in worki working ng enviro environm nment ent (tempe (temperat rature ure or humidity, for example) Tired or poorly trained labor
5.
The 5 steps are: 1. Coll Collec ectt 20 to 25 samp sample les, s, often often of n = 4 or 5 each; compute the mean and range of each sample. 2. Compu Compute te the overal overalll means ( x and and R ), set appropriate control limits, usually at the 99.73% level, and calculate the preliminary preliminary upper and lower lower control control limits. If the process is not currently stable, use the desired mean, µ , instead of x to calculate limits. 3. Graph Graph the sample means means and ranges ranges on their respective respective control charts and determine whether they fall outside the acceptable limits. 4. Investig Investigate ate points points or patterns patterns that indicate indicate the process process is out of control. control. Try to assign causes causes for the variation variation,, address the causes, and then resume the process. 5. Collect additional additional samples samples and, if necessary, necessary, revalidate revalidate the control limits using the new data.
6. Text Text list includes includes machine machine wear, wear, misadjus misadjusted ted equipme equipment, nt, fatigued or untrained workers, new batches of raw materials, etc. Others might be bad measuring device, workplace lighting, other ergonomic conditions etc. 7. Two sigma covers only 95.5% of all natural variation; even in the absence of assignable cause, points will fall outside the control limits 4.5% of the time. 8. The desired mean is used when the mean of a process being observed is unknown or out of control or when there is an established or known known µ , provided by the manufacturer or designer of the equipment or process.
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9. Yes; “out of control” means that the process has changed. If we are doing something “too well,” then the process has changed from the norm. We want to find out what we are doing “too well” so that we can do the same thing in the future and for other products. 10. Control charts are designed for specific sample sizes because the sample standard deviation or range is dependent on the sample size. The control charts presented here should not be used if the sample size varies. 11. C pk , the process capabilit capability y index, index, is one way to express process capability. It measures the proportion of natural variation (3σ ) betw betwee een n the the cent center er of the the proc proces esss and and the the near neares estt specification limit. C p is the process capability ratio and determines if the process meets design specifications.
12.
A “run of 5” implies that assignable variation is present.
13. The AQL is the quality level of a lot considered to be good. The LTPD is the quality level of a lot we consider bad. These are combi combine ned d with with risk risk level levelss to deter determin minee an accep acceptan tance ce sampling plan. 14. A run test is is used to help spot abnormalities in a control chart chart process. process. It is used if points points are not individua individually lly out of control, but form a pattern above or below the nominal (center) line. 15.
Managerial issues include:
Selecting places in a process that need SPC Deciding which type of control charts best fit Setting rules for workers to follow if certain points or patterns emerge
16. An OC curve is a graph showing the probability of accepting a lot given a certain quality (percentage of defective). 17. The purpose of acceptance sampling is to determine a course of action action (accept or reject) reject) regardin regarding g the dispositio disposition n of a lot without inspecting each item in a lot. Acceptance sampling does not estimate the quality of a lot. 18. The two risks when acceptance sampling is used are type I error: rejecting a good lot; type II error: accepting a bad lot. 19. A process that has a capability index of one or greater—a “capable” process—produces small percentages of unacceptable items. The capability formula is built around an assumption of exactly exactly one, those those parts parts that are more than three sigma from center are unacceptable; they are 0.00135 of all output. If the capability index is greater than one, that fraction falls.
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SUPPLEMENT 6
STATISTICAL
P ROCESS
Active Model Eercises ACTIVE MODEL S6.1: p-Chart—with data
C ONTROL
END!O"!SUPPLEMENT P RO#LEMS S6.1
1.
Has the the process been in control? Samples 3 and 19 were “too good,” and sample 16 was out of control. 2. Suppose we use a 95% p-chart. What are the upper and lower control limits? Has the process gotten more out of control? .074 and .0008. It is the same process but sample 13 is also out of control. 3. Suppose that the sample size used was actually 120 instead of the 100 that it was supposed to be. How does this affect the chart? The overall percentage of defects drops and, in addition, the UCL and LCL get closer to the center line and each other.
σ
=
µ
= 14 oz.
n
5. What happens to the chart as we reduce the percentage defects? The chart gets “tighter.” The UCL and LCL get closer to the center line and each other.
2. Increase the standard deviation. At what value will the curve cross the upper specification? About .9
4. What happens to the actual alpha and beta as the LTPD is increased? Alpha remains the same and beta decreases. 5. What happens to the actual alpha and beta as the sample size, n, is increased? Alpha increases and beta decreases. 6. What happens to the actual alpha and beta as the critical value, c, is increased? Alpha decreases and beta increases.
6
= 0.0167
= 5 X = 50
= 1.72 = 3 1.72 UCL = 50 + 3 = 52.31 5 1.72 LCL = 50 − 3 = 47.69 5
n
S6.3
The relevant constants are:
σ
A2 = 0.419 D4 = 1.924 D3 = 0.076
The control limits are: (a)
UCL x LCL x
(b)
UCL R LCL R
S6.4
1. What is the designed value for alpha? What is the actual value for alpha? designed = .05; actual = .0492
3. What happens to the appropriate sampling plan as the AQL is increased? Both the sample size a nd critical value increase.
0.1
S6.2
ACTIVE MODEL S6.3: Acceptance Sampling/OC Curve
2. What is the designed value for beta? What is the actual value for beta? designed = .10; actual = .0525
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=
= 14 + 0.05 = 14.05 oz. LCL = 14 − 3σ x = 14 − 3(0.0167) = 13.95 oz.
ACTIVE MODEL S6.2: Process Capability 1. How far can the mean shift to the right before the curve extends over the upper specification? How does this affect the C p and C pk ? Approximately 1.45; C p is not affected; C pk is about 1.0.
=
UCL = 14 + 3σ x = 14 + 3(0.0167)
4.
What happens to the chart as we reduce the z-value The chart gets “tighter.” The UCL and LCL get closer to the center line and each other.
0.1
x
σ
S6.5
= 57.75 + 0.419(1.78) = 58.496 = 57.75 − 0.419(1.78) = 57.004 = 1.924(1.78) = 3.4272 = 0.076(1.78) = 0.135
= 420. So σ LCL = x – Z = 420 – 4 n σ UCL = x + Z = 420 + 4 n
Target of
x
UCL x =
440
calories
LCL x =
400
calories
= 400. 25 25 = 440 . Thus, 25
25
From Table S6.1, A2 = 0.308, D4 = 1.777, D3 = 0.223 UCL x
= x + A2 × R = 705 + 0.308 × 6 = 706.848
UCL R
= D4 × R = 1.777 × 6 = 10.662
LCL x
= x − A2 × R = 705 − 0.308 × 6 = 703.152
LCL R
= D3 × R = 0.223 × 6 = 1.338
SUPPLEMENT 6
STATISTICAL
P ROCESS
C ONTROL
6$
S6.6 %o&r 1 2 3 4 5 6 7 8
X
R
%o&r
X
R
%o&r
X
R
3.25 3.10 3.22 3.39 3.07 2.86 3.05 2.65
0.71 1.18 1.43 1.26 1.17 0.32 0.53 1.13
9 10 11 12 13 14 15 16
3.02 2.85 2.83 2.97 3.11 2.83 3.12 2.84
0.71 1.33 1.17 0.40 0.85 1.31 1.06 0.50
17 18 19 20 21 22 23 24
2.86 2.74 3.41 2.89 2.65 3.28 2.94 2.64
1.43 1.29 1.61 1.09 1.08 0.46 1.58 0.97
Average X = 2.982, Average R = 1.02375, n = 4. From Table S6.1, A2 = 0.729, D4 = 2.282, D3 = 0.0.
UCL X
= X + A2 × R = 2.982 + 0.729 × 1.024 = 3.728
LCL X
= X − A2 × R = 2.982 − 0.729 × 1.024 = 2.236
UCL R
= D4 × R = 2.282 × 1.024 = 2.336
LCL R
= D3 × R = 0 × 1.024 = 0
The smallest sample mean is 2.64, the largest 3.39. Both are well within the control limits. Similarly, the largest sample range is 1.61, also well within the control limits. We can conclude that the process is presently within control. However, the first five values for the mean are above the expected mean; this may be the indication of a problem in the early stages of the process.
S6.7 X = R
=
156.9 + 153.2 + 153.6 + 155.5 + 156.6 155.16 mm = 5 4.2 + 4.6 + 4.1 + 5.0 + 4.5 4.48 mm = 5
X -chart: X = 155.16 mm from the sample data
UCL x = X
+ A2R = 155.16 + (0.308 × 4.48) = 156.54 mm
LCL x = X
− A2 R = 155.16 − (0.308 × 4.48) = 153.78 mm.
