Quality And Performance.pptx

5 – 1
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Quality And Performance
5
For Operations Management, 9e by
Krajewski/Ritzman/Malhotra
© 2010 Pearson Education
Homework: 5, 13, 14,
15

5 – 2
Costs of Quality
 A failure to satisfy a customer is
considered a defect
 Prevention costs
 Appraisal costs
 Internal failure costs
 External failure costs
 Ethics and quality

5 – 3
Total Quality Management
Figure 5.1 – TQM Wheel
Customer
satisfaction

5 – 4
 Customer satisfaction
 Conformance to specifications
 Value
 Fitness for use
 Support
 Psychological impressions
 Employee involvement
 Cultural change
 Teams

5 – 5
 Continuous improvement
 Kaizen
 A philosophy
 Not unique to quality
 Problem solving process

5 – 6
The Deming Wheel
Plan
Do
Study
Act
Figure 5.2 – Plan-Do-Study-Act Cycle

5 – 7
Six Sigma
X X
X
X
X
X
X
X X
X
XX
X
X X
X
X
Process average OK;
too much variation
Process variability OK;
process off target
Process
on target with
low variability
Reduce
spread
Center
process
X
X
X
X
X
X
X
X
X
Figure 5.3 – Six-Sigma Approach Focuses on Reducing Spread and Centering the Process

5 – 8
SPC, Causes of Variation
 Common causes
 Random, unavoidable sources of variation
 Assignable causes (out of statistical control)
 Can be identified and eliminated
 Change in the mean, spread, or shape

5 – 9
Control Charts
 Time-ordered diagram of process
performance

5 – 10
Control Chart Factors
TABLE 5.1 | FACTORS FOR CALCULATING THREE-SIGMA LIMITS FOR
| THE x-CHART AND R-CHART
Size of
Sample (n)
Factor for UCL and
LCL for x-Chart (A2)
Factor for LCL for
R-Chart (D3)
Factor for UCL for
R-Chart (D4)
2 1.880 0 3.267
3 1.023 0 2.575
4 0.729 0 2.282
5 0.577 0 2.115
6 0.483 0 2.004
7 0.419 0.076 1.924
8 0.373 0.136 1.864
9 0.337 0.184 1.816
10 0.308 0.223 1.777

5 – 11
Using x- and R-Charts
EXAMPLE 5.1
The management of West Allis Industries is concerned about
the production of a special metal screw used by several of the
company’s largest customers. The diameter of the screw is
critical to the customers. Data from five samples appear in the
accompanying table. The sample size is 4. Is the process in
statistical control?

5 – 12
Data for the x- and R-Charts: Observation of Screw Diameter (in.)
Observation
Sample
Number
1 2 3 4 R x
1 0.5014 0.5022 0.5009 0.5027 0.0018 0.5018
2 0.5021 0.5041 0.5024 0.5020 0.0021 0.5027
3 0.5018 0.5026 0.5035 0.5023 0.0017 0.5026
4 0.5008 0.5034 0.5024 0.5015 0.0026 0.5020
5 0.5041 0.5056 0.5034 0.5047 0.0022 0.5045
Average 0.0021 0.5027

5 – 13
Figure 5.10 – Range Chart from the OM Explorer x and R-Chart Solver for the
Metal Screw, Showing That the Process Variability Is in Control

5 – 14
Figure 5.11 – The x-Chart from the OM Explorer x and R-Chart Solver for the
Metal Screw, Showing That Sample 5 is out of Control

5 – 15
Application 5.1
Webster Chemical Company produces mastics and caulking for
the construction industry. The product is blended in large
mixers and then pumped into tubes and capped.
Webster is concerned whether the filling process for tubes of
caulking is in statistical control. The process should be
centered on 8 ounces per tube. Several samples of eight tubes
are taken and each tube is weighed in ounces.
Tube Number
Sample 1 2 3 4 5 6 7 8 Avg Range
1 7.98 8.34 8.02 7.94 8.44 7.68 7.81 8.11 8.040 0.76
2 8.23 8.12 7.98 8.41 8.31 8.18 7.99 8.06 8.160 0.43
3 7.89 7.77 7.91 8.04 8.00 7.89 7.93 8.09 7.940 0.32
4 8.24 8.18 7.83 8.05 7.90 8.16 7.97 8.07 8.050 0.41
5 7.87 8.13 7.92 7.99 8.10 7.81 8.14 7.88 7.980 0.33
6 8.13 8.14 8.11 8.13 8.14 8.12 8.13 8.14 8.130 0.03
Avgs 8.050 0.38

5 – 16
Application 5.1
Assuming that taking only 6 samples is sufficient, is the process
in statistical control?
UCLR = D4R =
LCLR = D3R =
1.864(0.38) = 0.708
0.136(0.38) = 0.052
The range chart is out of control since sample 1 falls outside the
UCL and sample 6 falls outside the LCL. This makes the x
calculation moot.
Conclusion on process variability given R = 0.38 and n = 8:
Too much variability in the process!

5 – 17
Process Capability
 Process capability refers to the ability of
the process to meet the design
specification for the product or service
 Design specifications are often expressed
as a nominal value and a tolerance
 Three-sigma quality, Ratio>1
 Four-sigma quality, Ratio>1.33
 Six-sigma quality, Ratio>2

5 – 18
Process Capability Index
 The process capability index measures
how well a process is centered and
whether the variability is acceptable
Cpk = Minimum of ,
x – Lower specification
3σ
Upper specification – x
3σ
where
σ = standard deviation of the process distribution
If Cpk is less than the critical value, either the process average is too close to one of
the tolerance limits and is generating defective output, or the process variability is too
large.

5 – 19
Process Capability Ratio
 The process capability ratio tests whether
process variability is the cause of
problems
Cp =
Upper specification – Lower specification
6σ

5 – 20
In Stat Control, but meeting Design
Specs?
If a process is in statistical control, we may also want to determine if the
process is meeting a design specification or standard.
To do this first determine Cpk, if this metric is okay (1, 1.33, 2), the process is
capable of meeting design specs to a high level of quality (three, four, or six-
sigma).
Otherwise, if Cpk is small (less than 1, 1.33, or 2), determine Cp.
If the Cp ratio is not okay, we conclude the process has too much variability.
If the Cp is okay (and Cpk is not), we conclude the process mean is not
centered appropriately, i.e. to design spec.

5 – 21
Assessing Process Capability
EXAMPLE 5.5
 The intensive care unit lab process has an average
turnaround time of 26.2 minutes and a standard deviation of
1.35 minutes
 The nominal value for this service is 25 minutes with an
upper specification limit of 30 minutes and a lower
specification limit of 20 minutes
 The administrator of the lab wants to have four-sigma
performance for her lab
 Is the lab process capable of this level of performance?

5 – 22
SOLUTION
The administrator began by taking a quick check to see if the
process is capable by applying the process capability index:
Lower specification calculation = = 1.53
26.2 – 20.0
3(1.35)
Upper specification calculation = = 0.94
30.0 – 26.2
3(1.35)
Cpk = Minimum of [1.53, 0.94] = 0.94
Since the target value for four-sigma performance is 1.33, the
process capability index told her that the process was not
capable. However, she did not know whether the problem was
the variability of the process, the centering of the process, or
both. The options available to improve the process depended
on what is wrong.

5 – 23
She next checked the process variability with the process
capability ratio:
The process variability did not meet the four-sigma target of
1.33. Consequently, she initiated a study to see where
variability was introduced into the process. Two activities,
report preparation and specimen slide preparation, were
identified as having inconsistent procedures. These procedures
were modified to provide consistent performance. New data
were collected and the average turnaround was now 26.1
minutes with a standard deviation of 1.20 minutes.
Cp = = 1.23
30.0 – 20.0
6(1.35)

5 – 24
She now had the process variability at the four-sigma level of
performance, as indicated by the process capability ratio:
However, the process capability index indicated additional
problems to resolve:
Cp = = 1.39
30.0 – 20.0
6(1.20)
Cpk = , = 1.08
(26.1 – 20.0)
3(1.20)
(30.0 – 26.1)
3(1.20)
Minimum of

5 – 25
Application 5.4
Webster Chemical’s nominal weight for filling tubes of caulk
is 8.00 ounces ± 0.60 ounces. The target process capability
ratio is 1.33, signifying that management wants 4-sigma
performance. The current distribution of the filling process is
centered on 8.054 ounces with a standard deviation of
0.192 ounces. Compute the process capability index and
process capability ratio to assess whether the filling process
is capable and set properly.

5 – 26
Application 5.4
Recall that a capability index value of 1.0 implies that the firm
is producing three-sigma quality (0.26% defects) and that the
process is consistently producing outputs within
specifications even though some defects are generated. The
value of 0.948 is far below the target of 1.33. Therefore, we
can conclude that the process is not capable. Furthermore,
we do not know if the problem is centering or variability.
Cpk = Minimum of ,
3σ
3σ
= Minimum of = 1.135, = 0.948
8.054 – 7.400
3(0.192)
8.600 – 8.054
3(0.192)
a. Process capability index:

5 – 27
Application 5.4
b. Process capability ratio:
Cp =
6σ
= = 1.0417
8.60 – 7.40
6(0.192)
Recall that if the Cpk is greater than the critical value (1.33 for
four-sigma quality) we can conclude that the process is
capable. Since the Cpk is less than the critical value, either the
process average is close to one of the tolerance limits and is
generating defective output, or the process variability is too
large. The value of Cp is less than the target for four-sigma
quality. Therefore we conclude that the process variability must
be addressed first, and then the process should be retested.

5 – 28
Solved Problem 1
The Watson Electric Company produces incandescent
lightbulbs. The following data on the number of lumens for 40-
watt lightbulbs were collected when the process was in control.
Observation
Sample 1 2 3 4
1 604 612 588 600
2 597 601 607 603
3 581 570 585 592
4 620 605 595 588
5 590 614 608 604
a. Calculate control limits for an R-chart and an x-chart.
b. Since these data were collected, some new employees were
hired. A new sample obtained the following readings: 570,
603, 623, and 583. Is the process still in control?

5 – 29
Solved Problem 1
Sample R
1 601 24
2 602 10
3 582 22
4 602 32
5 604 24
Total 2,991 112
Average x = 598.2 R = 22.4
x
604 + 612 + 588 + 600
4
= 601
x =
R = 612 – 588 = 24
SOLUTION
a. To calculate x, compute the mean for each sample. To
calculate R, subtract the lowest value in the sample from the
highest value in the sample. For example, for sample 1,

5 – 30
Solved Problem 1
The R-chart control limits are
b. First check to see whether the variability is still in control
based on the new data. The range is 53 (or 623 – 570), which
is outside the UCL for the R-chart. Since the process
variability is out of control, it is meaningless to test for the
process average using the current estimate for R. A search
for assignable causes inducing excessive variability must be
conducted.
2.282(22.4) = 51.12
0(22.4) = 0
UCLR = D4R =
LCLR = D3R =
The x-chart control limits are
UCLx = x + A2R =
LCLx = x – A2R =
598.2 + 0.729(22.4) = 614.53
598.2 – 0.729(22.4) = 581.87

5 – 31
Solved Problem 4
Pioneer Chicken advertises “lite” chicken with 30 percent fewer
calories. (The pieces are 33 percent smaller.) The process
average distribution for “lite” chicken breasts is 420 calories,
with a standard deviation of the population of 25 calories.
Pioneer randomly takes samples of six chicken breasts to
measure calorie content.
a. Design an x-chart using the process standard deviation.
b. The product design calls for the average chicken breast to
contain 400 ± 100 calories. Calculate the process
capability index (target = 1.33) and the process capability
ratio. Interpret the results.

5 – 32
Solved Problem 4
SOLUTION
a. For the process standard deviation of 25 calories, the
standard deviation of the sample mean is




n
x calories
10.2
6
25

UCLx = x + zσx =
LCLx = x – zσx =
420 + 3(10.2) = 450.6 calories
420 – 3(10.2) = 389.4 calories

5 – 33
Solved Problem 4
Because the process capability ratio is 1.33, the process should
be able to produce the product reliably within specifications.
However, the process capability index is 1.07, so the current
process is not centered properly for four-sigma performance.
The mean of the process distribution is too close to the upper
specification.
The process capability ratio is
b. The process capability index is
Cpk = Minimum of ,
3σ
3σ
= Minimum of = 1.60, = 1.07
420 – 300
3(25)
500 – 420
3(25)
Cp =
6σ
= = 1.33
500 – 300
6(25)

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