Operations Management- Case 2
Topic 4
Quality Management
Dr. Bin Jiang
Department of Management
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Dichotomy of Quality
MGT 504 by Dr. Bin Jiang
Quality is the ability of a product or service to consistently meet or exceed customer expectations.
Red Side of Quality:
Dimensions of Quality
- Performance – main characteristics of the product
- Reliability – consistency of performance
- Durability – useful life of the product
- Safety – risk of injury
- Aesthetics – appearance, feel, etc
- Service After Sale
- Perceived Quality – e.g. reputation
MGT 504 by Dr. Bin Jiang
Blue Side of Quality:
Process Control Charts
Control Charts show sample data plotted on a graph with Center Line (CL), Upper Control Limit (UCL), and Lower Control Limit (LCL).
MGT 504 by Dr. Bin Jiang
Control Chart
- Control Chart
- Purpose: to monitor process output to see if its variation is random
- A time ordered plot representative sample statistics obtained from an on going process (e.g. sample means)
- Upper and lower control limits define the range of acceptable variation
MGT 504 by Dr. Bin Jiang
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Types of Control Charts
- Control chart for variables are used to monitor characteristics that can be measured, e.g. length, weight, diameter, time, etc.
- Control charts for attributes are used to monitor characteristics that have discrete values and can be counted, e.g. % defective, number of flaws in a shirt, number of broken eggs in a box, etc.
MGT 504 by Dr. Bin Jiang
Empirical Rule
MGT 504 by Dr. Bin Jiang
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-3
-1
-2
+1
+2
+3
68%
95%
99.7%
Upper Limit
Lower Limit
Expected Output
*
Six Sigma
MGT 504 by Dr. Bin Jiang
Mean
Lower limits
Upper limits
±6
3.4 defects/million
2,700 defects/million
±3
Control Charts for Variables
- Are named according to the statistics being plotted, i.e., X bar, R
- Have a center line that is the overall average
- Have limits above and below the center line at ± 3 standard deviations (usually)
MGT 504 by Dr. Bin Jiang
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Center line
Lower Control Limit (LCL)
Upper Control Limit (UCL)
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Control Charts for Variables
- Mean control chart: X bar chart
- Used to monitor the central tendency of a process.
Range control chart: R chart
- Used to monitor the process dispersion
MGT 504 by Dr. Bin Jiang
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Variables Data Charts
- Process Centering
- X bar chart
- X bar is a sample mean
- Process Dispersion (consistency)
- R chart
- R is a sample range
MGT 504 by Dr. Bin Jiang
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X chart
- Center line is the grand mean (X double bar)
- Points are X bars
MGT 504 by Dr. Bin Jiang
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-OR-
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Example 1
MGT 504 by Dr. Bin Jiang
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Standard Deviation is 0.02
| Sample 1 | Sample 2 | Sample 3 | Sample 4 | Sample 5 | |
| O1 | 12.11 | 12.15 | 12.09 | 12.12 | 12.09 |
| O2 | 12.10 | 12.12 | 12.09 | 12.10 | 12.14 |
| O3 | 12.11 | 12.10 | 12.11 | 12.08 | 12.13 |
| O4 | 12.08 | 12.11 | 12.15 | 12.10 | 12.12 |
| X bar | 12.10 | 12.12 | 12.11 | 12.10 | 12.12 |
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Factors for Calculating Three-Sigma Limits
For x-Chart and R-Chart
Example 1
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MGT 504 by Dr. Bin Jiang
Approach 1
Approach 2
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Example 1
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MGT 504 by Dr. Bin Jiang
UCL=12.14
LCL=12.08
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R Chart
- Center line is the grand mean (R bar)
- Points are R
- D3 and D4 values are tabled according to n (sample size)
MGT 504 by Dr. Bin Jiang
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Example 1
MGT 504 by Dr. Bin Jiang
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UCL = 0.105
LCL = 0
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X and Range Charts
MGT 504 by Dr. Bin Jiang
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UCL
Does not
reveal increase
UCL
LCL
LCL
R-chart
Reveals increase
(process variability is increasing)
Sampling
Distribution
x-Chart
*
X and Range Charts
MGT 504 by Dr. Bin Jiang
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UCL
LCL
UCL
LCL
R-chart
x-Chart
Detects shift
Does not
detect shift
(process mean is
shifting upward)
Sampling
Distribution
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Example 2
Processing new accounts at a bank is intended to average 10 minutes each. Five samples of four observations each have been taken. Use the sample data in the following table to construct upper and lower control limits for both a mean chart and a range chart. Do the results suggest that the process is in control?
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MGT 504 by Dr. Bin Jiang
| Sample1 | Sample2 | Sample3 | Sample4 | Sample5 |
| 10.2 | 10.3 | 9.7 | 9.9 | 9.8 |
| 9.9 | 9.8 | 9.9 | 10.3 | 10.2 |
| 9.8 | 9.9 | 9.9 | 10.1 | 10.3 |
| 10.1 | 10.4 | 10.1 | 10.5 | 9.7 |
| 10.0 | 10.1 | 9.9 | 10.2 | 10.0 |
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Example 2
MGT 504 by Dr. Bin Jiang
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Step 1:
Step 2:
For n = 4: A2 = 0.73, D4 = 2.28, D3 = 0.
Step 3:
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Example 2
Step 4:
MGT 504 by Dr. Bin Jiang
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x-Chart
R-chart
10.04
10.42
9.66
1.19
0.52
0
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Control Chart Checks
MGT 504 by Dr. Bin Jiang
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Any point outside of the control limits
A run of 7 points all above or all below the nominal line
A run of 7 points up or down (trends)
Any other obviously non-random pattern
Control Charts for Attributes
P-Charts & C-Charts
- Use P-Charts for quality characteristics that are discrete and involve yes/no or good/bad decisions
- Percent of leaking caulking tubes in a box of 48
- Percent of broken eggs in a carton
- Use C-Charts for discrete defects when there can be more than one defect per unit
- Number of flaws or stains in a carpet sample cut from a production run
- Number of complaints per customer at a hotel
MGT 504 by Dr. Bin Jiang
Constructing a P-Chart:
A Production manager for a tire company has inspected the number of defective tires in five random samples with 20 tires in each sample. The table below shows the number of defective tires in each sample of 20 tires.
MGT 504 by Dr. Bin Jiang
| Sample | Sample Size (n) | Number Defective |
| 1 | 20 | 3 |
| 2 | 20 | 2 |
| 3 | 20 | 1 |
| 4 | 20 | 2 |
| 5 | 20 | 1 |
Step 1:
Calculate the Percent defective of Each Sample and the Overall Percent Defective (P-Bar)
MGT 504 by Dr. Bin Jiang
| Sample | Number Defective | Sample Size | Percent Defective |
| 1 | 3 | 20 | .15 |
| 2 | 2 | 20 | .10 |
| 3 | 1 | 20 | .05 |
| 4 | 2 | 20 | .10 |
| 5 | 1 | 20 | .05 |
| Total | 9 | 100 | .09 |
Step 2: Calculate the Standard Deviation of P.
MGT 504 by Dr. Bin Jiang
Step 3: Calculate CL, UCL, LCL
MGT 504 by Dr. Bin Jiang
Center line (p bar):
Control limits for ±3σ limits:
Step 4: Draw the Chart
MGT 504 by Dr. Bin Jiang
Constructing a C-Chart:
The number of weekly customer complaints are monitored in a large hotel. Develop a three sigma control limits For a C-Chart using the data table On the right.
MGT 504 by Dr. Bin Jiang
| Week | Number of Complaints |
| 1 | 3 |
| 2 | 2 |
| 3 | 3 |
| 4 | 1 |
| 5 | 3 |
| 6 | 3 |
| 7 | 2 |
| 8 | 1 |
| 9 | 3 |
| 10 | 1 |
| Total | 22 |
Calculate CL, UCL, LCL
MGT 504 by Dr. Bin Jiang
Center line (c bar):
Control limits for ±3σ limits:
Process Capability
MGT 504 by Dr. Bin Jiang
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Product Specifications
Preset product or service dimensions, tolerances: bottle fill might be 16 oz. ±.2 oz. (15.8oz.-16.2oz.)
Based on how product is to be used or what the customer expects
Process Capability – Cp and Cpk
Assessing capability involves evaluating process variability relative to preset product or service specifications
Cp assumes that the process is centered in the specification range
Cpk helps to address a possible lack of centering of the process
Relationship between Process Variability and Specification Width
- Three possible ranges for Cp
- Cp = 1, as in Fig. (a), process
variability just meets specifications
- Cp ≤ 1, as in Fig. (b), process not capable of producing within specifications
- Cp ≥ 1, as in Fig. (c), process
exceeds minimal specifications
- One shortcoming, Cp assumes that the process is centered on the specification range
- Cp=Cpk when process is centered
MGT 504 by Dr. Bin Jiang
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Process Capability
800
1000
1200
MGT 504 by Dr. Bin Jiang
Nominal
value
Hours
Upper
specification
Lower
specification
Process distribution
Process Capability
Nominal
value
Hours
Upper
specification
Lower
specification
Process distribution
800
1000
1200
MGT 504 by Dr. Bin Jiang
MGT 504 by Dr. Bin Jiang
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Computing the Cp Value at Cocoa Fizz: 3 bottling machines are being evaluated for possible use at the Fizz plant. The machines must be capable of meeting the design specification of 15.8-16.2 oz. with at least a process capability index of 1.0 (Cp≥1).
Example of Cp
The table below shows the information gathered from production runs on each machine. Are they all acceptable?
Machine A:
MGT 504 by Dr. Bin Jiang
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Example of Cpk
Design specifications call for a target value of 16.0 ±0.2 OZ.
(USL = 16.2 & LSL = 15.8)
Observed process output has now shifted and has a µ of 15.9 and a
σ of 0.1 oz.
Cpk is less than 1, revealing that the process is not capable
What if we don’t know ?
MGT 504 by Dr. Bin Jiang
Sample Size
d2
2
1.128
3
1.693
4
2.059
5
2.326
6
2.534
10
3.078
n
X
X
n
i
i
å
=
=
1
)
min(
)
max(
i
i
X
X
R
-
=
x
z
X
UCL
s
+
=
n
x
/
s
s
=
x
z
X
LCL
s
-
=
m
X
X
m
j
j
å
=
=
1
R
A
X
UCL
2
+
=
R
A
X
LCL
2
-
=
11
.
12
5
5
4
3
2
1
=
+
+
+
+
=
x
x
x
x
x
x
046
.
0
5
5
4
3
2
1
=
+
+
+
+
=
R
R
R
R
R
R
14
.
12
046
.
0
73
.
0
11
.
12
2
=
´
+
=
+
=
R
A
x
UCL
08
.
12
046
.
0
73
.
0
11
.
12
2
=
´
-
=
-
=
R
A
x
LCL
14
.
12
4
02
.
0
3
11
.
12
=
÷
ø
ö
ç
è
æ
+
=
+
=
x
z
x
UCL
s
08
.
12
4
02
.
0
3
11
.
12
=
÷
ø
ö
ç
è
æ
-
=
-
=
x
z
x
LCL
s
10
.
12
4
=
x
12
.
12
2
=
x
10
.
12
1
=
x
12
.
12
5
=
x
11
.
12
3
=
x
11
.
12
=
x
R
D
UCL
4
=
R
D
LCL
3
=
046
.
0
5
5
4
3
2
1
=
+
+
+
+
=
R
R
R
R
R
R
105
.
0
046
.
0
28
.
2
4
=
´
=
=
R
D
UCL
0
046
.
0
0
3
=
´
=
=
R
D
LCL
046
.
0
=
R
52
.
0
,
04
.
10
=
=
R
x
0
)
52
.
0
(
0
19
.
1
)
52
.
0
(
28
.
2
66
.
9
)
52
.
0
(
73
.
0
04
.
10
42
.
10
)
52
.
0
(
73
.
0
04
.
10
3
4
2
2
=
=
=
=
=
=
=
-
=
-
=
=
+
=
+
=
R
D
LCL
R
D
UCL
R
A
x
LCL
R
A
x
UCL
R
R
x
x
p
p(1-p)(.09)(.91)
σ===0.064
n20
CLp.09
==
(
)
(
)
p
p
UCLpz
σ.093(.064).282
LCLpz
σ.093(.064).1020
=+=+=
=-=-=-=
#complaints22
CL2.2
# of samples10
===
UCLcc2.232.26.65
LCLcc2.232.22.250
z
z
=+=+=
=-=-=-=
s
6
LSL
USL
C
p
-
=
ú
ú
û
ù
ê
ê
ë
é
-
-
=
s
s
3
,
3
x
US
LS
x
Min
C
pk
Sample Size
d
2
2
1.128
3
1.693
4
2.059
5
2.326
6
2.534
10
3.078
2
ˆ
d
R
@
s