quantitative review iii. chapter 6 and 6s statistical process control
TRANSCRIPT
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Quantitative Review III
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Chapter 6 and 6SStatistical Process Control
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Control Charts for Variable Data(length, width, etc.)
• Mean (x-bar) charts– Tracks the central tendency (the average
or mean value observed) over time• Range (R) charts:
– Tracks the spread of the distribution over time (estimates the observed variation)
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x-Bar Computationsx-bar = sample average
xx
xx
x
n
zxLCL
zxUCL
nk
xxxx
...21
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x = standard deviation of the sample means
z = standard normal variable or the # of std. deviations desired to use to develop the control limits
nx
“n” here equals # of observations in each
sample
k
xxxx
n...21
“k” here = # of samples
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Assume the standard deviation of the process is given as 1.13 ouncesManagement wants a 3-sigma chart (only 0.26% chance of alpha error)
Observed values shown in the table are in ounces. Calculate the UCL and LCL.
0
Sample 1 Sample 2 Sample 3
Observation 1 15.8 16.1 16.0
Observation 2 16.0 16.0 15.9
Observation 3 15.8 15.8 15.9
Observation 4 15.9 15.9 15.8
Sample means 15.875 15.975 15.9
A. 18.56, 16.32
B. 16.22, 18.56
C. 17.62, 14.23
D. 19.01, 12.56
E. 18.33, 14.28
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ouncesxLCL
ouncesxUCL
x
x
x
23.14565.39167.15
62.17565.39167.15
565.2
13.1
4
13.1
9167.15
3
9.15975.15875.15
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Inspectors want to develop process control charts to measure the weight of crates of wood. Data (in pounds) from three samples are:
Sample Crate 1 Crate 2 Crate 3 Sample Means
1 18 38 22 26
2 26 24 28 26
3 26 26 26 26
What are the upper and lower control limits for this process?
x
Z=3 = 4.27
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6.19135.2326
4.32135.2326
135.22
27.4
4
27.4
26
3
262626
xLCL
xUCL
x
x
x
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Range or R Chart
RDLCL,RDUCL 3R4R
k
RR
k = # of sample ranges
Range Chart measures the dispersion or variance of the process while The X Bar chart measures the central tendency of the process. When selecting D4 and D3 use sample size or number of observations for n.
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Control Chart Factors
D3 D4
2 0.00 3.273 0.00 2.574 0.00 2.285 0.00 2.116 0.00 2.007 0.08 1.928 0.14 1.869 0.18 1.82
10 0.22 1.7811 0.26 1.7412 0.28 1.7213 0.31 1.6914 0.33 1.6715 0.35 1.65
Factors for R-ChartSample Size (n)
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Example
Sample 1 Sample 2 Sample 3
Observation 1 15.8 16.1 16.0
Observation 2 16.0 16.0 15.9
Observation 3 15.8 15.8 15.9
Observation 4 15.9 15.9 15.8
Sample means 15.875 15.975 15.9
Sample ranges 16.0-15.8=0.2 16.1-15.8=0.3 16.0-15.8=0.2
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R-chart Computations(Use D3 & D4 Factors: Table 6-1)
00233.
531.28.2233.
233.3
2.03.02.0
3
4
DRLCL
DRUCL
R
R
R
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Ten samples of 5 observations each have been taken form aSoft drink bottling plant in order to test for volume dispersionin the bottling process. The average sample range was foundTo be .5 ounces. Develop control limits for the sample range.
A. .996, -.320B. 1.233, 0C. 1.055, 0D. .788, 1.201E. 1.4, 0
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05.0
055.15.11.2
3
4
DRLCL
DRUCL
R
R
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(P) Fraction Defective Chart
• Used for yes or no type judgments (acceptable/not acceptable, works/doesn’t work, on time/late, etc.)
• p = proportion of nonconforming items
p = average proportion of nonconforming items
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(P) Fraction Defective Chart
pppp
p
zpzp
n
pp
p
LCL,UCL
)1(
,)n"times"k"sampled(" units ofnumber total
defects ofnumber total
n = # of observations in each sample
z = standard normal variable or the # of std. deviations desired to use to develop the control limits
K = number of samples
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P-Chart Example: 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.
Z= 3. Calculate the control limits.
Sample Number of
Defective Tires
Number of Tires in each
Sample
Proportion 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
• Solution:
0.1023(.064).09σzpLCL
.2823(.064).09σzpUCL
0.06420
(.09)(.91)
n
)p(1pσ
.09100
9
Inspected Total
Defectives#p
p
p
p
Note: Lower control limit can’t be negative.
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Number-of-Defectives or C Chart
Used when looking at # of defects
c = # of defects
c = average # of defects per sample
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Number-of-Defectives or C Chart
cccc
c
zczc
cc
LCL,UCL
,(k) samples ofnumber
observed incidents ofnumber
= standard normal variable or the # of std. deviations desired to use to develop the control limits
z
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C-Chart Example: The number of weekly customer complaints are monitored in a large hotel using a
c-chart. Develop three sigma control limits using the data table below. Z=3.Week Number of
Complaints1 3
2 2
3 3
4 1
5 3
6 3
7 2
8 1
9 3
10 1
Total 22
• Solution:
02.252.232.2ccLCL
6.652.232.2ccUCL
2.210
22
samples of #
complaints#c
c
c
_
z
z
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Process Capability Index
• Can a process or system meet it’s requirements?
6
LSL - USL
system production theof deviations standard 6
rangeion specificatdesign sproduct'pC
Cp < 1: process not capable of meeting design specsCp ≥ 1: process capable of meeting design specs
One shortcoming, Cp assumes that the process is centered on the specification range
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Process Capability• Product Specifications
– Preset product or service dimensions, tolerances– e.g. 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 process6σ
LSLUSL
width process
width ionspecificatCp
3σ
LSLμ,
3σ
μUSLminCpk
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3σ
LSLμ,
3σ
μUSLminCpk
min = minimum of the two
= mu or mean of the process
If pkC Is less than 1, then the process is not capable or
is not centered.
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Capability Indexes
• Centered Process (Cp):
• Any Process (Cpk):
6 widthprocess
ion widthspecificat LSLUSLC p
3
;3
minLSLUSL
C pk
Cp=Cpk when process is centered
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Example
• Design specifications call for a target value of 16.0 +/-0.2 ounces (USL = 16.2 & LSL = 15.8)
• Observed process output has a mean of 15.9 and a standard deviation of 0.1 ounces
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Computations
• Cp:
• Cpk:
66.06.0
4.0
1.06
8.152.16
6
LSLUSL
C p
33.033.0or 1min3.0
1.0or
3.0
3.0min
1.03
8.159.15or
1.03
9.152.16min
3or
3min
LSLUSLC pk
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Chapter 3Project Mgt. and Waiting Line
Theory
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Critical Path Method (CPM)• CPM is an approach to scheduling and controlling
project activities.
• The critical path is the sequence of activities that take the longest time and defines the total project completion time.
• Rule 1: EF = ES + Time to complete activity
• Rule 2: the ES time for an activity equals the largest EF time of all immediate predecessors.
• Rule 3: LS = LF – Time to complete activity
• Rule 4: the LF time for an activity is the smallest LS of all immediate successors.
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Example
Activity DescriptionImmediate
PredecessorDuration (weeks)
A Develop product specifications None 4B Design manufacturing process A 6C Source & purchase materials A 3D Source & purchase tooling & equipment B 6E Receive & install tooling & equipment D 14F Receive materials C 5G Pilot production run E & F 2H Evaluate product design G 2I Evaluate process performance G 3J Write documentation report H & I 4K Transition to manufacturing J 2
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CPM Diagram
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Add Activity Durations
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Identify Unique Paths
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Calculate Path Durations
• The longest path (ABDEGIJK) limits the project’s duration (project cannot finish in less time than its longest path)
• ABDEGIJK is the project’s critical path
Paths Path durationABDEGHJK 40ABDEGIJK 41ACFGHJK 22ACFGIJK 23
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Adding Feeder Buffers to Critical Chains
• The theory of constraints, the basis for critical chains, focuses on keeping bottlenecks busy.
• Time buffers can be put between bottlenecks in the critical path• These feeder buffers protect the critical path from delays in non-
critical paths
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B(6) D(6)
A(4)
C(3)F(5)
E(14)
G(2)
I(3)
H(2)
J(4) K(2)
ES=0EF=4LS=0LF=4
ES=4EF=10LS=4LF=10
ES=10EF=16LS=10LF=16
ES=16EF=30LS=16LF=30
ES=32EF=34LS=33LF=35 ES=35
EF=39LS=35LF=39
ES=39EF=41LS=39LF=41
ES=32EF=35LS=32LF=35
ES=30EF=32LS=30LF=32ES=7
EF=12LS=25LF=30
ES=4EF=7LS=22LF=25
Critical Path
E Buffer
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Some Network Definitions
• All activities on the critical path have zero slack• Slack defines how long non-critical activities can be
delayed without delaying the project• Slack = the activity’s late finish minus its early finish (or
its late start minus its early start)• Earliest Start (ES) = the earliest finish of the immediately
preceding activity• Earliest Finish (EF) = is the ES plus the activity time• Latest Start (LS) and Latest Finish (LF) depend on
whether or not the activity is on the critical path
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B(6) D(6)
A(4)
C(3)F(5)
E(14)
G(2)
I(3)
H(2)
J(4) K(2)
ES=0EF=4LS=0LF=4
ES=4+6=10EF=10LS=4LF=10
ES=10EF=16
ES=16EF=30
ES=32EF=34
ES=35EF=39
ES=39EF=41
ES=32EF=35LS=32LF=35
ES=30EF=32LS=30LF=32
ES=7EF=12LS=25LF=30
ES=4EF=7LS=22LF=25
Calculate EarlyStarts & Finishes
Latest EF= Next ES
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B(6) D(6)
A(4)
C(3)F(5)
E(14)
G(2)
I(3)
H(2)
J(4) K(2)
ES=0EF=4
ES=4EF=10LS=4LF=10
ES=10EF=16LS=10LF=16
ES=16EF=30LS=16LF=30
ES=32EF=34LS=33LF=35 ES=35
EF=39LS=35LF=39
ES=39EF=41LS=39LF=41
ES=32EF=35LS=32LF=35
ES=30EF=32LS=30LF=32ES=7
EF=12LS=25LF=30
ES=4EF=7LS=22LF=25
Calculate LateStarts & Finishes
Earliest LS= Next LF
39-4=35
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Activity Slack Time
TES = earliest start time for activity
TLS = latest start time for activity
TEF = earliest finish time for activity
TLF = latest finish time for activity
Activity Slack = TLS - TES = TLF - TEF
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Calculate Activity Slack
ActivityLate
FinishEarly Finish
Slack (weeks)
A 4 4 0B 10 10 0C 25 7 18D 16 16 0E 30 30 0F 30 12 18G 32 32 0H 35 34 1I 35 35 0J 39 39 0K 41 41 0
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Waiting Line Models
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Arrival & Service Patterns
• Arrival rate:– The average number of customers arriving
per time period
• Service rate:– The average number of customers that can
be serviced during the same period of time
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Infinite Population, Single-Server, Single Line, Single Phase Formulae
systemincustomersofnumberaverageL
nutilizatiosystemaverage
rateservicemeanmu
ratearrivalmeanlambda
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Infinite Population, Single-Server, Single Line, Single Phase Formulae
lineinwaitingspenttimeaverageWW
serviceincludingsystemintimeaverageW
lineincustomersofnumberaverageLL
Q
Q
1
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Example
• A help desk in the computer lab serves students on a first-come, first served basis. On average, 15 students need help every hour. The help desk can serve an average of 20 students per hour.
• Based on this description, we know:– Mu = 20 (exponential distribution)– Lambda = 15 (Poisson distribution)
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Average Utilization
%7575.020
15or
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Average Number of Studentsin the System
studentsL 31520
15
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Average Number of StudentsWaiting in Line
studentsLLQ 25.2375.0
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Average Time a Student Spends in the System
1520
11
W
.2 hours or 12 minutes
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Average Time a StudentSpends Waiting (Before
Service)
minutes9
hours15.02.075.0
or
WWQ
Too long?After 5 minutes peopleget anxious
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Consider a single-line, single-server waiting line system. Suppose that customers arrive
according to a Poisson distribution at an average rate of 60 per hour, and the average
(exponentially distributed) service time is 45 seconds per customer. What is the average
number of customers in the system?
A. 2.25B. 3C. 4D. 3.25E. 4.25
60/80-60= 3
L