ISQS 3344 Quantitative Review #3. Control Charts for Measurements of Quality Example Usage: number...
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Transcript of ISQS 3344 Quantitative Review #3. Control Charts for Measurements of Quality Example Usage: number...
ISQS 3344Quantitative Review #3
Control Charts for Measurements of Quality
• Example Usage: number of ounces per bottle; diameters of ball bearings; lengths of screws
• 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 (largest - smallest) over time
X-Bar Chart Computations
1. First, find “xbar-bar”, the average of the averages
2. Now find , where σ is the standard deviation and n is the
size of each sample
3). Find the upper control limit (UCL) and lower control limit (LCL) using the following formulas:
k
xxxx
n...21
nx
Number of sample averages
nzxLCL
nzxUCL
Z is the number of sigma limits specified in the problem. For “3 sigma limits” use z = 3, for example.
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.
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
222.144
13.13917.15
612.174
13.13917.15
3z ;4 problem); the(from 13.1
917.153/)9.15975.15875.15(
LCL
UCL
n
x
Range or R Chart
RDLCL,RDUCL 3R4R
k
RR
k = # of sample ranges
A range chart measures the variability of the process using the averageof the sample ranges (range = largest – smallest)
The values of D3 and D4 are special constants whose values depend on the sample size. These constants will be given to youin a chart.
Range 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)
First Example Revisited
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 0.2 0.3 0.2
0)2333.0(0
5319.0)2333.0(28.2
28.2 ;0 so ,4
2333.03/)2.03.02.0(
43
LCL
UCL
DDn
R
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.
0)5.0(0
055.1)5.0(11.2
11.2 ;0 so ,5
problem) the(from 5.0
43
LCL
UCL
DDn
R
P Fraction Defective Chart
• “Proportion charts”• Used for yes-or-no type judgments
(acceptable/not acceptable, works/doesn’t work, on time/late, etc.)
• p = proportion of nonconforming items• Control limits are based on
p = average proportion of nonconforming items
P-Charts
1). Find p-bar:
2). Compute
3). Compute UCL and LCL using the formulas:
P-Chart Computations
)n"times"k"sampled(" units ofnumber total
defects ofnumber totalp
n
ppp
)1(
Number of observations per sample
n
ppzpLCL
n
ppzpUCL
)1(
)1(
As with the X-Bar chart, z is the number of sigma limits specified in the problem
If LCL turns out to be negative, set it to 0 (lower limit can’t be negative—why?)
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
0.282 UCL
0 LCL
are limits The 0. LCLset negative, is LCL Since
.1023(.064).09σzpLCL
.2823(.064).09σzpUCL
0.06420
(.09)(.91)
n
)p(1pσ
.09100
9
Inspected Total
Defectives#p
p
p
p
• “Count charts”• Used when looking at # of defects• Control limits are based on average number of
defects, c
C-Charts
Number-of-Defectives or C Chart
C-Chart Computations
1). Compute c-bar:
2). Compute
3). Compute LCL and UCL using the formulas:
takensamples ofnumber
observed defects ofnumber c
cc
czcLCL
czcUCL
As with the X-Bar chart, z is the number of sigma limits specified in the problem
As with the P-Bar chart, if the LCL turns out to be negative, set LCL to 0 (LCL can’t be negative, why?
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
6.65 UCL
0 LCL
thenare limits The
0. toLCLset negative, is LCL Since
2.25.2.232.2ccLCL
6.652.232.2ccUCL
2.210
22
samples of #
complaints#c
c
c
_
z
z
• “Capability” : Can a process or system meet its requirements?
Limit"ion SpecificatLower "LSL
Limit"ion SpecificatUpper "
6
LSL - USL
system production theof deviations standard 6
rangeion specificatdesign sproduct'
USL
C p
Cp < 1: process not capable of meeting design specsCp ≥ 1: process capable of meeting design specs
Cp assumes that the process is centered on the specification range, which may not be the case!
To see if a process is centered, we use Cpk:
Process Capability
3σ
LSLμ,
3σ
μUSLminCpk
3σ
LSLμ,
3σ
μUSLminCpk
min = “minimum of the two”
= mean of the process
A value of Cpk < 1 indicates that the process is notcentered.
Cpk
Cp=Cpk when process is centered
Example
Design specifications call for a target value of 16.0 +/-0.2 ounces. Observed process output has a mean of 15.9 and a standard deviation of 0.1 ounces. Is the process capable?
LSL = 16-0.2 = 15.8 USL =16 + 0.2 = 16.2
capable.not is process so ,1 and CpBoth
3333.0)333.0,1min(
)1(.3
8.159.15,
)1(.3
9.152.16min
667.0)1(.6
8.152.16
Cpk
Cpk
Cp
Chapter 3Project Mgt. and Waiting Line
Theory
Critical Path Method (CPM)• CPM is an approach to scheduling and controlling
project activities.
• The critical path: Longest path through the process
• 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.
Critical Path Method (CPM)
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
Step 1: Draw the Diagram
Step 2: Add Activity Durations
Step 3: Identify All Unique Paths And Path Durations
Path Duration = Sum of all task times along the path
Path Duration
ABDEGHJK 40
ABDEGIJK 41
ACFGHJK 22
ACFGIJK 23
Critical path
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
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=30E=32
ES=7EF=12LS=25LF=30
ES=4EF=7LS=22LF=25
Critical Path
E Buffer
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
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
Strategy: Find all the ES’s and EF’s first by moving left to right (start to finish).Then find LF and LS by working backward (finish to start)
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
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
If an item is on the critical path, there is no slack!!!!
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
The critical path was ABDEGIJK
Notice that the slack for these task times is 0.
Waiting Line Models
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
• Arrival rate and service rate must be in the same units!!
Infinite Population, Single-Server, Single Line, Single Phase Formulae
systemincustomersofnumberaverageL
nutilizatiosystemaverage
rateservicemeanmu
ratearrivalmeanlambda
Infinite Population, Single-Server, Single Line, Single Phase Formulae
lineinwaitingspenttimeaverageWW
serviceincludingsystemintimeaverageW
lineincustomersofnumberaverageLL
Q
Q
1
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:– µ = 20– λ= 15
• Note that both arrival rate and service rate are in hours, so we don’t need to do any conversion.
Average Utilization
%7575.020
15or
Average Number of Studentsin the System
31520
15
L
Average Number of StudentsWaiting in Line
studentsLLQ 25.2375.0
Average Time a Student Spends in the System
1520
11
W
.2 hours or 12 minutes
Average Time a StudentSpends Waiting (Before
Service)
minutes9
hours15.02.075.0
or
WWQ
Too long?After 5 minutes peopleget anxious
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?Convert to hours first!
36080
60
,1
customers 80
1hr
secs 3600
secs 45
customer 1
L
Nowhr