### ES=30

```Quantitative Review III
Chapter 6 and 6S
Statistical Process Control
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)
x-Bar Computations
x-bar = sample average
x1  x 2  ... x n
x
k
UCLx  x  z x
LCLx  x  z x
x 

n
x
x 
= standard deviation of the sample means

n
“n” here equals # of
observations in each
sample
x1  x 2  ... x n
x
k
z
“k” here = # of samples
= standard normal variable or the # of std. deviations
desired to use to develop the control limits
Assume the standard deviation of the process is given as 1.13 ounces
Management 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
A.
B.
C.
D.
E.
18.56, 16.32
16.22, 18.56
17.62, 14.23
19.01, 12.56
18.33, 14.28
0
15.875 15.975 15.9
x
3
x  15.9167
1.13 1.13
x 

 .565
2
4
UCL  15.9167 3x.565  17.62ounces
LCL  15.9167 3x.565  14.23ounces
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?
Z=3

x
= 4.27
26  26  26
x
3
x  26
4.27
4.27
x 

 2.135
2
4
UCL  26  3x 2.135  32.4
LCL  26  3x 2.135  19.6
Range or R Chart
UCLR  D4 R, LCLR  D3 R
k = # of sample ranges
R

R
k
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.
Control Chart Factors
Sample Size (n)
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Factors for R-Chart
D4
D3
0.00
3.27
0.00
2.57
0.00
2.28
0.00
2.11
0.00
2.00
0.08
1.92
0.14
1.86
0.18
1.82
0.22
1.78
0.26
1.74
0.28
1.72
0.31
1.69
0.33
1.67
0.35
1.65
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
R-chart Computations
(Use D3 & D4 Factors: Table 6-1)
0 .2  0 .3  0 .2
R
 .233
3
UCLR  RD4  .2332.28  .531
LCLR  RD3  .2330   0
Ten samples of 5 observations each have been taken form a
Soft drink bottling plant in order to test for volume dispersion
in the bottling process. The average sample range was found
To be .5 ounces. Develop control limits for the sample range.
A.
B.
C.
D.
E.
.996, -.320
1.233, 0
1.055, 0
.788, 1.201
1.4, 0
UCLR  RD4  2.11.5  1.055
LCLR  RD3  0.5  0
(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
(P) Fraction Defective Chart
totalnumber of defects
p
,
totalnumber of units sampled("k"times"n")
p 
p(1  p)
n
n = # of observations in each sample
UCL p  p  z p , LCL p  p  z p
z
= standard normal variable or the # of std. deviations
desired to use to develop the control limits
K = number of samples
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
p(1 p)
(.09)(.91)
σp 

 0.064
n
20
5
1
20
.05
Total
9
100
.09
UCLp  p  zσ   .09  3(.064) .282
• Solution:
p
# Defectives
9

 .09
T otalInspected 100
LCLp  p  zσ   .09  3(.064) .102  0
Note: Lower control limit can’t be negative.
Number-of-Defectives or C
Chart
Used when looking at # of defects
c = # of defects
c = average # of defects per sample
Number-of-Defectives or C
Chart
number of incidentsobserved
c
, c  c
number of samples(k)
UCLc  c  z c , LCLc  c  z c
z
= standard normal variable or the # of std. deviations
desired to use to develop the control limits
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
Complaints
1
3
2
2
3
3
4
1
5
3
6
3
7
2
8
1
9
3
10
1
Total
22
• Solution:
_
# complaints 22
c

 2.2
# of samples 10
UCLc  c  z c  2.2  3 2.2  6.65
LCLc  c  z c  2.2  3 2.2  2.25  0
Process Capability Index
• Can a process or system meet it’s
requirements?
product's design specification range
USL - LSL
Cp 

6 standarddeviationsof theproductionsystem
6
Cp < 1: process not capable of meeting design specs
Cp ≥ 1: process capable of meeting design specs
One shortcoming, Cp assumes that the process
is centered on the specification range

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
spe cificat
ion width USL  LSL
Cp

proce ss width
6σ
– Cpk helps to address a possible lack of centering of the process
 USL  μ μ  LSL 
C pk  min
,

3σ 
 3σ
 USL  μ μ  LSL 
C pk  min
,

3σ 
 3σ
min = minimum of the two

If
= mu or mean of the process
C pk
Is less than 1, then the process is not capable or
is not centered.
Capability Indexes
• Centered Process (Cp):
specification width USL  LSL
Cp 

processwidth
6
• Any Process (Cpk):
 USL     LSL 
C pk  min
;

3 
 3
Cp=Cpk when process is centered
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
Computations
• C p:
• Cpk:
USL  LSL 16.2  15.8 0.4
Cp 


 0.66
6
60.1
0.6
C pk
  LSL 
 USL  
 min
or

3 
 3
 16.2  15.9 15.9  15.8 

 min
or
30.1 
 30.1
 0. 3 0. 1 
 min
or
  min1 or 0.33  0.33
 0.3 0.3 
Chapter 3
Project Mgt. and Waiting Line
Theory
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.
Example
Activity
A
B
C
D
E
F
G
H
I
J
K
Description
Develop product specifications
Design manufacturing process
Source & purchase materials
Source & purchase tooling & equipment
Receive & install tooling & equipment
Pilot production run
Evaluate product design
Evaluate process performance
Write documentation report
Transition to manufacturing
Immediate
Predecessor
None
A
A
B
D
C
E&F
G
G
H&I
J
Duration
(weeks)
4
6
3
6
14
5
2
2
3
4
2
CPM Diagram
Identify Unique Paths
Calculate Path Durations
Paths
ABDEGHJK
ABDEGIJK
ACFGHJK
ACFGIJK
Path duration
40
41
22
23
• 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
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 noncritical paths
ES=10 ES=16
EF=16 EF=30
LS=16
LS=10
E Buffer
LF=16 LF=30
ES=4
EF=10
LS=4
LF=10
B(6)
D(6)
E(14)
A(4)
ES=0
EF=4
LS=0
LF=4
ES=32
EF=34
LS=33
LF=35
H(2)
G(2)
C(3)
ES=4
EF=7
LS=22
LF=25
F(5)
ES=7
EF=12
LS=25
LF=30
ES=30
EF=32
LS=30
LF=32
I(3)
ES=32
EF=35
LS=32
LF=35
ES=35
EF=39
LS=35
LF=39
ES=39
EF=41
LS=39
LF=41
J(4)
K(2)
Critical Path
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
ES=4+6=10
EF=10
LS=4
LF=10
ES=16
EF=30
ES=10
EF=16
B(6)
D(6)
E(14) Latest EF H(2)
= Next ES
ES=35
EF=39
ES=39
EF=41
G(2)
J(4)
K(2)
A(4)
ES=0
EF=4
LS=0
LF=4
C(3)
ES=4
EF=7
LS=22
LF=25
ES=32
EF=34
F(5)
ES=7
EF=12
LS=25
LF=30
ES=30
EF=32
LS=30
LF=32
I(3)
ES=32
EF=35
LS=32
LF=35
Calculate Early
Starts & Finishes
ES=16
EF=30
LS=16
LF=30
ES=10
EF=16
LS=10
LF=16
ES=4
EF=10
LS=4
LF=10
B(6)
D(6)
E(14)
H(2)
39-4=35
ES=35
EF=39
LS=35
LF=39
Earliest LS
G(2) = Next LF J(4)
A(4)
ES=0
EF=4
ES=32
EF=34
LS=33
LF=35
C(3)
ES=4
EF=7
LS=22
LF=25
F(5)
ES=7
EF=12
LS=25
LF=30
ES=30
EF=32
I(3)
LS=30
LF=32 ES=32
EF=35
LS=32
LF=35
ES=39
EF=41
LS=39
LF=41
K(2)
Calculate Late
Starts & Finishes
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
Calculate Activity Slack
Activity
A
B
C
D
E
F
G
H
I
J
K
Late
Finish
4
10
25
16
30
30
32
35
35
39
41
Early
Finish
4
10
7
16
30
12
32
34
35
39
41
Slack
(weeks)
0
0
18
0
0
18
0
1
0
0
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
Infinite Population, Single-Server,
Single Line, Single Phase Formulae
  lam bda m eanarrival rate
  m u  m eanservice rate

   averagesystem utilization


L
 averagenum berof custom ersin system
 
Infinite Population, Single-Server,
Single Line, Single Phase Formulae
LQ  L  averagenum berof custom ersin line
1
W
 averagetim e in system includingservice
 
WQ  W  averagetim e spent waiting  in  line
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)
Average Utilization
 15
 
 0.75 or 75%
 20
Average Number of Students
in the System

15
L

 3 students
   20  15
Average Number of Students
Waiting in Line
LQ  L  0.753  2.25 students
Average Time a Student
Spends in the System
1
1
W

   20  15
.2 hours or 12 minutes
Average Time a Student
Spends Waiting (Before
Service)
WQ  W  0.750.2  0.15 hours
or 9 minutes
Too long?
After 5 minutes people
get anxious
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.
B.
C.
D.
E.
2.25
3
4
3.25
4.25
L

 
60/80-60
=3
```