Quantitative Review III
Chapter 6 & 6S
Managing Quality
&
Statistical Process Control (SPC)
How do we ensure quality of manufacturing and
service delivery processes?
Using Statistical Process Control (SPC) to help identify and
eliminate unwanted causes of variation in the process
To monitor the degree of product conformance to a specification
on continuous scale of measurement such as length, weight,
and time (continuous metric), we use X-bar and R-charts.
To monitor the noncontinous scale of measurement (discrete
metric)such as from data that are counted (e.g. no. of compete
orders, no. of good service experiences), we use p- or c-charts
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)
Control Charts for Variable Data
(length, width, etc.)
K: Number of samples
n: Sample size
x i : The mean for each sample (sample average)
R: the range for each sample
x : The overall mean of all the sample means
R : The average range of all the sample ranges
x 
x 1  x 2  ... x n
k
Upper Control Limit:
Lower Control Limit:
UCL
LCL
x
x
 x  z
 x  z
x
x
;

x


n
standard deviation of the sample means
standard deviation of the process
x 
x

“n” = # of observations in each sample
n
x 1  x 2  ... x n
k
z
“k” = # of samples
= standard normal variable or the # of std. deviations
desired to use to develop the control limits
Example: X Bar-Chart
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
x 
15 . 875  15 . 975  15 . 9
3
We have 3 samples here
x  15 . 9167

x

1 . 13
4

1 . 13
2
 . 565
We have 4 observations in each sample
UCL  15 . 9167  3 x . 565  17 . 62 ounces
LCL  15 . 9167  3 x . 565  14 . 23 ounces
Example: X Bar-Chart
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
Crate 4
1
18
38
22
26
2
26
30
28
20
3
26
26
26
26
What are the upper and lower x control limits for this process?
Z=3

= 4.27
We first need to calculate x-bar for each sample
x1
=
18  38  22  26
= 26
4
x2
=
26  30  28  20
= 26
4
x3
=
26  26  26  26
4
= 26
26  26  26
x 
 26
3
We have 3 samples

x

4 . 27
4

4 . 27
 2 . 135
2
We have 4 observations in each sample
UCL  26  3 x 2 . 135  32 . 4
LCL
 26  3 x 2 . 135  19 . 6
Range or R Chart
UCL
R
 D4R,
R

R 
LCL
R
 D3R
k = # of sample ranges
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 from the table, 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: R-Chart
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
Sample ranges = highest observation – lowest observation
R-chart Computations
(Use D3 & D4 Factors: Table 6-1)
R 
0 .2  0 .3  0 .2
3
 . 233
D4 from control chart factors when n=4
UCL
R
 R D 4  . 233  2 . 28   . 531
LCL
R
 R D 3  . 233 0   0
D3 from control chart factors when n=4
Example: R-Chart
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.
R = 0.5, n = 5
UCL R  R D 4  ( 0 . 5 )( 2 . 11 )  1 . 055
LCL R  R D 3  0 . 5 0   0
P- Chart
Many quality characteristics assumes only two values,
such as good or bad, pass or fail. The proportion of
nonconforming items can be monitored using a control
chart called a p-chart, where p is the proportion of
nonconforming items found in a sample.
(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
total number
p
total number

p

UCL
z
p (1  p )
n
p
 p  z p ,
of defects
of units sampled("
,
k" times" n" )
K = # of samples
n = # of observations in each sample
LCL
p
 p  z
p
= standard normal variable or the # of std. deviations
desired to use to develop the control limits
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:
p
# Defectives
Total Inspected
σp 
p (1  p )
n


9
 .09
100
(.09)(.91)
 0.064
20
UCL p  p  z σ   .09  3(.064)  .282
LCL p  p  z σ   .09  3(.064)   .102  0
Note: Lower control limit can’t be negative, if LCLp is less than zero, a value
of zero is used.
C-Chart
A p-chart monitors the proportion of nonconforming items,
but a nonconforming item may have more than one
conformance. E.g. a customer’s order may have several
errors, such as wrong item, wrong quantity, wrong price.
To monitor the number of nonconformance per unit, we
use a c-chart. It is used extensively in service applications.
Number-of-Defectives or C-Chart
Used when looking at # of nonconformances
c = # of nonconformances
c = average # of nonconformances per sample
Number-of-Defectives or C Chart
c
number of incidents observed
number of samples (k)
,
c 
c
UCL c  c  z  c  c  z c
LCL c  c  z  c  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:
_
c
# complaints
# of samples

22
 2.2
10
UCL c  c  z c  2.2  3 2.2  6.65
LCL c  c  z c  2.2  3 2.2   2.25  0
Process Capability
The natural variation in a process
that results from common causes
Process Capability Index
A measure that quantify the relationship
between the natural variation and specifications
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
Cp 
specificat ion width
process width

USL  LSL
6σ
– Cpk helps to address a possible lack of centering of the process
 USL  μ μ  LSL 
C pk  min 
,

3σ
3σ


Process Capability
LSL
USL
Spec Width
-3δ -2δ -1δ μ 1δ 2δ 3δ
Cp 
specificat ion width

USL  LSL
process width
C pk
 USL  μ μ  LSL 
 min 
,

3σ
3σ


6σ
If Cp is ≥ 1; process capable of meeting
design specs
If Cpk is ≥ 1; process capable of meeting
design specs
Process Capability Index
• Can a process or system meet it’s
requirements?
Cp 
product' s design specificat ion range
6 standard deviations
of the production

USL - LSL
system
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

 USL  μ μ  LSL 
C pk  min 
,

3σ
3σ


min = minimum of the two
 = mu or mean of the process
If C pk is less than 1, then the process is not capable or
is not centered.
Capability Indexes
• Centered Process (Cp):
Cp 
specificat ion width
process width

USL  LSL
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:
Cp 
USL  LSL
6

16 . 2  15 . 8
6 0 . 1 

0 .4
 0 . 66
0 .6
Cp < 1: process not capable of meeting design specs
• Cpk:
  LSL 
 USL  
C pk  min 
or

3
3


 16 . 2  15 . 9
15 . 9  15 . 8 

 min 
or
3 0 . 1  
 3 0 . 1 
0 .1 
 0 .3
 min 
or
  min 1 or 0 . 33   0 . 33
0 .3 
 0 .3
C pk <1, then this measure also indicates that process is not capable
Chapter 3
Project Management
Project Planning, Scheduling and
Controlling
Calculate Path Durations
P a th s
ABD EG H JK
A B D E G IJ K
AC FG H JK
A C F G IJ K
P a th d u ra tio n
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
ES=10 ES=16
ES=4
Project
Network
EF=16 EF=30
EF=10
LS=16
LS=10
E
Buffer
LS=4
LF=30
LF=16
LF=10
B(6)
A(4)
D(6)
ES=32
EF=34
LS=33
LF=35
E(14)
H(2)
G(2)
ES=35
EF=39
LS=35
LF=39
ES=39
EF=41
LS=39
LF=41
J(4)
K(2)
ES=0
ES=30
C(3)
EF=4
F(5)
EF=32
I(3)
LS=0
LS=30
ES=4
LF=4
ES=7
LF=32 ES=32
EF=7
Critical Path: the sequence of activities that
takes the longest time &
EF=12
LS=22defines the total project completion
EF=35 time
LS=25
LF=25
LS=32
LF=30
LF=35
Calculate Early Starts & Finishes
ES=10
EF=16
ES=4
EF=10
B(6)
D(6)
E(14) Latest EF H(2)
= Next ES
A(4)
G(2)
ES=0
EF=4
LS=0
LF=4
ES=32
EF=34
ES=16
EF=30
C(3)
ES=4
EF=7F=25
F(5)
ES=7
EF=12
LS=25
LF=30
J(4)
ES=30
EF=32
LS=30
LF=32
ES=35
EF=39
I(3)
ES=32
EF=35
LS=32
LF=35
ES=39
EF=41
K(2)
Calculate Late Starts & Finishes
ES=10
EF=16
LS=10
LF=16
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
ES=16
EF=30
LS=16
LF=30
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
ES=35
EF=39
LS=35
LF=39
J(4)
I(3)
ES=32
EF=35
LS=32
LF=35
ES=39
EF=41
LS=39
LF=41
K(2)
What is the critical path?
E(5)
B(3)
G(4)
A(4)
C(5)
D(2)
F(6)
1.
2.
3.
4.
ABEGH
ACEGH
ACFGH
ADFGH
H(3)
What is ES, EF for G?
E(5)
B(3)
G(4)
A(4)
C(5)
D(2)
F(6)
1.
2.
3.
4.
14, 18
15, 19
16, 20
17, 21
Latest EF
= Next ES
H(3)
What is LS, LF for C?
E(5)
B(3)
G(4)
A(4)
C(5)
D(2)
F(6)
1.
2.
3.
4.
4, 9
5, 10
3, 8
6, 11
H(3)
ES 9 LS 10
EF 14 LF 15
ES 4 LS 7
EF 7 LF 10
ES 19 LS 19
EF 22 LF 22
E(5)
B(3)
ES 15 LS 15
EF 19 LF 19
ES 4 LS 4
EF 9 LF 9
ES 0
EF 4
H(3)
G(4)
C(5)
A4
D2
ES 4 LS 7
EF 6 LF 9
F(6)
ES 9 LS 9
EF 15 LF 15
Latest EF
= Next ES
Earliest LS
= Previous LF
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
Ac tiv ity
A
B
C
D
E
F
G
H
I
J
K
L a te
E a rly
S la c k
F in is h
4
10
25
16
30
30
32
35
35
39
41
F in is h
4
10
7
16
30
12
32
34
35
39
41
(w e e k s )
0
0
18
0
0
18
0
1
0
0
0
Path Slack
Duration of Critical Path (41)
- Path Duration
= Path Slack
P ath s
A B D E G H JK
A B D E G IJK
A C F G H JK
A C F G IJK
P ath d u ratio n
40
41
22
23
Path Slack
1
0
19
18
Consider the following project information.
Activity
Activity Time
(weeks)
Immediate
Predecessor(s)
A
3
none
B
5
A
C
2
B
D
5
B
E
12
C, D
F
3
E
G
6
E
H
5
G, F
What is the critical path?
A (3)
B (5)
It is ABDEGH
C (2)
E (12)
D (5)
F (3)
G (6)
H(5)
A (3)
B (5)
C (2)
E (12)
D (5)
F (3)
G (6)
H(5)
What is ES and EF for E?
ES for E: 3+5+5=13
EF for E: 13+12 =25
What is LS and LF for C?
LF for C: = ES for E =13
LS for C: 13-2=11
What is the activity slack for C?
ES for C: 3+5 =8
Activity slack for C = LS-ES=11-8=3
EF for C: 8+2=10
If as a project manager you see that Activity F in the project above is
going to take 3 extra weeks, what should you do?
Watchful waiting
Waiting Line Models
How long you wait in line depends on a number
of Factors:
1. the number of people served before you
2. the number of servers working
3. the amount of time it takes to serve each individual
customer
Wait time is affected by the design of the waiting
line system. A waiting line system (or queuing
system) is defined by two elements:
• The customer population (people or objects to be
processed)
• The process or service system
Waiting Line System Performance Measures
• The average number of customers waiting in
queue
• The average number of customers in the system
• The average waiting time in queue
• The average time in the system
• The system utilization rate (% of time servers
are busy)
Infinite Population, Single-Server, Single Line,
Single Phase Formulae
  lambda  mean arrival rate
  mu  mean service rate
 
L


 average system utilizatio n

 
 average number of customers
in system
Infinite Population, Single-Server, Single Line,
Single Phase Formulae
L Q   L  average number of customers
W 
1
 
 average time in system
in line
including
service

W Q   W  average time spent waiting  in  line
Pn  1    
n
 probabilit y of n customers
Do not confuse L from LQ ; or W from WQ
in the system
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. Assess the performance measures in this
case.
Based on this description, we know:
– μ= 20 (exponential distribution)
– λ= 15 (Poisson distribution)
- Average System Utilization
 



15
 0 . 75 or 75 %
20
- Average Number of Students in the System
L

 

15
20  15
 3 students
- Average Number of Students Waiting in Line
L Q   L  0 . 75 3   2 . 25 students
- Average Time a Student Spends in the System
W 
1
 

1
20  15
= .2 hours or 12 minutes
- Average Time a Student Spends Waiting (Before Service)
W Q   W  0 . 75 0 . 2   0 . 15 hours
or 9 minutes
- Probability of n Students in the Line
P0  1   
P1  1   
0
1
 1  0 . 75  1  0 . 25
 1  0 . 75  0 . 75  0 . 188
P2  1   
2
 1  0 . 75  0 . 75
2
 0 . 141
P3  1   
3
 1  0 . 75  0 . 75
3
 0 . 105
P4  1   
4
 1  0 . 75  0 . 75
4
 0 . 079
Example
Consider a single-server queuing system. If the arrival
rate is 40 units per hour, and each customer takes 60
seconds on average to be served, what is the average #
of customers in the line? Answer to 2 decimal places.
λ = 40 units/ hr; μ = (60 *60)/60 = 60 units / hr
Average number of customers in the line
LQ = ρL = (λ /μ) * (λ /μ-λ) = (40/60)*(40/20) =1.33
Example
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 36
seconds per customer. What is the average number of
customers in the system? Should they add servers?
λ = 60 customers/hour
Service rate is 36 seconds for each customer,
so μ =60*60 / 36 = 100 customers / hour
The average number of customers in the system
L = λ / (μ-λ) = 60 / (100-60) =1.5, no need to add servers
Example
Consider a single-server queuing system. If the
arrival rate is 25 customers per hour and it takes 3
minutes on average to serve a customer, what is the
average waiting time in the waiting line in minutes?
λ= 25 customers/hr,
μ = 60/3 =20 customers/hr
WQ = ρW=(λ/μ)(1/μ-λ) =(25/20)(1/20-25) =-.25
Answer: the waiting time will continuously increase
Example
Consider a single-line, single-server waiting line
system. The arrival rate lambda is 80 people per
hour, and the service rate µ is 120 people per
hour. What is the probability of having 3 units in
the system?
ρ = λ/μ= 80/120 =.66
Probability Pn = (1- ρ)ρn =(1-.66) (.66^3)
=(.33)(.287) = .0947 =9.47%