ISQS 3344
Quantitative 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
x
x 1  x 2  ... x n
Number of sample
averages
k
2. Now find
x 

, where σ is the standard deviation and n is the
n
size of each sample
3). Find the upper control limit (UCL) and lower control limit (LCL) using the
following formulas:

UCL  x  z
Z is the number of sigma
limits specified in the
problem. For “3 sigma limits”
use z = 3, for example.
n
LCL  x  z

n
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  15 . 917
  1 . 13 (from the problem);
UCL  15 . 917  3
1 . 13
n  4; z  3
 17 . 612
4
LCL  15 . 917  3
1 . 13
4
 14 . 222
Range or R Chart
UCL
R
 D4R,
LCL
k = # of sample ranges
R
 D3R
R 
R
k
A range chart measures the variability of the process using the average
of 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 you
in a chart.
Range 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
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
R  ( 0 . 2  0 . 3  0 . 2 ) / 3  0 . 2333
n  4 , so D 3  0 ; D 4  2 . 28
UCL  2 . 28 ( 0 . 2333 )  0 . 5319
LCL  0 ( 0 . 2333 )  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.
R  0 . 5 (from the problem)
n  5 , so D 3  0 ; D 4  2 . 11
UCL  2 . 11 ( 0 . 5 )  1 . 055
LCL  0 ( 0 . 5 )  0
P-Charts
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-Chart Computations
1). Find p-bar:
total number
p
total number
2). Compute

p

of defects
of units sampled("
k" times" n" )
p (1  p )
n
Number of observations
per sample
3). Compute UCL and LCL using the formulas:
UCL  p  z
p (1  p )
As with the X-Bar chart, z is the number of
sigma limits specified in the problem
n
LCL  p  z
p (1  p )
n
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.
Sampl
e
Number
of
Defectiv
e Tires
Number of
Tires in
each
Sample
Proportio
n
Defective
1
3
20
.15
2
2
20
.10
p
# Defectives

Total Inspected
p (1  p )
9
 .09
100
(.09)(.91)
3
1
20
.05
σp 
4
2
20
.10
5
1
20
.05
UCL p  p  z σ   .09  3(.064)  .282
Total
9
100
.09

n
 0.064
20
LCL p  p  z σ   .09  3(.064)   .102
Since LCL is negative,
LCL  0
UCL  0.282
set LCL  0. The limits are
C-Charts
• “Count charts”
• Used when looking at # of defects
• Control limits are based on average number of
defects, c
Number-of-Defectives
or
C
C-Chart Computations
Chart
1). Compute c-bar:
c
number
number
2). Compute
c 
of defects observed
of samples taken
c
3). Compute LCL and UCL using the formulas:
UCL  c  z c
LCL  c  z c
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
Complaints
1
3
2
2
_
3
3
c
4
1
5
3
6
3
7
2
LCL c  c  z c  2.2  3 2.2   2.25.
8
1
Since LCL is negative,
9
3
10
1
Total
22
# complaints

# of samples
22
 2.2
10
UCL c  c  z c  2.2  3 2.2  6.65
The limits
are then
LCL  0
UCL  6.65
set LCL to 0.
Process Capability
• “Capability” : Can a process or system
meet its requirements?
Cp 
product' s design specificat
6 standard
deviations
ion range
of the production
USL  " Upper Specificat
ion Limit"
LSL  " Lower Specificat
ion Limit"

USL - LSL
system
6
Cp < 1: process not capable of meeting design specs
Cp ≥ 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:

 USL  μ μ  LSL 
Cpk  min 
,

3σ
3σ


Cpk
 USL  μ μ  LSL 
Cpk  min 
,

3σ
3σ


min = “minimum of the two”

= mean of the process
A value of Cpk < 1 indicates that the process is not
centered.

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
Cp 
16 . 2  15 . 8
 0 . 667
6 (. 1)
 16 . 2  15 . 9 15 . 9  15 . 8 

Cpk  min 
,
3 (. 1)
3 (. 1)


 min( 1, 0 . 333 )  0 . 3333
Both Cp and Cpk  1, so process is not capable.
Chapter 3
Project Mgt. and Waiting Line
Theory
Critical Path Method (CPM)
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.
Example
Ac tiv ity
A
B
C
D
E
F
G
H
I
J
K
D e s c rip tio n
D e ve lo p p ro d u c t s p e c ific a tio n s
D e s ig n m a n u fa c tu rin g p ro c e s s
S o u rc e & p u rc h a s e m a te ria ls
S o u rc e & p u rc h a s e to o lin g & e q u ip m e n t
R e c e ive & in s ta ll to o lin g & e q u ip m e n t
R e c e ive m a te ria ls
P ilo t p ro d u c tio n ru n
E va lu a te p ro d u c t d e s ig n
E va lu a te p ro c e s s p e rfo rm a n c e
W rite d o c u m e n ta tio n re p o rt
T ra n s itio n to m a n u fa c tu rin g
Im m e d ia te
D u ra tio n
P re d e c e s s o r
N one
A
A
B
D
C
E & F
G
G
H & I
J
(w e e k s )
4
6
3
6
14
5
2
2
3
4
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 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
E=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=10
EF=16
B(6)
D(6)
ES=16
EF=30
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=32
EF=34
F(5)
ES=30
EF=32
LS=30
LF=32
ES=4
ES=7
EF=7
EF=12
LS=22
LS=25
Strategy: Find
all
the
ES’s
and EF’s first
LF=25
LF=30
by moving left to right (start to finish).
Then find LF and LS by working backward
(finish to start)
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
If an item is on the critical path, there is
no slack!!!!
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
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
  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

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
 



15
 0 . 75 or 75 %
20
Average Number of Students
in the System
L

 

15
20  15
3
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
 

.2 hours or 12 minutes
1
20  15
Average Time a Student
Spends Waiting (Before
Service)
W Q   W  0 . 75 0 . 2   0 . 15 hours
or 9 minutes
Too long?
After 5 minutes people
get 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!
1 customer

45 secs
3600 secs
1hr
1hr
Now ,
L

 

60
80  60

80 customers
3