1. Quantitative Review III

2. Chapter 6 and 6S Statistical Process Control

3. 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)

4. x-Bar Computations x-bar = sample average

5. “n” here equals # of observations in each sample
= standard deviation of the sample means
“n” here equals # of observations in each sample
“k” here = # of samples
= standard normal variable or the # of std. deviations
desired to use to develop the control limits

6. Assume the standard deviation of the process is given as 1
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
15.9
Observation 3
Observation 4
Sample means
15.875
15.975
18.56, 16.32
16.22, 18.56
17.62, 14.23
19.01, 12.56
18.33, 14.28

8. Inspectors want to develop process control charts to measure the weight of crates of wood. Data (in pounds) from three samples are:
Sample Crate Crate 2 Crate 3 Sample Means
What are the upper and lower control limits for this process?
Z=3
= 4.27

10. Range or R Chart 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.

11. Control Chart Factors D3 D4 2 0.00 3.27 3 2.57 4 2.28 5 2.11 6 2.00 7
0.08
1.92 8
0.14
1.86 9
0.18
1.82 10
0.22
1.78 11
0.26
1.74 12
0.28
1.72 13
0.31
1.69 14
0.33
1.67 15
0.35
1.65
Factors for R-Chart
Sample Size (n)

12. Example Sample 1 Sample 2 Sample 3 Observation 1 15.8 16.1 16.0
15.9
Observation 3
Observation 4
Sample means
15.875
15.975
Sample ranges
=0.2
=0.3

13. R-chart Computations (Use D3 & D4 Factors: Table 6-1)

14. 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.
.996, -.320
1.233, 0
1.055, 0
.788, 1.201
1.4, 0

16. (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
= average proportion of nonconforming items

17. (P) Fraction Defective Chart
n = # of observations in each sample
= standard normal variable or the # of std. deviations
desired to use to develop the control limits
K = number of samples

18. Note: Lower control limit can’t be negative.
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
.10
.05 4 5
Total 9
100
.09
Solution:
Note: Lower control limit can’t be negative.

19. Number-of-Defectives or C Chart
Used when looking at # of defects
c = # of defects
= average # of defects per sample

20. Number-of-Defectives or C Chart
= standard normal variable or the # of std. deviations
desired to use to develop the control limits

21. 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 4 5 6 7 8 9 10
Total 22
Solution:

22. Process Capability Index
Can a process or system meet it’s requirements?
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

23. 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 process

24. min = minimum of the two
= mu or mean of the process If
Is less than 1, then the process is not capable or
is not centered.

25. Capability Indexes Centered Process (Cp): Any Process (Cpk):
Cp=Cpk when process is centered

26. Example
Design specifications call for a target value of /-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

27. Computations
Cp:
Cpk:

28. Chapter 3 Project Mgt. and Waiting Line Theory

29. 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.

30. Example

31. CPM Diagram

32. Add Activity Durations

33. Identify Unique Paths

34. 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

35. 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

36. ES=16
EF=30
LS=16
LF=30
ES=32
EF=34
LS=33
LF=35
ES=10
EF=16
LS=10
LF=16
ES=4
EF=10
LS=4
LF=10
E Buffer
ES=35
EF=39
LS=35
LF=39
ES=39
EF=41
LS=39
LF=41
D(6)
E(14)
H(2)
B(6)
A(4)
K(2)
G(2)
J(4)
ES=0
EF=4
LS=0
LF=4
C(3)
ES=30
EF=32
LS=30
LF=32
F(5)
I(3)
Critical Path
ES=4
EF=7
LS=22
LF=25
ES=7
EF=12
LS=25
LF=30
ES=32
EF=35
LS=32
LF=35

37. 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

38. ES=4+6=10
EF=10
LS=4
LF=10
ES=16
EF=30
ES=32
EF=34
ES=10
EF=16
D(6)
E(14)
H(2)
B(6)
Latest EF
= Next ES
ES=35
EF=39
ES=39
EF=41
A(4)
K(2)
G(2)
J(4)
ES=0
EF=4
LS=0
LF=4
C(3)
ES=30
EF=32
LS=30
LF=32
F(5)
I(3)
Calculate Early
Starts & Finishes
ES=4
EF=7
LS=22
LF=25
ES=7
EF=12
LS=25
LF=30
ES=32
EF=35
LS=32
LF=35

39. ES=16 EF=30 LS=16 LF=30 ES=32 EF=34 LS=33 LF=35 ES=10 EF=16 LS=10
39-4=35
ES=35
EF=39
LS=35
LF=39
ES=39
EF=41
LS=39
LF=41
D(6)
E(14)
H(2)
B(6)
A(4)
Earliest LS
= Next LF
K(2)
G(2)
J(4)
ES=0
EF=4
C(3)
ES=30
EF=32
LS=30
LF=32
F(5)
I(3)
Calculate Late
Starts & Finishes
ES=4
EF=7
LS=22
LF=25
ES=7
EF=12
LS=25
LF=30
ES=32
EF=35
LS=32
LF=35

40. Activity Slack = TLS - TES = TLF - TEF
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

41. Calculate Activity Slack

42. Waiting Line Models

43. 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

44. Infinite Population, Single-Server, Single Line, Single Phase Formulae

45. Infinite Population, Single-Server, Single Line, Single Phase Formulae

46. 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)

47. Average Utilization

48. Average Number of Students in the System

49. Average Number of Students Waiting in Line

50. Average Time a Student Spends in the System
.2 hours or 12 minutes

51. Average Time a Student Spends Waiting (Before Service)
Too long?
After 5 minutes people
get anxious

52. Consider a single-line, single-server waiting line system
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?
2.25 3 4
3.25
4.25
60/80-60
= 3