Quantitative Review III

Chapter 6 and 6SStatistical 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 Computationsx-bar = sample average

xx

xx

x

n

zxLCL

zxUCL

nk

xxxx

...21

x = standard deviation of the sample means

z = standard normal variable or the # of std. deviations desired to use to develop the control limits

nx

“n” here equals # of observations in each

sample

k

xxxx

n...21

“k” here = # of samples

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.

0

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. 18.56, 16.32

B. 16.22, 18.56

C. 17.62, 14.23

D. 19.01, 12.56

E. 18.33, 14.28

ouncesxLCL

ouncesxUCL

x

x

x

23.14565.39167.15

62.17565.39167.15

565.2

13.1

4

13.1

9167.15

3

9.15975.15875.15

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?

x

Z=3 = 4.27

6.19135.2326

4.32135.2326

135.22

27.4

4

27.4

26

3

262626

xLCL

xUCL

x

x

x

Range or R Chart

RDLCL,RDUCL 3R4R

k

RR

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.

Control 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)

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)

00233.

531.28.2233.

233.3

2.03.02.0

3

4

DRLCL

DRUCL

R

R

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.

A. .996, -.320B. 1.233, 0C. 1.055, 0D. .788, 1.201E. 1.4, 0

05.0

055.15.11.2

3

4

DRLCL

DRUCL

R

R

(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

pppp

p

zpzp

n

pp

p

LCL,UCL

)1(

,)n"times"k"sampled(" units ofnumber total

defects ofnumber total

n = # of observations in each sample

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

5 1 20 .05

Total 9 100 .09

• Solution:

0.1023(.064).09σzpLCL

.2823(.064).09σzpUCL

0.06420

(.09)(.91)

n

)p(1pσ

.09100

9

Inspected Total

Defectives#p

p

p

p

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

cccc

c

zczc

cc

LCL,UCL

,(k) samples ofnumber

observed incidents ofnumber

= standard normal variable or the # of std. deviations desired to use to develop the control limits

z

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

• Solution:

02.252.232.2ccLCL

6.652.232.2ccUCL

2.210

22

samples of #

complaints#c

c

c

_

z

z

Process Capability Index

• Can a process or system meet it’s requirements?

6

LSL - USL

system production theof deviations standard 6

rangeion specificatdesign sproduct'pC

Cp < 1: process not capable of meeting design specsCp ≥ 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

– Cpk helps to address a possible lack of centering of the process6σ

LSLUSL

width process

width ionspecificatCp

3σ

LSLμ,

3σ

μUSLminCpk

3σ

LSLμ,

3σ

μUSLminCpk

min = minimum of the two

= mu or mean of the process

If pkC Is less than 1, then the process is not capable or

is not centered.

Capability Indexes

• Centered Process (Cp):

• Any Process (Cpk):

6 widthprocess

ion widthspecificat LSLUSLC p

3

;3

minLSLUSL

C pk

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

• Cp:

• Cpk:

66.06.0

4.0

1.06

8.152.16

6

LSLUSL

C p

33.033.0or 1min3.0

1.0or

3.0

3.0min

1.03

8.159.15or

1.03

9.152.16min

3or

3min

LSLUSLC pk

Chapter 3Project 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 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

CPM Diagram

Add Activity Durations

Identify Unique Paths

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

Paths Path durationABDEGHJK 40ABDEGIJK 41ACFGHJK 22ACFGIJK 23

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=30EF=32LS=30LF=32ES=7

EF=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

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

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

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

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:– Mu = 20 (exponential distribution)– Lambda = 15 (Poisson distribution)

Average Utilization

%7575.020

15or

Average Number of Studentsin the System

studentsL 31520

15

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

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. 2.25B. 3C. 4D. 3.25E. 4.25

60/80-60= 3

L