ISQS 3344Quantitative 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

2. Now find , where σ is the standard deviation and n is the

size of each sample

3). Find the upper control limit (UCL) and lower control limit (LCL) using the following formulas:

k

xxxx

n...21

nx

Number of sample averages

nzxLCL

nzxUCL

Z is the number of sigma limits specified in the problem. For “3 sigma limits” use z = 3, for example.

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.

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

222.144

13.13917.15

612.174

13.13917.15

3z ;4 problem); the(from 13.1

917.153/)9.15975.15875.15(

LCL

UCL

n

x

Range or R Chart

RDLCL,RDUCL 3R4R

k

RR

k = # of sample ranges

A range chart measures the variability of the process using the averageof 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 youin a chart.

Range 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)

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

0)2333.0(0

5319.0)2333.0(28.2

28.2 ;0 so ,4

2333.03/)2.03.02.0(

43

LCL

UCL

DDn

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.

0)5.0(0

055.1)5.0(11.2

11.2 ;0 so ,5

problem) the(from 5.0

43

LCL

UCL

DDn

R

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-Charts

1). Find p-bar:

2). Compute

3). Compute UCL and LCL using the formulas:

P-Chart Computations

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

defects ofnumber totalp

n

ppp

)1(

Number of observations per sample

n

ppzpLCL

n

ppzpUCL

)1(

)1(

As with the X-Bar chart, z is the number of sigma limits specified in the problem

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.

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

0.282 UCL

0 LCL

are limits The 0. LCLset negative, is LCL Since

.1023(.064).09σzpLCL

.2823(.064).09σzpUCL

0.06420

(.09)(.91)

n

)p(1pσ

.09100

9

Inspected Total

Defectives#p

p

p

p

• “Count charts”• Used when looking at # of defects• Control limits are based on average number of

defects, c

C-Charts

Number-of-Defectives or C Chart

C-Chart Computations

1). Compute c-bar:

2). Compute

3). Compute LCL and UCL using the formulas:

takensamples ofnumber

observed defects ofnumber c

cc

czcLCL

czcUCL

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

Complaints1 3

2 2

3 3

4 1

5 3

6 3

7 2

8 1

9 3

10 1

Total 22

6.65 UCL

0 LCL

thenare limits The

0. toLCLset negative, is LCL Since

2.25.2.232.2ccLCL

6.652.232.2ccUCL

2.210

22

samples of #

complaints#c

c

c

_

z

z

• “Capability” : Can a process or system meet its requirements?

Limit"ion SpecificatLower "LSL

Limit"ion SpecificatUpper "

6

LSL - USL

system production theof deviations standard 6

rangeion specificatdesign sproduct'

USL

C p

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

Process Capability

3σ

LSLμ,

3σ

μUSLminCpk

3σ

LSLμ,

3σ

μUSLminCpk

min = “minimum of the two”

= mean of the process

A value of Cpk < 1 indicates that the process is notcentered.

Cpk

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

capable.not is process so ,1 and CpBoth

3333.0)333.0,1min(

)1(.3

8.159.15,

)1(.3

9.152.16min

667.0)1(.6

8.152.16

Cpk

Cpk

Cp

Chapter 3Project Mgt. and Waiting Line

Theory

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.

Critical Path Method (CPM)

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

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 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=30E=32

ES=7EF=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

Strategy: Find all the ES’s and EF’s first by moving left to right (start to finish).Then find LF and LS by working backward (finish to start)

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

If an item is on the critical path, there is no slack!!!!

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

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

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:– 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

%7575.020

15or

Average Number of Studentsin the System

31520

15

L

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

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!

36080

60

,1

customers 80

1hr

secs 3600

secs 45

customer 1

L

Nowhr