Why learn Six Sigma, 4,28,15

©Product Ventures, Inc., March, 2015 1

Why Should I learn Six Sigma?(A Hypothetical Office Simulation to Demonstrate the Merits of Statistical Analysis in Continuous Improvement)

Jim McCarthy - 4/28/15


Agenda1. CEO wants customer quoting issues resolved fast2. I am selected to get it done3. Initial known facts & data4. Conventional thinking & methods5. Capability Study6. Analysis of Variance (ANOVA) & Std. Dev. Test 7. Charter: Deliverables, Team, Gantt Chart, Scope,

Dates, Report-Out/s, Process Analyses, etc.8. Design of Experiments (DOE)9. Possible solutions


How It Started

CEO receives phone calls from customers:“You are not providing quotes in a timely manner!”

Customers threaten to go to a competitor

Global Customer Base

(All Time Zones)


I Get The Job to “Fix Quoting”

All I know is: • Customers unhappy with quoting lead time• CEO wants this fixed “Fast”• No other direction is given


Initial Investigation (Discussion with Department Supervisor)

1. Quoting Department composed of six stations2. Stations serve all customers at random (not

region specific)3. Supervisor compiled data - last 1,200 quotes

(200 from each of six stations) to check lead time4. Company Lead Time policy expectation has been

72 hours – (24/7 basis) 5. Supervisor is adamant that policy is being met!


How Did Supervisor Know?1. Compiled a bar chart depicting the

performance over the last 1,200 quotes2. Average of all six stations = 68.1185 Hrs.3. “See I told you we were meeting the company

policy!” (Said the supervisor)4. We are 5.4% better than the 72 hr. policy!5. Supervisor carried out to four decimal places

just to show how precise the data was!


Station FStation EStation DStation CStation BStation A

70

60

50

40

30

20

10

0

Lead

Tim

e Q

uotin

g H

ours

(24

/7 b

asis)

68.532666.0853

68.5282 69.5961 68.870567.0983

Average Quoting Lead Times, Jan.- March, 2015 - 200 Quotes/Station


Where Do I Go From Here?

1. Customer complaints (Why?–averages look good)2. How to respectfully challenge data?3. Supervisor is taking it personal 4. Need specific data on quotes associated with

customer complaints, but data is not available.

I am Very Nervous! &

the Clock is Ticking!


Can Six Sigma Help?1. Averages are only one way to measure2. What are the variations around average/s?3. How stable/predictable is the process?4. Are all quoting stations statistically similar?5. Does system “Capability” = 72 hours or less?

A Typical “Normal Distribution”

Looks Like This


Data Shown = 24/7 Lead Times (in hours) for (Jan. – March, 2015)1,200 quotes (200 per Station) Data are “Time Ordered” per

individual Station (42.0 Hrs. is first completed quote for Station A)


How Bad could It be from Customer’s View?

• Capability study - compare to 72 hr. max. • What is the % of late quotes• Analyze all six stations as “one system” (1st)


Choose “Continuous”



Normality Test

Results FailP-value 0.005

(Anderson-Darling)

Customer RequirementsLower Spec Target Upper Spec

* * 72

% Out of spec (expected) 39.92 PPM (DPMO) (observed) 393333 PPM (DPMO) (expected) 399203

Total N 1200Mean 68.118Standard deviation 15.197

Capability statistics Pp * Ppk 0.09 Z.Bench 0.26 % Out of spec (observed) 39.33

Process Characterization

14012010080604020

Quoting Lead time (24/7 Hrs.)

USL

measures represent long-term performance, may not apply.time. Therefore, the usual interpretation, that the capabilitysources of variation that may appear over a longer period ofHowever, the data collection method used may not capture allThe capability measures use the overall standard deviation.

HistogramAre the data below the limit?

Normality Plot

Points should be close to line.

Comments

Capability Snapshot for Stations A-F (24/7 Quoting Lead Time, Jan.- Mar., 2015)Summary Report

Natural Process width (NPW) = 6 x Std. Dev = 6 x 15.197= 91.98

39.9% Ouch!

Ppk = Difference between mean & closest spec. ÷ 1/2 NPW (Long Term)

Z Bench = metric of normal distribution with same PPM level as that shown (close to

sigma level) - Closest spec. ÷ Std. dev. (ST)

Why?


10

5

01

51

02

52

03

53

02 04 06 08 001 021 04

68.53 12.84 20066.09 14.02 20068.53 12.92 20069.60 13.28 20068.87 12.41 20067.10 22.90 200

Mean StDev N

D

tnecreP

ata

SelbairaV

F noitatSE noitatSD noitatSC noitatSB noitatSA noitat

H lamroN

... ,D noitatS ,C noitatS ,B noitatS ,A noitatS fo margotsi


1501251007550250

99.9

99

95

90

80706050403020

10

5

1

0.1

68.53 12.84 200 0.357 0.45266.09 14.02 200 0.531 0.17368.53 12.92 200 0.404 0.35369.60 13.28 200 0.251 0.73868.87 12.41 200 0.242 0.76967.10 22.90 200 0.266 0.687

Mean StDev N AD P

Data

Perc

ent

68.5

66.1

68.5

69.6

68.9

67.1

50

Station AStation BStation CStation DStation EStation F

Variable

Probability Plot of Station A, Station B, Station C, Station D, ...Normal - 95% CI


What Else Do We Know? Quoting is a 1st shift, Monday-Friday operation 1,200 completed quotes (Jan.– March, 2015) 90 days & 68 hours (2.833 calendar days - lead Time) (TH)

From Little’s Law: Cycle Time (24/7) 2.833 Days/Quote x 13.33 Quotes/Day 90 Calendar days = 65 work days:


Need to Be “Data Driven”!1. What is present incoming rate from customer?2. Visual daily metrics & “stand-up” meetings (10

minutes) for: Cycle Time, Throughput, & WIP3. Can’t improve (or manage) what we don’t measure!


Let’s Dig DeeperAnalysis of Variance (ANOVA)

• 6 Sigma tool (comparison-multiple data sets)• ANOVA tries to determine if all data sets are from the same

population• Looks at both the mean (average) & variation of all data sets• Develops a ratio of the “between” data groups to “within”

each data group. • Calculates an “F” ratio & compares it to a critical value to

determine if the null hypothesis (data is statistically the same) can be rejected

• 1st – ANOVA assumes “Normal” data – need to check


ANOVA

• Between-Groups Variance (BGV)• Within Groups Variance (WGV)

• If the “F” ratio is larger than “F critical” then we reject the null hypothesis because we cannot be sure that all data sets are from the same population.

• ANOVA requires significant calculations and is best done using software.


We have 6 samples to compare so

use “One Way Anova”

A Hypothesis test is testing to

see if all data are of the same

population.Null hypothesis assumes all data

are from the same population



Which means differ?

1 Station B2 Station A3 Station C4 Station F None Identified5 Station E6 Station D

# Sample Differs from

Differences among the means are not significant (p > 0.05).

Yes No

0 0.05 0.1 > 0.5

P = 0.161

Station D

Station E

Station F

Station C

Station A

Station B

7270686664

not differ significantly.• Comparison Chart: Blue intervals indicate that the means dodifferences among the means at the 0.05 level of significance.• Test: There is not enough evidence to conclude that there are

Do the means differ?

Means Comparison ChartBlue indicates there are no significant differences. Comments

One-Way ANOVA for Stations A-F (Lead Time data J an.-Mar. 2015)Summary Report



Red dashed lines = α2 = 0.052 significance level (95% probability)

α = .05 significance level (critical value) 95% probability- 2 tailed(.025 per each tail)


100

50

0

100

50

0

100

50

0

Station A Station B

Station C Station D

Station E Station F

14012010080604020

Station A

Station B

Station C

Station D

Station E

Station F

Data in Worksheet OrderInvestigate any outliers (marked in red).

Distribution of DataCompare the location and spread.

One-Way ANOVA for Station A, Station B, Station C, Station D,...Diagnostic Report


difference of 7?What sample sizes are required to detect a

129 60.3 - 94.7%156 70.1 - 97.9%191 80.1 - 99.4%245 90.1 - 99.9%

All Samples Power

Statistics

Station A 200 68.533 12.841 (66.742, 70.323)Station B 200 66.085 14.020 (64.130, 68.040)Station C 200 68.528 12.917 (66.727, 70.329)Station D 200 69.596 13.279 (67.744, 71.448)Station E 200 68.870 12.409 (67.140, 70.601)Station F 200 67.098 22.905 (63.905, 70.292)

Sample SizeSample

Mean DeviationStandard

95% CI for MeanIndividual

difference of 7 ranges from 82.21% to 99.58%.Based on your samples and α level (0.05), the chance of detecting a

Difference

Power< 40% 60% 90% 100%

What is the chance of detecting a difference of 7?

increasing the sample sizes.Power is a function of the sample sizes and the standard deviations. If the power is not satisfactory, consider

One-Way ANOVA for Station A, Station B, Station C, Station D,...Power Report

β = 1- Power (Power is probability of being able to detect a 7 hour differenceβ = Probability of making “Type 2 error”(not being able to detect 7 hr. difference β =17.79% (Worst) to 0.42% (Best) – Power goes up with increase in sample size


Station F

Station E

Station D

Station C

Station B

Station A

28262422201816141210

P-Value 0.000

P-Value 0.000

Multiple Comparisons

Levene’s Test

Test for Equal Variances: Stations A-F - Customer Service Lead TimeMultiple comparison intervals for the standard deviation, α = 0.05

If intervals do not overlap, the corresponding stdevs are significantly different.

If “P” value is low, the null Hypothesis is

rejected(We can’t say that variances are the same)


Observations, Questions, & Next Steps

1. Something is different about Station “F”2. Although Station F has a mean of 67.098 hrs.,

it has large variability3. What do the individual station outputs look

like with respect to 72 hours ( What is the yield?)


Normality Test

Results PassP-value 0.687

(Anderson-Darling)


* * 72





14012010080604020

Quoting Lead Time (24/7 Hours)

USL



Normality Plot


Comments

Capability Snapshot - Quoting Station F (24/7 Lead Time, Jan, - Mar., 2015)Summary Report


Normality Test

Results PassP-value 0.494

(Anderson-Darling)


* * 72





112.5100.087.575.062.550.037.525.0

Quoting Lead Time (24/7 Hours)

USL



Normality Plot


Comments

Capability Snapshot - Quoting Stations A-E (24/7 Lead Time, Jan.- Mar. 2015)Summary Report


What Next?1. Station “F” has high variability & is not capable (41.53% failure rate

for 72 hr. Spec.)2. “A-F” have a 39.92% (collective) failure rate (not capable)3. “A-E” have 38.97 (collective) failure rate (not capable)4. This requires a systemic change to reduce common cause

variability & reduce “mean” across all stations.5. Write a charter, form a team, get understanding & agreement from

management , & write project plan 6. Spend some time investigating Station “F” variance in order to

better understand process. (DOE)7. Do “Data Tagging” & value stream map (I.D. waste)8. I.D. working hours, shifts, RFQ input rate, departments, etc.9. Design & review daily metrics charts: WIP, Throughput, 24/7 Cycle

Time, & Quality Issues


Touch Time

White Space

Working on Quotes 2.9% of time! : 029 x 100=2.9%

0.15 hrs. 1.0 hrs. 0.75 hrs. 0.10 hrs.

Station A-F12 hrs.

Engineering24 hrs.

Purchasing18 hrs

Station A-F 14 hrs.

Total Lead Time = 68 Hours (24/7- Average)

Work on 97% or 3% to Reduce Overall Time?

Total Touch Time = 2.0 Hrs. (Ave.)

4 Operations in Our Quoting Example[Initial guesses (no confirmed data) shown below]


Station “F” Design of Experiments

Observations About Station “F”:

1. Two, operators, not one, work in station “F”- usually separately, but sometimes simultaneously- one experienced & one new

2. They both use two separate processes “at will” – manual or computerized (no standardized pattern)

3. The station gets incoming requests for quotes via email (local) or Fax (remote)


Design of Experiments (DoE)

Controlled Experiments for Improvement


What is “Design of Experiments”

• A controlled experiment in which the experimenter is interested in both quantifying and optimizing the effect of a number of variables on one or more variables

• Input variables causing the variation = “factors”• Output variables being measured = “responses”


When Should it Be Used?

1. To identify factors which most affect responses

2. To identify those factors which act together on a particular response - these are called interactions

3. To determine the optimum settings of factors to satisfy a number of required responses


DOE Basics

• Follows the 6 sigma transfer function:

Which means that “Y” (output or response) is equal to a function of the “X” input/s (factors)


A Typical DoE Application

• Determine how to get the best gas mileage (response) based on factors (input variables) of tire pressure & speed

Y


What are Factor levels?• Factors have agreed upon settings based on experience &

/or observation• Settings typically “high” & “low”

• For gas mileage (MPG) example:Tire pressure ; low = 25 Psi

Tire pressure ; high = 35 PsiSpeed ; low = 45 MPHSpeed ; High = 65 MPH

Output (Response) = Miles Per Gallon (MPG)

Speed

Pressure

MPG


One Factor at a Time (OFAT)

• OFAT experiments change one thing at a time• This seems reasonable, but it is costly and does

not allow us to see interactions between inputsDOE Enables:1. logical Experiments2. Minimal trials3. Visible Interactions


Common Types of DOEs• Fractional Factorials = 4 to 15 variables

Screening designs when product knowledge is low – does not test all combinations

• Full Factorials = 2 to 5 Input variablesUsed when necessary to fully understand effects

of interactions – not one factor at a time (OFAT)• Response Surface Methods (RSM) = 2 to 4

input variablesTypical when response surface has significant

curvature


DOE Notation

• General notation used to designate a full factorial design is shown as:

• Where k is the number of input variables or factors

• 2 = the number of “levels” that will be used for each factor


Levels & Coding• Typically uses coding: -1, +1 for low & high

respectively

LevelFactor 1 Pressure

(PSI)

Factor 1Coding

Factor 2Speed(MPH)

Factor 2Coding

Low 25 -1 45 -1

High 35 +1 65 +1


DOE - Full Factorial With 2 Factors & 2 Levels)

Pressure (PSI)(Factor 1)

25 PSI

35 PSI

Speed (MPH)(Factor 2)

45 65

(-1,+1) (+1,+1)

(-1,-1) (+1,-1)


DOE Full Factorial Combinations

Number of factors;

In this case we have 2 factors “Pressure” &

“Speed”

Number of Levels;

In this case we have 2 levels

“high” & “low”

Combinations

©Product Ventures, Inc., March, 2015

Full Factorial Test Combinations“Standard Order” – (not randomized & no replicates)

47

Test Run No.

Factor 1 Pressure (PSI)

Factor 2Speed(MPH)

Factor 1, Factor 2

Test Coding

1 25 45 -1, -1

2 35 45 +1, -1

3 25 65 -1, +1

4 35 65 +1, +1


3 Factors & 2 LevelsLet’s add one more input variable (Fuel Type)

Type A & Type B

LevelFactor

1 Pres-sure (PSI)

Factor 1

Coding

Factor 2

Speed(MPH)

Factor 2

Coding

Factor 3

Fuel

Factor 3

Coding

Low 25 -1 45 -1 A -1

High 35 +1 65 +1 B +1


DOE - Full Factorial With 3 Factors & 2 Levels

Pressure (PSI)(Factor 1 = Y)

25 (-)

35 (+)

Speed (MPH)(Factor 2 = X)

45(-) 65 (+)

(-1,+1,-1)

(+1,+1,+1)

(-1,-1,-1) (+1,-1,-1)

Fuel (Factor 3 = Z)

A (-)

B (+)(+1,-1,+1)(-1,-1,+1)

(+1,+1,-1)

(-1,+1,+1)X,Y,Z (Respectively)


3 Factors & 2 Levels – (8 combinations)Std. Run Order

(Yates Order)

Tire Pressure25 PSI = -135 PSI = +1

Speed45 MPH= -165 MPH= +1

FuelA=-1B=+1

Response(MPG)

1 -1 -1 -1 45

2 +1 -1 -1 553 -1 +1 -1 334 +1 +1 -1 325 -1 -1 +1 486 +1 -1 +1 497 -1 +1 +1 258 +1 +1 +1 33


DOE - Full Factorial - 3 Factors & 2 Levels (MPG Responses Shown)

Pressure (PSI)(Factor 1 = Y)

25 (-)

35 (+)

Speed (MPH)(Factor 2 = X)

45(-) 65 (+)

(55 MPG)

(33 MPG)

(45 MPG) (33 MPG)

Fuel (Factor 3 = Z)

A (-)

B (+)(25 MPG)(48 MPG)

(32 MPG)

(49 MPG)


Back to Station “F” Quoting IssueFactors & levels shown below

LevelFactor 1 Operator

Factor 2Process

Type

Factor 3Input

Method

Low New Computerized Fax

High Experienced Manual Email





Minitab Populates

Columns C1-C7After Design is

Completed

Experimenter Populates the “Response” Column (C8)

After Tests are Done


Then Select “Factorial plots” to get Main Effects & Interaction Plots. Select “Cube Plot” to Get Same.


Term

AB

AC

ABC

BC

B

C

A

76543210

A OperatorB ProcessC Input

Factor Name

Standardized Effect

2.306

Pareto Chart of the Standardized Effects(response is Lead Time (24/7 Hours), α = 0.05)

Factors & Interactions

(A,C,B, & BC) are Statistically Significant


ExperiencedNew

100

90

80

70

60

ManualComputerized EmailFax

Operator

Mea

n of

Lea

d Ti

me

(24/

7 H

ours

)

Process Input

Main Effects Plot for Lead Time (24/7 Hours)-Station "F" 4/28/15Fitted Means


100

75

50

ExperiencedNew

100

75

50

ManualComputerized

Operator * Process

Operator * Input

Operator

Process * Input

Process

ComputerizedManual

Process

FaxEmail

Input

Mea

n o

f Le

ad T

ime

(24/

7 H

ours

)

Interaction Plot for Lead Time (24/7 Hours) - Station "F" 4/28/15Fitted Means


Email

Fax

Manual

Computerized

ExperiencedNew

Input

Process

Operator

64.0

39.570.0

104.0

69.0

63.5111.0

105.5

Cube Plot (fitted means) for Lead Time (24/7 Hours)


Observations & Next Steps1. Standardize “Computerized” process2. Train operators to one (computerized) process3. Do Data tagging - find out issues with the Fax4. Populate value stream map to show waste5. Show Value Added vs. Non Value Added time 6. Design Standard work for flow (all stations)7. Consider “runner” to deliver FAXes or localize8. Plot WIP, Lead Time, Rework & Throughput daily 9. Design a visual Queue to work FIFO & make it easy

to see “work in temporal jeopardy”


Lessons LearnedSIX Sigma Enabled:

1. Me to “get project started” with little data2. My frustration level to be lowered significantly3. Me to logically formulate “next steps” during project4. Revelation of a “Systemic Problem”- didn’t lead the

team down wrong path chasing “Special Causes” 5. All involved to “rally around” logical process analyses

& not emotional “gut-feel” or “finger-pointing”6. Us to see our system from the customer’s viewpoint

Documents

Why learn Six Sigma, 4,28,15