Systems Thinking and the Theory of Constraints

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Systems Thinking and the Theory of Constraints




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Presentations text content in Systems Thinking and the Theory of Constraints

Slide1

Systems Thinking and the Theory of Constraints

A Statement for Quality Goes Here

These sides and note were prepared using

Managing

Business Flow

processes.

Anupindi

, Chopra,

Deshmukh

, Van

Mieghem

, and

Zemel.Pearson

Prentice

Hall.

F

ew of the graphs of the slides of Prentice Hall for this book, originally prepared by professor

Deshmukh

.

Slide2

Introduction ~

The Garage Door ManufacturerAccording to

the sales manager of a high-tech manufacturer of garage doors, while the company has 15% of market share, customers are not satisfiedDoor

Quality

in terms of safety,

durability, and ease

of use

High Price compared competitors’ process

Not on-time orders

Poor After Sales Service

We can not rely of subjective statements and opinions

Collect and analyze

concrete data –facts-

on performance

measures that drive customer satisfaction

I

dentify

,

correct, and

prevent sources of future problems

Slide3

9.1

Performance VariabilityAll internal and

external performance measures display vary from tome to time.

External Measurements - customer satisfaction, product rankings, customer complaints.

Internal Measurements - flow units cost, quality,

and time.

No two

cars rolling off an assembly line

have identical cost. No two customers for identical transaction spend the same time in a bank. The same meal you have had in two different

occasions in a restaurant do not taste exactly the same.

Sources of Variability

Internal: imprecise equipment, untrained workers, and lack of standard operating

procedures.

External: inconsistent raw materials,

supplier

delays,

consumer taste change, and

changing economic

conditions.

Slide4

9.1

Performance VariabilityA discrepancy between the actual and the expected performance often leads to cost↑, flow time↑, quality↓

 dissatisfied customers. P

rocesses with greater variability are judged less satisfactory than those with consistent, predictable performance.

What is the base of the customer judgment the exact unit of product or service s/he gets, not how the average product performs. Customers

perceive any variation in their product or service from what they expected as a

loss in value.

In general, a product is classified as defective if its cost, quality, availability or flow time differ significantly from their expected values, leading to dissatisfied customers.

Slide5

Quality Management Terms

Quality of Design

. How well product specifications aim to meet customer requirements (what we promise consumers ~ in terms of what the product can do).

Quality

Function Deployment (QFD

)

is

a

conceptual

framework for translating customers’ functional requirements (such as ease of operation of a door or its durability) into concrete design specifications (such as the door weight should be between 75 and 85 kg

.)

Quality of C

onformance

. How

closely the actual product conforms to the chosen design

specifications. Ex.

# defects per car, fraction of output that meets

specifications. Ex.

irline

conformance can be measured in terms of the percentage of flights delayed for more than 15 minutes OR the number of reservation errors made in a specific period of time.

Slide6

9.2

Analysis of VariabilityTo analyze and improve variability there are diagnostic tools to help us:

Monitor the actual process performance over time

Analyze variability in the process

Uncover root causes

Eliminate those causes

Prevent them from recurring in the future

Slide7

9.2.1

Check Sheetscheck Sheet is simply a tally of the types and frequency of problems with a product or a service experienced by customers

.Pareto Chart is a bar chart of frequencies

of occurrences

in non-increasing order. The

80-20 Pareto principle states that 20% of problem types account for 80% of all occurrences.

Slide8

9.2.3

HistogramsCollect data on door weight – Ex. one door, five times a day, 20 days, total of 100 door weight.

Histogram

is a bar plot that displays the frequency distribution of an observed performance characteristic.

Ex. 14% of the doors weighed about 83 kg, 8% weighed about 81 kg, and so forth.

Slide9

9.2.4

Run ChartsRun

chart is a plot of some measure of process performance monitored over time.

Slide10

9.2.5

Multi-Vari Charts

Multi-vari chart is a plot of high-average-low values of performance measurement sampled over time.

Slide11

Comparison

Pareto Chart. The importance of each item. Quality was the most important item. Quality

was then defined as finish, ease of use, and durability. Ease of use and durability which are subjective, must be translated into some thing measurable. We translate them into weight. If weight is high, it cannot operate easily, if weight is low, it will not be durable. A high quality door, based on engineering design must weight 82.5 lbs. Histogram.

Shows the tendency (mean) and the standard deviation. Ex. For door weight.

Run Chart.

Can show trend.

Multi-

Vari

Chart. Shows average and variability inside the samples and among the samples.

Slide12

Process Management

Two aspects to process management; Process planning’s

goal is to produce and deliver products that satisfy targeted customer needs.Structuring the process

Designing operating

procedures

Developing key competencies such as process capability, flexibility, capacity, and cost

efficiency

Process control’s

goal

is to ensure that actual performance conforms to the planned performance.

Tracking

deviations between the actual and the planned performance and taking corrective actions to identify and eliminate sources of these variations.

There could be various reasons behind variation in performance.

Slide13

9.3.1

The Feedback Control PrincipleProcess performance management is based on the general principle of feedback control of dynamical systems.

Applying the feedback control principle to process control.

“involves periodically monitoring the actual process performance (in terms of cost, quality, availability, and response time), comparing it to the planned levels of performance, identifying causes of the observed discrepancy between the two, and taking corrective actions to eliminate those causes.”

Slide14

Plan-Do-Check-Act (PDCA)

Process planning and process control are similar to the Plan-Do-Check-Act (PDCA) cycle. Performed

continuously to monitor and improve the process performance.Problems in Process ControlPerformance variances are determined by comparison of the current and previous period’s performances.

Decisions are based on results of this comparison.

Some variances may be due to factors beyond a worker’s control

.

According to W. Edward Deming, incentives based on factors that are beyond a worker’s control is like rewarding or punishing workers according to a lottery.

Slide15

Two categories of performance variability

Normal Variability. Is statistically predictable and includes both structural variability and stochastic

variability. Cannot be removed easily. Is not in worker’s control. Can

be removed only by process re-design, more precise equipment, skilled workers,

better material,

etc.

Abnormal

variability.

Unpredictable

and disturbs the state of statistical equilibrium of the process by changing parameters of its distribution in an unexpected way. Implies that one or more performance affecting factors may have

changed due

to

external

causes or

process tampering.

Can

be identified and removed easily therefore

is worker’s

responsibility.

Slide16

Process Control

If observed performance variability is

Normal - due to random causes - process is in controlAbnormal - due to assignable causes - process is out of control

The short run goal is:

Estimate normal stochastic variability.

Accept it as an inevitable and avoid tampering

Detect presence of abnormal variability

Identify and eliminate its sources

The long run goal is to reduce normal variability by improving process.

Slide17

9.3.3

Control Limit PolicyHow to decide whether observed variability is normal or abnormal?

Control Limit PolicyControl band - A range within which any variation in performance is interpreted as normal due to causes that cannot be identified or eliminated in short run.Variability outside this range is abnormal.

Lower limit of acceptable mileage, control band for house temperature.

Slide18

Process Control

Process control is useful to control any type of process.Application of control limit policy

Managing inventory, process capacity and flow time.Cash management - liquidate some assets if cash falls below a certain level.Stock trading - purchase a stock if and when its price drops to a specific level.

Control limit policy has usage in a wide variety of business in form of critical threshold for taking action

Slide19

9.3.4 Statistical Process Control

Statistical process control involves setting a “range of acceptable variations” in the performance of the process, around its mean.

If the observed values are within this range:

Accept the variations as “normal”

Don’t make any adjustments to the process

If

the observed values are

outside

this range

:

The process is

out of control

Need to investigate what’s causing the problems – the

assignable cause

Slide20

9.3.4 Process Control Charts

Let

 be the expected value and

be

the standard

deviation of

the

performance. Set

up

an

Upper Control Limit

(UCL) and a

Lower Control

Limit

(LCL).

LCL

= 

- z

UCL =

+ z

D

ecide

how

tightly

to monitor and control

the

process. The

smaller the

z

,

the tighter the control

Slide21

9.3.4 Process Control Charts

If observed data within

the control limits and does not show any systematic pattern

Performance variability is normal

.

Otherwise

Process is

out

of

control

Type

I

error (

 error).

Process is in control, its statistical parameters have not changed, but

data

falls outside

the

limits. Type II

error (

 error)

Process is

out of

control, its statistical parameters have

changed, but data falls

inside

the limits.

Slide22

9.3.4 Control Charts … Continued

Optimal Degree of Control

depends on 2 things:How

much variability in the performance measure we consider

acceptable

How frequently we monitor the process performance.

Optimal

frequency of monitoring

is a balance between the costs and

benefits

I

f

we set ‘z’ to be

too small

:

We’ll

end up doing unnecessary

investigation. Incur

additional

costs.

I

f

we set ‘z’ to be

too large

:

We’ll

accept a lot more variations as

normal

.

W

e

wouldn’t look for problems in the process – less costly

Slide23

9.3.4 Control Charts … Continued

In

practice, a value of

z = 3

is

used.

99.73

%

of all measurements will fall within the “normal” range

Slide24

We have collected 20 samples, each of size 5, n=5, of our variable of interest X – the door weight in our example. We have 100 pieces of data. We can simple use excel to compute the average and standard deviation of this data.

Overall average weight

Variance

Standard deviation

A higher value of the average indicates a shift in the entire distribution to the right, so that all doors produced are consistently heavier. An increase in the value of the standard deviation means a wider spread of the distribution around the mean, implying that many doors are much heavier or lighter than the overall average weight.

Slide25

X Bar Chart

Copyright © 2013 Pearson Education Inc. publishing as Prentice Hall

If we compute the average of the

random variable X, in each sample of n, in our example 5, and show it by

Slide26

X Bar Chart

Copyright © 2013 Pearson Education Inc. publishing as Prentice Hall

Therefore, if we compute the average weight door

68.26

% of all doors will weigh within 82.5 

+

(1

)(1.88),

95.44% of doors will weight within 82.5 

+

(2

)(1.88),

and 99.73% of door weights will be within 82.5 

+

(3

)(1.88), or between and 76.86 and 88.14 .

UCL

LCL

Slide27

Process Control and Improvement

LCL

UCL

Out of Control

In

Control Improved

Copyright © 2013 Pearson Education Inc. publishing as Prentice Hall

Slide28

R Chart

Copyright © 2013 Pearson Education Inc. publishing as Prentice Hall

UCL = 10.1+3(3.5) = 20.6 , LCL

=

10.1-3(3.5) = -0.4

= 0

UCL

LCL

Process Is “

In Control

” (i.e., variation is stable)

Slide29

Number of Defects (c) Chart

Discrete Quality Measurement: D = Number of “defects” (errors) per unit of work

Examples: Number of typos/page, errors/thousand transactions,

equipment breakdowns/shift, bags lost/thousand flown,

power outages/year, customer complaints/month, defects/car.......

If n = No. of opportunities for defects to occur, and

p = Probability of a defect/error occurrence in each

then D ~ Binomial (n, p) with mean

np

, variance

np

(1-p)

Poisson (m) with m = mean = variance =

np

, if

n is large (≥ 20) and p is small (≤ 0.05)

With m =

np

= average number of defects per unit,

Control limits = m + 3 √m

Copyright © 2013 Pearson Education Inc. publishing as Prentice Hall

Slide30

Performance Variation

Trend Cyclical Shift

Copyright © 2013 Pearson Education Inc. publishing as Prentice Hall

Stable Unstable

Slide31

Performance Variation

Stable

Unstable

Trend

Cyclical

Shift

Copyright © 2013 Pearson Education Inc. publishing as Prentice Hall

Slide32

Performance Variation

Stable

Unstable

Trend

Cyclical

Shift

Copyright © 2013 Pearson Education Inc. publishing as Prentice Hall

Slide33

Performance Variation

Stable

Unstable

Trend

Cyclical

Shift

Copyright © 2013 Pearson Education Inc. publishing as Prentice Hall

Slide34

Performance Variation

Stable

Unstable

Trend

Cyclical

Shift

Copyright © 2013 Pearson Education Inc. publishing as Prentice Hall

Slide35

X Bar Chart

Copyright © 2013 Pearson Education Inc. publishing as Prentice Hall

If

the door weight distribution was Normal, 68.26% of all doors will weigh within 82.5 + (1)(4.2), 95.44% of doors will weight within 82.5 + (2)(4.2), and 99.73% of door weights will be within 82.5 + (3)(4.2). This would be the distribution of the weight of each individual door.

Slide36

9.3.4 Control Charts … Continued

Average and Variation Control Charts

    

Let z = 3 Sample Averages

UCL =

A

+ zs/

n =

82.5 + 3 (4.2) /

5 =

88.13

LCL =

A

- zs/

n =

82.5 – 3 (4.2) /

5 =

76.87

Slide37

9.3.4 Control Charts … Continued

Average and Variation Control Charts

    

Let z = 3 Sample Variances

UCL =

V

+ z

s

V

=

10.1 + 3 (3.5) =

20.6

LCL =

V

- zs

s

V

=

10.1 – 3 (3.5) =

- 0.4

Slide38

9.3.4 Control Charts … Continued

Extensions

    

Continuous Variables

:

Garage Door Weights

Processing Costs

Customer Waiting Time

Use Normal distribution

Discrete Variables

:

Number of Customer Complaints

Whether a Flow Unit is Defective

Number of Defects per Flow Unit Produced

Use Binomial or Poisson distribution

Slide39

9.3.5 Cause-Effect Diagrams

Cause-Effect Diagrams

    

Now what?!!

Answer 5 “WHY” Questions !

Sample

Observations

Plot

Control Charts

Abnormal

Variability !!

Brainstorm Session!!

Slide40

9.3.5 Cause-Effect Diagrams … Continued

Why…? Why…? Why…? (+2)

    

Our famous “Garage Door” Example:

1. Why are these doors so heavy?

Because the Sheet Metal was too ‘thick’.

2. Why was the sheet metal too thick?

Because the rollers at the steel mill were set incorrectly.

3. Why were the rollers set incorrectly?

Because the supplier is not able to meet our specifications.

4. Why did we select this supplier?

Because our Project Supervisor was too busy getting the product out – didn’t have time to research other vendors.

5. Why did he get himself in this situation?

Because he gets paid by meeting the production quotas.

Slide41

9.3.5 Cause-Effect Diagrams … Continued

Fishbone Diagram

    

Slide42

9.3.6 Scatter Plots

The Thickness of the Sheet Metals

    

Change Settings on Rollers

Measure the Weight of the Garage Doors

Determine Relationship between the two

Plot the results on a graph:

Scatter Plot

Slide43

9.4 Process Capability

Ease of external product measures (door operations and durability) and internal measures (door weight) Product specification limits vs. process control limits

Individual units, NOT sample averages - must meet customer specifications.Once process is in control, then the estimates of μ (82.5kg) and σ (4.2k) are reliable. Hence we can estimate the process capabilities.

Process capabilities - the ability of the process to meet customer specifications

Three measures of process capabilities:

9.4.1 Fraction of Output within Specifications

9.4.2 Process Capability Ratios (Cpk and Cp)

9.4.3 Six-Sigma Capability

Slide44

9.4.1 Fraction of Output within Specifications

The fraction of the process output that meets customer specifications. We can compute this fraction by:

- Actual observation (see Histogram, Fig 9.3) - Using theoretical probability distribution

Ex. 9.7:

- US: 85kg; LS: 75 kg (the range of performance variation that customer is willing to accept)

See figure 9.3 Histogram: In an observation of 100 samples, the process is 74% capable of meeting customer requirements, and

26% defectives!!!

OR:

Let W (door weight): normal random variable with mean = 82.5 kg and standard deviation at 4.2 kg,

Then the proportion of door falling within the specified limits is:

Prob (75 ≤ W ≤ 85) = Prob (W ≤ 85) - Prob (W ≤ 75)

Slide45

9.4.1 Fraction of Output within Specifications cont…

Let Z = standard normal variable with μ = 0 and σ = 1, we can use the standard normal table in Appendix II to compute: AT US:

Prob (W≤ 85) in terms of:

Z = (W-μ)/ σ

As Prob [Z≤ (85-82.5)/4.2] = Prob (Z≤.5952) = .724 (see Appendix II)

(In Excel: Prob (W ≤ 85) = NORMDIST (85,82.5,4.2,True) = .724158)

AT LS:

Prob (W ≤ 75)

= Prob (Z≤ (75-82.5)/4.2) = Prob (Z ≤ -1.79) = .0367 in Appendix II

(In Excel: Prob (W ≤ 75) = NORMDIST(75,82.5,4.2,true) = .037073)

THEN:

Prob (75≤W≤85)

= .724 - .0367 = .

6873

Slide46

9.4.1 Fraction of Output within Specifications cont…

SO with normal approximation, the process is capable of producing 69% of doors within the specifications, or delivering 31% defective doors!!!

Specifications refer to INDIVIDUAL doors, not AVERAGES.

We cannot comfort customer that there is a

30% chance

that they’ll get doors that is either TOO LIGHT or TOO HEAVY!!!

Slide47

9.4.2 Process Capability Ratios (C pk and Cp)

2nd measure of process capability that is easier to compute is the process capability ratio (Cpk)

If the mean is 3σ above the LS (or below the US), there is very little chance of a product falling below LS (or above US).

So we use:

(US- μ)/3σ (.1984 as calculated later)

and (μ -LS)/3σ (.5952 as calculated later)

as measures of how well process output would fall within our specifications.

The higher the value, the more capable the process is in meeting specifications.

OR take the smaller of the two ratios [aka (US- μ)/3σ =.1984] and define a single measure of process capabilities as:

Cpk = min[(US-μ/)3σ, (μ -LS)/3σ] (.1984, as calculated later)

Slide48

9.4.2 Process Capability Ratios (C pk and Cp)

Cpk of 1+- represents a capable process Not too high (or too low)

Lower values = only better than expected quality Ex: processing cost, delivery time delay, or # of error per transaction process

If the

process is properly centered

Cpk is then either:

(US- μ)/3σ or (μ -LS)/3σ

As both are equal for a centered process.

Slide49

9.4.2 Process Capability Ratios (C pk and Cp) cont…

Therefore, for a correctly centered process, we may simply define the process capability ratio as:Cp = (US-LS)/6σ (.3968, as calculated later)

Numerator = voice of the customer / denominator = the voice of the process Recall: with normal distribution:

Most process output is 99.73% falls within +-3σ from the μ.

Consequently, 6σ is sometimes referred to as the

natural tolerance of the process.

Ex: 9.8

Cpk = min[(US- μ)/3σ , (μ -LS)/3σ ]

= min {(85-82.5)/(3)(4.2)], (82.5-75)/(3)(4.2)]}

= min {.1984, .5952}

=.1984

Slide50

9.4.2 Process Capability Ratios (C pk and Cp)

If the process is correctly centered at μ = 80kg

(between 75 and 85kg), we compute the process capability ratio as Cp = (US-LS)/6σ

= (85-75)/[(6)(4.2)] = .3968

NOTE: Cpk = .1984 (or Cp = .3968) does not mean that the process is capable of meeting customer requirements by 19.84% (or 39.68%), of the time. It’s about 69%.

Defects are counted in parts per million (ppm) or ppb, and the process is assumed to be properly centered. IN THIS CASE, If we like no more than

100 defects per million

(.01% defectives), we SHOULD HAVE the probability distribution of door weighs so closely concentrated around the mean that the standard deviation is 1.282 kg, or Cp=1.3 (see Table 9.4) Test: σ = (85-75)/(6)(1.282)] = 1.300kg

Slide51

Table 9.4

Slide52

9.4.3 Six-Sigma Capability

The 3rd process capability

Known as Sigma measure, which is computed as S = min[(US- μ /σ), (μ -LS)/σ] (= min(.5152,1.7857) = .5152 to be calculated later)

S-Sigma process

If process is correctly centered at the middle of the specifications,

S = [(US-LS)/2σ]

Ex: 9.9

Currently the sigma capability of door making process is

S=min(85-82.5)/[(2)(4.2)] = .5952

By centering the process correctly, its sigma capability increases to

S=min(85-75)/[(2)(4.2)] = 1.19

THUS, with a 3σ that is correctly centered, the US and LS are 3σ away from the mean, which corresponds to Cp=1, and 99.73% of the output will meet the specifications.

Slide53

9.4.3 Six-Sigma Capability cont…

SIMILARLY, a correctly centered six-sigma process has a standard deviation so small that the US and LS limits are 6σ from the mean each. Extraordinary high degree of precision.

Corresponds to Cp=2 or 2 defective units per billion produced!!! (see Table 9.5)

In order for door making process to be a six-sigma process, its standard deviation must be:

σ = (85-75)/(2)(6)] = .833kg

Adjusting for Mean Shifts

Allowing for a shift in the mean of +-1.5 standard deviation from the center of specifications.

Allowing for this shift, a six-sigma process amounts to producing an average of 3.4 defective units per million. (see table 9.5)

Slide54

Table 9.5

Slide55

9.4.3 Six-Sigma Capability cont…

Why Six-Sigma? See table 9.5Improvement in process capabilities from a 3-sigma to 4-sigma = 10-fold reduction in the fraction defective (66810 to 6210 defects)

While 4-sigma to 5-sigma = 30-fold improvement (6210 to 232 defects)While 5-sigma to 6-sigma = 70-fold improvement (232 to 3.4 defects, per million!!!).

Average companies deliver about 4-sigma quality, where best-in-class companies aim for six-sigma.

Slide56

9.4.3 Six-Sigma Capability cont…

Why High Standards? The overall quality of the entire product/process that requires ALL of them to work satisfactorily will be significantly lower.

Ex: If product contains 100 parts and each part is 99% reliable, the chance that the product (all its parts) will work is only (.99)100 = .366, or

36.6%!!!

Also, costs associated with each defects may be high

Expectations keep rising

Slide57

9.4.3 Six-Sigma Capability cont…

Safety capability - We may also express process capabilities in terms of the desired margin [(US-LS)-zσ] as safety capability

- It represents an allowance planned for variability in supply and/or demand - Greater process capability means less variability

- If process output is closely clustered around its mean, most of the output will fall within the specifications

- Higher capability thus means less chance of producing defectives

- Higher capability = robustness

Slide58

9.4.4 Capability and Control

So in Ex. 9.7: the production process is not performing well in terms of MEETING THE CUSTOMER SPECIFICATIONS. Only 69% meets output specifications!!! (See 9.4.1: Fraction of Output within Specifications)

 Yet in example 9.6, “the process was in control!!!”, or WITHIN US & LS LIMITS.

 

Meeting customer specifics: indicates internal stability and statistical predictability of the process performance.

 

In control (aka within LS and US range): ability to meet external customer’s requirements.

Observation of a process in control ensures that the resulting estimates of the process mean and standard deviation are reliable so that our measurement of the process capability is accurate.

Slide59

9.5 Process Capability Improvement

Shift the process meanReduce the variabilityBoth

Slide60

9.5.1 Mean Shift

Examine where the current process mean lies in comparison to the specification range (i.e. closer to the LS or the US)Alter the process to bring the process mean to the center of the specification range in order to increase the proportion of outputs that fall within specification

Slide61

Ex 9.10

MBPF garage doors (currently) -specification range: 75 to 85 kgs -process mean: 82.5 kgs

-proportion of output falling within specifications: .6873The process mean of 82.5 kgs was very close to the US of 85 kgs (i.e. too thick/heavy)To lower the process mean towards the center of the specification range the supplier could change the thickness setting on their rolling machine

Slide62

Ex 9.10 Continued

Center of the specification range: (75 + 85)/2 = 80 kgsNew process mean: 80 kgsIf the door weight (W) is a normal random variable, then the proportion of doors falling within specifications is: Prob (75 =< W =< 85)Prob (W =< 85) – Prob (W =< 75)

Z = (weight – process mean)/standard deviationZ = (85 – 80)/4.2 = 1.19Z = (75 – 80)/4.2 = -1.19

Slide63

Ex 9.10 Continued

[from table A2.1 on page 319] Z = 1.19 .8830 Z = -1.19 (1 - .8830) .1170Prob (W =< 85) – Prob (W =< 75) =

.8830 - .1170 = .7660By shifting the process mean from 82.5 kgs to 80 kgs, the proportion of garage doors that falls within specifications increases from .6873 to .7660

Slide64

9.5.2 Variability Reduction

Measured by standard deviationA higher standard deviation value means higher variability amongst outputsLowering the standard deviation value would ultimately lead to a greater proportion of output that falls within the specification range

Slide65

9.5.2 Variability Reduction Continued

Possible causes for the variability MBPF experienced are: -old equipment -poorly maintained equipment

-improperly trained employeesInvestments to correct these problems would decrease variability however doing so is usually time consuming and requires a lot of effort

Slide66

Ex 9.11

Assume investments are made to decrease the standard deviation from 4.2 to 2.5 kgsThe proportion of doors falling within specifications: Prob (75 =< W =< 85)Prob (W =< 85) – Prob (W =< 75)Z = (weight – process mean)/standard deviation

Z = (85 – 80)/2.5 = 2.0Z = (75 – 80)/2.5 = -2.0

Slide67

Ex 9.11 Continued

[from table A2.1 on page 319] Z = 2.0 .9772 Z = -2.0 (1 - .9772) .0228Prob (W =< 85) – Prob (W =< 75) =

.9772 - .0228 = .9544By shifting the standard deviation from 4.2 kgs to 2.5 kgs and the process mean from 82.5 kgs to 80 kgs, the proportion of garage doors that falls within specifications increases from .6873 to .9544

Slide68

9.5.3 Effect of Process Improvement on Process Control

Changing the process mean or variability requires re-calculating the control limitsThis is required because changing the process mean or variability will also change what is considered abnormal variability and when to look for an assignable cause

Slide69

9.6 Product and Process Design

Reducing the variability from product and process design -simplification -standardization

-mistake proofing

Slide70

Simplification

Reduce the number of parts (or stages) in a product (or process) -less chance of confusion and errorUse interchangeable parts and a modular design -simplifies materials handling and inventory control

Eliminate non-value adding steps -reduces the opportunity for making mistakes

Slide71

Standardization

Use standard parts and procedures -reduces operator discretion, ambiguity, and opportunity for making mistakes

Slide72

Mistake Proofing

Designing a product/process to eliminate the chance of human error -ex. color coding parts to make assembly easier -ex. designing parts that need to be connected with perfect symmetry or with obvious asymmetry to prevent assembly errors

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9.6.2 Robust Design

Designing the product in a way so its actual performance will not be affected by variability in the production process or the customer’s operating environmentThe designer must identify a combination of design parameters that protect the product from the process related and environment related factors that determine product performance

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QUESTIONS

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