Ch. 9 : Managing Flow Variability Chapter 9: Managing Flow Variability Sections 9.3.4 to End Team 10...

Ch. 9 : Managing Flow Variability

Chapter 9: Managing Flow VariabilitySections 9.3.4 to End

Team 10Alex Ichiroku

Vivian Ramos

Hamid Orandi

(and…Shehzad Khan)


9.3.4 Control Charts

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


9.3.4 Control Charts … Continued

Let be the expected value of the performance

Set up a “control band” around

» UCL = Upper Control Limit

» LCL = Lower Control Limit

Calculate the standard deviation,

Decide how “tightly” we want to monitor and control the process

The smaller the value of “z”, the tighter the control



The Upper and Lower Control Bands:

LCL = - z UCL = + z

Process Control Chart:



If observed data within the control band:

Performance variability is normal

If observed data outside the control band:

Process is “out of control”

Data Misinterpretation

Type I error, : Process is “in control”, but data outside the Control Band

Type II error, : Process is “out of control”, but data inside the Control Band



Optimal Degree of Control

Depends on 2 things:

1. How much variability in the performance measure we consider acceptable

2. How frequently we monitor the process performance.

The value of “z” determines how tightly we control the process

LCL = - z UCL = + z

A small value of “z”: Narrower Control Band -- Tighter Control

Optimal frequency of monitoring is a balance between the costs and benefits




If we set ‘z’ to be too small:

We’ll end up doing unnecessary investigation

Incur additional costs

If we set ‘z’ to be too large:

We’ll accept a lot more variations as normal

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



In practice, a value of z = 3 is used

99.73% of all measurements will fall within the “normal” range




Average and Variation Control Charts

Monitor process performance by taking random samples

For each Sample:

Calculate the average value, A1, A2….AN

Calculate the variance of each sample, V1, V2….VN

Sample Averages:

Normally distributed Mean of A

Standard Deviation of A




A = /n (n = sample size)

LCL = - z/n and UCL = + z/n

Estimate by the overall average of all the sample averages, A

A = (A1+ A2+…+AN) / N (N = # of samples)

Also estimate by the standard deviation of all N x n observations, S

Take it one step further:




Our New, Improved equations for Upper and Lower Control Limits are:

LCL = A - zs/n and UCL = A + zs/n

CalculateV -- the average variance of the sample variances

V = (V1+ V2+…+VN) / N (N = # of samples)

Also calculate SV -- the standard deviation of the variances

We can do the same calculations with the Sample Variances:




The New equations for Variance Control Limits are:

LCL = V - z sV and UCL = V + z sV

If observed variations fall within this range:

Process Variability is stable

If not:

Need to investigate the cause of abnormal variations




Garage Door Example revisited…

Ex: A1 = (81 + 73 + 85 + 90 + 80) / 5 = 81.8 kg

Ex: V1 = (90 - 73) = 17 kg

Std. Dev. of Door Weights: s = 4.2 kgStd. Dev. of Sample Variances: sV = 3.5 kg




Average Weights of Garage Door Samples:




Average Weight Control Chart

74

76

78

80

82

84

86

88

90

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Days

Ave

rag

e W

t. (

Kg

)

`

UCL = 88.13

LCL = 76.87

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




Let z = 3 Sample Variances

UCL = V + z sV = 10.1 + 3 (3.5) = 20.6

LCL = V - zs sV = 10.1 – 3 (3.5) = - 0.4

Variance Control Chart

0

5

10

15

20

25

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Days

Var

ian

ce (

ran

ge)

of

Wt.

(K

g)

UCL = 20.6

LCL = 0



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


9.3.5 Cause-Effect Diagrams

Cause-Effect Diagrams

Now what?!!

Answer 5 “WHY” Questions !

Sample

Observations

Plot

Control Charts

Abnormal

Variability !!

Brainstorm Session!!


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.


9.3.5 Cause-Effect Diagrams … Continued

Fishbone Diagram


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:

Roller Settings & Garage Door Weights

0

5

10

15

20

0 1 2 3 4 5 6 7 8 9 10 11 12 13

Roller Setting (mm)

Do

or

Wei

gh

t (

Kg

)

Scatter Plot


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


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)


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


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!!!


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)



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.


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



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


Table 9.4

Table 9.4 Relationship Between Process Capability Ratio and Proportion DefectiveDefects (ppm) 10000 1000 100 10 1 2 ppbCp 0.86 1 1.3 1.47 1.63 2


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: 9.9Currently the sigma capability of door making process is S=min(85-82.5)/[(2)(4.2)] = .5952By centering the process correctly, its sigma capability increases toS=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.


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)


Table 9.5

Table 9.5 Fraction Defective and Sigma MeasureSigma S 3 4 5 6Capability Ratio Cp 1 1.33 1.667 2Defects (ppm) 66810 6210 233 3.4



Why Six-Sigma?

– See table 9.5

– Improvement 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.



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



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


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.


9.5 Process Capability Improvement

Shift the process mean Reduce the variability Both


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


Ex 9.10

MBPF garage doors (currently)

-specification range: 75 to 85 kgs

-process mean: 82.5 kgs

-proportion of output falling within specifications: .6873 The 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


Ex 9.10 Continued

Center of the specification range: (75 + 85)/2 = 80 kgs New process mean: 80 kgs If 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 deviation Z = (85 – 80)/4.2 = 1.19 Z = (75 – 80)/4.2 = -1.19


Ex 9.10 Continued

[from table A2.1 on page 319]

Z = 1.19 .8830

Z = -1.19 (1 - .8830) .1170 Prob (W =< 85) – Prob (W =< 75) =

.8830 - .1170 = .7660 By 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


9.5.2 Variability Reduction

Measured by standard deviation A higher standard deviation value means higher variability amongst outputs Lowering the standard deviation value would ultimately lead to a greater proportion

of output that falls within the specification range


9.5.2 Variability Reduction Continued

Possible causes for the variability MBPF experienced are:

-old equipment

-poorly maintained equipment

-improperly trained employees Investments to correct these problems would decrease variability however doing so

is usually time consuming and requires a lot of effort


Ex 9.11

Assume investments are made to decrease the standard deviation from 4.2 to 2.5 kgs

The 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 = 2.0 Z = (75 – 80)/2.5 = -2.0


Ex 9.11 Continued

[from table A2.1 on page 319]

Z = 2.0 .9772

Z = -2.0 (1 - .9772) .0228 Prob (W =< 85) – Prob (W =< 75) =

.9772 - .0228 = .9544 By 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


9.5.3 Effect of Process Improvement on Process Control

Changing the process mean or variability requires re-calculating the control limits This 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


9.6 Product and Process Design

Reducing the variability from product and process design

-simplification

-standardization

-mistake proofing


Simplification

Reduce the number of parts (or stages) in a product (or process)

-less chance of confusion and error Use interchangeable parts and a modular design

-simplifies materials handling and inventory control Eliminate non-value adding steps

-reduces the opportunity for making mistakes


Standardization

Use standard parts and procedures

-reduces operator discretion, ambiguity, and opportunity for making mistakes


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


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 environment

The designer must identify a combination of design parameters that protect the product from the process related and environment related factors that determine product performance


QUESTIONS

???

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Ch. 9 : Managing Flow Variability Chapter 9: Managing Flow Variability Sections 9.3.4 to End Team 10...

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