Quality Management09. lecture

Statistical process control

Variability of process

• Random variation – uncontrollable, caused by chance, centered around a mean with a consistent amount of dispersion

• Non-random variation – has a systematic cause, shift in process mean

• Process stability – only random variation exist

Sampling methods

• Less expensive• Take less time• Less intrusive

• 100% sampling – during acceptance sampling or work-in-process inspection

• Random sample: equal chance to be inspected, independence among observations

• Systematic samples: according to time or sequence• Rational subgroup: logically homogeneous, if we not

separate these groups, non-random variation can biased results

Process control chart

• Tools for monitoring process variation

• Continuous variable

• Attribute – either or situation

• Weight will be variable, while number of defective items will be attributes

Steps

• Identify critical operations

• Identify critical product characteristics

• Determine whether variables or attributes

• Select the proper control charts

• Determine control limits and improve the process

• Update the limits

Control limits

• UCL – Upper Control limit

• CL – Central line

• LCL – lower Control limit

• Control limits comes from the process and are very different from specification limits.

Distribution

• Central limit theorem:

• If the samples number is high (above 30) than the mean of the samples will follow normal distribution

Hypothesis test

• H0: μ=11 cm• H1: μ≠11 cm• 95% (z=1,96) rejection limit• If σ=0,001 (n=10), than the rejection limits:• 11+1,96*0,001 and 11-1,96*0,001 • (11,00196;10,99804)• The sample mean μ=10,998 falls between the

rejection limits, we accept the null hypothesis• Then we accept that a process is in control

Errors during hypothesis test

Decision

In control Out of control

Reality In control OK Error type of one (risk of supplier)

Out of control

Error type of two (risk of customer)

OK

X mean and R control chart

• Mean chart monitor the average of the process

• Range chart monitor the dispersion of the process

• K>25, n=4 or 5

• Sample mean

• Range of sample

• n is number of observations

• Average of sample means

• Average of ranges

• k is number of samples

n

xxxx n

......21

minmax xxR

k

x....xxx

k21

k

R......RRR k21

Counting of control limits

RDUCLR 4

RDLCLR 3 RAxLCLx 2

RAxUCLx 2

A2, D3, D4 comes from factor for control limits table

Exercise

Obs1 Obs2 Obs3 Obs4

day1 6 6 5 7

day2 8 6 6 7

day3 7 6 6 6

day4 6 7 5 4

x-bar chart

024

68

1 2 3 4

day

centim

eter

Means Cl x-bar LCL x-bar UCL x-bar

Rchart

0

2

4

6

1 2 3 4

Day

Cen

tim

eter

Sample Range R-bar UCL R

X and MR (moving range) chart

• If it is not possible to draw samples• Only one or two units per day are produced• Central limit theorem doesn’t apply you make

sure that the datas normally distributed. If the distribution is not normal– Use g chart of h chart

• X – individual observation from a population 3 std dev limit is a natural variation X chart limits ate not control limits. They are natural limits.

)(2 RMExLCLx

)(2 RMExUCLx

)(4 RMDUCLMR

0MRLCL

Exercise

• The following table shows the daily trips. The trucks generally take 6,5 hours to make the daily trip. The owner want to know whether there is any other reason of the increasing delivery time, or it just depend on the traffic.

• Use X chart and MR chart to determine.

Travel Time MR

6,2 -

6,1 0,1

6,5 0,4

7,2 0,7

6,8 0,4

7,7 0,9

Solution

• Xmean=6,75• MRmean=0,5• UCLx=6,75+2,66*0,5=8,08• CLx=6,75• LCLx=6,75-2,66*0,5=5,42• UCLMR=3,268*0,5=1,634• CLMR=0,5• LCLMR=0

-0,1

0,1

0,3

0,5

0,7

0,9

1,1

1,3

1,5

1,7

1 1,5 2 2,5 3 3,5 4 4,5 5

Median chart

• If counting average takes too much time or effort

• Number of observations (n) is better to be odd number, (3,5,7)

• 20<k<25

• In sum the number of observations must reach 100

RAxLCLx 2

~~ RAxUCLx 2

~~

Example

• The table below contains observations of a process. Use median chart and determine, whether the process is in control.

Obs 1 Obs 2 Obs 3 Obs 4 Obs 5

1,1 1,2 1,4 1,5 1,6

1 1,02 1,5 1,6 1,6

1,2 1,4 1,4 1,4 1,5

1,3 1,3 1,3 1,5 1,6

Solution

• CLx=1,4

• LCLx=1,4-0,691*0,425=1,1063

• UCLx=1,4+0,691*0,425=1,693

Obs 1 Obs 2 Obs 3 Obs 4 Obs 5 Median Range

1,1 1,2 1,4 1,5 1,6 1,4 0,5

1 1,02 1,5 1,6 1,6 1,5 0,6

1,2 1,4 1,4 1,4 1,5 1,4 0,3

1,3 1,3 1,3 1,5 1,6 1,3 0,3

s and X mean chart

• Concerned about the dispersion of a process, than R chart is not sufficently precise

• Use std dev chart, when variationis small (high tech industry)

• New formula must be used for compute limits of x mean chart

• Si – the dtd dev for sample i• K number of samples• B3 and A3 factors

sBUCLs 4

sBLCLs 3 k

ss i

)(3 sAxLCLx

)(3 sAxUCLx

Example

• Determine useing s chart whether the process is in control, we have 4 samples with n=3.

  Obs1 Obs2 Obs3

Sample 2,0000 1,9995 2,0002

1 1,9998 2,0003 2,0002

2 1,9997 2,0000 2,0004

3 2,0003 1,9998 1,9997

4 2,0004 2,0001 2,0000

Sample Mean Std.dev

1 1,99990 0,000361

2 2,00010 0,000265

3 2,00003 0,000351

4 1,99993 0,000321

5 2,00017 0,000208

sum 10,00013 0,00151

• UCLs=2,568*(0,00151/5)=0,000775

• CLs=0,000302• LCLs=0• UCLx=2,000026+1,954*0,0003

02=2,00061• CLx=2,000026• LCLx=2,000026-

1,954*0,000302=1,99943

1,99985

1,99990

1,99995

2,00000

2,00005

2,00010

2,00015

2,00020

1 1,5 2 2,5 3 3,5 4 4,5 5

0

0,0001

0,0002

0,0003

0,0004

0,0005

0,0006

0,0007

0,0008

1 1,5 2 2,5 3 3,5 4 4,5 5

Process capability

• If the process is in control, than there is only non-random variation in the process. But it doesn’t mean that the products produced by the process meet the specifications or defect-free.

• Process capability refers to the ability of a process to produce a product that meet the specifications.

Specification limit

• USL – Upper specification limit

• LSL – lower specification limit

• Specification limit comes from outside, determined by engineers or administration, and not calculated from the process.

Population capability

• If there are no subgroups, calculate population capability, where

- population mean• - population process std.dev

);min{ PplPpuPpk

3

)( USL

Ppu

3

)( LSLPpl

1

)( 2

n

xxi

Capability index• 1. select critical operation• 2. select k sample of size n

– 19<k<26– n>50 (if n binomial)– 1<n<11 (measurement)

• Use control chart whether it is stable

• Compare process natural tolerance limit with specification limits

• Compute capability indexes: Cpl, Cpu, Cpk

• - computed population process mean

• - estimated process std.dev

);min{ CplCpuCpk

2

ˆd

R

ˆ3

)ˆ( USL

Cpu

ˆ3

)ˆ( LSLCpl

6LSLUSL

Cp

USL

6σ

LSL

Cp=1

Cpk=1

Exercise

• For an overhead projector, the thickness of component is specified to be between 30 and 40 millimeters. Thirty samples of components yield a grand mean ( ) of 34 millimeters with a standard deviation ( ) of 3,5. Calculate process capability index. If the process is not capable, what proportion of a product will not conform?

x

Solution

• Cpu=(40-34)/3*3,5=0,57• Cpl=(34-30)/3*3,5=0,38• Cpk=0,38• The process is not capable.• To determine the proportion of product that not

conform, we need to use normal distribution table.• Z=(LSL- )/ =(30-34)/3,5=-1,14• Z=(USL- )/ =(40-34)/3,5=1,71• 0,1271+0,0436=0,1707 17,07% will not conform

x

x

Thank you for your attention!