# Quality Management 09. lecture Statistical process control.

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Quality Management 09. 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 controlOut of control RealityIn control OKError 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

Counting of control limits A 2, D 3, D 4 comes from factor for control limits table

Exercise Obs1Obs2Obs3Obs4 day16657 day28667 day37666 day46754

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.

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 TimeMR 6,2- 6,10,1 6,50,4 7,20,7 6,80,4 7,70,9

Solution X mean =6,75 MR mean =0,5 UCL x =6,75+2,66*0,5=8,08 CL x =6,75 LCL x =6,75-2,66*0,5=5,42 UCL MR =3,268*0,5=1,634 CL MR =0,5 LCL MR =0

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 { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/13/3878034/slides/slide_20.jpg", "name": "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

Example The table below contains observations of a process. Use median chart and determine, whether the process is in control. Obs 1Obs 2Obs 3Obs 4Obs 5 1,11,21,41,51,6 11,021,51,6 1,21,4 1,5 1,3 1,51,6

Solution CL x =1,4 LCL x =1,4-0,691*0,425=1,1063 UCL x =1,4+0,691*0,425=1,693 Obs 1Obs 2Obs 3Obs 4Obs 5MedianRange 1,11,21,41,51,61,40,5 11,021,51,6 1,50,6 1,21,4 1,51,40,3 1,3 1,51,61,30,3

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

Example Determine useing s chart whether the process is in control, we have 4 samples with n=3. Obs1Obs2Obs3 Sample2,00001,99952,0002 11,99982,00032,0002 21,99972,00002,0004 32,00031,99981,9997 42,00042,00012,0000

SampleMeanStd.dev 11,999900,000361 22,000100,000265 32,000030,000351 41,999930,000321 52,000170,000208 sum10,000130,00151 UCL s =2,568*(0,00151/5)=0,00 0775 CL s =0,000302 LCL s =0 UCL x =2,000026+1,954*0,0003 02=2,00061 CL x =2,000026 LCL x =2,000026- 1,954*0,000302=1,99943

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

Capability index 1. select critical operation 2. select k sample of size n –1950 (if n binomial) –1 { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/13/3878034/slides/slide_31.jpg", "name": "Capability index 1. select critical operation 2.", "description": "select k sample of size n –1950 (if n binomial) –1

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?

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

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