Making A Quality Product

Making a Quality Product Product ?????? What is required to make a product?A REVIEW / INTRODUCTION OF PROBLEM SOLVING TOOLS FOR ACHIEVING PROCESS CONTROL AND WASTE REDUCTION please contact mrdrking@gmail.com for an animated PowerPoint presentation Copyright ISandR

Making a Quality Product Product RawMaterial Processing Cell The process needs: the raw materials ... the equipment to produce the product ... Is that all? Copyright ISandR

Making a Quality Product Process Control Process Control Chart Product RawMaterial Processing Cell The process also needs ... regulation or control using ... A limited amount of the raw materials ... - how much raw material can be processed at one time? A limited range on the control factors ... - temperature: how hot or cold? - time: what duration? Monitoring of materials and parameters ... Is this enough to always make a good product? Copyright ISandR

Making a Quality Product Process Control Process Control Chart Product RawMaterial Processing Cell Sure! Why not?! So start the process and make product. Copyright ISandR

Making a Quality Product Process Control Process Control Chart Product RawMaterial Processing Cell The customer expects uniformity. Does all the product behave the same and conform to the manufacturing specifications? Copyright ISandR

Making a Quality Product Process Control Process Control Chart Product RawMaterial Processing Cell Wait a second! What’s this? This product is different! The customer won’t accept this part! So this product gets trashed. Copyright ISandR

Making a Quality Product Process Control Process Control Chart Product RawMaterial Processing Cell And there is more trash, and more ... $ and more ... $Hey, this is getting expen$ive!!How can this be improved? TRASH Copyright ISandR

Making a Quality Product Process Control Process Control Chart Product RawMaterial Processing Cell Tell the operator when bad Feedback $ product is made and to watch the process better. But the operator claims all process parameters are ? $ being maintained! What else can be done? TRASH Copyright ISandR

Making a Quality Product Process Control Process Control Chart Product RawMaterial Processing Cell Feedback Data Find out what conditions produce SPC Chart very good or bad product. Control Inspection establishes data on the Charts normal output of all product. It would be easiest to monitor all output and look at what conditions existed when a deviation from normal occurs. Data is easily organized and interpreted with a Control Chart. Copyright ISandR

Making a Quality Product Process Control Process Control Chart Product RawMaterial Processing Cell Data has several uses ... Feedback Feedback Data Control Charts produce SPC Chart improvements by comparing Control typical and unusual data Charts Design of Efficient experiments produce data that results in an Experiments improved process yielding a better product . (DOE) Process Data is used to estimate the ability of the process to produce Capability conforming product Copyright ISandR

Making a Quality Product Process Control Process Control Chart Product RawMaterial Processing Cell Feedback Feedback Data Design of Experiments Engineering Analysis Control Charts SPC Chart Optimize Output Process Capability ID out-of-control events Reduce Variation Cp > 2.0 TYPES Factorial Design Cpk > 1.5 Variable (measurable) Conventional & Taguchi Attribute (yes/no, on/off) Copyright ISandR

Making a Quality Product We will look at Process ControlHow to identify an out-of-control Process Control Chart Product Raw process withProcessingMaterial statistical process Cell control (SPC). How to predict the amount of Feedback Data Feedbacknon-conforming product from the Design of Experiments Engineering Analysis Control Charts process data. SPC Chart Optimize Output Process Capability ID out-of-control events Reduce Variation Cp > 2.0 How to improve the process by Factorial Design Conventional & Taguchi Cpk > 1.5 TYPES Variable (measurable)conducting efficient experiments. Attribute (yes/no, on/off) Copyright ISandR

Making a Quality Product SPC and DOE Reduce Variation in a Process SPC ChartControl Charts - reduce special (non-random) causes. They are used by the operator as a feedback mechanism to correct problems shown by the control chart.Engineering Analysis - compares the process capability to process tolerance. Scrap is reduced when parts are processed through areas capable of holding tolerance. Design of Experiments - analyze the influence of factors that cause variation. Factors are deliberately changed in an controlled and organized fashion so that their effects can be analyzed and then optimized to reduce output variation. Copyright ISandR

Making a Quality ProductHere is an example of making and testing bullets to illustrate: control charts design of experiments engineering analysisThe test of a well made bullet is to hit the target bull’s eye • This is what the customer and manufacturer wants! Is this always produced? Copyright ISandR

Making a Quality Product Of course we can’t expect every bullet to be identical. •• • •• •So we will look the process of making a bullet and show: process control - How are factors controlled in the manufacturing? control charts - Why do weed need control charts? Show the measure of good performance. Show when the process has poor performance. engineering analysis - Predict the amount of scrap. design of experiments - Show how to improve process performance. Copyright ISandR

Making a Quality ProductSo what are the input factors to be controlled in the manufacture. Let’s assume only three factors require monitoring for process control. A heavier weight projectile is slower so it hits the target lower than a lighter and faster projectile, but too little weight and the wind affects the path. The path of a smaller diameter projectile is erratic since the projectile wobbles, but too large and it Projectile doesnt fit the barrel. More powder weight makes the projectile faster and less makes it slower. Powder No case factors influence bullet quality. Here, this was chosen for convenience, but acquiring from an approved vendor could reduce monitoring. case Copyright ISandR

Making a Quality Product We need process control to monitor the input variables The projectile has manufacturing limitations: a maximum and minimum weight a maximum and minimum diameter Projectileweight and diameter The powder has manufacturing limitations: a maximum and minimum weight Powder weight So let’s look at the process control or “rainbow” charts for several of the most recent lots of bullets. Copyright ISandR

Making a Quality Product Process control monitors the input variables Here are the “rainbow” charts for the lots 980701 through 980707 Operation Characteristic: WEIGHT of PROJECTILE DATE 980701 980702 980703 980704 980705 980706 980707 TIME MAX PROJECTILE WEIGHT MIN INITIALS OK NOTES Operation Characteristic: DIAMETER of PROJECTILE DATE 980701 980702 980703 980704 980705 980706 980707 Projectile TIME MAXweight and diameter PROJECTILE DIAMETER OK MIN INITIALS NOTES Operation Characteristic: WEIGHT of POWDER DATE 980701 980702 980703 980704 980705 980706 980707 TIME MAX Powder weight MIN POWDER WEIGHT OK INITIALS NOTES Let’s look at the testing of these lots. Copyright ISandR

Making a Quality Product Control Charts monitor the output variablesTo measure the quality of the product, a few of the bullets from lot mustbe tested; this is called a sample. A sample is used because you can’tuse the entire lot in testing or there would be nothing left to sell.The quality of the lot is determined by the spread of the hole pattern and the distance the center of the spread is to the center of the bull’s eye . • •• • • • Here is the testing of lot 980701.Let’s look closer at this pattern and put the results into a control chart. Copyright ISandR

Making a Quality ProductSo we will look the process of making a bullet and show: Let’s look at process control - How are factors controlled in the manufacturing? control charts - Why do weed need control charts? Show the measure of good performance. Show when the process has poor performance. DISCUSSION Copyright ISandR

Making a Quality Product Control Charts monitor the output variables • •• •• • Test pattern of lot 980701The diameter of the blue circle around the pattern is 7 inches indiameter. This circle represents the pattern spread and is a measureof variation.This distance from the center of the pattern to the center of the bull’seye is 6.5 inches. This is the a location measurement which comparesthe output to the desired or true value.A proper evaluation requires a variation and a location measurement.Control charts plot both location and variation output measurements. Copyright ISandR

Making a Quality Product Variable Control Charts Variable Control Chart (Average and Range) Part Number Chart No.Part Name (Product) Operation (Process) Specification LimitsOperator Machine Gage Unit of Measure Zero EqualsDATETIME 1 2 3 4 5SUMAVERAGERANGENOTES 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 Control charts plot both location and variation output measurements. On this control chart the location is called the average and the variation is called the range. Control charts also have boundaries called UCL and LCL which stands for upper and lower control limits. These boundaries represent values that a stable process should not exceed. When the control boundaries are exceeded, the operator needs look for something that may be wrong with the process. Let’s fill out the chart with the results from 980701.Any change in people, equipment, materials, methods or environment to be noted on the reverse side; the notes will help to make corrections / improvements when indicated by the control chart. Copyright ISandR

The chart is provided with previously established process control limits. First fill in the information required intoQuality Product Making a the header. Part Number Chart No. Part Name (Product) Variable Control Chart (Average and Range) Operation (Process) 123 Specification Limits 29 Big Bullet Final Test See Customer Spec Operator Machine Gage Unit of Measure Zero Equals Kim Tester #7 Tester #7 inch 0.0 DATE 01 TIME 1 2 3 4 5 SUM AVERAGE 6.5 this is the distance of the pattern from the bull’s eye - the location of the sample data RANGE 7.0 this is the diameter of the pattern - the variation of the sample data NOTES 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26UCL 8 •average 6 4LCL 2 15 range UCL 10 5 • Let’s look at the tests for the remaining lots. Any change in people, equipment, materials, methods or environment to be noted on the reverse side; the notes will help to make corrections / improvements when indicated by the control chart. Copyright ISandR

Making a Quality ProductOh good! We are just in time to see the tests of lots 980702 thruough 980707. • • •••• • •••• •• • •• • • that is 980702 to 980704 • • • •• • ••• ••• • • • • • • that is 980705 to 980707 Record the patterns of location and variation from the targets and then plot them on the control chart. Copyright ISandR

Making a Quality Product980702 980703 980704 Fill in the table in with the variation and location results. • • •••• • •••• •• • •• • lot variation location • 980702 6.5 2.56.5, 2.5 7.0, 3.0 5.0, 5.5 980703 7.0 3.0 • 980704 5.0 5.5 • • •• • •• • • 980705 7.0 1.5 • •• • • • •7.0, 1.5 7.5, 1.0 • 13.5, 1.5 980706 7.5 1.0 980707 13.5 1.5980705 980706 980707 Use this table to fill in the control chart. Copyright ISandR

Making a Quality Product Part Number Chart No. Variable Control Chart (Average and Range) 123 29 Fill in the information for lots Part Name (Product) Big Bullet Operation (Process) Final Test 980702 to 980704. Spec See Customer Specification Limits Operator Machine Gage Unit of Measure Zero Equals Kim Tester #7 Tester #7 inch 0.0 DATE 01 02 03 04 TIME 1 2 3 4 5 SUM AVERAGE 6.2 2.5 3.0 5.5 RANGE 7.0 6.5 7.0 5.0 Now plot the points on the average and range graphs NOTES 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26UCL 8 lot variation location •average 6 980702 6.5 2.5 • 980703 7.0 3.0 4 • 980704 5.0 5.5LCL 2 • 980705 7.0 1.5 980706 7.5 1.0 15 range UCL 980707 13.5 1.5 • • • • 10 5 Let’s look at this before finishing. Any change in people, equipment, materials, methods or environment to be noted on the reverse side; the notes will help to make corrections / improvements when indicated by the control chart. Copyright ISandR

Making a Quality Product Part Number Chart No. Variable Control Chart (Average and Range) 123 29 Part Name (Product) Big Bullet Operation (Process) Final Testthe location and variation All of See Customer Spec Specification Limits Operator Machine Gage Unit of Measure Zero Equals Kim Tester #7 Tester #7 inch 0.0 DATE 01 02 03 04 data looks normal so the process TIME 1 is behaving as expected. 2 3 None of the new values exceed 4 5 the dotted lines which are the SUM control limits that signal when to AVERAGE 6.2 2.5 3.0 5.5 RANGE 7.0 6.5 7.0 5.0 look for problems within the NOTES 1 2 3 4 5 6 7 8 9 10 11 12 13 process. 14 15 16 17 18 19 20 21 22 23 24 25 26UCL 8 lot variation location •average 6 980702 6.5 2.5 • 980703 7.0 3.0 4 • 980704 5.0 5.5LCL 2 • 980705 7.0 1.5 980706 7.5 1.0 15 range UCL 980707 13.5 1.5 10 5 • • • • Let’s continue. Any change in people, equipment, materials, methods or environment to be noted on the reverse side; the notes will help to make corrections / improvements when indicated by the control chart. Copyright ISandR

Making a Quality Product Part Number Chart No. Variable Control Chart (Average and Range) 123 Ok, you know there is something 29 Part Name (Product) Big Bullet Operation (Process) Final Testwith the remaining data.Spec wrong See Customer Specification Limits Operator Machine Gage Unit of Measure Zero Equals Kim Tester #7 Tester #7 inch 0.0 DATE TIME 01 02 03 04 05 06 07 Think about where the data 1 2 becomes unusual and what to do. 3 4 5 SUM AVERAGE 6.2 2.5 3.0 5.0 1.5 2.0 2.5 RANGE 7.0 7.5 7.0 5.0 7.0 7.5 13 NOTES 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26UCL 8 lot variation location •average 6 980702 6.5 2.5 • 980703 7.0 3.0 4 980704 5.0 5.5 • •LCL 2 • • 980705 7.0 1.5 • 980706 7.5 1.0 15 • range UCL 980707 13.5 1.5 10 5 • • • • • • Do you see a problem? Any change in people, equipment, materials, methods or environment to be noted on the reverse side; the notes will help to make corrections / improvements when indicated by the control chart. Copyright ISandR

Making a Quality Product Part Number Chart No. Variable Control Chart (Average and Range) 123 OK. There are some hints here! 29 Part Name (Product) Big Bullet Operation (Process) Final Test See Customer Spec Specification Limits Operator Kim Machine Tester #7 • DidTester #7 Gage you think the red location inch 0.0 Unit of Measure Zero Equals DATE 01 02 03 04 05 06 07 value was a problem? TIME 1 • Did you think the blue variation 2 3 value was a problem? 4 SUM 5 • Are both a problem? AVERAGE 6.2 2.5 3.0 5.0 1.5 2.0 2.5 • Maybe neither are a problem. RANGE 7.0 7.5 7.0 5.0 7.0 7.5 13 NOTES 1 2 3 4 5 6 7 8 9 10 11 12 13 Do both values have to exceed a 14 15 16 17 18 19 20 21 22 23 24 25 26UCL 8 limit at the same time to act? •average 6 • What do you think and why? 4 •LCL 2 • • • • 15 • range UCL 10 Take a minute to think. 5 • • • • • • Any change in people, equipment, materials, methods or environment to be noted on the reverse side; the notes will help to make corrections / improvements when indicated by the control chart. Copyright ISandR

So what do you think? Variable Control Chart (Average and Range) Part Name (Product) Big Bullet Operation (Process) Oh-Oh!? Operator DATE TIME Kim 01 02 03 04 05 06 07 Machine Tester Thinking The red dot, the blue dot,... 1 2 Both, neither,... 3 4 5 Maybe it’s a trick and SUM AVERAGE 6.2 2.5 3.0 5.0 1.5 2.0 2.5 it’s all the above. RANGE 7.0 7.5 7.0 5.0 7.0 7.5 13 NOTES 1 2 3 4 5 6 7 8 9 10 11UCL 8 •average 6 • 4 • • •LCL 2 • • 15 • range UCL 10 5 • • • • • • Any change in people, equipment, materials, methods or environment to be noted on the reverse OK. Here’s the answer and why.

Making a Quality Product Part Number Chart No. Variable Control Chart (Average and Range)Certainly you would stop and look if 123 29 Part Name (Product) Big Bullet Operation (Process) Final Test Specification Limits the location upper controlCustomer Spec See limit was Operator Machine Gage Kim Tester #7 exceeded. #7 That means Zerothe0.0 Tester Unit of Measure inch Equals hole DATE 01 02 03 04 05 06 07 pattern has shifted a large distance away from the bull’s eye and that is TIME 1 2 3 bad. 4 5 But the location has exceeded the SUM lower control limit (LCL). AVERAGE 6.2 2.5 3.0 5.0 1.5 2.0 2.5 RANGE 7.0 7.5 7.0 5.0 7.0 7.5 13 That would mean that the hole pattern NOTES 1 2 3 4 5 6 7 8 9 10 11 12 13 14 was close to the 20 15 16 17 18 19 bull’s 22 23 and that’s 21 eye 24 25 26UCL 8 good. Why tell anyone if the process is better than what is expected? •average 6 Well if the process got better perhaps we can figure out why the process is • better. So always look at what is 4 happening to the process when any • control limit is exceeded.LCL 2 • • • • A special note. The variation limit has not been exceeded at the same time as the location value. This 15 • range UCL means that this may be a rare 10 • • • • exception when a limit is exceeded 5 • • although the process is okay. Any change in people, equipment, materials, methods or environment to be noted on the reverse side; the notes will help to make corrections / improvements when indicated by the control chart. Copyright ISandR

Making a Quality Product Part Number Chart No. Variable Control Chart (Average and Range)Now if a variation and a location 123 29 Part Name (Product) Big Bullet Operation (Process) Final Test limit are exceeded at Spec control Specification Limits See Customer the Operator Machine Gage Unit of Measure Zero Equals Kim Tester #7 same Tester #7 time there is usually a real inch 0.0 DATE 01 02 03 04 05 06 07 problem. TIME 1 2 But the variation limit has been 3 exceeded by itself. Does this mean 4 5 there is a probelm? SUM AVERAGE 6.2 2.5 3.0 5.0 1.5 2.0 2.5 YES! RANGE 7.0 7.5 7.0 5.0 7.0 7.5 13 NOTES A “well behaved” process will usually 1 2 3 4 5 6 7 8 9 10 11 12 13 14 have 16 17 18variation. When variation 15 stable 19 20 21 22 23 24 25 26UCL 8 changes there is a good chance that something has definitely influenced •average 6 the process. • 4 When any control limit is exceeded, assume there is a problem and look • • • • for a source that influences theLCL 2 • variation and/or the location value. 15 • range UCL 10 5 • • • • • • How are these problems identified? Any change in people, equipment, materials, methods or environment to be noted on the reverse side; the notes will help to make corrections / improvements when indicated by the control chart. Copyright ISandR

Control Charts and ProbabilitySPC Chart It would be valuable to know when a process is producing parts that meet a desirable outcome (like high reliability or yield) and if it was not producing, why not? Control charts are used to visually show when a process is producing parts within specification and when is it not Come on snake producing parts within specification. eyes! We want to build parts that would be identical, but we know all parts are not the same. The parts vary. Probability relates the possibility of meeting and not meeting a desirable outcome. The discussion of control charts requires some understanding of probability. Copyright ISandR

Control Charts and ProbabilitySPC Chart Just as in gambling we cannot predict what will be the outcome of an event before it happens, for instance rolling a two with a pair of dice, we can know how frequently we should expect that event to occur. Come on snake eyes! When we make an item we can also predict how frequently the part should be out of some desirable range. When the frequency gets too high then we should look for the source that causes the part to vary too much so it is unacceptable. We can pictorially represent the shape of how frequently events occur. Copyright ISandR

Probability Come on snake eyes! What is the probability of rolling a “one” with one die? A = the number of ways an event can happen B = the number of way an event fails to happen A + B = the total number of all possibilities Probability is calculated by dividing A by the sum of A and B A 1 Probability = = A+B 1+5 1 way to Probability = 16.6% get a one5 ways fail to get a one What is the probability of a “head” with a coin toss? Copyright ISandR

Probability Come on snake eyes! What is the probability of a “head” on a coin toss? A = the number of ways an event can happen B = the number of way an event fails to happen A + B = the total number of all possibilities Probability is calculated by dividing A by the sum of A and B A 1 TAILS Probability = = A+B 1+11 way to fail to Probability = 50% 1 way to get a head get a headWhat is the probability of tossing two coins and both are “heads”? Copyright ISandR

Probability Come on snake eyes! What is the probability of tossing two coins and both are “heads”? What are all the HH HT combinations? TH TT HT A 1 TH TT Probability = = A+B 1+3 1 way to get3 ways to fail to get two heads two heads Probability = 25% What is the probability of tossing coins five consecutive times and getting “heads”? Copyright ISandR

Probability Come on snake eyes! What is the probability of tossing coins five consecutive times and getting “heads”? HHHHH HHHHT HHHTH HHHTT HHTHH What are all the combinations? HHTHT HHTTH HHTTT 31 ways to fail to get HTHHH HTHHT five heads HTHTH HTHTT HTTHH 1 way to get five heads HTTHT HTTTH HTTTT THHHH THHHT THHTH A 1 THHTT THTHH Probability = = THTHT THTTH0 1 1 A+B 1 + 31 THTTT TTHHH1 5 5 TTHHT2 10 10 Probability = 3.125% TTHTH TTHTT3 10 10 12 TTTHH4 5 5 TTTHT5 1 1 Note how some outcomes are more 10 TTTTH TTTTT likely and some are less likely and how this 8 influences the shape of the distribution. 6 4 What is the probability of rolling a “two” with a pair of dice? 2 0 Copyright ISandR 0 1 2 3 4 5

Probability Come on snake eyes! What is the probability of rolling a “two” with a pair of dice?? What are all the outcomes from 2 dice? 36 total combinations 1st die 2nd die 1 way to get a two 6 1,2,3,4,5,6 35 ways to fail to get a two 5 1,2,3,4,5,6 4 1,2,3,4,5,6 3 1,2,3,4,5,6 A 1 2 1,2,3,4,5,6 Probability = = 1 1,2,3,4,5,6 A+B 1 + 35 Probability = 2.78%The graphical2 1 1 113presentation 2 2 12 4 5 3 4 3 4 31 41 8 The developing shape is similar to the 8 6 5 5 51 6 4 “Normal Distribution Curve”. 6 4 7 6 6 61 8 5 5 62 2 2 9 4 4 63 0 010 3 3 64 2 3 4 5 6 7 8 9 10 11 1211 2 2 65 Copyright ISandR

Making a Quality Producto ooo o o o o o o o oThis is precise but This is accurate butnot accurate. not precise. Copyright ISandR

Probability The Characteristics of a Normal Distribution Curve When we make an item the location (mean/average) is not a zero value as shown here. There is an actual length or weight or whatever is important mean enough to be measured. All items do not have the same value; this is the variation. The shape of the curve results from the fact that most items will have a value at the peak of the curve and other items will have other values, but these will occur less frequently.0 -15 -10 -5 0 5 10 15 variationmaybe a histogram of parts being measured would help more Copyright ISandR

Probability The Characteristics of a Normal Distribution Curve location X The Normal Distribution Curve has a mean = 0 location and a variation value which describes the entire shape of the curve. Literally these are the essential variables of the mathematical equation The location value is called the mean. The variation value is called the standard deviation.0 -15 -10 -5 0 5 10 15 variation S uo standard deviation = +/- 5 Copyright ISandR

Probability Note how changes in location and variation affect the characteristics of a Normal Distribution Curve Horizontally the graphs show changes in variation The standard deviation is, from left to right, 3, 5, and 9 3 5 9 As the standard deviation gets 0 bigger, the curves gets wider and lower. 0 3 -8The change in location moves -8 the curve left 3 and right 5 5 Vertically the graphs show changes in location The mean is, from top to bottom, 0, -8, and 5 Copyright ISandR

Probability The Characteristics of a Normal Distribution Curve 100% of all possibilities are within the curve! +/- S INSIDE OUTSIDE 1 1 1 68.25% 32.75% 2 2 2 95.44% 4.56% 3 3 3 99.73% ???% This describes the possibilities of obtaining an outcome for any process that is totally random0 3 2 1 +/- S 1 2 3 axis marked in units of std. dev. Copyright ISandR

Probability The Characteristics of a Normal Distribution Curve How are the location (X) and variation (S) values determined? Gather a sample from the group to be evaluated. Measure the response (length, time, pressure, ...). Calculate the mean, X, by adding all the measured values and divide by the number of measurements added together. find X of 5 measurements: 2, 4, 5, 8, 9 (2+4+5+8+9) = 28 S = ?? 28 / 5 = 5.6 Calculate the standard deviation, S, by summing the square obtained from subtracting each measured value from the average, divide this sum by the number of measurements minus 1, and then0 take the square root of that number. X = ?? find S of same 5 measurements 2, 4, 5, 8, 9 (5.6-2)2+(5.6-4)2+(5.6-5)2+(5.6-8)2+(5.6-9)2 = 33.2 33.2 / (5-1) = 8.3 (8.3)1/2 = 2.88 Copyright ISandR

Probability The Characteristics of a Normal Distribution Curve What is variation and of what is it composed?Variation is composed of common and special sources.Common cause of variation - is the stable random pattern caused bynatural or inherent conditions of a process. Performance is predictableand is a state of statistical control. This is the type of variation handledby probability and depicted with the Normal Distribution Curve.Special cause of variation - is a source of variation that is intermittent,unpredictable unstable; sometimes called assignable causes. This istool wear, a balance missing a weight, a misread gage.

Process Capability Cpk = X - nearest limit 3sCpk = 1says the manufacturing tolerance is equal to 6sigma and is evenly centered about the processcapability Copyright ISandR

Process CapabilityYou are a car salesperson.You want to sell your customer a new SUV (Sports Utility Vehicle).Assume the width of the car represents the capability of the process (that’s whatyour selling) and the width of the garage door represents the customer’sspecifications (they are limited to what can be bought) To get the SUV through the door is easiest when the door is much wider than the car. It is easiest to meet requirements when the customer’s specification is big compared to what the process delivers. Customer Specification Process Capability

Process CapabilityComparison of Cp (Process Capability) and Cpk (where the Process Capability is k centered with respect to the specifications) Process Process Capability Customer Specification Disruption Cp < 1 Cp = 1 Cp > 1 Cpk < Cp Cpk = Cp Cpk < Cp Cpk << Cp Copyright ISandR

Process Capabilitywhen Cp > or = 1 then it it starts to get easier to get the car through the garage doorto get a calculation of process capabilityremove all assignable causes - this is done with the control chartonce all random events achieved in the processget x bar and std devcalculate Cp and Cpkcalculate process yield Copyright ISandR

Process Capability z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09This is called a “z” table. 0.0 0.5000 0.4960 0.4920 0.4880 0.4840 0.4801 0.4761 0.4721 0.4681 0.4641 0.1 0.4602 0.4562 0.4522 0.4483 0.4443 0.4404 0.4364 0.4325 0.4286 0.4247The table is used to find the 0.2 0.4207 0.4168 0.4129 0.4090 0.4052 0.4013 0.3974 0.3936 0.3897 0.3859probability that events will 0.3 0.4 0.3821 0.3446 0.3783 0.3409 0.3745 0.3372 0.3707 0.3336 0.3669 0.3300 0.3632 0.3264 0.3594 0.3228 0.3557 0.3192 0.3520 0.3156 0.3483 0.3121occur. 0.5 0.3085 0.3050 0.3015 0.2981 0.2946 0.2912 0.2877 0.2843 0.2810 0.2776 0.6 0.2743 0.2709 0.2676 0.2643 0.2611 0.2578 0.2546 0.2514 0.2483 0.2451 0.7 0.2420 0.2389 0.2358 0.2327 0.2296 0.2266 0.2236 0.2206 0.2177 0.2148In the next slide we will look 0.8 0.2119 0.2090 0.2061 0.2033 0.2005 0.1977 0.1949 0.1922 0.1894 0.1867 0.9 0.1841 0.1814 0.1788 0.1762 0.1736 0.1711 0.1685 0.1660 0.1635 0.1611up 1.81 because we want to 1.0 0.1587 0.1562 0.1539 0.1515 0.1492 0.1469 0.1446 0.1423 0.1401 0.1379know what is the possibility 1.1 0.1357 0.1335 0.1314 0.1292 0.1271 0.1251 0.1230 0.1210 0.1190 0.1170 1.2 0.1151 0.1131 0.1112 0.1093 0.1075 0.1056 0.1038 0.1020 0.1003 0.0985of an event occurring 1.81 1.3 0.0968 0.0951 0.0934 0.0918 0.0901 0.0885 0.0869 0.0853 0.0838 0.0823 1.4 0.0808 0.0793 0.0778 0.0764 0.0749 0.0735 0.0721 0.0708 0.0694 0.0681standard deviations away 1.5 0.0668 0.0655 0.0643 0.0630 0.0618 0.0606 0.0594 0.0582 0.0571 0.0559from the mean. 1.6 1.7 0.0548 0.0446 0.0537 0.0436 0.0526 0.0427 0.0516 0.0418 0.0505 0.0409 0.0495 0.0401 0.0485 0.0392 0.0475 0.0384 0.0465 0.0375 0.0455 0.0367 1.8 0.0359 0.0351 0.0344 0.0336 0.0329 0.0329 0.0314 0.0307 0.0301 0.0294This illustrates how to look 1.9 0.0287 0.0281 0.0274 0.0268 0.0262 0.0256 0.0250 0.0244 0.0239 0.0233 2.0 0.0228 0.0222 0.0217 0.0212 0.0207 0.0202 0.0197 0.0192 0.0188 0.0183up 1.81 and see that it 2.1 0.0179 0.0174 0.0170 0.0166 0.0162 0.0158 0.0154 0.0150 0.0146 0.0143 2.2 0.0139 0.0136 0.0132 0.0129 0.0125 0.0122 0.0119 0.0116 0.0113 0.0110represents 0.0351 or 3.51% 2.3 0.0107 0.0104 0.0102 0.0099 0.0096 0.0094 0.0091 0.0089 0.0087 0.0084 2.4 0.0082 0.0080 0.0078 0.0075 0.0073 0.0071 0.0069 0.0068 0.0066 0.0064probability an event will 2.5 0.0062 0.0060 0.0059 0.0057 0.0055 0.0054 0.0052 0.0051 0.0049 0.0048occur. 2.6 0.0047 0.0045 0.0044 0.0043 0.0041 0.0040 0.0039 0.0038 0.0037 0.0036 2.7 0.0035 0.0034 0.0033 0.0032 0.0031 0.0030 0.0029 0.0028 0.0027 0.0026 2.8 0.0026 0.0025 0.0024 0.0023 0.0023 0.0022 0.0021 0.0021 0.0020 0.0019 2.9 0.0019 0.0018 0.0018 0.0017 0.0016 0.0013 0.0015 0.0015 0.0014 0.0014Careful when using these; the table 3.0 0.0013 0.0013 0.0013 0.0012 0.0012 0.0011 0.0011 0.0011 0.0010 0.0010can be single or double tailed. This 3.1 0.0010 0.0009 0.0009 0.0009 0.0008 0.0008 0.0008 0.0008 0.0007 0.0007one is single tailed; the probability is 3.2 0.0007 0.0007 0.0006 0.0006 0.0006 0.0006 0.0006 0.0005 0.0005 0.0005 for one tail. 3.3 0.0005 0.0005 0.0005 0.0004 0.0004 0.0004 0.0004 0.0004 0.0004 0.0003 3.4 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0002 Copyright ISandR

Process Capability Using the Z Table (Using a single tailed table)Z tables are used to determine the percent probability of an event in thetail of a distribution (a variation in an input or an output variable).This is a look up table for the % probability between two events, themean (x bar) and another event, the distance between them given instandard deviation units. 1.81 S = 0.0351 Normalized Gaussian120 or 3.51% of all100 events within one tail80 at 1.81 standard6040 deviations units and20 beyond 0 -4 -3.6 -3.2 -2.8 -2.4 -2 -1.6 -1.2 -0.8 -0.4 0 0.4 0.8 1.2 1.6 2 2.4 2.8 3.2 3.6 4 std dev units Remember that these predictions work if the distribution is Normal / Gaussian. If the data is not Normal then use control charts to find the assignable causes.

X = 1.7 Process CapabilityS = 0.010 Calculate the out-of-spec parts for this process.UCL = 1.715 Calculate the control limits in standard deviation unitsLCL = 1.670 UCL = (1.715 - 1.7) / 0.010 = (0.015) / 0.010 = 1.50 LCL = (1.670 - 1.7) / 0.010 = (0.030) / 0.010 = 3.00 look up the “z” fraction beyond the points 1.50 and 3.00 Z1.50 = 0.0668 and Z3.00 = 0.00135 or add together and make it a percent: 6.815% out-of-spec Calculate the UCL & LCL in z 0.0 0.00 0.5000 0.01 0.4960 units found in a standard 0.1 0.2 0.3 0.4602 0.4207 0.3821 0.4562 0.4168 0.3783 Normal Distribution table 0.4 0.5 0.6 0.3446 0.3085 0.2743 0.3409 0.3050 0.2709 0.7 0.2420 0.2389 1.4 0.0808 0.8 0.2119 0.2090 0.9 0.1841 0.1814 the second curve is 1.5 0.0668 0.0 1.0 0.1587 0.1562 1.6 0.0548 0 1.1 0.1357 0.1335 centered on zero by 1.2 1.3 1.4 0.1151 0.0968 0.0808 0.1131 0.0951 0.0793 subtracting the 2.9 0.0019 1.5 1.6 1.7 0.0668 0.0548 0.0446 0.0655 0.0537 0.0436 3.0 0.0013 0. average value 3.1 0.0010 1.8 1.9 0.0359 0.0287 0.0351 0.0281 2.0 0.0228 0.0222 2.1 0.0179 0.0174 2.2 0.0139 0.0136 2.3 0.0107 0.0104 the third curve has 2.4 2.5 0.0082 0.0062 0.0080 0.0060 2.6 0.0047 0.0045 the variation scaled 2.7 2.8 0.0035 0.0026 0.0034 0.0025 2.9 0.0019 0.0018 in whole units of 3.0 3.1 0.0013 0.0010 0.0013 0.0009 3.2 0.0007 0.0007 standard deviation 3.3 3.4 0.0005 0.0003 0.0005 0.0003 Copyright ISandR

Process CapabilityMeasurement adds variationAdjust machines to get x bar in the center of UCL and LCL so Cpkbecomes as large as possibleWhat happen when Cpk produces yields of 0.98, 0.95, and 0.92? (0.90) * (0.95) * (0.92) = 0.7866When variation improves, get smaller, then yields improve. Copyright ISandR

Process CapabilityAssume a product tolerance of 1.00 +/- 0.01 1.00 x = 1.003 Cp = 0.02 / 0.0009486 = 0.35 1.02 s = 0.009486 Cpk = (1.10 - 1.003) / (3 * 0.009486) 1.00 LCL = 0.99 = 0.007 / 0.028458 0.99 UCL = 1.01 = 0.245 1.01 Z statistics 0.99 (1.003 - 0.99) / 0.009486 = 1.37 1.00 1.37 = 0.0853 or 8.53% 1.01 1.00 (1.003 - 1.01) / 0.009486 = 0.7379 0 0.7379 (about .74) = .2296 or 22.96% 1.01 total of 31.46% Out of Tolerance Copyright ISandR

Process CapabilityAssume a product tolerance of 2.5 + / - 0.05 2.49 x = 2.509 Cp = 0.10 / (6 * 0.02331) = 0.71 2.50 s = 0.02331 Cpk = (2.509 - 2.55) / (3 * 0.02331) 2.54 LCL = 2.45 = 0.041 / 0.06993 2.50 UCL = 2.55 = 0.586 2.47 Z statistics 2.49 (2.509 - 2.55) / 0.02331= 1.76 2.51 1.76 = 0.0392 or 3.92% 2.52 2.54 (2.509 - 2.45) / 0.02331= 2.53 0 2.53 2.53 = 0.0057 or 0.57% total of 3.44% Out of Tolerance Copyright ISandR

Control Charts All Data Yes / No, Good / Bad, Pass / Fail Measurable Attribute Variable Defects Defectives X bar & R individual & Unlimited limited moving x bar C u p np mixed sample short run size production fixed variable variable fixedsample sample sample sample size size size size Best to use variable Copyright ISandR

Control ChartsP CHARTWhen variable data cannot be obtainedWhen charting fraction rejected as non-conformingWhen screening multiple characteristics for potential control chartsWhen tracking the quality level of a process before (how? By counting the number of defective items from asample and then plotting the percent defective)Conditionsto be of help: there should be some rejects in each observed samplethe higher the quality level, the larger the sample size needs to be, since needs rejects. For example, 20% of aproduct is rejectable......................................................................................................needed. However, a sample of 1000 will give a ......................................................................................sample if 0.1% of the product is rejectableUCL - pbar + ( 3(pbar (1-pbar)/n)1/2 .... LCL want a normal dist UCL 7 LCL are calc calc pbar on n = 20 ( also based on a small LCL and UCL) sample size; LCL 0 usually = 0; if LCL = -3s then part is bad < 0.0135% Copyright ISandR

Variable Control Charts Part Number Chart No. Variable Control Chart (Average and Range)Part Name (Product) Operation (Process) Specification LimitsOperator Machine Gage Unit of Measure Zero EqualsDATETIME 1 2 3 4 5SUMAVERAGERANGENOTES 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26Any change in people, equipment, materials, methods or environment to be noted on the reverse side; the notes will help to make corrections / improvements when indicated by the control chart. Copyright ISandR

Variable Control Charts Worksheet Fac t o rs fo r Co n t ro l lim it sRANGES CHART n d2 D3 D. A2 A2Subgroups Included: ,________________________ 2 1 .1 2 8 0 3 .2 6 7 1 .8 8 1 .8 8Sum of Ranges = S = 3 1 .6 9 3 0 2 .5 7 5 1 .0 2 3 1 .1 8 7Number of Subgroups = K = 4 2 .0 5 9 0 2 .2 8 2 0 .7 2 9 0 .7 9 6Subgroup Size = n = 5 2 .3 2 6 0 2 .1 1 5 0 .5 7 7 0 .6 9 1Average Range = Rbar = S/K = ________ 6 2 .5 3 4 0 2 .0 0 4 0 .4 8 3 0 .5 4 8D3 factor = 7 2 .7 0 4 0 .0 7 6 1 .9 2 4 0 .4 1 9 0 .5 0 8 8 2 .8 4 7 0 .1 3 6 1 .8 6 4 0 .3 7 3 0 .4 3 3 9 2 .9 7 0 .1 8 4 1 .8 1 6 0 .3 3 7 0 .4 1 2LCLR = D3Rbar = ( )( ) = ________ 10 3 .0 7 8 0 .2 2 3 1 .7 7 7 0 .3 0 8 0 .3 6 2D4 factor =UCLR = D4Rbar = ( )( ) = ________AVERAGES CHARTSubgroups Included: ,________________________Sum of Averages = S =Number of Subgroups = K =Grand Average = X = S /K= ________A2 factor =UCLX= X + A2R = +( )( ) = ________LCLX= X - A2R = -( )( ) = ________ Copyright ISandR

Control Charts X bar and R Chartworks for even for non-normal distribution because of central limittheorem. The examples are from the Boeing Manual for Supplier’s. Xbar Copyright ISandR

Pareto Charts Problem Solving looking for largest contributor; 20% 0f items cause 80% of the problem 100% Other Copyright ISandR

Problem Solving Fishbone diagrams - to brainstorm process improvement Machine Operator PM Shift Temp Effect Season Storage PreparationWeather Raw Materials Copyright ISandR

Problem Solving Fishbone diagrams - to brainstorm process improvementthis is a way of analyzing problem is also known as Ishikowa Diagramthe effect is usually negative - a problemthe problem should be specific and clearly statedthe ideas are generated by using brainstormingremind the group to ask themselves “what would cause the problem?”keep attention on the effect of the problemthe goal is to find as many sources for variation as possible that cause the problem Measurement Machine Operator s PM Shift Temp Effect Season Storage Preparation Methods Environment Raw Materials Copyright ISandR

Problem SolvingRANK CAUSElet each person rank the problems, tally all votes, rank causesSCATTER PLOTSCV - coefficient of Variation, of +1 and -1 for strong proportional andinverse proportional Copyright ISandR

Process CapabilityExample: If a box of capacitors has an mean (xbar) = 100 uF andthe standard deviation (s) = 2 uF, what is the probability whenpulling out 1 capacitor that ... • 1 the cap is greater than 106 uF; (106-100)2 = 3; z(3) = 0.0013; 1.3% • 98 uF < X < 102 uF; (100-98)/2 and (102-100)/2; z is 2(.1587); 100 - 31.74 = 68.26% • 100 uF < X < 102 uF; (100-100)/2 and (102-100)/2; z is .50 + .1587 or 100-65.87 =34.13% • 102 uF < X < 104 uF; (102-100)/2 and (104-100)/2; z is .1587 - .0228 or 13.59% • <95 uF; (100-95)/2 = 2.5; z is 0.0062; .62% 0 Copyright ISandR

Process Capability 2 .5 0 0 0 2 .5 0 1 0 These parts are tolerance as 2.500” +/- 0.005”. 2 .4 9 9 0 2 .5 0 4 0 What is Cp, Cpk and how many parts will be made out-of-spec? 2 .5 0 2 0 2 .4 9 8 0 2 .5 0 1 0 2 .5 0 0 0 2 .5 0 5 0 2 .5 0 4 0 2 .5 0 1 0 2 .5 0 0 0 2 .5 0 3 0 2 .5 0 1 0 2 .5 0 3 0 2 .5 0 2 0 2 .5 0 2 0 2 .5 0 3 0 2 .4 9 9 0 2 .5 0 1 5 xbar0 .00 1 8 9 6 std dev Copyright ISandR

Control Charts Construct a Control Chart for Individual and Moving Average Specification 10.000” +/- 0.010” Individual and Moving Range Part Number Chart No. ind ivid |MR| Variable Control Chart (Average and Range) Part Name (Product) Operation (Process) Specification Limits 1 1 0 .0 0 0 0 Operator Machine Gage Unit of Measure Zero Equals 2 9 .9 9 3 0 0 .0 0 7 0 3 9 .9 9 9 0 0 .0 0 6 0 DATE TIME 4 1 0 .0 0 7 0 0 .0 0 8 0 1 5 1 0 .0 0 9 0 0 .0 0 2 0 2 3 6 1 0 .0 0 1 0 0 .0 0 8 0 4 5 7 1 0 .0 0 0 0 0 .0 0 1 0 SUM 8 1 0 .0 1 1 0 0 .0 1 1 0 AVERAGE RANGE 9 1 0 .0 0 0 0 0 .0 1 1 0 NOTES 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 10 1 0 .0 0 3 0 0 .0 0 3 0 11 9 .9 9 7 0 0 .0 0 6 0 12 9 .9 9 8 0 0 .0 0 1 0 13 9 .9 9 8 0 0 .0 0 0 0 14 1 0 .0 0 2 0 0 .0 0 4 0 15 9 .9 8 9 0 0 .0 1 3 0 16 1 0 .0 0 1 0 0 .0 1 2 0 17 9 .9 9 9 0 0 .0 0 2 0 18 1 0 .0 0 4 0 0 .0 0 5 0 19 1 0 .0 1 0 0 0 .0 0 6 0 20 9 .9 9 0 0 0 .0 2 0 0 a v g 1 0 .0 0 0 6 0 .0 0 6 6s t d d e v 0 .0 0 5 9 UCL 1 0 .0 1 8 2 LCL 9 .9 8 2 9 Any change in people, equipment, materials, methods or environment to be noted on the reverse side; the notes will help to make corrections / improvements when indicated by the control chart.UCL = Xbar + (2.66)(MRbar)LCL = Xbar - (2.66)(MRbar) Copyright ISandR

Gage Repeatability and Reproducibility (Gage R&R)• Gages have an error distribution• R&R are the two different ways of characterizing gage measurements• The total variation is a function of the parts and the gage.• Look at how much of the variation is due to the gage variation – Rule of thumb - gage tolerance is a tenth of the minimum tolerance unit. Copyright ISandR

Gage Repeatability and Reproducibility (Gage R&R)• Repeatability of a gage - one person measures one part with the same instrument Accuracy “true reading” Repeatability NIST traceable• Reproducibility of a gage - different operators measure one part with the same instrument B C SB = sC - sA A ?????????? Reproducibility Copyright ISandR

Design of Experiments Objectives of Design of Experiments (DOE)DOE is a systematic approach to correlate the response betweenprocess inputs and outputs (independent and dependent variables). tool pressure yield PROCESS temperature tolerance The objective is to optimize for highest yield and best tolerance.A good experiment doe not change only one variable since variables can interact (i.e.. temperature and wind or gasoline and air).Consider the interaction of temperature and wind velocity in the wind chill factor; more wind makes it feel colder and an engine requires both gasoline and air.The experiment that changes variables one-at-a-time (OAAT) requires more experiments and cannot evaluate interactions; it is inefficient at best. Copyright ISandR

Design of Experiments SPC vs. DOESPC - find special effects Control Charts: capability studies, sampling frequency, Cp, Cpk Passive measure of the performance of a process and alerts when the process is out-of-control Eliminate process variation by eliminating assignable or special causesDOE - reduces the magnitude of randomness Balanced orthogonal arrays, ANOVA, main effects, interacting variables Actively manipulate factors in order to measure their effects on the process Copyright ISandR

Design of Experiments A Simple ExperimentPerform an experiment to determine if temperature affects yield.Change temperature and follow the percent yield.T1 has yields of 82 88 86 93 89T2 has yields of 89 98 88 91 92Find mean of each of the yields at each temperaturePlot mean and range then connect the means Have to be careful about the conclusions. There could still be another factor not held constant that could influence the variation. Duplicate the experiment to increase confidence in the result. T1 T2 Copyright ISandR

Design of Experiments A Two Factor ExperimentIf you want to know the affects from two INPUT RESULT P T Yield factors (i.e. temperature and pressure), P1 T1then vary both T and P and record the P1 T2 results of the dependent variable (i.e. yield). P2 T1 Don’t just change one factor at a time. P2 T2Run an experiment. Change the pressure between 50 and 100 psi andthe temperature between 25 and 75 degrees F; record the yield results. INPUT RESULT P T Yield 50 25 94 50 75 92 100 25 96 100 75 88Calculate the mean yield for each factor 96 94 T1 = 95; T2 = 91; P1 = 93; P2 = 93 92Plot the response. From the main effects, 90 pressure has little influence on the yield, 88while temperature has a larger effect on P1 P2 T1 T2the yield.This experiment evaluates all possible combinations, but one cannotalways run full factorial experiments, so use designed experiments. Copyright ISandR

Design of ExperimentsThis experiment looks at the affects of temperature and pressure on the yield of a process.Determine the number of levels for each variable (we will choose only two).Determine all combinations of the variables. Two variables at two levels creates four combinations Run the experiment in duplicate to confirm the results INPUT OUTPUT Temp Press Result1 Result2 T1 P1 90 92 T1 P2 84 88 T2 P1 94 96 T2 P2 88 90Average the results of the runs at each variable level 96 Average at T1 = (90 + 92 + 84 + 88) / 4 = 88.5 94 92 Average at T2 = (94 + 96 + 88 + 90) / 4 = 92 90 Average at P1 = (90 + 92 + 94 + 96) / 4 = 93 88 Average at P2 = (84 + 88 + 88 + 90) / 4 = 87.5 86 P1 P2 T1 T2This is a main effects experiment using a full factorial orthogonal balanced array Copyright ISandR

Design of Experiments Three Factor Full Factorial Experiment INPUT OUTPUT Factor A Factor B Factor C Result 1 Result 2 A1 B1 C1 A1 B1 C2This is a three factor experiment (A, B, C) A1 B2 C1 A1 B2 C2 and the effect on the process (yield, A2 B1 C1 tolerance, etc...). A2 B1 C2 A2 B2 C1 A2 B2 C2Get engineering recommendations for each level of A1, A2, B1, B2, C1, and C2.Determine the combination of factors. The formula for all combinations (number of levels) (factors) = (2)3 = (2)(2)(2) = 8 This is a full factorial arrayRun the experiment, preferably obtaining more than one result.Calculate the mean result for each factor at each level.Plot the main effects.Run the experiment again to confirm the results. Copyright ISandR

Design of Experiments Three Factor Full Factorial Experiment INPUT OUTPUT Factor A Factor B Factor C Result 1 Result 2Run the experiment and A1 B1 C1 84 90gather the results A1 B1 C2 86 92 A1 B2 C1 90 88 A1 B2 C2 98 88 A2 B1 C1 90 96Calculate the mean for each A2 B1 C2 84 82factor at each level A2 B2 C1 90 98 A2 B2 C2 98 94Plot the main effectsRun the experiment again toconfirm the results A1 = (84 + 90 + 86 + 92 + 90 + 88 + 98 + 88) / 8 = 89.5 A2 = (90 + 96 + 84 + 82 + 90 + 98 + 98 + 94) / 8 = 91.5 B1 = (84 + 90 + 86 + 92 + 90 + 96 + 84 + 82) / 8 = 88.0Next, compare a partial and B2 = (90 + 88 + 98 + 88 + 90 + 98 + 98 + 94) / 8 = C1 = (84 + 90 + 90 + 88 + 90 + 96 + 90 + 98) / 8 = 93.0 90.7full factorial array by using C2 = (86 + 92 + 98 + 88 + 84 + 82 + 98 + 94) / 8 = 90.2the same experimental data 94(we will compare the results 92in this full factorial to an array 90called an “L4” with Taguchi 88nomeclature) 86 84 A1 A2 B1 B2 C1 C2 Copyright ISandR

Design of Experiments Three Factor Full vs. Partial Factorial DOE INPUT OUTPUTThis is a comparison of a full factorial Factor A Factor B Factor C Result 1 Result 2 1 A1 B1 C1 84 90array to a partial factorial for the same A1 B1 C2 86 92experimental data. A1 B2 C1 90 88 2 A1 B2 C2 98 88Run the experiment. A2 B1 C1 90 96 Instead of running the experiment, 3 A2 B1 C2 84 82 compare the data from the identical 4 A2 B2 C1 90 98 A2 B2 C2 98 94 rows in each array. INPUT OUTPUTCalculate the mean for each factor at Factor A Factor B Factor C Result 1 Result 2each level. 1 A1 B1 C1 84 90Plot the main effects. 2 A1 B2 C2 98 88 3 A2 B1 C2 84 82 Note the difference between a full 4 A2 B2 C1 90 98 (solid line) and a partial factorial (broken line) in the main effects A1 = (84 + 90 + 98 + 88) / 4 = 90.7 responses; B is similar, but C has a A2 = (84 + 82 + 90 + 98) / 4 = 88.5 B1 = (84 + 90 + 84 + 82) / 4 = 85.0 larger influence, and A is less. B2 = (98 + 88 + 90 + 98) / 4 = 93.5 There is a price to pay for running a C1 = (84 + 90 + 90 + 98) / 4 = 94.0 C2 = (98 + 88 + 84 + 82) / 4 = 90.2 less trials; it is information loss. 94The information, in either a full orpartial factorial, could be influenced 92by an omitted variable and the 90response could be different than a 88straight line (only plotted 2 points). A 86partial factorial could be influenced 84by an interaction between factors. A1 A2 B1 B2 C1 C2 Copyright ISandR

Design of Experiments 1 3 2This is a Taguchi L4 array and a linear diagram where theinteraction (called confounding) between factors occur.The actual array can be set up to acquire information oninteractions. Here 3 represents the column where interaction, ifany, between factors 1 and 2 occurs.Suppose we ran an experiment to determine the result fromchanging the levels for each factor. The proof of interactionwould be found from calculating the mean of the response andplotting the results.Plot the results and determine if there is an interaction. A1 = (90 + 85) / 2 INPUT OUTPUT A2 = (92 + 88) / 2 A A (AxB) Result B1 = (98 + 92) / 2 A1 B1 (AxB)1 90 B2 = (85 + 88) / 2 A1 B2 (AxB)2 85 (AxB)1 = (98 + 88) / 2 A2 B1 (AxB)2 92 (AxB)2 = (85 + 92) / 2 A2 B2 (AxB)1 88 Copyright ISandR

Design of Experiments INPUT OUTPUT Interaction of A and B A B (AxB) R A1 A2 95 A1 B1 (AxB)1 90 90 A1 B2 (AxB)2 85 85 A2 B1 (AxB)2 92 92 90 B1 A2 B2 (AxB)1 88 88 87.5 90 92 85 B2 Interaction of A and B A1 A2 90 95 b1 90 92 88 85 88 80 NO 90 b2 A1 A2 86 84 INTERACTION 85 87.5 90 82 A1 A2 80 A1 A2 A1 A2The upper left plot shows there is little interaction between factorsA and B because the lines representing (AxB) do not intersect. Themore non-parallel the lines, the greater is the interaction.The upper right plot is the main effects. Here we see that A and Bboth contribute to the output result. A1 = (90 + 85) / 2 = 87.5 INPUT OUTPUT A2 = (92 + 88) / 2 = 90 A A (AxB) Result B1 = (90 + 92) / 2 = 91 A1 B1 (AxB)1 90 B2 = (85 + 88) / 2 = 86.5 A1 B2 (AxB)2 85 (AxB)1 = (90 + 88) / 2 = 89 A2 B1 (AxB)2 92 (AxB)2 = (85 + 92) / 2 = 88.5 A2 B2 (AxB)1 88 Copyright ISandR

Design of Experiments DOE: Full or Partial Factorial ArraysThis is an orthogonal Taguchi array called an L8. The symmetry or balance allows the evaluation of each factor independently of other factors varied at the same time. How does this pattern work? FACTOR trial A B C D E F G 1 1 1 1 1 1 1 1 2 1 1 1 0 0 0 0 3 1 0 0 1 1 0 0 4 1 0 0 0 0 1 1 5 0 1 0 1 0 1 0 6 0 1 0 0 1 0 1 7 0 0 1 1 0 0 1 8 0 0 1 0 1 1 0Consider factor A. While factor A is “1”, factor B has 2 levels at “0” and “1”, as does C, D, E, F, and G.; and while factor A is “0” , factor B has 2 levels at “0” and “1”, as does C, D, E, F, and G.The same symmetry is true for each of the other factors as they are tested at each level of “0” and “1”.This is a “balanced array”. Each factor can be tested with a balanced influence from every other factor. Up to 7 factors can be tested, but confounding will occur. Copyright ISandR

Design of ExperimentsHere is the L8 array with the linear diagram(s) which provides the columnswhere the interactions/confounding occurs. Columns A, B, D, and G will befree of confounding, while C, E, and F will provide information on interactionsor have information mixed with that from factors from the “ends” of each line. FACTOR trial A B C D E F G 1 1 1 1 1 1 1 1 2 1 1 1 0 0 0 0 3 1 0 0 1 1 0 0 4 1 0 0 0 0 1 1 5 0 1 0 1 0 1 0 6 0 1 0 0 1 0 1 7 0 0 1 1 0 0 1 8 0 0 1 0 1 1 0 A A G G C E C E B F D B F DNote that not all columns need be used in an experiment. Only the columnpatterns that minimize confounding might be used (you could chose only A, B,D, and G). However, all of the trials must be conducted (use all of the rows). Copyright ISandR

Design of Experiments• This is where you do an experiment• You will set up an experiment to determine what influences the time a pendulum takes to make one complete period of an arc swing. A2• Consider weight, length of radius, and the start angle A1 prior to beginning the arc Time swing. L1 Start• Label these independent Stop variables W, L, A and t2 determine if there is any W2 interaction with W and L. The L2 dependent variable, time, is labeled t. Time• Use two levels for each factor. Start Stop W1 t1

Design of ExperimentsThis is one of the assignments of the factors weight (W), length (L), interactionof weight and length (WxL), and the angle (A). Other columns could be used aswell to eliminate undesirable interaction (i.e. B, D, F, and G, respectively). W L (WxL) A FACTOR trial A B C D E F G A 1 1 1 1 1 1 1 1 G 2 1 1 1 0 0 0 0 3 1 0 0 1 1 0 0 E C 4 1 0 0 0 0 1 1 5 0 1 0 1 0 1 0 6 0 1 0 0 1 0 1 7 0 0 1 1 0 0 1 B F D 8 0 0 1 0 1 1 0 Copyright ISandR

Analysis of Variance ANOVA stands for ANalysis Of VArianceThe technique assigns the percent contribution to the result from each factor, including the noise. Copyright ISandR

Analysis of VarianceFirst, define the terms of an ANOVA table, then evaluate a sample experiment.Column 1 - this is the degrees of freedom for a factor and is the number of levels minus 1. The degrees of freedom for the entire matrix, DOFTotal, is equal to DOFTotal = (number of runs) x (number of repetitions) - 1 DOFError = (DOFTotal ) - (the sum of all factors’ DOF)Column 2 - this is the sum of squares , SS (demonstrated in an example)Column 3 - this is the mean SS, which equals SS divided by DOF for that row.Column 4 - the F Ratio is the mean SS for the factor divided by the mean SS for the error.Column 5 - the percent contribution is the SS for the factor divided by the SSTotal DOF Factor (Degrees of SS (Sum of SS (Mean Percent Freedom) 1 Squares) 1 of SS) 3 F Ratio 4 Contribution 5 A B C Error Total Copyright ISandR

Analysis of VarianceUse this T and P experimental data for RUN P INPUT T Result1 OUTPUT Result1calculation and to fill out the ANOVA table 1 50 25 98 95 2 50 75 92 96Column 1 - this is the degrees of freedom 3 10 25 89 88 0The degrees of freedom for a factor 4 10 75 98 87is the number of levels minus 1. runs 0 repetitionsDOFA = 2 - 1 = 1DOFB = 2 - 1 = 1 ... there is no factor CTotal degrees of freedom equals the number of runs, times the number of repetitions, minus 1DOFTotal = (4) x (2) - 1 = 7The degrees of freedom for the error is DOFTotal minus DOF from each of the factors A and BDOFError = 7 - 1 - 1 = 5 Fill in the table DOF Factor (Degrees of SS (Sum of SS (Mean Percent Freedom) 1 Squares) 1 of SS) 3 F Ratio 4 Contribution 5 A 1 B 1 C Error 5 Total 7 Copyright ISandR

Analysis of Variance INPUT OUTPUT RUN P T Result1 Result1 1 50 25 98 95 2 50 75 92 96Column 2 - this is the sum of squares 3 10 25 89 88 0 4 75 98A correction factor (CF) must be calculated 10 87 0CF = (98 + 95 + 92 + 98 + 96 + 89 + 88 + 98 + 87)2 / 8 = 66866.1SSA = [(98 + 95 + 92 + 96)2 /4] - [(89 + 88 + 98 + 87)2 / 4] - CF = 45.125SSB = [(98 + 95 + 89 + 88)2 /4] - [(92 + 96 + 98 + 87)2 / 4] - CF = 1.125SSTotal = (98)2 + (95)2 + (92)2 + (98)2 + (96)2 + (89)2 + (88)2 + (98)2 + (87)2 - CF = 72.875 Error makes up the difference between the total and the contribution from each factorSSError = SSTotal - SSA - SSB = 72.875 - 45.125 - 1.125 = 26.575 Fill in the table DOF Factor (Degrees of SS (Sum of SS (Mean Percent Freedom) 1 Squares) 1 of SS) 3 F Ratio 4 Contribution 5 A 1 45.125 B 1 1.123 C Error 5 26.575 Total 7 72.875 Copyright ISandR

Analysis of Variance INPUT OUTPUT RUN P T Result1 Result1 1 50 25 98 95 2 50 75 92 96Column 3 - this is the mean SS 3 10 25 89 88 which equals SS divided by DOF 4 0 75 98 87 10 0SSA = 45.125 / 1 = 45.125SSB = 1.125 / 1 = 1.125SSError = 26.575 / 5 = 5.315SSTotal = 72.875 / 7 = 10.41 Enter these numbers DOF Factor (Degrees of SS (Sum of SS (Mean Percent Freedom) 1 Squares) 1 of SS) 3 F Ratio 4 Contribution 5 A 1 45.125 45.125 B 1 1.123 1.125 C Error 5 26.575 5.315 Total 7 72.875 10.41 Copyright ISandR

Analysis of Variance INPUT OUTPUT RUN P T Result1 Result1 1 50 25 98 95 2 50 75 92 96Column 4 - the F Ratio is the mean SS for the factor 3 10 0 25 89 88 4 75 98 divided by the mean SS for the error 10 87 0FA = 46.126 / 5.315 = 8.49FB = 1.126 / 5.315 = 0.211 Fill in the Table DOF Factor (Degrees of SS (Sum of SS (Mean Percent Freedom) 1 Squares) 1 of SS) 3 F Ratio 4 Contribution 5 A 1 45.125 45.125 8.49 B 1 1.123 1.125 0.211 C Error 5 26.575 5.315 Total 7 72.875 10.41 Copyright ISandR

Analysis of Variance INPUT OUTPUT RUN P T Result1 Result1 1 50 25 98 95Column 5 - the percent contribution is the SS for the 2 50 75 92 96 3 10 25 89 factor divided by the total SS 4 0 75 98 88 87 10 0%A = (45.125 / 72.825) x 100 = 61.9%%B = (1.125 / 72.825) x 100 = 1.5%%Error = (26.575 / 72.825) x 100 = 36.5% Fill in the Table DOF Factor (Degrees of SS (Sum of SS (Mean Percent Freedom) 1 Squares) 1 of SS) 3 F Ratio 4 Contribution 5 A 1 45.125 45.125 8.490 61.9 B 1 1.123 1.125 0.211 1.5 C Error 5 26.575 5.315 36.5 Total 7 72.875 10.41 Copyright ISandR

• the end• THE END• THIS IS REALLY THE END•THE END! please contact mrdrking@gmail.com for an animated PowerPoint presentation Copyright ISandR

Making A Quality Product

by Thomas King

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