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South Dakota School of Mines & Technology Introduction to Probability & Statistics Industrial Engineering South Dakota School of Mines & Technology Applications in Quality Industrial Engineering South Dakota School of Mines & Technology Acceptance Sampling Industrial Engineering Quality Methods Quality Engineering Methods 100 Percent of Application 75 Acceptance Sampling Process Control 50 25 Design of Experiments 0 0 1 2 3 4 5 6 7 Time Quality Design & Process Variation Lower Spec Upper Spec Limit Limit Acceptance Sampling 60 80 100 120 140 Statistical Process Control 60 140 Experimental Design 60 140 Acceptance Sampling Suppose we buy parts in lot sizes of 5,000 and we wish to inspect to ensure that the incoming quality is sufficient. Obviously, we would not like to do 100% inspection. Idea: Select a portion, n, of the lot at random and determine the number of defects in the sample. If the number of defects, d, surpasses some critical value, c, assume the entire lot is defective and ship it back. Example Let N = lot size = 5,000 n = sample size = 50 c = critical value = 1 d = number of defectives found in sample size n If d 1 , accept the entire lot d 2 , reject the entire lot Theory We assume that each part in the lot is either defective or “successful”. The probability of a defective part is constant with at some proportionate rate p. In our example we assume that an acceptable p is given by c 1 pa .02 2% n 50 We call this the Acceptable Quality Level or AQL Theory (cont.) Now, if p is constant from trial to trial in a sample of n, then the probability of accepting a lot given that the lot has acceptable quality level is given by Pa = P{d c} c Pa n p d (1 p ) n d d 0 d Theory (cont.) Then, 1 Pa 50 .02 d (.98) 50 d d 0 d 50! .02 0 (.98) 50 50! .02 1 (.98) 49 50!0! 49 !1! = .7358 Risks Decision Ho HA no Ho error Truth no HA error P{reject lot | lot is good} P{accept lot | lot is bad} Risks (cont.) P{reject lot | lot is good} =1 - P(accept lot | lot is good} = 1 - Pa = 1 - .7358 = .2642 P{accept lot | lot is bad} Problem: What constitutes a bad lot? Computing Let’s suppose that we consider a lot with 4% (p=.04) to be a bad lot. This is Lot Tolerance Percent Defective, LTPD. Then P{accept lot | lot is bad} =P{d < c | pb } 1 50 .04 d (.96) 50 d d 0 d Computing Then 50! .04 0 (.96) 50 50! .04 1 (.96) 49 50!0! 49 !1! = .4005 Suppose we wish to evaluate for a number of different LTPD values (pb). Operating Characteristic Operating Characteristic Curve 1.0 0.9 0.8 0.7 P{accept lot} 0.6 0.5 0.4 0.3 0.2 0.1 0.0 0.00 0.02 0.04 0.06 0.08 0.10 Pb Acceptance Sampling Uses when testing is destructive cost of 100% inspection is high inspection error rate is high cost of passing limited number of defective products is low the process is stable Acceptance Sampling Limitations does not estimate lot quality does not provide for quality control risks associated with accepting bad lots and rejecting good lots Today’s Trends high quality small lots FRO’s find Acceptance Sampling of little use South Dakota School of Mines & Technology Introductory SPC The “Magnificent 7” Industrial Engineering Quality Design & Process Variation Lower Spec Upper Spec Limit Limit Acceptance Sampling 60 80 100 120 140 Statistical Process Control 60 140 Experimental Design 60 140 Statistical Process Control u1 u2 u 3 u4 Input Output x1 x2 x3 x4 ui = random noise, unassignable causes of variation xi = assignable causes of variation Statistical Process Control u1 u2 u3 u4 Input Output x1 x2 x3 x4 Objective: remove all assignable causes of variation smallest variation in output Deming Wheel P A D C The Magnificent “7” Problem Solving Useful Tools Step Understand Mess Flow Charts Find Facts Check Sheets Identify Problems Pareto Diagrams Histograms Generate Ideas Cause-and-Effect Develop Solutions Scatter Diagrams Implementation Control Charts Problem Identification/Analysis Problem Identification Problem Analysis Flow Chart Histogram Pareto Chart Scatter Diagram Check Sheet Cause & Effect Process Capability Nominal Group Technique Stratification Control Chart Brainstorming Run Chart Force Field Analysis Flow Charting Mix Dry Mix Wet Mix Wet Fold 10 Ingredients Ingredients & Dry Minutes Extrude Fold 2 Add Chips Cut Dough Dough Minutes OK Bake 10 Qual. Pack and Minutes Ship Poor Feed to Hogs Check Sheet; Cookies Type Check Total Burnt 11 Crumbly 5 Too Few Chips 14 Poor Taste 6 Other 3 Histograms Defects by Category 15 10 Frequency 5 0 Burnt Crumbly Few Chips Poor Taste Other Type Defect Pareto Chart Pareto Analysis 1.0 0.8 Percentage 0.6 0.4 0.2 0.0 Too Few Burnt Poor Taste Crumbly Other Chips Category Histogram; Class Problem Now that we’ve decided that too few chips is our quality problem. We need to determine how many chips are in each cookie. Count your chips. Histogram; Class Problem Chips per Cookie 8 Frequency 6 4 2 0 10 12 14 16 18 20 No. Chips Cause & Effect Mix Chips Automate in Dry Fold Longer Too Few Chips Place Chips on Drop Liquid by hand Chips Cause & Effect People Methods Handling Effect Design Tools Idea; Cookies As a first step, let’s try Increasing Folding Time Scatter Diagram Chips vs Fold Time 25 20 # Chips 15 10 5 5 10 15 20 25 Fold Time New Idea; Cookies Try Mixing Chips in With Dry Ingredients Flow Charting Mix Dry Mix Wet Mix Wet Add Chips Ingredients Ingredients & Dry Extrude Fold 2 Fold 10 Cut Dough Minutes Dough Minutes OK Bake 10 Qual. Pack and Minutes Ship Poor Feed to Hogs Evaluate; Control Chart Avg Chips 18 15 Time 12 South Dakota School of Mines & Technology Control Charts Industrial Engineering Control Charts We assume that the underlying distribution is normal with some mean and some constant but unknown standard deviation . Let n xi x i 1 n Distribution of x Recall that x is a function of random variables, so it also is a random variable with its own distribution. By the central limit theorem, we know that x N ( , x ) where, x x n Control Charts x x x x Control Charts UCL x x LCL UCL & LCL Set at 3 x Problem: How do we estimate & ? Control Charts m x i x i 1 m m R R i 1 f ( ) m Control Charts m x i x i 1 m m R R i 1 f ( ) m UCLx x A2 R UCLR D4 R LCLx x A2 R LCLR D3 R Example Suppose specialized o-rings are to be manufactured at .5 inches. Too big and they won’t provide the necessary seal. Too little and they won’t fit on the shaft. Twenty samples of 2 rings each are taken. Results follow. Control Charts Part Measurements R Chart X Chart No. 1 2 x R UCL R UCL LCL Xbar 1 0.502 0.504 0.503 0.002 0.0077 0.002 0.5052 0.4964 0.503 2 0.495 0.497 0.496 0.002 0.0077 0.002 0.5052 0.4964 0.496 3 0.492 0.496 0.494 0.004 0.0077 0.004 0.5052 0.4964 0.494 4 0.501 0.498 0.500 0.003 0.0077 0.003 0.5052 0.4964 0.500 5 0.507 0.508 0.508 0.001 0.0077 0.001 0.5052 0.4964 0.508 6 0.504 0.504 0.504 0.000 0.0077 0.000 0.5052 0.4964 0.504 7 0.497 0.496 0.497 0.001 0.0077 0.001 0.5052 0.4964 0.497 8 0.493 0.496 0.495 0.003 0.0077 0.003 0.5052 0.4964 0.495 9 0.502 0.501 0.502 0.001 0.0077 0.001 0.5052 0.4964 0.502 10 0.498 0.500 0.499 0.002 0.0077 0.002 0.5052 0.4964 0.499 11 0.505 0.507 0.506 0.002 0.0077 0.002 0.5052 0.4964 0.506 12 0.502 0.499 0.501 0.003 0.0077 0.003 0.5052 0.4964 0.501 13 0.495 0.497 0.496 0.002 0.0077 0.002 0.5052 0.4964 0.496 14 0.499 0.496 0.498 0.003 0.0077 0.003 0.5052 0.4964 0.498 15 0.503 0.507 0.505 0.004 0.0077 0.004 0.5052 0.4964 0.505 16 0.507 0.509 0.508 0.002 0.0077 0.002 0.5052 0.4964 0.508 17 0.503 0.501 0.502 0.002 0.0077 0.002 0.5052 0.4964 0.502 18 0.497 0.493 0.495 0.004 0.0077 0.004 0.5052 0.4964 0.495 19 0.504 0.508 0.506 0.004 0.0077 0.004 0.5052 0.4964 0.506 20 0.505 0.503 0.504 0.002 0.0077 0.002 0.5052 0.4964 0.504 Avg = 0.501 0.002 x R Std. = 0.0047 Control Charts X-Bar Chart 0.510 0.505 x 0.500 0.495 0.490 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Xbar charts can identify special causes of variation, but they are only useful if the process is stable (common cause variation). Control Limits for Range UCL = D4R = 3.268*.002 = .0065 LCL = D3 R = 0 R Chart 0.010 0.008 Range 0.006 0.004 0.002 0.000 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Observation The P-Chart UCL p 3 p LCL p 3 p where m di p (1 p ) p i 1 p nm n Example; P-Chart Operators of a sorting machine must read the zip code on a letter and diver the letter to the proper carrier route. Over one month’s time, 25 samples of 100 letters were chosen, and the number of errors was recorded. Error counts for each of the 25 days follows. Example Day Error %Defect Day Error %Defect 1 3 0.03 14 2 0.02 2 1 0.01 15 3 0.03 3 0 0.00 16 3 0.03 4 0 0.00 17 2 0.02 5 2 0.02 18 0 0.00 6 3 0.03 19 1 0.01 7 5 0.05 20 5 0.05 8 2 0.02 21 4 0.04 9 4 0.04 22 3 0.03 10 3 0.03 23 2 0.02 11 0 0.00 24 4 0.04 12 1 0.01 25 3 0.03 13 4 0.04 Example p .03.01.00. . . .03 25 Example p .03.01.00. . . .03 25 p (1 p ) UCL p p 3 n .024(1 .024) .024 3 100 0.070 Example p .03.01.00. . . .03 25 p (1 p ) UCL p p 3 n .024(1 .024) .024 3 100 0.070 p (1 p ) LCL p p 3 .022 0.0 n Example; P-Chart P-Chart; Mail Sort 0.08 0.06 % Errors 0.04 0.02 0.00 -0.02 0 10 20 30 Day South Dakota School of Mines & Technology Process Capability Industrial Engineering Process Capability Precise but not Accurate 7 6 5 Frequency 4 3 2 1 0 1 2 3 4 5 6 7 8 9 Value Process Capability Accurate but not Precise 7 6 5 Frequency 4 3 2 1 0 1 2 3 4 5 6 7 8 9 Value Process Capability Precise and Accurate 7 6 5 Frequency 4 3 2 1 0 1 2 3 4 5 6 7 8 9 Value Process Capability LSL USL x x USL LSL Cp 6 x Process Capability LSL USL x x USL LSL C pk min , 3 x 3 x Cookie Example Suppose we have the cookie example and it is desired to have 15 + 3 chips. Obs. Chips Obs. Chips 1 12 11 17 2 18 12 16 3 19 13 19 4 16 14 13 5 13 151 15 6 11 16 14 7 16 17 17 8 14 18 11 9 10 19 16 10 12 20 15 Average 14.1 Std. Dev. 3.0 Cookie Example 18 14.1 14.1 12 Cp min , 3(3) 3(3) min(0.433, 0.233) 0.233 South Dakota School of Mines & Technology Funnel Experiment Industrial Engineering Funnel Experiment TheFunnel Experiment is a famous experiment first developed by Dr. Deming. It was designed to show primarily two things: 1. All processes have statistical variation 2. Improving processes to reduce variation is infinitely better than trying to control process variation Funnel Experiment Funnel Experiment Funnel Experiment Rule 1: Do nothing Funnel Experiment Rule 2: Move the funnel in an equal but opposite direction Rule 3: Move the funnel in an equal Funnel Experiment but opposite direction from (0,0) Rule 4: Move the funnel to the last Funnel Experiment position where the last hit. The Funnel Experiment Rule 1: (LTDTA) Set the funnel over the target at (0,0) and leave the funnel fixed through all 50 drops. Funnel Experiment; Rule 1 4 3 2 1 Yk 0 -4 -3 -2 -1 -1 0 1 2 3 4 -2 -3 -4 Xk The Funnel Experiment Rule 2: Move the funnel in an equal but opposite direction from where the last marbel hit. The funnel should be moved the distance (-xk,-yk) from its last resting point. Funnel Experiment; Rule 2 4 3 2 1 Yk 0 -4 -3 -2 -1 -1 0 1 2 3 4 -2 -3 -4 Xk The Funnel Experiment Rule 2: Move the funnel in an equal but opposite direction from where the last marbel hit. The funnel should be moved the distance (-xk,-yk) from its last resting point. Machine/Process adjustment rules using control chart limits. Automatic compensating machinery The Funnel Experiment Rule 3: Move the funnel in an equal but opposite direction from target (0,0) to where the last marble hit. Rule 3 states that the funnel should be moved a distance (-xk,-yk) from the target (0,0). Funnel Experiment; Rule 3 4 3 2 1 Yk 0 -4 -3 -2 -1 -1 0 1 2 3 4 -2 -3 -4 Xk The Funnel Experiment Rule 3: Move the funnel in an equal but opposite direction from target (0,0) to where the last marble hit. Rule 3 states that the funnel should be moved a distance (-xk,-yk) from the target (0,0). Management by Objectives Committees The Funnel Experiment Rule 4: Move the funnel to the last position where the last marble hit. Rule 4 states that the funnel should be moved to the resting point, (xk,yk). Funnel Experiment; Rule 4 4 3 2 1 Yk 0 -4 -3 -2 -1 -1 0 1 2 3 4 -2 -3 -4 Xk The Funnel Experiment Rule 4: Move the funnel to the last position where the last marble hit. Rule 4 states that the funnel should be moved to the resting point, (xk,yk). Operator matches color from batch to batch without reference to original swatch. Operator attempts to make every piece like the last one. On the job training. Rule 2: Deadband Controlling } Xn o e X1 (o X n ) o X2 X3 o e 1. Deviation |Xn - o| is random; do nothing 2. Deviation |Xn - o| is shift; adjust by (o - Xn) Infinitely Large Deadband Xn X1 X2 X3 Xn ~ N(,) Infinitesimally Small Deadband Xn X1 } ( X n ) X2 X3 n n 1 X n n 1 ( n 1 Zn ) Infinitesimally Small Deadband Xn X1 } ( X n ) X2 X3 Xn ~ N(, 22) Process Variation Process Variance vs Deadband 2 Variance 1.5 1 0 1 2 3 4 5 K South Dakota School of Mines & Technology Loss Functions Industrial Engineering Cost of Quality Traditional View Costs Total Cost Cost of Control Cost of Failure Quality Level Cost of Quality Contemporary View Costs Total Cost Cost of Control Cost of Failure Quality Level Traditional Loss Function LSL USL x T LSL T USL Taguchi Loss Function x T Taguchi Loss Function x T x T Example (Sony, 1979) Comparing cost of two Sony television plants in Japan and San Diego. All units in San Diego fell within specifications. Japanese plant had units outside of specifications. Loss per unit (Japan) = $0.44 Loss per unit (San Diego) = $1.33 How can this be? Sullivan, “Reducing Variability: A New Approach to Quality,” Quality Progress, 17, no.7, 15-21, 1984. Example LSL USL x T U.S. Plant (2 = 8.33) Japanese Plant (2 = 2.78) Taguchi Loss Function L(x) k(x - T)2 x T L(x) = k(x - T)2 Estimating Loss Function Suppose we desire to make pistons with diameter D = 10 cm. Too big and they create too much friction. Too little and the engine will have lower gas milage. Suppose tolerances are set at D = 10 + .05 cm. Studies show that if D > 10.05, the engine will likely fail during the warranty period. Average cost of a warranty repair is $400. Estimating Loss Function L(x) 400 10.05 10 400 = k(10.05 - 10.00)2 = k(.0025) Estimating Loss Function L(x) 400 10.05 10 400 = k(10.05 - 10.00)2 = k(.0025) k = 160,000 Expected Loss Expected Loss: Piston Diameter Probability Probability Diameter Process A Process B 9.925 0.000 0.025 9.950 0.200 0.075 9.975 0.200 0.200 10.000 0.200 0.400 10.025 0.200 0.200 10.050 0.200 0.075 10.075 0.000 0.025 Expected Loss Expected Loss: Piston Diameter Probability Probability Diameter Loss Process A Process B 9.925 900 0.000 0.025 9.950 400 0.200 0.075 9.975 100 0.200 0.200 10.000 0 0.200 0.400 10.025 100 0.200 0.200 10.050 400 0.200 0.075 10.075 900 0.000 0.025 Expected Loss Expected Loss: Piston Diameter Probability Weighted Probability Diameter Loss Process A Loss Process B 9.925 900 0.000 0.0 0.025 9.950 400 0.200 80.0 0.075 9.975 100 0.200 20.0 0.200 10.000 0 0.200 0.0 0.400 10.025 100 0.200 20.0 0.200 10.050 400 0.200 80.0 0.075 10.075 900 0.000 0.0 0.025 Expected Loss Expected Loss: Piston Diameter Probability Weighted Probability Weighted Diameter Loss Process A Loss Process B Loss 9.925 900 0.000 0.0 0.025 22.5 9.950 400 0.200 80.0 0.075 30.0 9.975 100 0.200 20.0 0.200 20.0 10.000 0 0.200 0.0 0.400 0.0 10.025 100 0.200 20.0 0.200 20.0 10.050 400 0.200 80.0 0.075 30.0 10.075 900 0.000 0.0 0.025 22.5 Expected Loss Expected Loss: Piston Diameter Probability Weighted Probability Weighted Diameter Loss Process A Loss Process B Loss 9.925 900 0.000 0.0 0.025 22.5 9.950 400 0.200 80.0 0.075 30.0 9.975 100 0.200 20.0 0.200 20.0 10.000 0 0.200 0.0 0.400 0.0 10.025 100 0.200 20.0 0.200 20.0 10.050 400 0.200 80.0 0.075 30.0 10.075 900 0.000 0.0 0.025 22.5 Exp. Loss = 200.0 145.0 Aside L( x) k ( x T ) 2 If we replace T by , L(x) looks similar to the variance function. Indeed, we can show that L( x) k ( ( x T ) ) 2 2 Example (Sony) LSL USL x T U.S. Plant (2 = 8.33) Japanese Plant (2 = 2.78) E[LUS(x)] = 0.16 * 8.33 = $1.33 E[LJ(x)] = 0.16 * 2.78 = $0.44 Tolerance (Pistons) Recall, L(x) 400 10.05 10 400 = k(10.05 - 10.00)2 = k(.0025) k = 160,000 Tolerance L(x) Suppose repair for an engine which will fail during warranty can be made for 400 only $200 200 10.05 LSL 10 USL Tolerance L(x) Suppose repair for an engine which will fail during warranty can be made for 400 only $200 200 10.05 200 = 160,000(tolerance)2 LSL 10 USL Tolerance L(x) Suppose repair for an engine which will fail during warranty can be made for 400 only $200 200 10.05 200 = 160,000(tolerance)2 LSL 10 USL tolerance = (200/160,000)1/2 = .0354

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posted: | 5/16/2012 |

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