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Unformatted text preview: Quality Quality Control (QC)
MGMT 46000 Operations Management (OM)
Acceptance sampling Process control V  Quality Management  II
1. 1. 2. Input Output 3. Process control: During process** Process capability Acceptance sampling: For incoming material Control
ChV:QualityManagement‐II‐Sp10 feedback ** Also called Statistical Process control (SPC)
1 ChV:QualityManagement‐II‐Sp10 2 MGMT 36100 MGMT 36100 Process Variability Operation: from a rod about 6’ in length, we want to cut small pieces of 3” length. Tolerance limits: 3.00 ± 0.015” length. 2.997 3.002 3.007 2.983 3.003 Process Variability •Assignable variation: caused by material, tools, worker related problems. •Common variation (random variation): caused by type of process, equipment etc. If we eliminate assignable variation, we will bring the process within control. The process itself may still produce the variation due to common causes. Back to statistics again
Products from any process have variability and a dimension of the product we are interested in may have certain probability distribution with mean μ and standard deviation σ. We estimate these parameters (μ and σ) using random samples. From samples we can calculate sample mean (xbar) and standard (xdeviation (s). ChV:QualityManagement‐II‐Sp10 3 ChV:QualityManagement‐II‐Sp10 4 MGMT 36100 MGMT 36100 Control charts
Used for controlling process variability during production. There are many types of control charts (XBar, R, C, p); (Xwe will study Xbar and R charts. There are two stages. 1. Establishing stable control limits:
• • • • Collect random samples. Determine (trial) control limits. Determine out of control point. Revise limits – repeat.
UCL: Upper Control Limit
•• • • • • • • • Here we have collected 20 samples of size 5 each.
1 1 2 3 4 30 30 34 33 32 2 55 51 47 33 48 3 19 25 21 44 28 4 36 36 37 35 38 17 27 27 26 29 27 18 29 32 34 35 34 19 30 34 34 35 36 m:
20 30 31 29 29 29 n: 5 CL: CL: Center Line •• • • Lower LCL: Now we calculate sample mean for 20 samples. X bar 31.8 46.8 27.4 36.4 27.2 32.8 Now we calculate sample range for 20 samples. Range 4 22 25 3 3 6 33.8 6 29.6 2 Control Limit 2. Use stable limits to maintain
process control. Out of control Now we calculate the grand mean and average range. We will mainly consider stage 1 in class.
ChV:QualityManagement‐II‐Sp10 5 X = (31.8 + 46.8 + … + 33.8 + 29.6) / 20 = 32.17 R = (4 + 22 + … + 6 + 2) / 20 = 9.00 Actual observations in row 1: 2.030, 2.055, 2.019, 2.036 …..
ChV:QualityManagement‐II‐Sp10 6 MGMT 36100 MGMT 36100 ChV:QualityManagementIISp10 1 To establish control limits: Collect m samples of size n each. Calculate trial control limits using appropriate sample statistics (different ones are used for different charts). X Chart:
We start with m samples of size n each. We are dealing with the distribution of the average X , (normally distributed). Most observations should be within 3 std. dev. from the mean. So we set up control limits at these points. We turn around this figure by 900 before plotting.
3 • •• • • Θ • • • • • • • • • out out of control UCL: Upper Control Limit CL: Center Line UCL 3 2 1 0 1 2 3 CL LCL LCL CL CL UCL
UCL UCL LCL: Lower Control Limit LCL 2 1 0 out out of control Plot points and check if the process is in control. Eliminate out of control points after determining reasons. Repeat till stable limits are established.
ChV:QualityManagement‐II‐Sp10 7 1 2 3 Sample Number CL LCL
8 ChV:QualityManagement‐II‐Sp10 MGMT 36100 MGMT 36100 X Chart: CL = UCL = LCL = Limits for this chart are given by X Chart: Example 1
Sample # Mean Range Sample # Mean 1 2 3 4 m = 20 samples of size n = 5 each. each. We generally plot graphs μX μ X +3σX μ X 3σ X ≈X ≈ X +A2 R ≈ X  A2 R 5 6 7 8 9 10 31.8 46.8 27.4 36.4 34.8 34.6 32.6 42.8 4 11 22 12 25 13 3 14 5 15 9 16 15 17 6 18 26.6 31.2 5 19 6 20 33.6 20.8 32.6 25.8 28.2 13 2 12 9 5 34 27.2 32.8 22 3 6 33.8 29.6 6 2 Use A2 values from the table.
n A2 11 2 3 4 5 6 7 8 9 10 Range X = 32.17 1.88 1.023 0.729 0.577 0.483 12 13 14 15 16 0.419 0.373 0.337 0.308 17 18 19 20 R = 9.00 For n = 5, A2 = 0.577 CL = 32.17 = X 0.285 0.266 0.249 0.235 0.223 0.212 0.203 0.196 0.187 0.180 UCL = 32.17 + 0.577 * 9 = 37.36 LCL = 32.17  0.577 * 9 = 26.98
9 ChV:QualityManagement‐II‐Sp10 Which points outside UCL? Which points outside LCL?
10 ChV:QualityManagement‐II‐Sp10 MGMT 36100 MGMT 36100 Example 1
X bar chart: trial control limits Example 1: Revision 1, m = 15 Revision
X = 32.04 R = 9.07 CL = 32.04 = X UCL = 32.04 + 0.577 * 9.07 = 37.27 42.55 37.36 32.17 26.98 21.79 16.60 1 3 5 7 9 11 13 15 17 19 LCL = 32.04  0.577 * 9.07 = 26.81
Revised X bar chart 37.27 32.04 26.81 1 3 5 7 9 11 13 15 17 19 Points 2, 8, 9, 12, 14 are outside limits. Find assignable cause and eliminate these points. Then find new control limits.
ChV:QualityManagement‐II‐Sp10 11 All points are within limits. The process is now in control and the limits become stable.
ChV:QualityManagement‐II‐Sp10 12 MGMT 36100 MGMT 36100 ChV:QualityManagementIISp10 2 R chart We have m samples of size n each and plot sample range values. So we are dealing with the distribution of R. Distribution of R is not normal. LCL D3 R
5 0 6 0 7 0.076 Example 2: R chart chart
n = 5, D3 = 0.0, D4 = 2.115 X = 32.17 R = 9.00 CL = 9.00, UCL = 2.115 * 9.00 = 19.04, LCL = 0.00 * 9.00 = 0.00 9.00, 19.04,
R chart: trial control limits We need control limits. Estimates for limits are:
n D3 D4 11 0.256 1.744 2 0 3 0 4 0 CL
R
8 UCL D4 R
9 10 0.136 0.184 0.223 1.816 1.777 19 20 0.404 0.414 1.596 1.586 19.04 9.00 1.04 1 3 5 7 9 11 13 15 17 19 3.268 2.574 2.282 2.115 12 13 14 15 0.284 0.308 0.329 0.348 1.716 1.692 1.671 1.652 2.004 1.924 16 17 0.364 0.379 1.636 1.621 1.864 18 0.392 1.608 Points 2, 3 and 16 are outside limits. Find assignable causes; eliminate points; find new control limits. ChV:QualityManagement‐II‐Sp10 13 ChV:QualityManagement‐II‐Sp10 14 MGMT 36100 MGMT 36100 Example 2: Revision 1, m = 17 Revision
R = 6.53 CL = 6.53, UCL = 13.80, LCL = 0
R chart: revised control limits Example 2: Revision 2, m = 16 Revision
R = 6.00 CL = 6.00, UCL = 12.69, LCL = 0
R chart: control limits (revision 2) Revision 3, m = 15 R = 5.53 CL = 5.53, UCL = 11.7, LCL = 0 21.07 13.80 6.53
0.0 0.74 R chart: control limits (revision 3) 12.69 6.00 11.7 5.53 1 3 5 7 9 11 13 15 17 19 0.69
Point 7 is outside limits. Perform revision 2 0 5 10 15 20 25 0.64 1 3 5 7 9 11 13 15 17 19 Point 11 is outside limits. Perform revision 3
ChV:QualityManagement‐II‐Sp10 15 ChV:QualityManagement‐II‐Sp10 Point 13 is outside limits. Perform revision 4
16 MGMT 36100 MGMT 36100 Example 2: Revision 4, m = 14
R = 5.07 Using X bar and R charts simultaneously.
X bar chart: trial control limits CL = 5.07, UCL = 10.73, LCL = 0
42.55 37.36 R chart: trial control limits R chart: revised control limits (revision 4) 19.03 9.00 1.03 1
1 3 5 7 9 11 13 15 17 19 10.73 32.17 26.98 21.79 5.07 16.60 3 5 7 9 11 13 15 17 19 0.59 1 3 5 7 9 11 13 15 17 19 All points are in control now! What if we lose too many points before we get stable limits?
ChV:QualityManagement‐II‐Sp10 17 Out of UCL LCL Control Trial 37.36 26.98 2,8,9,12,14 19.04 0 2,8,9,12,1419.04 Rev. 1 36.20 28.30 4,15,17 14.48 0 Rev. 2 36.79 28.72 14.81 0 UCL LCL Out of Eliminate Control 2,3,16 2,3,8,9,12,14,16 7 4,7,15,17  ChV:QualityManagement‐II‐Sp10 18 MGMT 36100 MGMT 36100 ChV:QualityManagementIISp10 3 Control charts
Stage 2: To maintain control limits: Keep on taking new samples as per established procedure and plot new points. If process goes out of control, take corrective action (but do not recalculate control limits). Revise limits whenever major changes occur. Determination of out of control •
Points outside control limits •• • • • • • UCL • • • • • • •
LCL UCL CL Trial Control limis
Stable Control limits
0.39 0.21 0.04
0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 30 32 34 36 38 40 42 44 46 48 50 Other measures: • Seven consecutive points above / below CL. CL. • Cycle pattern repeating • All points close to CL (hugging). • A run of 2 or 3 points outside 2sigma, 4 or 5 2outside 1sigma 119 ChV:QualityManagement‐II‐Sp10 •• • • •• • • •• •• CL ChV:QualityManagement‐II‐Sp10 LCL
MGMT 36100 20 MGMT 36100 Process Capability
Suppose we are making a component using certain process and we need to produce at least 99.5% good components. components. Here is the first question we want to answer. Is the process capable? Answer to the first question is expressed using a measure called ]”. “process capability ratio [ CP ]”. If the answer is no, we need to find a better process. no If the answer is yes, we start using the process and periodically yes ask the second question. How is the process doing at any given time? Answer to the second question is expressed using a measure called “process capability index” [ Cpk ]. If the answer to the second question reveals lower capability, we take some corrective action.
ChV:QualityManagement‐II‐Sp10 21 Process Capability Analysis:
The methodology used for finding Cp and CPK is called process capability analysis. Here we make certain assumptions • • • We are measuring only one critical dimension X. X is normally distributed. Process is stable. μ The specification for this dimension is A ± B where “A + B” is the Upper Tolerance Limit [UTL] “A” is the nominal value and AB A A+B “A – B” is the Lower Tolerance Limit [LTL] If the dimension of the component produced falls between LTL and UTL, we accept the component. If it falls outside, we reject. When we do the analysis, we collect random samples to estimate mean and standard deviation and to ensure that the process is stable.
ChV:QualityManagement‐II‐Sp10 22 MGMT 36100 MGMT 36100 Process Capability Ratio (CP):
The formula for the process capability ratio is given by LTL) Cp = (UTL – LTL) / (6σ). Here are the specifications A ± B Now we will superimpose normal distribution on the specifications. It is assumed that the process mean is at the nominal value “A”. Red area shows probability of rejection. If standard deviation (σ) of the process becomes smaller, CP becomes higher and probability of rejection becomes smaller.
ChV:QualityManagement‐II‐Sp10 Ex 1: Process capability ratio: Cp = (UTL – LTL) / (6σ). LTL)
Stable Stable process, normally distributed. Tolerance limits for critical dimension: 2.500 ± 0.0035”. Estimated std dev = 0.001”. 1: Calculate CP UTL = 2.5035” and LTL = 2.4965” AB A A+B CP = (2.5035 – 2.4965)/(6 * 0.001) = (0.007)/(0.006) = 1.167
AB A A+B AB A A+B
23 2: Calculate acceptance probability. Prob. Prob. of acceptance = Pr (LTL ≤ X ≤ UTL) with μ = 2.500 = Pr(2.4965 ≤ X ≤ 2.5035) = Pr[ (2.4965 – 2.5000) / 0.001 ≤ Z ≤ (2.5035  2.5000)/0.001] = Pr(3.5 ≤ Z ≤ 3.5) Pr(= 0.9996 So ChV:QualityManagement‐II‐Sp10 99.96% components will be accepted. 24
MGMT 36100 MGMT 36100 ChV:QualityManagementIISp10 4 Ex 2: Process capability ratio: Cp = (UTL – LTL) / (6σ). LTL)
Stable process, normally distributed. Tolerance limits for critical dimension: 4.000 ± 0.003”. Estimated std dev = 0.002”. What is the significance of Cp?
Here is a table of probabilities for different values of CP Cp 1/3 0.5 2/3 1.0 1.584 2.0 Probability of For many products, CP = 1.0 acceptance rejection may be acceptable. 0.68269 0.31731 However, for many other products, this means 0.86637 0.13363 rejection of 0.0027 or 2700 0.95450 0.04550 parts per million (ppm). parts per million ). This 0.99730 0.00270 This may not be acceptable. 0.999998 0.000002 Companies strive for Cp = 1.584 (rejection of 2 ppm). ppm). 1.000000 0.000000 1: 1: Calculate CP UTL = 4.003” and LTL = 3.997” AB A A+B CP = (4.003 – 3.997)/(6 * 0.002) = (0.006)/(0.012) = 0.5 2: Calculate acceptance probability. Prob. Prob. of acceptance = Pr (LTL ≤ X ≤ UTL) with μ = 4.000 = Pr(3.997 ≤ X ≤ 4.003) = Pr[ (3.997 – 4.000) / 0.002 ≤ Z ≤ (4.003 – 4.000)/0.002] = Pr(1.5 ≤ Z ≤ 1.5) Pr(= 0.8664 So 86.64% components will be accepted.
ChV:QualityManagement‐II‐Sp10 25 The ideal is CP = 2 with rejection of about 2 parts per billion. For CP = 2, UTL and LTL are at a distance of 6 σ from the nominal value.
ChV:QualityManagement‐II‐Sp10 26 MGMT 36100 MGMT 36100 Process Process Capability Index (CPK):
The formula for the process capability index is given by CPK = min {(μ – LTL) / (3σ), (UTL – μ) / (3σ)}. Here are the specifications A ± B We will superimpose normal distribution on the specifications. However, now μ may not be equal to nominal value “A” Ex 1: continued continued Tolerance limits: 2.500 ± 0.0035”. σ = 0.001”. From parts 1 and 2: CP = 1.167, acceptance probability = 0.9996 2: 3: If the current μ is 2.501” calculate Cpk. If 2.501”
= min{ CPK = min {(μ – LTL) / (3σ), (UTL – μ) / (3σ)}. min (2.501 − 2.4965 ) (2.5035 − 2.501) , }= min {0.833, 1.500) = 0.833 (3 * 0.001) (3 * 0.001) AB A μ A+B Red area shows probability of rejection. As μ moves away from “A” while running the process, CPK will become smaller and rejection probability will increase. increase. 4: Prob. of accepting = Pr (LTL ≤ X ≤ UTL) with μ = 2.501 Prob. = Pr(2.4965 ≤ X ≤ 2.5035) Pr(2.4965 = Pr[ (2.4965 – 2.501) / 0.001 ≤ Z ≤ (2.5035  2.501)/0.001] = Pr( 4.5 ≤ Z ≤ 2.5) Pr(= 0.9938 When μ = nominal value, CPK = CP otherwise CPK < CP. otherwise
27 ChV:QualityManagement‐II‐Sp10 28 ChV:QualityManagement‐II‐Sp10 MGMT 36100 MGMT 36100 Ex 2: continued continued Tolerance limits: 4.000 ± 0.003”. σ = 0.002”. From parts 1 and 2: CP = 0.5, acceptance probability = 0.8664 2: General comments: comments:
Cp 1/3 0.5 2/3 1.0 1.584 2.0 LTL Probability of acceptance 0.68269 0.86637 0.95450 0.99730 0.999998 1.000000 UTL rejection 0.31731 0.13363 0.04550 0.00270 0.000002 0.000000 σ CP CPK Probability of acceptance 3: If the current μ is 4.001” calculate Cpk. If 4.001”
( 4.001 − 3.997) ( 4.003 − 4.001) = min{ , } (3 * 0.002) (3 * 0.002) CPK = min {(μ – LTL) / (3σ), (UTL – μ) / (3σ)}. min = min {0.667, 0.333) = 0.333 {0.667, Please note the Cp table to the left. When Cp = 2/3, acceptance probability is always 0.9545. This is not true for C. See the table below. 4: Prob. of accepting = Pr (LTL ≤ X ≤ UTL) with μ = 4.001 Prob. = Pr(3.997 ≤ X ≤ 4.003) Pr(3.997 = Pr[ (3.997 – 4.001) / 0.002 ≤ Z ≤ (4.003 – 4.001)/0.002] 4.001)/0.002] = Pr( 2.0 ≤ Z ≤ 1) Pr(= 0.8115 Textbook problem 13, page 202. a. Find Cpk b. Should the company use this machine? Why?
ChV:QualityManagement‐II‐Sp10 29 Current μ 98.0 102.0 100.0 97.5 102.5 100.5 97.0 103.0 101.0 96.0 105.0 101.0
ChV:QualityManagement‐II‐Sp10 1 2/3 2/3 1 5/6 2/3 1 1 2/3 2 3/4 2/3 0.9545 0.9759 0.9772 0.9710
30 MGMT 36100 MGMT 36100 ChV:QualityManagementIISp10 5 How is process capability used?
• First we select target process capability ( based on customer requirement / our own standard / industry standard). • Using standard deviation estimate, we determine process capability ratio (Cp). If this exceeds target value, we use the process and periodically calculate Cpk. If Cpk falls below target value, we need some simple adjustments (sharpen the tool, adjust the setting, etc). • If Cp value is less than the target value, we won’t be able to use the process; we need a long term solution – Use better process, train operators, use better quality raw material. Some times tolerance limits can be relaxed, – You need to reduce variability (std. deviation). The effect of reduced variability is shown in the next slide. When σ is large, you produce high % defectives. When μ is shifted, you produce even more % defectives. 2.988 3.000 3.012 2.988 3.000 3.012 See the effect of small σ even when μ shifts. 2.988 3.000 3.012
ChV:QualityManagement‐II‐Sp10 31 ChV:QualityManagement‐II‐Sp10 2.988 3.000 3.012
32 MGMT 36100 MGMT 36100 Acceptance Sampling Acceptance Sampling
Acceptance sampling Check number of items ( based on statistical principles), randomly selected from the entire lot. Various sampling plans have been developed. If the inspection involves counting, we have plans involving attributes. If the inspection involves measurements we have If the inspection involves measurements, we have plans plans involving variables. Types of plans: Single – double – sequential sampling plans. Standards: MIL standard, ISO, … Suppose our supplier has sent us a “lot” containing 1000 items. How do we ensure quality of this incoming lot? 1. 2. 3. 4. Accept the lot w ithout inspection (counting, cursory visual th inspection). inspection). Check 100% items. Check some predetermined number from top layer. Check number of items ( based on statistical principles), randomly selected from the entire lot. ChV:QualityManagement‐II‐Sp10 33 ChV:QualityManagement‐II‐Sp10 34 MGMT 36100 MGMT 36100 Single Sampling Plan
Receive 1000 “N”
Random Sample: 80 n Start inspecting Accept if defective ≤ 3 C Sampling Plan Characteristics:
Sampling plans are based on many assumptions. The The customer is willing to accept agreed % defective. Due Due to sampling, both consumer and supplier face some risk. Consumer’s Consumer’s risk is accepting bad lot because the sample happens to be good. Supplier’s Supplier’s risk is rejection of acceptable lot because the sample happens to contain more defective units. Probability Probability distribution for defective units is assumed.
Second Sample: 40 n2 Accept if total defective ≤ 4 Double Sampling Plan
Receive 1000 N
Random Sample: 60 60 n1 Defective ≤ 1:C1 1:C Defective 4:C Defective > 4:C2 Accept Reject Based Based on % defective acceptable, agreed risk levels and probability distributions for defectives, sampling plans have been prepared. For For different lot sizes, you can find plans giving the values of n, c, etc.
35 ChV:QualityManagement‐II‐Sp10 36 Which plan Single or double sampling? planWhat is done when a lot is rejected?
ChV:QualityManagement‐II‐Sp10 MGMT 36100 MGMT 36100 ChV:QualityManagementIISp10 6 ...
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This note was uploaded on 12/14/2010 for the course MGMT 361 taught by Professor Panwalker during the Spring '10 term at Purdue UniversityWest Lafayette.
 Spring '10
 panwalker
 Management

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