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XBAR and R Charts - 平均值-极差控制图


http://www.sytsma.com/tqmtools/xbar-r.html

Theoretical Control Limits for XBAR Charts

Although theoretically possible, since we do not know either the population process mean or standard deviation, these formulas cannot be used directly and both must be estimated from the process itself. First the R chart is constructed. If the R chart validates that the process variation is in statistical control, the XBAR chart is constructed.

Steps in Constructing an R Chart

  1. Select k successive subgroups where k is at least 20, in which there are n measurements in each subgroup. Typically n is between 1 and 9. 3, 4, or 5 measurements per subgroup is quite common.
  2. Find the range of each subgroup R(i) where R(i)=biggest value - smallest value for each subgroup i.
  3. Find the centerline for the R chart, denoted by

  4. Find the UCL and LCL with the following formulas: UCL= D(4)RBAR and LCL=D(3)RBAR with D(3) and D(4) can be found in the following table:
    Table of D(3) and D(4)
    n    D(3)    D(4)       n    D(3)    D(4)
    2     0      3.267      6     0      2.004
    3     0      2.574      7    .076    1.924
    4     0      2.282      8    .136    1.864
    5     0      2.114      9    .184    1.816
  5. Plot the subgroup data and determine if the process is in statistical control. If not, determine the reason for the assignable cause, eliminate it, and the subgroup(s) and repeat the previous 3 steps. Do NOT eliminate subgroups with points out of range for which assignable causes cannot be found.
  6. Once the R chart is in a state of statistical control and the centerline RBAR can be considered a reliable estimate of the range, the process standard deviation can be estimated using:

    d(2) can be found in the following table:

    n  d(2)     n   d(2)
    2  1.128    6   2.534
    3  1.693    7   2.704
    4  2.059    8   2.847
    5  2.326    9   2.970

Steps in Constructing the XBAR Chart

  1. Find the mean of each subgroup XBAR(1), XBAR(2), XBAR(3)... XBAR(k) and the grand mean of all subgroups using:

  2. Find the UCL and LCL using the following equations:

    A(2) can be found in the following table:
    n   A(2)     n   A(2)
    2   1.880    6   .483
    3   1.023    7   .419
    4    .729    8   .373
    5    .577    9   .337
  3. Plot the LCL, UCL, centerline, and subgroup means
  4. Interpret the data using the following guidelines to determine if the process is in control:
    a.  one pount outside the 3 sigma control limits
    b.  eight successive points on the same side of the centerline
    c.  six successive points that increase or decrease
    d.  two out of three points that are on the same side of the centerline, 
        both at a distance exceeding 2 sigmas from the centerline
    e.  four out of five points that are on the same side of the centerline, 
        four at a distance exceeding 1 sigma from the centerline
    f.  using an average run length (ARL) for determining process anomolies

Example:

The following data consists of 20 sets of three measurements of the diameter of an engine

shaft.

n       meas#1  meas#2  meas#3          Range   XBAR
1       2.0000  1.9998  2.0002          0.0004  2.0000
2       1.9998  2.0003  2.0002          0.0005  2.0001
3       1.9998  2.0001  2.0005          0.0007  2.0001
4       1.9997  2.0000  2.0004          0.0007  2.0000
5       2.0003  2.0003  2.0002          0.0001  2.0003
6       2.0004  2.0003  2.0000          0.0004  2.0002
7       1.9998  1.9998  1.9998          0.0000  1.9998
8       2.0000  2.0001  2.0001          0.0001  2.0001
9       2.0005  2.0000  1.9999          0.0006  2.0001
10      1.9995  1.9998  2.0001          0.0006  1.9998
11      2.0002  1.9999  2.0001          0.0003  2.0001
12      2.0002  1.9998  2.0005          0.0007  2.0002
13      2.0000  2.0001  1.9998          0.0003  2.0000
14      2.0000  2.0002  2.0004          0.0004  2.0002
15      1.9994  2.0001  1.9996          0.0007  1.9997
16      1.9999  2.0003  1.9993          0.0010  1.9998
17      2.0002  1.9998  2.0004          0.0006  2.0001
18      2.0000  2.0001  2.0001          0.0001  2.0001
19      1.9997  1.9994  1.9998          0.0004  1.9996
20      2.0003  2.0007  1.9999          0.0008  2.0003
        
RBAR CHART LIMITS:      
                                                
RBAR  = 0.0005                                          

UCL=D(4)*RBAR = 2.574 * .0005 = 0.001287                
LCL=D(3)*RBAR = 0.000 * .0005 = 0.000           


XBAR CHART LIMITS:                                                      
        
XDBLBAR  = 2.0000                                       
        
UCL  =  XDBLBAR + A(2)*RBAR  =  2.000+1.023*.0005 = 2.0005115
LCL  =  XDBLBAR - A(2)*RBAR  =  2.000-1.023*.0005 = 1.9994885

R - Chart:

XBAR - Chart:







XBAR and s Charts - 平均值-标准偏差控制图


http://www.sytsma.com/tqmtools/xbar-s.html
Theoretical Control Limits for XBAR Charts

Although theoretically possible, since we do not know either the population process mean or standard deviation, these formulas cannot be used directly and both must be estimated from the process itself. First the s chart is constructed. If the s chart validates that the process variation is in statistical control, the XBAR chart is constructed.

Steps in Constructing an s Chart

  1. Select k successive subgroups where k is at least 20, in which there are n measurements in each subgroup. Typically n is between 1 and 9. 3, 4, or 5 measurements per subgroup is quite common.
  2. Find the sample standard deviation of each subgroup s(i).
  3. Find the centerline for the s chart, denoted by

  4. Find the UCL and LCL with the following formulas: UCL= B(4)SBAR and LCL=B(3)SBAR with B(3) and B(4) can be found in the following table:
    Table of B(3) and B(4)
    n    B(3)    B(4)       n    B(3)    B(4)
    2     0      3.267      6    .03     1.970
    3     0      2.568      7    .118    1.882
    4     0      2.266      8    .185    1.815
    5     0      2.089      9    .239    1.761
  5. Plot the subgroup data and determine if the process is in statistical control. If not, determine the reason for the assignable cause, eliminate it, and the subgroup(s) and repeat the previous 3 steps. Do NOT eliminate subgroups with points out of range for which assignable causes cannot be found.
  6. Once the s chart is in a state of statistical control and the centerline SBAR can be considered a reliable estimate of the range, the process standard deviation can be estimated using:

    c(4) can be found in the following table:

    n  c(4)     n   c(4)
    2  .7979    6   .9515
    3  .8862    7   .9594
    4  .9213    8   .9650
    5  .9400    9   .9693

Steps in Constructing the XBAR Chart

  1. Find the mean of each subgroup XBAR(1), XBAR(2), XBAR(3)... XBAR(k) and the grand mean of all subgroups using:

  2. Find the UCL and LCL using the following equations:

    A(3) can be found in the following table:
    n   A(3)     n   A(3)
    2   2.659    6   1.287
    3   1.954    7   1.182
    4   1.628    8   1.099
    5   1.427    9   1.032
  3. Plot the LCL, UCL, centerline, and subgroup means
  4. Interpret the data using the following guidelines to determine if the process is in control:
    a.  one pount outside the 3 sigma control limits
    b.  eight successive points on the same side of the centerline
    c.  six successive points that increase or decrease
    d.  two out of three points that are on the same side of the centerline, 
        both at a distance exceeding 2 sigmas from the centerline
    e.  four out of five points that are on the same side of the centerline, 
        four at a distance exceeding 1 sigma from the centerline
    f.  using an average run length (ARL) for determining process anomolies

Example:

The following data consists of 20 sets of three measurements of the diameter of an engine

shaft.

n	meas#1	meas#2	meas#3		StdDev	XBAR		
1	2.0000	1.9998	2.0002		0.0002	2.0000		
2	1.9998	2.0003	2.0002		0.0003	2.0001		
3	1.9998	2.0001	2.0005		0.0004	2.0001		
4	1.9997	2.0000	2.0004		0.0004	2.0000		
5	2.0003	2.0003	2.0002		0.0001	2.0003		
6	2.0004	2.0003	2.0000		0.0002	2.0002		
7	1.9998	1.9998	1.9998		0.0000	1.9998		
8	2.0000	2.0001	2.0001		0.0001	2.0001		
9	2.0005	2.0000	1.9999		0.0003	2.0001		
10	1.9995	1.9998	2.0001		0.0003	1.9998		
11	2.0002	1.9999	2.0001		0.0002	2.0001		
12	2.0002	1.9998	2.0005		0.0004	2.0002		
13	2.0000	2.0001	1.9998		0.0002	2.0000		
14	2.0000	2.0002	2.0004		0.0002	2.0002		
15	1.9994	2.0001	1.9996		0.0004	1.9997		
16	1.9999	2.0003	1.9993		0.0005	1.9998		
17	2.0002	1.9998	2.0004		0.0003	2.0001		
18	2.0000	2.0001	2.0001		0.0001	2.0001		
19	1.9997	1.9994	1.9998		0.0002	1.9996		
20	2.0003	2.0007	1.9999		0.0004	2.0003		
								
SBAR CHART LIMITS:							
SBAR  =	0.0002						

UCL   =	 B(4)*SBAR   =	2.568*.0002 =	0.0005136		
LCL   =  B(3)*SBAR   =    0 * .0002 =	0.00		

XBAR CHART LIMITS:							
XDBLBAR  =		2.0000		
			
UCL  =	XDBLBAR + A(3)*SBAR   =	2.000+1.954*.0002  = 2.0003908
LCL   =	XDBLBAR - A(3)*SBAR   =	2.000-1.954*.0002  = 1.9996092

s - Chart:

XBAR - Chart:






Median Charts - 中值控制图


http://www.sytsma.com/tqmtools/median.html
Preparing Median Charts

The primary reason for using medians is that it is easier to do on the shop floor because no arithmetic must be done. The person doing the charting can simply order the data and pick the center element. For simplicity, odd numbers of samples are chosen 3, 5, 7, etc. The major disadvantage of using a median chart is that it is less sensitive (powerful) in detecting process changes when extreme values occur.

Traditionally, all subgroup values are plotted, and only the median values are connected by line segments. One must be careful when interpreting the chart that the "out of control" rules are only applied to the median elements.

Steps in Constructing a Median Chart

  1. Either an R chart or s chart is developed as shown on the respective XBAR-r or XBAR-s charts and the process variation is shown to be in statistical control.
  2. If an R chart was used, the control limits are as follows:

  3. If an s chart was used, the control limits are as follows:

    Table of A(6) and A(7)
    
    n    A(6)    A(7)       n    A(6)    A(7)
    
    2    1.880   1.880      6    .549    .580
    3    1.187   1.067      7    .509    .521
    4     .796    .796      8    .434    .477
    5     .691    .660      9    .412    .444
    
    The centerline is XDBLBAR.  Note that it is the subgroup MEANS that determine
    both the centerline and the control limits.
    
  4. Plot the centerline XDBLBAR, LCL, UCL, and the subgroup medians.
  5. Interpret the data using the following guidelines to determine if the process is in control:
    a.  one pount outside the 3 sigma control limits
    b.  eight successive points on the same side of the centerline
    c.  six successive points that increase or decrease
    d.  two out of three points that are on the same side of the centerline, 
        both at a distance exceeding 2 sigmas from the centerline
    e.  four out of five points that are on the same side of the centerline, 
        four at a distance exceeding 1 sigma from the centerline
    f.  using an average run length (ARL) for determining process anomolies

Example:

The following data consists of 20 sets of three measurements of the diameter of an engine shaft. An R-Chart will be used to examine variability followed by a Median Chart.

n       meas#1  meas#2  meas#3          Range   XBAR    Median  
1       2.0000  1.9998  2.0002          0.0004  2.0000  2.0000  
2       1.9998  2.0003  2.0002          0.0005  2.0001  2.0002  
3       1.9998  2.0001  2.0005          0.0007  2.0001  2.0001  
4       1.9997  2.0000  2.0004          0.0007  2.0000  2.0000  
5       2.0003  2.0003  2.0002          0.0001  2.0003  2.0003  
6       2.0004  2.0003  2.0000          0.0004  2.0002  2.0003  
7       1.9998  1.9998  1.9998          0.0000  1.9998  1.9998  
8       2.0000  2.0001  2.0001          0.0001  2.0001  2.0001  
9       2.0005  2.0000  1.9999          0.0006  2.0001  2.0000  
10      1.9995  1.9998  2.0001          0.0006  1.9998  1.9998  
11      2.0002  1.9999  2.0001          0.0003  2.0001  2.0001  
12      2.0002  1.9998  2.0005          0.0007  2.0002  2.0002  
13      2.0000  2.0001  1.9998          0.0003  2.0000  2.0000  
14      2.0000  2.0002  2.0004          0.0004  2.0002  2.0002  
15      1.9994  2.0001  1.9996          0.0007  1.9997  1.9996  
16      1.9999  2.0003  1.9993          0.0010  1.9998  1.9999  
17      2.0002  1.9998  2.0004          0.0006  2.0001  2.0002  
18      2.0000  2.0001  2.0001          0.0001  2.0001  2.0001  
19      1.9997  1.9994  1.9998          0.0004  1.9996  1.9997  
20      2.0003  2.0007  1.9999          0.0008  2.0003  2.0003  
                                                                
RBAR CHART LIMITS:                                                      
RBAR  = 0.0005                                          

UCL   =  D(4)*RBAR   =  2.574*.0005 =   0.001287                
LCL   =  D(3)*RBAR   =  0 * .0005   =   0.00            

XBAR CHART LIMITS:                                                      
XDBLBAR  = 2.0000                                       
                                                                
UCL  =  XDBLBAR + A(6)*RBAR   = 2.000+1.187*.0005  = 2.0005935
LCL   = XDBLBAR - A(6)*RBAR   = 2.000-1.187*.0005  = 1.9994065

R Chart:

Median - Chart:








Individuals Charts - 单值控制图


http://www.sytsma.com/tqmtools/individ.html
Preparing Individuals Charts

An Individuals Chart is used when the nature of the process is such that it is difficult or impossible to group measurements into subgroups so an estimate of the process variation can be determined. This occurs frequently in low volume production situations and in situations in chich continuously varying quantities within the process are process-related variables.

The solution is to artificially create subgroups from the data and then calculate the range of each subgroup. This is done by creating rolling groups (most often pairs) of data through time and using the pairs to determine the range R. The resulting ranges are called moving ranges.

Steps in Constructing an Individuals Chart

  1. A moving range average is calculated by taking pairs of data (x1,x2), (x2,x3), (x3,x4),..., (xn-1,xn), taking the sum of the absolute value of the differences between them and dividing by the number of pairs (one less than the number of pieces of data). This is shown mathematically as:

  2. An estimate of the process standard deviation is given by

    and the three sigma control limits become:

  3. Plot the centerline, XBAR, LCL, UCL, and the process measurement X(i).
  4. Interpret the data using the following guidelines to determine if the process is in control:
    a.  one pount outside the 3 sigma control limits
    b.  eight successive points on the same side of the centerline
    c.  six successive points that increase or decrease
    d.  two out of three points that are on the same side of the centerline, 
        both at a distance exceeding 2 sigmas from the centerline
    e.  four out of five points that are on the same side of the centerline, 
        four at a distance exceeding 1 sigma from the centerline
    f.  using an average run length (ARL) for determining process anomolies

Some Additional Notes:

If the process has been brought into statistical control, the sample standard deviation, s/c(4), can well be used as an estimate of the process standard deviation sigma. This works well unless there are trends in the data (which cause an inflated value of s). When one is certain that no trends exist in the data, s/c(4), will provide considerable more power than MRBAR/1.128. This means that one is more likely to detect out-of-control situations when they exist.

Example:

The following data consists of 20 sets of three measurements of the diameter of an engine shaft. An R-Chart will be used to examine variability followed by a Median Chart.

n	meas#1		MR					
1	2.0000		0.0002					
2	1.9998		0.0000					
3	1.9998		0.0001					
4	1.9997		0.0006					
5	2.0003		0.0001					
6	2.0004		0.0006					
7	1.9998		0.0002					
8	2.0000		0.0005					
9	2.0005		0.0010					
10	1.9995		0.0007					
11	2.0002		0.0000					
12	2.0002		0.0002					
13	2.0000		0.0000					
14	2.0000		0.0006					
15	1.9994		0.0005					
16	1.9999		0.0003					
17	2.0002		0.0002					
18	2.0000		0.0003					
19	1.9997		0.0006					
20	2.0003							
								
INDIVIDUALS CHART LIMITS:							
	
XBAR =	2.0000						
MRBAR  =	0.0004						
								
UCL  =	XBAR + 2.66*MRBAR   =	2.000+2.66*.0004  =	2.001064
LCL   =	XBAR - 2.66*MRBAR   =	2.000-2.66*.0004  =	1.997872

Individuals Chart:

 



Exponentially Weighted Moving Average (EWMA) Charts - 指数权重移动均值控制图


http://www.sytsma.com/tqmtools/ewma.html
An EWMA (Exponentially Weighted Moving Average) Chart is used when it is desirable to detect out-of-control situations very quickly. EWMA charts have a built in mechanism for incorporating information from all previous subgroups, weighting the information from the closest subgroup with a higher weight. Thus, the control/out-of-control decision is made with information from previous subgroups as well as the current subgroup. The chief advantage of EWMA charts is that they detect out-of-control conditions more quickly than XBAR charts and that this detection can be done by using only one rule...being within or outside the 3-sigma limits. The chief disadvantage is the EWMA chart is that it is more difficult to construct. When the subgroup size is n>1, the EWMA chart is an alternative to an XBAR chart; when the subgroup size is n=1, the EWMA chart is an alternative to the Individuals Chart.

Steps in Constructing EWMA Charts

  1. Estimate the process standard deviation, sigma, using RBAR/d(2) or SBAR/c(4) if n1 or by using MRBAR/1.128 if n=1.
  2. Determine a weighting constant, lambda, that weights past and current information. If, for example, lambda=.3, 70% of the weight will be given to past information and 30% to current information. Typically a lambda between .1 and .3 provides a reasonable balance between past and current information and .2 is very common in actual practice.
  3. Determine the points on the EWMA chart denoted by xhat(1), xhat(2),...,xhat(k)k and computed by using the equation xhat(i)=lambda*xhat(i)+(1-lambda)*xhat(i-1). This recursive formula begins by using as its initial value xhat(0)= XDBLBAR.
  4. The control limits are not straight lines in the early stages of the chart. The UCL increases and then stabilizes a fixed distance above the centerline while the LCL decreases and then stabilizes a fixed distance below the centerline.

  5. Plot the centerline, XDBLBAR, the LCL and UCL, and the process measurements XHAT(i).
  6. Interpret the data. The chart is out of control only if a point is outside the +/- sigma limits.

Example:

The following data consists of 20 sets of three measurements of the diameter of an engine shaft. Lambda has been chosen as .2.

Data and Preliminary Computations:
								lambda		
n	meas#1	meas#2	meas#3		Range	XBAR	XHAT	0.2		
0							2.0000	
1	2.0000	1.9998	2.0002		0.0004	2.0000	2.0000			
2	1.9998	2.0003	2.0002		0.0005	2.0001	2.0000			
3	1.9998	2.0001	2.0005		0.0007	2.0001	2.0001			
4	1.9997	2.0000	2.0004		0.0007	2.0000	2.0001			
5	2.0003	2.0003	2.0002		0.0001	2.0003	2.0001			
6	2.0004	2.0003	2.0000		0.0004	2.0002	2.0001			
7	1.9998	1.9998	1.9998		0.0000	1.9998	2.0001			
8	2.0000	2.0001	2.0001		0.0001	2.0001	2.0001			
9	2.0005	2.0000	1.9999		0.0006	2.0001	2.0001			
10	1.9995	1.9998	2.0001		0.0006	1.9998	2.0000			
11	2.0002	1.9999	2.0001		0.0003	2.0001	2.0000			
12	2.0002	1.9998	2.0005		0.0007	2.0002	2.0001			
13	2.0000	2.0001	1.9998		0.0003	2.0000	2.0000			
14	2.0000	2.0002	2.0004		0.0004	2.0002	2.0001			
15	1.9994	2.0001	1.9996		0.0007	1.9997	2.0000			
16	1.9999	2.0003	1.9993		0.0010	1.9998	2.0000			
17	2.0002	1.9998	2.0004		0.0006	2.0001	2.0000			
18	2.0000	2.0001	2.0001		0.0001	2.0001	2.0000			
19	1.9997	1.9994	1.9998		0.0004	1.9996	1.9999			
20	2.0003	2.0007	1.9999		0.0008	2.0003	2.0000			
										
ESTIMATE PROCESS STD DEVIATION:										
RBAR  =	0.0005								
SIGMA =	RBAR/d(2) =.0005/1.693 =0.000721501				
										
EWMA CHART LIMITS:										
XDBLBAR  =	2.0000							
UCL=XHAT(0)+((3*SIGMA)/SQRT(n))*SQRT((lambda/(1-lambda))(1-(1-lambda)^(2*i)))								
LCL=XHAT(0)-((3*SIGMA)/SQRT(n))*SQRT((lambda/(1-lambda))(1-(1-lambda)^(2*i)))								

Control Chart Computations:

Subgrp	CL	LCL		UCL		XHAT
1	2.0000	2.000406569	1.999656764	2.0000
2	2.0000	2.00046109	1.999602243	2.0000
3	2.0000	2.00051173	1.999551604	2.0001
4	2.0000	2.000541506	1.999521827	2.0001
5	2.0000	2.000559683	1.999503651	2.0001
6	2.0000	2.000570994	1.999492339	2.0001
7	2.0000	2.000578111	1.999485223	2.0001
8	2.0000	2.000582617	1.999480716	2.0001
9	2.0000	2.000585482	1.999477851	2.0001
10	2.0000	2.000587308	1.999476026	2.0000
11	2.0000	2.000588473	1.99947486	2.0000
12	2.0000	2.000589218	1.999474116	2.0001
13	2.0000	2.000589694	1.99947364	2.0000
14	2.0000	2.000589998	1.999473335	2.0001
15	2.0000	2.000590193	1.999473141	2.0000
16	2.0000	2.000590317	1.999473016	2.0000
17	2.0000	2.000590397	1.999472936	2.0000
18	2.0000	2.000590448	1.999472885	2.0000
19	2.0000	2.000590481	1.999472853	1.9999
20	2.0000	2.000590502	1.999472832	2.0000

EWMA - Chart:

 
 
Q: 为什么针对计量值的控制图一般都是2个?
A: 一个图监控数据点和数据点之间有多大变化; 一个图监控数据

posted on 2005-03-09 13:41  台风  阅读(2049)  评论(0)    收藏  举报