试验设计（DOE）
培训资料
试验设计
(Design Of Experiment)
指所有的有控制的輸入及有計划的結果分析的試驗。
试验输入因素与输出 (试验)
不同操作者
不同机器
不同班次
供货商/部件
时间
广告
过程
可控制的
输入
过程关键输出
确定
KPIV的工具
鱼骨图
原因与结果矩阵/失效模式及效果分析(FMEA)
多试验研究
短期能力分析
干扰的输入
(连续型的)
干扰的输入
(离散型的)
试验:
是连续变量数据还是
逻辑数据?
数据类型决定
分析方法!
DOE 目标
理解术语
建立简单的试验
解释基本结论
响应(Response): 试验输出的结果
因素(Factor): 试验过程中的不同输入变量
水平(Level): 试验中对因素的不同设定值.
干扰(Noise)：人不可控制的事物
Blocking:将干扰最小化的方法
主要影响(Main Effect)：对单个因素而言, 从一个水平到另一个水平的变化对输出的平均影响
术语
Interaction(交互作用)：两 个因素合起来对总输出的影响将高于两个单独的因素造成的影响
重复 (Replication)： 以随机次序重新做一次试验
随机化 (Randomization)：以一种无固定模式的次序做试验
试 验 策 略
定义问题
确定试验目的
试验输出方式(试验)
确定试验的限制条件
选择输入因素(试验输入变量)
试验因素的水平
选择试验方案
收集数据
分析数据
得出结论
在执行结论之前，做一次确认试验
达到试验目的
试验的目的是 更好地试验真实的世界 , 而不是试验试验数据。
William Diamond
IBM - Retired Statistician
我将有多少种组合方案 ？
总的方案数取决于
有多少种因素
每种因素有多少水平
如果知道因素的个数及每个因素的水平数，将各因素的水平数相乘即得到方案的个数。
方案 = 因素1的水平数 X 因素2的水平数 X 因素3的水平数 X …
例如:
因素 水平 方案数
3 2, 2, 2 2 X 2 X 2 = 8
3 2, 3, 3: 2 X 3 X 3 = 18
2 5, 5 5 X5 = 25
6 2, 2, 2, 2, 2, 2 2X2X2X2X2X2 = 64
确定试验限制条件
确定限制试验可采用方案数与试验次数的限制条件。
试验限制条件可以是试验，钱，人力资源，物质限制等。
决定你将做多少次试验。
结合你的试验目的，选择最佳试验设计
及你可以采用的最多的试验次数。
注意:
1. 不要在第一次试验中用完你所有的资源
2. 成功的DOE 不是一次试验就能试验的，需要有反复
曲奇饼DOE
烘烤时间（A）
烘烤温度（B）
目标：改进曲奇的口味和外观
试验设计- 例子1
DOE设计及结果
烘烤时间 烘烤温度
(min) (oF) 口味
A B
6 375 41
10 375 50
6 450 47
10 450 35
问题：
那一个因子比较重要？
如何设定重要的因子？
全因子实验示例1－曲奇饼DOE
DOE结果分析
实用性分析
图形分析
定量分析
全因子实验示例1－曲奇饼DOE
实用性分析
烘烤时间 烘烤温度
(min) (oF) 口味
A B
6 375 41
10 375 50
6 450 47
10 450 35
全因子实验示例1－曲奇饼DOE
图形分析
烘烤时间 烘烤温度
(min) (oF) 口味
A B
6(-) 375 (-) 41
10(+) 375 (-) 50
6(-) 450 (+) 47
10 (+) 450 (+) 35
全因子实验示例1－曲奇饼DOE
定量分析
开始分析之前，用（＋）、（－）号表示每一个水平
烘烤时间 烘烤温度
(min) (oF) 口味
A B
－ － 41
＋ － 50
－ ＋ 47
＋ ＋ 35
烘烤时间 烘烤温度
(min) (oF) 口味
A B
6 375 41
10 375 50
6 450 47
10 450 35
全因子实验示例1－曲奇饼DOE
定量分析
因子的影响＝高水平平均值－低水平平均值
在本例中，
A的主要影响＝ ＝ - 44 = -
B的主要影响＝ ＝ 41- = -
全因子实验示例1－曲奇饼DOE
交互作用 － 一个因子的影响取决于另一个因子的水平
烘烤时间 烘烤温度
(min) (oF) 口味
A B AB
6(-) 375 (-) ＋ 41
10(+) 375 (-) － 50
6(-) 450 (+) － 47
10 (+) 450 (+) ＋ 35
全因子实验示例1－曲奇饼DOE
交互作用 － 交互作用的水平可由因子的水平值相乘得出
烘烤时间 烘烤温度
(min) (oF) 口味
A B AB
6(-) 375 (-) ＋ 41
10(+) 375 (-) － 50
6(-) 450 (+) － 47
10 (+) 450 (+) ＋ 35
A x B = AB
1 - x - = +
2 + x - = -
3 - x + = -
4 + x + = +
交互作用的影响
AB= (41+35)/2 - (50+47)/2 = -
全因子实验示例1－曲奇饼DOE
DOE结果分析总结
因子 影响
时间
温度
时间×温度
试验设计- 例子2
假如你在看电视高尔夫节目中对所有声称能帮助你提高积分(通过增远击球的距离)的广告很有兴趣。你不确定这些球棒和球如何能提高球的距离，但是你急于提高自己的水平，于是你从你朋友处借来球和球棒：
1) 两种球棒 a) Ping b) Callaway
2) 两种球 a) Titliest b) Pinnacle
你平时在两个 草地打球，而这两个草地的风力不同。A草地四面环山，几乎没有什幺风；B草地位于草原上，因此有很大的风。
你拿到了工具。
你查明了天气。
你知道该怎幺做来提高你的成绩?
让我们来设计一个实验 . . .
试验策略
定义问题:
提高我的Golf分数。
确定目的:
增加击球的距离。
选择响应(输出) :
游戏结束的最终分数。
分数是我的(KPOV)
试验策略
选择因素的水平 :
试验次数/方案(Treatment)= 2*2*2=8
因素 水平 1 水平 2
球棒 Ping Callaway
球 Titleist Pinnacle
天气/场地 有风/B 无风/A
建立矩阵
游戏 球棒 球 风
1 Ping T球 有风
2 Ping T球 无风
3 Ping P球 有风
4 Ping P球 无风
5 Callaway T球 有风
6 Callaway T球 无风
7 Callaway P球 有风
8 Callaway P球 无风
无风
Titleist 球
Ping 球棒
大风
Titleist 球
Callaway 球棒
最终分数
(试验设计的结构)
无风
Titleist
Ping 球棒
Pinnacle
Titleist
Callaway 球棒
Pinnacle
1
2
3
4
5
6
7
8
有风
(245)
(210)
(215)
(205)
(200)
(240)
(220)
(215)
Avg. =
Avg. =
Avg. =
Avg. =
Titleist
Avg. =
Pinnacle
Avg. =
结论:
Graph >Boxplot Y - 距离
X - 球，球棒，風
Stat >Quality Tools>Run Chart Y - 距离
X - 球，球棒，風
生成主效应图 (STAT>ANOVA>Main Effects)
主效应图 - 球棒 和风的影响显得较重要.
什么是主效应图?
1) 对球棒 低水平-1時 所有距离值的平均數
2) 对球棒 高水平1時的所有距离值的平均數
主要影响 = 210+215+205+200=207
Ping 球棍 4
主要因素 =
Callaway
245+240+220+215= 230
4
因素: 球棒，球，风
响应: 距离
作相互作用图 (STAT>ANOVA>Interactions)
相互作用图
球棒与风速之间可能存在相互作用.
相互作用图从何而来
#1)对低水平球棒在低水平风速状态下所击出的所有距离取平均值。
205+200=
2
#2)对低水平球棒在高水平风速状态下所击出的所有距离重复相同计算。
210+215=
2
#3)对高水平球棒在低水平风速状态下所击出的所有距离重复相同计算。
220+215=
2
#4)对高水平球棒在高水平风速状态下所击出的所有距离重复相同计算。
245+240=
2
因素: 球棒 球 风
试验: 距离
分析数据
试验策略
用分析连续变量的原则
运行 ANOVA Procedure
Select Residuals and Fits in Storage
Distance
Balls Wind Clubs
P值<， 表明有统计上的显著性
ABC
AC
C
BC
AB
B
A
Pareto Chart of the Standardized Effects
(response is Distance, Alpha = .05)
A:
Clubs
B:
Wind
C:
Ball
Fractional Factorial Fit
Estimated Effects and Coefficients for Distance (coded units)
Term Effect Coef StDev Coef T P
Constant
Clubs
Wind
Ball
Clubs*Wind
Clubs*Ball
Wind*Ball
Clubs*Wind*
Analysis of Variance for Distance (coded units)
Source DF Seq SS Adj SS Adj MS F P
Main Effects 3
2-Way Interactions 3
3-Way Interactions 1
Residual Error 8
Pure Error 8
Total 15
How do the factors affect the response? 各因素怎样影响试验响应？
How do the combinations (interactions) of factors affect the response?各因素之间的相互作用怎样影响试验响应？
We can write the equation that answers these questions 可以用以下公式进行预测
Y = f (X1, X2, X3, …, Xn)
Y (Response) = Distance
Xi (Factors) = Club, Ball, Wind
Distance= Constant +
Club Effect + Ball Effect + Wind Effect +
Club*Ball Interaction Effect + Club*Wind
Interaction Effect + Ball*Wind Interaction Effect +
Club*Ball*Wind Interaction Effect
Fractional Factorial Fit
Estimated Effects and Coefficients for Distance (coded units)
Term Effect Coef StDev Coef T P
Constant
Clubs
Wind
Ball
Clubs*Wind
Clubs*Ball
Wind*Ball
Clubs*Wind*Ball
Analysis of Variance for Distance (coded units)
Source DF Seq SS Adj SS Adj MS F P
Main Effects 3
2-Way Interactions 3
3-Way Interactions 1
Residual Error 8
Pure Error 8
Total 15
Distance= + *Clubs + *Ball + *Wind +
*(Club*Ball) + *(Club*Wind) - *(Ball*Wind ) + * (Club*Ball*Wind )
Fractional Factorial Fit
Estimated Effects and Coefficients for Distance (coded units)
Term Effect Coef StDev Coef T P
Constant
Clubs
Wind
Ball
Clubs*Wind
Clubs*Ball
Wind*Ball
Clubs*Wind*Ball
Analysis of Variance for Distance (coded units)
Source DF Seq SS Adj SS Adj MS F P
Main Effects 3
2-Way Interactions 3
3-Way Interactions 1
Residual Error 8
Pure Error 8
Total 15
由于只有Clubs, Wind, Balls 的Main Effect 和 Clubs*Wind, Wind*Ball 的 Interaction Effect is Significant (p <)
Distance= + *Clubs + *Ball + *Wind + *(Club*Wind) - *(Ball*Wind )
球棒, 球和风以及球棒*风, 球*风之间的相互作用 对提高成绩有显著影响
Distance= + *Clubs + *Ball + *Wind
+ *(Club*Wind) - *(Ball*Wind )
Distance Max = + - + + +
=
为得到最佳成绩， 必须使用Callaway球棒、Titliest球在有风的场地上练习
Key
1 Callaway
-1 Ping
1 Windy
-1 No wind
1 Pinnacle
-1 Titliest
试验结论
试验策略
重复(Replication) : 以随机的次序把整个试验重做一次，而不是按同样的次序把试验再做一次。
随机化 (Randomization): 以一种无固定模式的次序来做试验。
试验需考虑的2个原则：
重复(Replication) 和
随机化(Randomization) :
试验设计(DOE)的R&R
有效试验的障碍
问题不清楚
试验目的不清楚
不理解DOE的策略和工具
试验结果不清楚
DOE 成本太高
DOE 浪费时间 -- 我们需要即时的结果
不理解DOE的好处
对 DOE缺乏信心
DOE 试验
Customers of CHI (Cellulose Helicopters Inc.) have been complaining about the limited flight time of CHI helicopters. 公司的客户 CHI 抱怨 公司直升机 的 巡航 飞行时间 太短
Management wants to increase flight time to improve customer satisfaction. 公司 决定 通过 提高直升机的 巡航飞行时间 来提高 客户 满意度 。
You are put in charge of this improvement project. 你 将 负责该项目
How would you approach this problem?
你将 怎样 来 解决 这 个 问题 呢 ？
The Standard Design
直升机的标准设置
别针处
Factor因子
Standard标准 (-)
Paper Clip 别针
无
Wing Width 机翼宽度
宽
Body Length机身长度
长
Wing Length机翼长度
长
可变化方式 (+)
裁开
折线 (1/3 处)
裁开
折线
1CM
有
窄一半
短3CM
短1CM
别针 机翼宽度 机身长度 机翼长度 飞行时间
-1 -1 -1 -1
1 -1 -1 -1
-1 1 -1 -1
1 1 -1 -1
-1 -1 1 -1
1 -1 1 -1
-1 1 1 -1
1 1 1 1
-1 -1 -1 1
1 -1 -1 1
-1 1 -1 1
1 1 -1 1
-1 -1 1 1
1 -1 1 1
-1 1 1 1
1 1 1 1
标准顺序
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
全因子实验结果记录
*
Reducing the Size of Experiments
减少试验次数
*
部分因子DOE学习目的
Recognize the need of DOE design to reduce number of experimental runs
认识减少DOE设计的试验次数的必要性
Define factional factorial DOE
定义部分因子DOE
Describe the generation of a half fractional factorial DOE design.
描述半因子DOE设计的产生
Define and explain “Confounding”
定义和解释“混淆”的概念
需要运行多少次实验...
对于有k个因素的 2 水平全因子试验试验次数=2k
Number of Factors
1
2
3
4
5
6
7
8
9
10
•
•
•
15
•
•
•
20
2
4
8
16
32
64
128
256
512
1024
•
•
•
32,768
•
•
•
1,048,576
The factorial strategy is an efficient approach to experimentation as compared to “one at a time.” 全因子的设计方法与“一次只考虑一个因素”的方法相比，更高效全面
This can result in a large number of runs, even with a relatively small number of factors.
即使因素相对较少，试验次数也很可观
Number of Runs
*
从2的阶乘实验中可以获得的信息
Number of
Factors
1
2
3
4
5
6
7
8
9
10
•
•
•
15
•
•
•
20
1
2
3
4
5
6
7
8
9
10
•
•
•
15
•
•
•
20
Main
Effects
2-way
Interactions
–
1
3
6
10
15
21
28
36
45
•
•
•
105
•
•
•
190
Higher Order
Interactions
–
–
1
5
16
42
99
219
466
968
•
•
•
32,647
•
•
•
1,048,365
因子数
主效应 两因素相互作用 更高的相互交互作用
*
可获得信息的例子...
...from a Full Factorial (4 Factors) 4个因子的全因子实验可获得的信息
Overall Average 平均值
Main effects主效应: A B C D
2-way interactions2因子相互作用: AB AC AD BC BD CD
3-way interactions3因子相互作用: ABC ABD ACD BCD
4-way interactions4因子相互作用: ABCD
Number
1
4
6
4
1
*
Reducing the Size of a Factorial Experiment 减少实验次数
–
+
–
+
–
+
–
+
A
1
2
3
4
5
6
7
8
Std.
Order
–
–
+
+
–
–
+
+
B
–
–
–
–
+
+
+
+
C
Equipment delays allow you time to run only 4 trials in the allotted time.
Which 4 trials will you choose?
如果只允许进行4次试验，你会怎样选择？
A
B
C
7
8
3
4
5
6
1
2
*
Choosing the Half Fraction 半因子设计
–
+
–
+
–
+
–
+
A
Std.
Order
–
–
+
+
–
–
+
+
B
–
–
–
–
+
+
+
+
C
1
2
3
4
5
6
7
8
We can select:
A
B
C
7
8
3
4
5
6
1
2
or
The 4 shaded trials4个有阴影的实验
+
-
–
+
A
–
+
–
+
B
–
–
+
+
C
Std.
Order
2
3
5
8
The 4 unshaded trials 4个无阴影的实验,
–
+
+
–
A
–
+
–
+
B
–
–
+
+
C
Std.
Order
1
4
6
7
Half Fraction designs use Half the runs of Full Factorial designs.
半因子设计只用全因子设计试验次数的一半
Design设计 Number of runs实验次数
Full Factorial全因子 2k = 23 = 8
Half Fraction 半因子 2k-1 = 23-1 = 22 = 4
*
Constructing a Half Fraction for Four Factors 4因子的半因子实验设计
A
B
–
+
–
+
–
+
–
+
–
–
+
+
–
–
+
+
–
–
–
–
+
+
+
+
C
–
+
+
–
+
–
–
+
D = ABC
Full factorial
for 3 factors 3因子的全因子设计
List the full factorial for three factors. This is called the base design. 列出3个因子的全因子试验
The fourth factor is assigned to the 3-factor interaction for the other three factors.第4个因子是其它3 个因子 的相互作用
Recall: a 3-factor interaction column is obtained by multiplying the three main effect columns. So D = ABC. 3个因子的相互作用可通过前3个因子的相乘获得
*
General Rule for Constructing a Half Fraction for k Factors
Define the base design as a full factorial for the first k-1 factors. 采用 k-1 个因子构成 试验的基础
Assign the kth factor to the interaction of the first k-1 factors from the base design.第k个因素 是 k-1个 因素的相互作用。
This interaction is found by multiplying the factor level settings together for the first k-1 factors from the base design. 用基础设计的前K-1个因子的因子水平设置相乘可获得它们的相互作用
实验次数计算Number of runs = N = 2k-1
*
Exercise
Half Fraction of a 25 Factorial 5个因子的半因子实验设计
Generate a half fraction experiment for
k = 5 factors, A, B, C, D, and E.
A
B
C
D
E
Factors
*
Answers
Half Fraction of a 25 Factorial
–
+
–
+
–
+
–
+
–
+
–
+
–
+
–
+
A
–
–
+
+
–
–
+
+
–
–
+
+
–
–
+
+
B
–
–
–
–
+
+
+
+
–
–
–
–
+
+
+
+
C
–
–
–
–
–
–
–
–
+
+
+
+
+
+
+
+
D
+
–
–
+
–
+
+
–
–
+
+
–
+
–
–
+
E
*
Trade-Offs
Between Full Factorial and Half Fraction Designs 比较全因子与半因子
Number of Effects Computed 可计算的实验效果的数量
Effects
Mean
Main Factors
2 Factor Int.
3 Factor Int.
4 Factor Int.
5 Factor Int.
Total Effects
Computed计算的总效果
Full Factorial
Half Fraction
1
5
10
10
5
1
32
1
5
10
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也是实验的次数
• Are the additional runs worth it? 额外的实验次数是否值得?
• What happens to the higher order interactions? 高次的相互作用如何?
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A 2-Factor Experiment with Confounded Effects 2因子实验中的混淆
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The effects of Factors A and B are confounded. A因素和B因素互相混淆
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Confounding is the combining of the effects of two or more factors into one resulting number such that the magnitude of the effects of the individual factors cannot be separated. 混淆是指两个或以上因素的作用组合成一个结果，因此单个因素的作用无法分开。
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Confounding in the Half Fraction半因子实验中的混淆
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For the half fraction, each letter must be present on one side of the equal sign or the other.在半因子实验中, 等式左右的因子作用混淆了, 所以可用一边的字母代替另一边字母的作用
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Effects Plot for the Full Factorial 全因子的效果图
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Effects from the Full Factorial
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Effects Plot for the Half Fraction半因子试验的 效果图
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Magnitude of effect
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Summary of the Half Fraction半因子试验的总结
The half fraction of a full factorial can often provide the same information as the full factorial, with only half the number of runs.半因子试验往往能够提供同样的信息 ， 却只需 进行 一半的试验
• Fewer runs saves time and money.经济
• More complicated to analyze (must understand confounding).分析更复杂 ， 必需理解存在混合效应
• In designs with few runs, important effects (such as 2-way interactions) are confounded.在实验次数较少的设计中, 有一些因素的效应混合了
Benefits
Costs
Note: Some texts (and Minitab) call confounding - aliasing.
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DOE 基础 — 小测验
1) T/F DOE is an iterative process....one DOE generally leads to another.
DOE是一个反复的过程… 一个DOE通常引导出另一个。
2) T/F The experimental design should be determined by the objective and constraints.
试验设计应该由目的和限制条件所决定。
3) The following is are conditions for a DOE:
下列是一个DOE的条件：
Account Manager 客户经理- Joe, Sue, Kelly, & Don
Sales Region销售区域 - North, South, East, West
Product Type产品类型 - Fluorescent, CFL
How many factors are there?
有多少个因素？
What are the factors?
什么因素？
How many levels are there for each factor?
每个因素有多少个水平？
What is the total # of treatment combinations?
要试验多少次？
4) T/F In order to run a DOE the response must be variables type data.
为了执行DOE，输出结果必须是连续变量。
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DOE is simply a method of ensuring that the data collected during a trial or test is properly controlled and analyzed.
Examples of DOE’s include:
A. Testing a coating weight of grams/bulb and grams/bulb and analyzing the impact on 100 hr. lumens of the two different coating weights.
B. Testing an exhaust head over ride spring with a k value of 86 N/m against a spring with a k value of 76 N/m and analyzing the impact on cracked flanges.
C. Testing the impact of changing the amount of epoxy in a base on the torsion required to remove the base.
D. Testing the coating weight, fill pressure, lehr temperature and E-mix weight all at once to determine the optimal combination.
B. When do we use a DOE?
1. During the improvement phase of 6 sigma
2. When we must decipher what variables are most important
C. Why do we use DOE
1. No human can decipher interactions especially beyond the 2nd order interactions. This needs to be statistically.
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These are the basic objectives of this training module
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Response:
Output(s) measured from the experiment. (examples:
shrinkage, defects such as drapes, no mercury and lens off set)
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Factor
: Input(s) varied during the experiment. (examples:
temperature, time, tension)
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Level:
Specific settings of the factor in the experiment.
(example: temperature, settings of 300 and 400 degrees)
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Noise -
Things you cannot control. (examples: humidity,
outside temperature)
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Blocking -
A method of minimizing interference. (example:
when testing the impact of a change in exhaust tubes on mercury
dosing, samples would be taken from a single head to eliminate
the noise created from head to head variation.)
TERMS
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This example is a demonstration on how to set-up the matrix and then will be used later on to explain main effects and interactions.
Have the class identify the factors and levels.
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This strategy slide describes takes us back, in concept, to the problem statement module. Recognize, though, that most Associates will not necessarily be involved in “Designing” the experiments . . . rather, they are likely to participate in “Running” an experiment already designed by a BB or GB.
If so, either:
1. remove this slide so as not to confuse the students
2. explain that this is not something for which they are responsible but you are sharing the background of what others will be doing.
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Ask the class what they would intuitively draw from this group of data and record the guesses at the bottom of the page and then come back to it after you have completed the Main Effects and Interactions.
Remember, in golf, LOW scores are good. This is not intuitive in our cultures where golf is not a common nor popular game.
Leading in to the next slide, Main Effects, have the students calculate the average scores, just for each pair of factors. For example, calculate the average of the scores for Ping clubs and then for Callaway clubs:
Ping = (76+78+78+79)/4 = 77
Callaway = (80+83+83+82)/4 = 82
do the same for the ball type and wind type pairs.
Conclusions that can be drawn:
1) Remind student’s that box plots are particularly useful for identifying data points which are outliers. From these Box Plots we don’t see any outliers.
2) The Horizontal line in the middle of the graph is the median. From the medians it looks like ball is not a KPIV but Club and Wind are.
3) Can’t draw any statistically significant conclusions yet...haven’t applied any statistics.
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In the Main Effect Plot, the more inclined the line, the more effect the factor has on the response variable. The more horizontal, the least effect.
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In the Interaction Plot, you look for parallelism between the lines to define the the importance of the interaction. The more parallel, the least interaction between the factors.. Consequently, the more perpendicular they are, the more interaction between the two factors.
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In the Introduction we defined some of the terms associated with DOE such as Levels, Factors and Response Variables. Before we move forward, let’s review another DOE Terms:
Replication: going back to the golf example we did in the introduction, this means going back and playing all the combinations we played. So now we will have two results per treatment.
Randomization: read the definition and mention that Minitab does this automatically. When you develop a DOE matrix using Minitab, randomization is a default. We will show you this later.
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There are barriers...the strategy forces you to blow through these barriers, the structure supports it.
Do not go through all of them. Choose the ones that most of the people relate to and move on.
Full factorial is time consuming and costly as the number of factors increase.
This chart shows what type of information is available from 2-level full factorial designs.
Note the huge number of higher order interactions as the number of factors increase.
A full factorial experiment will give you all the information on every possible main effect and interaction effect. For 20 factors you will get information on 20 + 190 + 1,048,365 = 1,048,575 effects. From the previous page, this relates to 1,048,576 runs!
The more effects you are evaluating the more runs you need to make.
Many of the higher order interactions do not exist. Most systems are dominated by some of the main effects and low-order ineractions.
Number of Runs = Number of effects + 1
For 4 factors this shows every possible effect (main and interaction) and the overall average. Evaluating the average is what the extra run (from the previous page) is used for.
If you can’t afford all the runs of a full factorial experiment, what can you do?
Which trials do you choose to run?
Choose the runs that cut through the cube the best. —Either pattern, the dark or the light circles.
Which runs would you choose if you only had
2 trials? Waste of time!
3 trials? Waste of time!
You don’t get enough coverage.
Other bad ideas:
1, 2, 3, 4 only covers one plane.
1, 2, 3, 5 only covers 1/6 of the design
space
There are only 2 good solutions to this problem: the dark or the light pattern. In practice the dark and light selections yield the same results.
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This shows which runs correspond to the light and dark patterns on the cube.
Bonus Question:
What happens if one of the variables is determined to be unimportant? You can run a full factorial design on the EXISTING data using only the two important variables.
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How many runs are there for a 4 factor half fraction design?
4 factors: k = 4
N = 2k-1 = 24-1 = 23 = 8
ABC is called the design generator. When this design is created using Minitab, it will be listed in the session window as “Design Generator = ABC.”
Generally higher level interactions are quite unlikely to exist.
If you do not have any reason to believe that there are high level interactions in your process, go with a half fraction design.
If you are interested in higher level interactions, go with a full factorial design.
You can’t tell if the change in response is due to A or B with this design.
Confounded effects are also referred to as Aliases. The meaning of confounded effects is explained with an example to follow.
When computing effects due to the factors, we take the difference between the average of the response at the high values and
the average of the response at the low values. For A, B, the effects due to the factors will be the same. That is, we cannot distinguish between the effect on the response due to factor A or factor B.
Can you determine what setting factor A, B should be set at in order to improve the average response from »90 to »180? Do we change A, B, or both?
When you use a half fraction design there are confounded effects. This will make the interpretation of your design results more difficult.
For each confounded set of effects, you will not be able to tell which effect the change in response is due to. You may be able to use your understanding of the process to guess which effect is significant.
Let’s look at an example. Suppose that the main effects A, B, C, & D are confounded with the effects of the 3-way interactions BCD, ACD, ABD & ABC respectively. Then, if you believe that there are no 3-way interactions present in your process, you can conclude that the response for the first four (A = BCD, B = ACD, C = ABD, D = ABC) are due to the main effects (A, B, C, D).
This is the result from the full factorial experiment.
The full factorial experiment confirms that A, B, and AB have a significant effect on diameter.
This is the result from the half fraction experiment.
The confounded effects are shown by a + sign.
Without running the full factorial experiment, would you know which effect produced the result? Is the greatest magnitude of effect due to A or BCD?
Main effects (A) are more likely to be significant than interaction effects (BCD). Therefore, it is more likely that A and B were significant (compared to BCD and ACD).
Since A and B are likely to be significant, then AB is more likely to be significant than CD.
Choose: A, B and AB. Does the full factorial experiment provide the same results?
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