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Economic Design of Charts Under Weibull Shock Models
X
TECHNOMETIRCS, November 1988, VOL. 30, NO.4 P.K. Banerjee and M.A. Rahim
組員： 914011 陳致翔 914015 張泰宏
914016 張裕鳳 914024 胡政宏
導 論 Introduction
Notations and preliminary results
Main results
Numerical illustration
Conclusion
Introduction
Design of control chart sample size sample frequency or interval between samples control limit for the chart
Why use economic design of control
charts？
目的 First proposed
“ The economic design of - control charts to control normal process means.” -- Duncan (1965)
The control limits : Sample size ： n （ Taken from the process every h hours. ）
The process mean may shift from to The design parameters of control charts are n, L and h.
The objective is to determine these parameters to minThe objective is to determine these parameters to minimize the expected total cost per unit timeimize the expected total cost per unit time
x
0 0
LX
背景 Duncan (1956)
Gibra (1975), Montgomery (1980), and Vance (1983) Baker (1971),
Heikes. Montgomery, and Yeung (1974) Montgomery and Heikes (1976)
Hu (1984)
Parkhideh, Behrooz; Case, Kenneth E. (1989)
Chung, Kun-Jen, Lin, Chuan-Neng (1993)
Assumptions
1. The time that the process remains in the in-control state follows a Weibull distribution, its pdf is given by
2. Drawing random samples of size n at times h1 , ( h1 + h2 ) , ( h1 + h2+ h3 ) ……………...
Duncan’s model assume h j =h for all j ( j=1,2,…. ).
3. The time to sample and chart one item is negligible.
4. Production ceases during the searches and repair.
0 ,1 ,0 }exp{)( )1( kttkttf kk
5. Make the following proposition ：h j are defined so as to keep the probability of a shift in an interval, given no shift up to its start constant for all intervals.
This can be achieved by defining the length of the sampling intervals h j ( j=1,2,……) in the following fashion ： h j=[ j1/k-(j-1)1/k ]h1
Note that hj satisfies the basic requirements; that is. (a) and (b) Further, for all j when k=1
.hhh 321 m
j jm h1
lim
1hh j
Assumptions
Define
Define pj ( j = 1, 2, … ) ：
The conditional probability that the unit used in the system will fail during the sampling interval j,
given that it was in the operating state at the beginning of the interval j
qj ： The probability that the unit will fail during the samp
ling interval j
0 ,,.........1 ,0 , 01
jhj
ijj
dttfqdttf
dttfp
j
j
j
j
j
jj and
1
1
1 )()(
)(
Definitions
The expected duration of the in-control period
within the sampling interval j, given that the
shock occurred during this sampling interval
jjj qdttftj
j
/)()(1
..... ,2 ,1j
The expectation for the time in control
during a sampling interval is defined as the
weighted average of ’s with qj ’s as the
respective weights
j
-)
1
11
/11()/1(j
jjj
jj qkq
兩種誤差
shift
alarm
Yes No
Yes
No
Notation as following
jh0h
0Z
n : sample size
: the length of the jth sampling interval (j=1,2,…; =0)
: expect search time associated with false alarm
1Z : expect time to discover assignable cause
2Z : expect time to repair process
0h
1
11121 )()1(
i
ii hZZ
)( 1TE
)( 10 TEZ
21 ZZ
The Expected Residual Times State E(residual cycle length) Probability
Out of control and alarm
Out of control but no alarm
In control and no alarm
In control and false alarm
)1( p
p
)1( p
)1)(1( p
State1. Out of control and alarm
1Z
2Z
sampling
1 2 3 j-10
shiftτ
j……
alarm
A cycle time
Residual time1h
State2. Out of control but no alarm
1Z
2Z
sampling
1 2 3 j-10
shiftτ
j……
alarm
A cycle time
Residual time1h
State3. In control and no alarm
sampling
1 2 3 j-10 j……
1h
State4. In control and false alarm
0Z
sampling
1 2 3 j-10 j……
alarm
A cycle time
1h
1
11121 )()1(
i
ii hZZ
)( 1TE
)( 10 TEZ
21 ZZ
The Expected Residual Times State E(residual cycle length) Probability
Out of control and alarm
Out of control but no alarm
In control and no alarm
In control and false alarm
)1( p
p
)1( p
)1)(1( p
Expected length of the production cycle
ppZ
pppAAph
ZZppAhTE
/)1(
)]1(/[)]1()()1[(
)1()(
0
1
211
Notation as following
0D
a : fixed sample cost
b : cost per unit sampled
L : control limit coefficient
Y : cost per false alarm
L : control limit coefficient
W : cost to locate and repair the assignable
cause : quality cost per hour while in control
: quality cost per hour while out of control
1D
)( 1110 hDDbna
The Expected Residual Costs State E(cost during the current period) E(residual cost)
Out of control and alarm
Out of control but no alarm
In control and no alarm
In control and false alarm
)( 1110 hDDbna
10hDbna
10hDbna
W
RW
)( 1CE
)( 1CEY
State1. Out of control and alarm
1Z
2Z
sampling
1 2 3 j-10
shift
τ
j……
alarm
A cycle time
1h Residual time
State2. Out of control but no alarm
1Z
2Z
sampling
1 2 3 j-10
shift
τ
j……
alarm
A cycle time
Residual time1h
State3. In control and no alarm
sampling
1 2 3 j-10 j……
1h
State4. In control and false alarm
0Z
sampling
1 2 3 j-10 j……
alarm
A cycle time
1h
)( 1110 hDDbna
The Expected Residual Costs State E(cost during the current period) E(residual cost)
Out of control and alarm
Out of control but no alarm
In control and no alarm
In control and false alarm
)( 1110 hDDbna
10hDbna
10hDbna
W
RW
)( 1CE
)( 1CEY
Expected cost per unit of time
)1(/)1(
)1()1(
)]1(/[)]1()()1[(
1
1)()(
12
1
10
11
10
pAhpDppY
pApphD
pppAAphD
WDDp
bnaCE
Numerical lustration
條件式
：equation following thehave wemodel,shock ), Weibull(kUnder the
)1/()]()1()1([h
)1(/)1()(
1
1021
pAppAp
ppAhppZZZTE
WpAphD
AppApD
pApphD
kDDppY
pbnaCEk
)1()-p-(1
)]()1()1([h
)1()1(
)/11()/1]([/)1(
]/1)1/()[()(
211
11
11
/110
)(
)(
TE
CE
Z0找出錯誤警報的平均時間 0.25 hr
Z1找出問題的平均時間 0.25 hr
Z2修復製程的平均時間 0.75 hr
a 固定抽樣成本 $20
b 檢驗一樣本的成本 $4.22
D0製程 in control 時每小時的製造成本 $50
D1製程 out control 時每小時的製造成本 $950
Y 錯誤警報時所需成本 $500
W 發生問題時所需的修復成本 $1100
製程偏移係數 0.5
製程 in control 時錯誤警報機率 L 與的函數製程 out control 沒有發出警報的機率
Example 1
當
抽樣時間為 Non-Uniform 時的最佳參數值如下
n 23
h1 10.04
L 1.56
0.118
0.2019
E(C) / E(T) $231.3
時且 5.0 )3,0002.0 ( ),( WeikWei
Example 2
當 Uniform 抽樣時間與 Non-Uniform 抽樣時間最大差異
5.0 )3,0002.0 ( ),( 且WeikWei
Uniform Non-Uniform
...... 2, 1,j )hexp(--1P k1j ]}h) )1((-h) ( [ exp{--1P kk
j jj
QZhhQZZTE 021 )1/()(
hDYQhQD
hDbna
WDDQbnaCE
10
1
10
11)(
)()1)(()(
hQk
where
k
)/11()/1(
)hjexp(-Q
/1
1j
kk
當
抽樣時間為 Uniform 時的最佳參數值如下
時且 5.0 )3,0002.0 ( ),( WeikWei
n 26
h 2.43 hr
L 1.54
0.1228
0.1571
E(C) / E(T) $259.85
Rahim V.S Duncan
Different Assumptions Rahim’s assumption – Weibull shock model Duncan’s assumption – Exponential shock model 、 Uniform Sampling
調整指數分配參數使得兩種模型的平均數相同可得
代入剛才所得之 E(C) 、 E(T) 中，可求出最佳參數值如下
0.06548902)]3/11()0002.0/1[()]/11()/1[( 13
11
1' kk
n 26
h 2.36 hr
L 1.56
0.1185
0.1614
E(C) / E(T) $261.18
與 Uniform Sampling 中 Weibull 模型並無明顯差異。
小結 In Rahim’s Weibull shock model
Various Sampling 所得的每小時平均花費顯著的小於 Uniform Sampling 中的平均花費
在 Rahim’s Weibull shock model 與 Duncan’s exponential shock model 中，且採用 Uniform Sampling 抽樣方式，則每小時平均花費兩者並無顯著差異
Sensitivity of the design and expected cost
何謂敏感性 ? keep the mean time to failure the same
Change one of the parameter values at a time
TABLE1
TABLE2 TABLE3
Cost of Misspecification in Weibull parameters
No misspecification: (j=1,2,….)are all equal
Misspecification : E(T) 與 E(C) 變得很困難去求得
小結論： The decision variable (n, L) are not sensitive to a moderate degree of
misspecification.
jp
Conclusion
Extensive computational experience indicates that the proposed non-uniform sampling scheme yields a lower cost than that of Hu(1984)uniform sampling scheme.
A fixed-sampling-interval control rule is widely used in practice, mainly because of
its administrative simplicity.
Extensive computational studies suggest that the lengths of the intervals can be rounded off to more convenient lengths . for example ,when k is close to 2, one can redefine
= , = /2 , = /3,and = for j=4,5,…
優點 :比 uniform sampling 有經濟利益缺點 :增加 the cost per hour
1h1h 2h 1h 3h 1h jh 3h
研究與發展 在韋伯分配下，其他管制圖之經濟設計
Rahim MA, Costa AFB (2000) – Xbar and R charts Yang SF, Rahim MA (2000) -- Xbar and S charts
時間序列模型 Ohta H, Kimura A, Rahim A (2002) – Time-varying control chart parameter
動態模型 Parkhideh and Case (1989) – Six decision variables in design methodology Ohta H, Rahim MA (1997) – Reduce to three decision variables
References
THE END
