monitoring high-yield processes monitoring high-yield processes cesar acosta-mejia june 2011

Monitoring High-yield processes

MONITORING HIGH-YIELD PROCESSES

Cesar Acosta-Mejia

June 2011


EDUCATION

– B.S. Catholic University of Peru

– M.A. Monterrey Tech, Mexico

– Ph.D. Texas A&M University

RESEARCH

– Quality Engineering - SPC, Process monitoring

– Applied Probability and Statistics – Sequential analysis

– Probability modeling – Change point detection, process surveillance


MOTIVATION

– High-yield processes

– Monitor the fraction of nonconforming units p

– Very small p (ppm)

– To detect increases or decreases in p

– A very sensitive procedure




ASSUMPTIONS

• Process is observed continuously

• Process can be characterized by Bernoulli trials

• Fraction of nonconforming units p is constant, but

may change at an unknown point of time


Hypothesis Testing

For (level ) two-sided tests

the region R is made up of two subregions R1 and R2

with limits L and U such that

P[X ≤ L] = / 2

P[X ≥ U] = / 2

L U


Hypothesis Testing

Consider testing the proportion p


Hypothesis Testing

The test may be based on different random variables

• Binomial (n, p)

• Geometric (p)

• Negative Binomial (r, p)

• Binomial – order k (n, p)

• Geometric – order k (p)

• Negative Binomial – order k (r, p)


Binomial tests

when p is very small


Test 1

• proportion p0 = 0.025 (25000 ppm)

• test H0 : p = 0.025

against

H1 : p 0.025

• X n. of nonconforming units in 500 items

0.0027


Test 1

Let X Binomial (500,p)

To test the hypothesis

H0 : p = 0.025 against H1 : p 0.025

the rejection region is

R = {x ≤ 2} {x ≥ 25}

since

P[X ≤ 2] = 0.000300 < 0.00135 = /2

P[X ≥ 25] = 0.001018 < 0.00135 = /2


Test 1

Plot of P[rejecting H0] vs. p is

0.0027

0.00000

0.00200

0.00400

0.00600

0.00800

0.01000

0.01200

5000 10000 15000 20000 25000 30000 35000 40000 45000

parts per million

prob

abili

ty

of

reje

ctin

g H

o


Hypothesis Testing

Now consider testing

p0 = 0.0001 (100 ppm)


Test 1

Let X Binomial (n = 500,p)

To test the hypothesis

H0 : p = 0.0001 against H1 : p 0.0001


R = {X ≥ 2}

since

P [X ≥ 2] = 0.0012

For n=500 there is no two-sided test for p = 0.0001.


Test 1

Binomial (n = 500, p = 0.025) Binomial (n = 500, p = 0.0001)


Test 1

For this test a plot of P[rejecting H0] vs. p is

0.0027

0

0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

20 40 60 80 100 120 140 160 180 200 220 240 260

parts per million

P [

rej

ectin

g H

o]


Consider a geometric test for p

when p is very small


Test 2

Let X Geo(p)

To test the hypothesis ( = 0.0027)

H0 : p = 0.0001 against H1 : p 0.0001


R = {X ≤ 13} {X ≥ 66075}

since

P[X ≤ 13] = 0.0013

P[X ≥ 66075] = 0.00135

An observation in {X ≤ 13} leads to conclude that p > 0.0001


0.00270

0.00000

0.00200

0.00400

0.00600

0.00800

0.01000

0.01200

50 100 150 200 250 300

p

P[r

eje

ctin

g H

o]

Test 2

For this test a plot of P[rejecting H0] vs. p is


Another performance measure

of a sequential testing procedure


Hypothesis Testing

Let X1, X2, … Geo(p) iid

Let T number of observations until H0 is rejected

Consider the random variables for j = 1,2,…

Aj = 1 if Xj R P[Aj = 0] = PR

Aj = 0 otherwise

then the probability function of T is

P[T= t] = P[A1 = 0] P[A2 = 0]… P[At-1 = 0] P[At = 1]

= PR [1-PR]t-1


Hypothesis Testing

therefore

T Geo(PR)

Let us consider E[T] = 1/PR as a performance measure

then

E[T] = 1/PR mean number of tests until H0 is rejected

when p = p0

E[T] = 1/


Test 2

Let X Geo(p) q = 1 - p

P [X ≤ x] = 1 – qx

Let the rejection region R = {X < L} {X > U}

then

PA = P [not rejecting H0]

= P [ L ≤ X ≤ U]

= 1 – qU – (1 – qL-1)

= qL-1 – qU

PR = 1 – (1- p )L-1 + (1 - p)U


Test 2

Let X Geo(p)

To test the hypothesis ( = 0.0027)

H0 : p = 0.0001 against H0 : p 0.0001


R = {X < 14} {X > 66074}

then P[rejecting H0] is

PR = 1 – (1 – p)13 + (1 – p)66074

E[T] = 1/PR

when p = p0 E[T] = 1/ = 370.4


Test 2

we want E[T] < 370.4 when p > 0.0001


Test 2

How can we improve upon this test ?

we want E[T] < 370.4 when p > 0.0001


run sum procedure


Geometric chart

A sequence of tests of hypotheses


THE RUN SUM – for the mean


THE GEOMETRIC RUN SUM


THE GEOMETRIC RUN SUM - DEFINITION

• Let us denote the following cumulative sums

SUt = SUt-1 + qt if Xt falls above the center line

= 0 otherwise

SLt = SLt-1 - qt if Xt falls below the center line

= 0 otherwise

where qt is the score assigned to the region in which Xt falls


THE GEOMETRIC RUN SUM - DEFINITION

• The run sum statistic is defined, for t = 1,2,…, by

St = max {SUt, -SLt}

with SU0 = 0, SL0 = 0

and limit sum L


THE GEOMETRIC RUN SUM - DESIGN

• Need to define

region limits (l1, l2, l3 and l5, l6, l7)

region scores (q1, q2, q3 and q4)

limit sum L



• Region limits above and below the center line are not symmetric around the center line.

• To define the region limits we use the cumulative probabilities of the distribution of X Geo (p0)

• Such probabilities were chosen to be the same as those of a run sum for the mean with the same scores


THE GEOMETRIC RUN SUM - EXAMPLE

• If X Geo (p0 = 0.0001)

the region limits are given by

0.00123 = P [X ≤ l1 ]

0.02175 = P [X ≤ l2 ]

0.15638 = P [X ≤ l3 ]

0.50000 = P [X ≤ l4 ]

0.84362 = P [X ≤ l5 ]

0.97825 = P [X ≤ l6 ]

0.99877 = P [X ≤ l7 ]



• If X Geo (p0 = 0.0001)

the region limits are given by

0.00123 = P [X ≤ 13 ]

0.02175 = P [X ≤ 220 ]

0.15638 = P [X ≤ 1701 ]

0.50000 = P [X ≤ 6932 ]

0.84362 = P [X ≤ 18554 ]

0.97825 = P [X ≤ 36280 ]

0.99877 = P [X ≤ 67007 ]



• Conclude H1: p p0 when St L

• Let T number of samples until H0 is rejected

• What is the distribution of T ?

• What is the mean and standard deviation?


RUN SUM (0,1,2,3) L = 5 - MODELING

• Markov chain

• States defined by the values that St can assume

• State space

= {-4,-3,-2,-1,0,1,2,3,4,C}

where

C ={n N | n = …,-6,-5,5,6,…}

is an absorbing state

• Transition probabilities



• Let p1 = P [ X ≤ l1 ]

p2 = P [ l1 ≤X ≤ l2]

p3 = P [ l2 ≤X ≤ l3 ]

p4 = P [ l3 ≤X ≤ l4]

p5 = P [ l4 ≤X ≤ l5]

p6 = P [ l5 ≤X ≤ l6]

p7 = P [ l6 ≤X ≤ l7]

p8 = P [ X > l8 ]

where X Geo (p0)



Transitions from St = 0



• Let T be the first passage time to state C

n. of observations until the run sum rejects H0

• Let Q be the sub matrix of transient states, then

P [T ≤ t] = e ( I – Qt ) J

G (s) = se ( I – s Q )-1 ( I – Q) J

E [T] = e ( I – Q )-1 J

e is a row vector defining the initial state {S0}


Geometric Run sum

For this chart a plot of E[T] vs. p is

0

100

200

300

400

500

600

20 30 40 50 60 70 80 90 100

110

120

130

140

150

160

170

180

ppm

aver

age

run

leng

th


Geometric Run sum

A comparison with Test 2

370.47

0

100

200

300

400

500

600

20 30 40 50 60 70 80 90 100

110

120

130

140

150

160

170

180

ppm

aver

age

run

leng

th


RUN SUM – FURTHER IMPROVEMENT

• Consider a geometric run sum

– No regions

– Center line equal to l4– Scores are equal to X

– Design – limit sum L


NEW GEOMETRIC RUN SUM - DEFINITION

• Let us denote the following cumulative sums

SUt = SUt-1 + Xt if Xt falls above the center line

= 0 otherwise

SLt = SLt-1 - Xt if Xt falls below the center line

= 0 otherwise


NEW GEOMETRIC RUN SUM - DEFINITION

• The run sum statistic is defined, for t = 1,2,…, by

St = max {SUt, -SLt}

with SU0 = 0, SL0 = 0

and limit sum L


NEW GEOMETRIC RUN SUM - MODELING

• Markov chain – not possible

– huge number of states

• Need to derive the distribution of T

• Can show that


NEW GEOMETRIC RUN SUM - MODELING


CONCLUSIONS

• The run sum is an effective procedure

for two-sided monitoring

• For monitoring very small p,

it is more effective than

a sequence of geometric tests

• If limited number of regions

it can be modeled by a Markov chain


TOPICS OF INTEREST

• Estimate (the time p changes – the change point)

• Bayesian tests

• Lack of independence (chain dependent BT)

• Run sum can be applied to other instances

- monitoring - arrival process

monitoring high-yield processes monitoring high-yield processes cesar acosta-mejia june 2011

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