[讨论]关于工程收率标准和lot size关系的研讨

这是我对公司lot size change后yield limit所作的研讨，下文是给我的国外的mbb
的请教信，所以是英文的，请见谅，希望各位多提意见。

our line has decreased lot size : from 430/lot to 150/lot
but yield limit has not been changed, it is 99.5%, then if 1 product failure will lead to low yield lot. summary data as followed:
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lot total q'ty which lot size is 150 is 29, but 9 lots is low yield lot, if transfer to big lot(430/lot), low yield lot is 0.
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we think it is not reasonable. followed is my study:

[]Cumulative Distribution study:[/]

if module ppm is 800, and lot size is 150.
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Cumulative Distribution Function

Binomial with n = 150 and p = 0.000800000

x P( X <= x )
2.00 0.9997


Cumulative Distribution Function

Binomial with n = 150 and p = 0.000800000

x P( X <= x )
1.00 0.9934


probability Density Function

Binomial with n = 150 and p = 0.000800000

x P( X = x )
0.00 0.8869
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if module ppm is 800, and lot size is 430.
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Cumulative Distribution Function

Binomial with n = 430 and p = 0.000800000

x P( X <= x )
2.00 0.9948
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>it means: base on yield limit, if lot size is 430 low yield lot ratio is 0.52%
if lot size is 150 low yield lot ratio is 11.31%
there is a great difference.

[]2 proportion research:[/]

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Test and CI for Two Proportions

Sample X N Sample p
1 146 150 0.973333
2 995 1000 0.995000

Estimate for p(1) - p(2): -0.0216667
95% CI for p(1) - p(2): (-0.0478167, 0.00448340)
Test for p(1) - p(2) = 0 (vs not = 0): Z = -2.81 P-Value = 0.005


Test and CI for Two Proportions

Sample X N Sample p
1 147 150 0.980000
2 995 1000 0.995000

Estimate for p(1) - p(2): -0.015
95% CI for p(1) - p(2): (-0.0378268, 0.00782678)
Test for p(1) - p(2) = 0 (vs not = 0): Z = -2.06 P-Value = 0.039


Test and CI for Two Proportions

Sample X N Sample p
1 148 150 0.986667
2 995 1000 0.995000

Estimate for p(1) - p(2): -0.00833333
95% CI for p(1) - p(2): (-0.0272019, 0.0105352)
Test for p(1) - p(2) = 0 (vs not = 0): Z = -1.22 P-Value = 0.221


Test and CI for Two Proportions

Sample X N Sample p
1 149 150 0.993333
2 995 1000 0.995000

Estimate for p(1) - p(2): -0.00166667
95% CI for p(1) - p(2): (-0.0154036, 0.0120703)
Test for p(1) - p(2) = 0 (vs not = 0): Z = -0.26 P-Value = 0.792

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[quote]>after compare with yield limit 99.5%, we can see there will observe difference clearly only
when defect count is more than 4 in one 150 lot.
and base on 95% CI, we can calculate low yield interval is (98%,100%)

[]my conclusion:[/]

yield limit for 150lot should be 99.3%, it means 1lot allow 1 product fail.
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yield limit low yield lot ratio
as is(150 lot) 99.5% 11.31%
to be (150 lot) 99.3% 0.66%
430 lot 99.5% 0.52%
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after change low yield lot ratio for 150lot is a little high than 430lot, but i think it is reasonable and acceptable.

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