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1
IEOR 130
Spring, 2009, Prof. Leachman
Solutions to HW 5
1. (a) We use the observed best yield to make a 3
σ
estimate of
Y
R
:
0.82 
Y
R
=
400
/
)
1
(
3
R
R
Y
Y
−
]
)[
400
/
9
(
64
.
1
6724
.
0
2
2
R
R
R
R
Y
Y
Y
Y
−
=
+
−
0
0225
.
1
6625
.
1
6724
.
0
2
=
+
−
R
R
Y
Y
7555
.
0
)
0225
.
1
(
2
75012
.
2
7639
.
2
6625
.
1
=
−
±
=
R
Y
The fatal defect density equivalent to this random yield satisfies
e
AD
=
Y
R
, where
A
= 0.5, i.e.,
D
= 0.56.
The inline inspections have found the following fatal defect density:
M1 (0.32)(0.38) =
0.1216
M2 (0.45)(0.50) =
0.225
Poly (0.25)(0.40) = 0.10

0.4466 in total
Thus 0.56 – 0.4466 = 0.1134 in fatal defect density remains to be found. This corresponds to a yield of exp(
0.5*0.1134) = 0.945, i.e., a yield loss of 5.5%.
(b) The overall systematic mechanismslimited yield is
.
794
.
0
7555
.
0
/
6
.
0
/
=
=
=
R
S
Y
Y
Y
The mechanisms identified to date account for (1 – 0.05 – 0.02)(0.98) = 0.9114
Thus about 0.117 in systematic yield loss remains to be explained.
2. c
1
= 4.5, c
2
= 0.5
t
p
t
1F(t)
denominator
numerator
G(t)
Nonavailability= G(t)/168
1
.15
.85
1.0
1.10
1.10
0.00655
2
.20
.65
1.85
1.90
1.027
0.00611
3
.25
.40
2.50
2.90
1.16
0.00690
4
.25
.15
2.90
3.90
1.34
0.00798
5
.15
0
3.05
4.50
1.48
0.00881
Replace O ring every 2 weeks.
Note G(t) above is expressed as hours of down time per weeks of cycle length,
so to express true nonavailability
we would have to convert units to hours of
down time divided by hours of cycle length, i.e., nonavailability equals
G(t)/168.
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3. (a) c
1
= 200, c
2
= 50
t
p
t
sum k*p
k
1F(t)
denominator
numerator
G(t)
1
.08
.08 .
92
1.0
62
62.0
2
.12
.32
.80
1.92
80
41.7
3
.16
.80
.64
2.72
104
38.2
4
.24
1.76
.40
3.36
140
41.7
t=3 gives the lowest cost rate.
(b) Let delta D denote the average defect density contributed by the machine in question. The production rate,
λ
=
10 wafers/day, A=0.5, DY = (0.5)*exp{A*(delta D)}
Avg. revenue per wafer = 2*100*(0.50)*exp{A*(delta D)}
Delta revenue = 100*[1  exp{A(delta D)}] per wafer
If we PM every day, delta D = 0.1 (average of 0 and 0.2). Hence
Delta revenue = 100*[1  exp{.05}] ~ 100*(.05)
If we PM every 3 days, delta D = 0.3 (average of 0 and 0.6). Hence
Delta revenue = 100*[1  exp{.15}] ~ 100*(.14)
We should modify the G(t) function to be
G*(t) = G(t) + {
λ
*{Delta revenue}*denominator} / denominator
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This note was uploaded on 03/16/2010 for the course ORMS 130 taught by Professor Leachman during the Spring '10 term at University of California, Berkeley.
 Spring '10
 Leachman

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