IEOR 130 Midterm Examination 2023_sol
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University of California, Berkeley *
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MISC
Subject
Industrial Engineering
Date
Jan 9, 2024
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11
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Solutions to IEOR Midterm Examination
Fall 2023, Prof. Leachman
Work all problems. 20 points for each problem, 100 points total. Your answers to the exam must be turned in via email to leachman@berkeley.edu and to jooseung_lee@berkeley.edu by 2pm Sunday Oct. 15. You can email me any questions you have. You also may talk to others in the class about the exam questions, but each student must write up his or her own answers. Grading will be based not just on the correctness of analysis and numerical results but also on the clarity of exposition. You don’t need to turn in spreadsheets of your calculations, but you should show all formulas that you used and the values of all input parameters to those formulas.
1. (20 points) The thickness of a film deposited on wafers at a particular process step is subject to statistical process control. The upper specification limit for the film thickness is 60 angstroms and the lower specification limit is 15 angstroms, i.e., wafers with film thickness more than 60 angstroms or less than 15 angstroms deposited on them are scrapped. At present, the process has considerable variability, with mean film thickness equal to 30 angstroms and standard deviation equal to 10 angstroms.
(a)
(3 points) What kind of control chart(s) should be used to track this parameter?
We should use X bar and R chart. The X bar is used to monitor the mean film thickness, and the R chart is used to monitor the standard deviation of the film thickness.
(b)
(5 points) What is the process performance index for this step?
?
??
= ?𝑖? (
??? − 𝜇
3𝜎
,
𝜇 − ???
3𝜎
) = min (
60 − 30
3 × 10
,
30 − 15
3 × 10
) = ?𝑖?(1, 0.5) = 0
.5 (c)
(7 points) Assume the only yield loss mechanism at this process step is out-of-spec film thickness. What is the yield of this process step?
? =
? − 𝜇
𝜎
?
𝐿𝑆𝐿
=
15 − 30
10
= −1.5
?
𝑈𝑆𝐿
=
60 − 30
10
= 3
Using the standard normal distribution table, we have Z
-1.5
= 0.0668 and Z
3
= 0.9986.
?𝑖??? = 0.9986 − 0.0668 = 0.9319 = 93
.18%
(d)
(5 points) To raise the yield of this step to 95%, what value for the process performance index must be achieved?
A yield of 0.9500 corresponds to about Z => ?
??
= 1.645.
2. (20 points) A stacked wafer map for a particular product with a die area of 0.5 sq cm has been prepared from 450 wafers believed not to have been involved in any defect excursions. The map shows the average yield by die site. Out of 300 total die sites, there are 250 die sites that seem to be free of edge loss or other spatial signatures. The best observed yield in the stack is 72% and this yield occurs at only two sites.
(a)
(8 points) What is the baseline random yield? What is the baseline defect density?
?? − ?
?
= ?𝜎 = 𝜙
−1
(1 −
?
?
)
√
?
?
(1 − ?
?
)
?
= 𝜙
−1
(1 −
2
300
)
√
?
?
(1 − ?
?
)
450
0.72 = ?
?
+ 2.475
√
?
?
(1 − ?
?
)
450
?
?
= 0
.665
= 66
.5%
?
0
= −
ln(?)
𝐴
= −
ln(0.665)
0.5
≈
0.816
???????
?
?
2
(b)
(2 points) The die yield for the product is 48%. What is the systematic mechanisms-limited yield?
?
?
=
?
?
?
=
0.48
0.665
≈ 0.722 = 72
.2%
(c)
(7 points) The following systematic mechanisms have been identified by the yield engineers:
Mechanism
Fraction of wafers
Fraction of dice
Kill rate
Parametric test fails
Metal deposition
0.02
1.00 of dice on wafer
1.00
defect excursions
0.05
0.7 of dice on wafer
0.5*
Edge loss
Contact layer
1.00
0.12 of dice on wafer
1.00
pattern fails
Implant layer
1.00
0.025 of dice on wafer
1.00
pattern fails
1.00
0.05 of dice on wafer
1.00
* 70% of the dice on wafers involved in metal deposition defect excursions experience particle contamination above baseline. For die experiencing such contamination, 50% of such die are found to fail at probe test.
What is the overall systematic mechanisms-limited yield is represented by the above mechanisms? How much systematic mechanisms yield loss remains to be discovered?
?𝑖???
???ℎ𝑎?𝑖??
= 1 − ?????𝑖??
𝑤𝑎????
× ?????𝑖??
?𝑖??
× ?𝑖??
?𝑎??
?𝑖???
?𝑎?𝑎????𝑖?
= 1 − 0.02 × 1.00 × 1.00 = 0.98
?𝑖???
???𝑎?
= 1 − 0.05 × 0.7 × 0.5 = 0.9825
?𝑖???
????
= 1 − 1.00 × 0.12 × 1.00 = 0.88
?𝑖???
????𝑎??
= 1 − 1.00 × 0.025 × 1.00 = 0.975
?𝑖???
𝑖???𝑎??
= 1 − 1.00 × 0.05 × 1.00 = 0.95
?
?_??𝑤
= 0.98 × 0.9825 × 0.88 × 0.975 × 0.95 ≈ 0.785 = 78
.5%
?𝑖???
????
= ?
s_new
− ?
?
= 0.785 − 0.722 = 0.063 = 6.3%
?𝑖??
?
????
= 6.3%
.
(d) (3 points) An engineer in the fab observed there are 2 points of yield loss at parametric test, 1.75 points from metal deposition defect excursions, 12 points from edge losses, 2.5 points from contact pattern fails, and 5 points from implant layer pattern fails, making for a total of 23.25 points of yield loss. Is that analysis incorrect? Why or why not?
No, the analysis is incorrect. The total yield loss should be the product of the individual yield losses, not the sum.
3. (20 points) Chemical mechanical polish (CMP) machines require periodic replacement of the slurry used in polishing because of a build-up of contaminating particles. While the slurry is being replaced, a CMP machine is down for an average 8 hours, including requalification time. All other down time on the machine totals to an average of 2 hours per day. The average process time per lot (
PT
) is 0.5 hours.
The CMP process engineer has studied the deposition of particles on wafers undergoing CMP processing. She has concluded that each processed lot since slurry change increases the particles deposited on the wafers in the subsequent lot by an average of 0.02 particles per sq cm of wafer surface. (The first lot processed after a slurry change does not experience any CMP particles.)
The yield engineer reports that, on average, 25% of particles deposited by CMP become fatal defects.
The fab has one product in production, as follows:
Product
Die size
Gross die per wafer
A
1 sq cm
600
(a)
(5 points) Suppose the baseline fatal defect density for the fabrication process excluding CMP particles is D
0
. Assuming a simple Poisson yield model, express the total good die output in the first n lots processed right after a slurry change. Assume lots contain 25 wafers and line yield is 100%. You don’t need to simplify the resulting expression.
?𝑖???
𝑖
= ?
−(𝐷
0
+0.02×0.25×(𝑖−1)
= ?
−(𝐷
0
+0.005(𝑖−1))
?????𝑖?
𝑖
= ?𝑖???
𝑖
× 25 × 600 = 15000?𝑖???
𝑖
?????𝑖?
???𝑎?
= ∑ ?????𝑖?
𝑖
?−1
𝑖=0
= 15000
?
−𝐷
0
∑
?
−0
.005
(𝑖
−1
)
?
𝑖
=1
(b)
(10 points) The CMP department is wondering how often they should schedule slurry changes. Suppose the frequencies under consideration are: once every 80 lots, once every 100 lots, or once every 120 lots. Assume CMP is the fab bottleneck and the minimum planned idle time for CMP is one hour per day. Which of these three frequencies maximizes good die output?
????
80
=
24 − 1 − 2
0.5 ∗ 80 + 8
=
21
48
????
100
=
24 − 1 − 2
0.5 ∗ 100 + 8
=
21
58
????
120
=
24 − 1 − 2
0.5 ∗ 120 + 8
=
21
68
?????𝑖
?
80
=
21
48
× 15000?
−𝐷
0
∑ ?
−0.005(𝑖−1)
80
𝑖=1
= 433788?
−𝐷
0
?????𝑖?
100
=
21
58
× 15000?
−𝐷
0
∑ ?
−0.005(𝑖−1)
100
𝑖=1
= 428458?
−𝐷
0
?????𝑖?
120
=
21
68
× 15000?
−𝐷
0
∑ ?
−0.005(𝑖−1)
= 419059?
−𝐷
0
120
𝑖=1
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