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2.830J / 6.780J / ESD.63J Control of Manufacturing Processes (SMA 6303)
Spring 2008
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SOLUTIONS
Massachusetts Institute of Technology
Department of Mechanical Engineering
Department of Electrical Engineering and Computer Science
2.830J/6.780J Control of Manufacturing Processes
Spring 2008
Quiz #1
Thursday – March 13, 2008
In all problems, please show your work and explain your reasoning. Statistical tables for the
cumulative standard normal distribution, percentage points of the
χ
2
distribution, percentage
points of the
t
distribution, percentage points of the
F
distribution, and factors for constructing
variables for control charts (all from Montgomery, 5
th
Ed.) are provided.
Problem 1
[35%]
Part (a)
*
[10%]
Control charts for
x
,
R
, and
s
are to be maintained for the threshold voltage of shortchannel
MOSFETs using sample sizes of
n
= 10. It is known that the process is normally distributed with
µ = 0.75 V and
σ
= 0.10 V. Find the centerline and ±3
σ
control limits for each of these charts.
All units in parts (a) and (b) are in V.
(1)
x
chart:
Center =
μ
= 0.75
3
σ
U
C
L
=
μ
+
=
0.845
3
LCL =
μ
−
=
0.655
n
(2)
R
chart:
Center =
R =
σ
·d
2
= 3.078 · 0.1 = 0.308
U
C
L
=
R
3
d
3
σ
=
D
4
R
=
0.547
LCL =
R
3
d
3
σ
=
D
3
R
=
0.069
(3)
s
chart:
Center =
s =
σ
·c
4
= 0.097
U
L
=
s
+
3
σ
1
−
4
2
=
B
6
σ
0.167
LCL =
s
−
3
σ
1
−
4
2
=
B
5
†
[10%]
Repeat part (a) assuming that µ and
σ
are unknown and that we have collected 50 observations of
sample size 10. These samples gave a grand average of 0.734 V, and average
s
i
of 0.125 V, and
an average
R
i
of 0.365 V.
(1)
x
chart:
There are two approaches here; in one, we use
R
to estimate variance, and in the
other we use
s
to estimate variance. Note that the resulting control limits are very
close in the two cases.
Using
R
U
s
i
n
g
s
Center =
x
=
0.734
C
e
n
t
e
r
=
x
= 0.734
s
1
UCL =
x
+
A
2
R
= 0.846
U
C
L
=
x
+
3
=
x
+
A
3
s
= 0.856
c
4
n
s
1
LCL =
x
−
A
2
R
= 0.622
LCL =
x
−
3
=
x
−
A
3
s
= 0.612
c
4
n
(2)
R
chart:
Center =
R
=
0.365
R
UCL =
R
+
3
d
3
=
D
4
R
=
0.649
d
2
R
LCL =
R
−
3
d
3
=
D
3
R
=
0.081
d
2
(3)
s
chart:
Center =
s
= 0.125
UCL =
s
+
3
s
1
−
c
4
2
=
B
4
s
=
0.215
c
4
LCL =
s
−
3
s
1
−
c
4
2
=
B
3
s
=
0.036
c
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This note was uploaded on 09/24/2010 for the course MECHE 2.830J taught by Professor Davidhardt during the Spring '08 term at MIT.
 Spring '08
 DavidHardt

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