q1_sol_2008

# q1_sol_2008 - MIT OpenCourseWare http/ocw.mit.edu 2.830J...

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MIT OpenCourseWare http://ocw.mit.edu 2.830J / 6.780J / ESD.63J Control of Manufacturing Processes (SMA 6303) Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .

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Name: 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 short-channel 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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q1_sol_2008 - MIT OpenCourseWare http/ocw.mit.edu 2.830J...

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