μ + = σ = D 4 σ = D 3 σ = σ = All units in parts (a) and (b) are in V. (1) x chart: Center = μ = 0.75 UCL =

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1 Our online Tutor are available 4*7 to provide Help with Proce control ytem Homework/Aignment or a long term Graduate/Undergraduate Proce control ytem Project. Our Tutor being experienced and proficient in Proce control ytem enure to provide high quality Proce control ytem Homework Help. Upload your Proce control ytem Aignment at Submit Your Aignment button or it to info@aignmentpedia.com. You can ue our Live Chat option to chedule an Online Tutoring eion with our Proce control ytem Tutor. In all problem, pleae how your work and explain your reaoning. Statitical table for the cumulative tandard normal ditribution, percentage point of the χ ditribution, percentage point of the t ditribution, percentage point of the F ditribution, and factor for contructing variable for control chart (all from Montgomery, 5 th Ed.) are provided. Problem 1 Part (a) * Control chart for x, R, and are to be maintained for the threhold voltage of hort-channel MOSFET uing ample ize of n = 10. It i known that the proce i normally ditributed with µ = 0.75 V and σ = 0.10 V. Find the centerline and ±3σ control limit for each of thee chart. All unit in part (a) and (b) are in V. (1) x chart: Center = μ = 0.75 UCL = 3σ μ + = n 3σ LCL = μ = n () R chart: Center = R = σ d = = UCL = R + 3d 3 σ = D 4 R = LCL = R 3d 3 σ = D 3 R = (3) chart: Center = = σ c 4 = UCL = + 3σ 1 c 4 = B 6 σ = LCL = 3σ 1 c 4 = B 5 σ =

2 Part (b) Repeat part (a) auming that µ and σ are unknown and that we have collected 50 obervation of ample ize 10. Thee ample gave a grand average of V, and average i of 0.15 V, and an average R i of V. (1) x chart: There are two approache here; in one, we ue R to etimate variance, and in the other we ue to etimate variance. Note that the reulting control limit are very cloe in the two cae. () R chart: (3) chart: Uing R Uing Center = x = Center = x = UCL = x + A R = UCL = 1 x + 3 = x + A 3 = c 4 n 1 LCL = x A R = 0.6 LCL = x 3 = x A 3 = 0.61 c 4 n Center = R = R UCL = R + 3d 3 = D 4 R = d R LCL = R 3d 3 = D 3 R = d Center = = 0.15 UCL = c 4 c 4 LCL = 3 1 c 4 c 4 = B 4 = 0.15 = B 3 = Note that a the ample ize get larger, the chart i omewhat more effective, a it ue all the data in the ample to etimate the within-ample variance, while the R chart only ue the extremal (max and min) value.

3 Part (c) The pecification limit for MOSFET threhold voltage in thi proce i x = 0.7 ± 0.05 V. What i the Cp and Cpk for thi proce? Auming the proce a pecified in part (a) i in control, what i the fraction of MOSFET that are nonconforming? If the proce in part (a) wa recentered uch that µ x i on target at 0.7 V, what would the fraction be of MOSFET that are nonconforming? We ee that USL = 0.75 V and LSL = 0.65 V. USL LSL (1) C p = = = = Note that thi i a very poor proce capability 6σ (and we will ee the impact on bad part in a moment). () C pk = min USL μ, μ LSL = 0 ince USL i equal to the actual proce mean. Thi i 3σ 3σ lower than C p ince the proce i not mean centered. (3) The proce i horribly off-center, a noted in (). All part above the proce mean are bad (above the USL), and an appreciably fraction of part till fall below LSL becaue the ditribution i o wide compared to the pecification limit. Thu USL μ μ LSL Pr(nonconforming, off-center) = 1 Φ + 1 Φ σ σ = 1 Φ Φ = [1 Φ(0)] + [1 Φ(1)] = (4) Since the proce wa o badly off-mean, perhap by recentering the proce we can improve the fraction of good part. When centered Pr(nonconf, centered) = 1 Φ USL T = 1 Φ = [1 Φ(0.5)] = σ 0.1 So we hardly reduce the fraction of bad part at all? Why? Becaue the inherent proce capability i o poor (C p = 0.167). The real problem i the ditribution i too wide compared to the pecification limit, not that we are off-center. 3

4 Problem Part (a) Baed on a fairly large hitorical ample of gate oxide thicknee coniting of 61 ample, an average oxide thickne of 96.5 Å i oberved, with a variance of 6 Å. What are your bet etimate, and 95% confidence interval, for the proce mean and proce variance? (1) Point etimate for mean: μˆ = x = 96.5 Å () 95% confidence interval for mean. Since n i large, the normal i a good approximation: Uing normal ditribution (large n): μ = x ± z α / 6 = 96.5 ±1.96 = 96.5 ± 0.61 Å n 61 or μ Å For more accuracy, ue t ditribution: μ = x ± t α /,n 1 6 = 96.5 ±.00 = 96.5 ± 0.63 Å n 61 or μ Å (3) Point etimate for variance: σˆ = = 6 Å (4) 95% confidence interval for variance: (n 1) σ (n 1) (60)6 o that σ (60)6 or 4.3 σ 8.89 Å Χ α /,n 1 Χ 1 α /,n Part (b) A change i propoed to increae the gate oxide thickne. Uing the modified proce, 11 ample are taken, with oberved ample mean of 100 Å and ample variance of 0 Å. What are your bet etimate, and 95% confidence interval, for the new proce mean and proce variance? (1) Point etimate for mean: μˆ = x = 100 Å () 95% confidence interval for mean: Now we need to ue the t ditribution for ure, ince n i 0 mall: μ = x ± t α /,n 1 = 96.5 ±.8 = 100 ± 3.00 Å or 97 μ 103 Å n 11 (3) Point etimate for variance: σˆ = = 0 Å (4) 95% confidence interval for variance: (n 1) σ (n 1) (10)0 o that σ (10)0 or 9.76 σ 61.6 Å Χ α /,n 1 Χ 1 α /,n

5 Part (c) To 95% confidence, ha the modification reulted in a proce with a different mean than in part (a)? To 95% confidence, ha the modification reulted in a proce with a different variance than in part (b)? (1) No becaue the 95% confidence interval for the old and new mean overlap, we don t have enough evidence to be 95% confident or better that the mean are different. One can alo attempt a direct hypothei tet on the difference in mean. Thi i more work than needed, given we already determined confidence interval. However, if a hypothei tet i ued, we hould be careful not to aume the variance are the ame (we have trong evidence they are not), and o we would then need to ue the omewhat complicated procedure decribed in Montgomery, pp () Ye here the 95% confidence interval for the new variance and old do not overlap. We are 95% confident or better that the chance ha reulted in a proce with larger variance. Here alo, one can formulate a hypothei tet a an alternative approach. Given that one variance appear to be o much larger than the other, we re bet off uing a one ided F- tet (that i to ay, the one ided tet will require even larger ratio to declare a larger variance, and thu the one ided tet i more conervative). So in thi cae, we would need to oberve an F > F 0, where F 0 = F 0.05,n 1,n 1 = F 0.05,10,60 = In our cae, we b a oberve F = b = 0 = 3.33 giving u better than 95% confidence that the econd a 6 variance i larger than the firt. Problem 3 You are in charge of a proce manufacturing part whoe key output parameter, x, i known to be normally ditributed. When the proce i in control, the mean value of x i known to be μ and it tandard deviation σ. Unfortunately, the upplier of an important input material for thi proce i rather unreliable, and every o often they inadvertently upply a large batch of material whoe propertie caue the mean of x to hift to μ + kσ, with the tandard deviation of x remaining unchanged. When thi mean hift occur, you want to be able to identify it a quickly a poible. Your metrology taff tell you that they can currently meaure at mot a fraction f of part produced, although you are free to chooe how thi meaurement budget i apportioned: i.e. you can dictate any n, where ample of n part are taken at interval of n/f part. 5

6 Part (a) Write down, in term of μ, σ and n, expreion for the upper and lower control limit for an x control chart monitoring thi proce. Anwer: μ ± 3σ/(n 0.5 ) Part (b) Write down an expreion for the probability, β, that the mean, x, of any given ample of ize n from an out-of-control proce with mean output x = μ + kσ, will not fall outide the control limit. Anwer: β = Φ(3 k n ) Φ( 3 k n ) Part (c) Derive, in term of n, f, and β, an expreion for the average number of part, P, that will be produced after a mean hift occur and before a ample fall outide the range (LCL, UCL). Anwer: P = n/[f(1 β)] We could ue the expreion that you have jut derived to chooe a value for n that minimize P for a given f (do not attempt to chooe n now). Sugget one other conideration that would feature in the election of ample ize and ample frequency. Anwer: The need to make the inter-ample interval longer than the correlation time of the output, o that ample are plauibly independent. 6

7 Problem 3, cont d. Part (d) Now uppoe that you are told that the quality lo function for the proce i defined a follow: c c q LSL: μ Δμ μ c = 0 USL: μ+δμ x Here, c i the cot penalty aociated with producing a part with a particular x. Write down an expreion for the expected cot penalty E k [c] for a part produced when the proce i x ~ N(μ + kσ, σ ). (Hint: your anwer hould be expreed a a um of integral; do not attempt to evaluate the integral.) Note: the probability denity function of x ~ N(μ, σ ) i given by Anwer: 1 1 x μ σ σ π e. c μ Δμ μ +Δμ q 1 x μ kσ 1 x μ kσ E k [c] = σ π exp 1 x μ kσ x μ dx + σ exp dx + Δμ μ Δμ σ exp dx μ+ Δμ σ 7

8 Problem 3, cont d. Part (e) With the above information in mind, you now want to invetigate whether you hould intall more metrology equipment o that the fraction of part meaured, f, can be increaed. Suppoe that meaurement cot can be completely decribed by an incremental cot c m per part meaured. Write down, and briefly explain, a cot expreion that hould be minimized when electing an optimal value for f. Your anwer hould be given in term of the following parameter: n ample ize c m cot of meauring one part f fraction of all part produced that are meaured E k [c] expected cot penalty for a part produced when the output i x ~ N(μ + kσ, σ ) E 0 [c] expected cot penalty for a part produced when the output i x ~ N(μ, σ ) β your anwer to part (b) D the frequency (in part 1 ), on average, with which the material upplier deliver a defective batch of material (e.g. if d = 10 5 part 1, the upplier would deliver a defective batch for every 10 5 part produced by the proce. Hopefully, not much of the batch will be ued before the problem i noticed! nd Anwer: Cot expreion: fc m + (E k [c] E 0 [c]) f (1 β ) We need to minimize the um of meaurement cot and the extra quality-related cot penalty that occur when proceing with defective material. The more frequently the upplier deliver the defective material, or the lower the meaurement cot per part, the larger the optimal f. What other factor hould be conidered when chooing f? Anwer: The defective material input may not be the only caue of mean hift that need to be detected. We may alo need to chooe f to allow u to detect particular tandard deviation change. Increaing f may low down the production line, even if more meaurement equipment were to be intalled: the metrology team may have other contraint. viit u at or u at info@aignmentpedia.com or call u at