Homework 3 - Student Caroline Uffman Instructor Peter Kelle...

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Unformatted text preview: Student: Caroline Uffman Instructor: Peter Kelle Assignment: HW 3: Ch S6 numerical . Date: 6/4/12 Course: ISDS 3115 Summer 2012 problems Time: 3:30 PM ‘ Book: Heizer/Render: Operations Management, 10eND 2. Refer to Table 56.1 — Factors for Computing Control Chart Limits (3 sigma) for this problem. Auto pistons at Yongpin Zhou's plant in Shanghai are produced in a forging process, and the diameter is a critical factor that must be controlled. From sample sizes of 10 pistons produced each day, the mean and the range of this diameter have been as follows: M . 5.9.3 E: RVWJB 0‘} WWSEID Mean 3; Range R 7 —_ New 05‘? Willie WW Day (mm) (mm) M pk n 4 {‘31}ng 1 154.9 4.0 AL, 9:5,. in; “ i M” ' 2 155.2 4.4 3 153.6 4.3 4 153.5 4.8 5 156.6 4.3 A a) What is the value of 97?», Mame away? 01> meml = 7738/ 5 : 34354-3412; . x = [1 mm (round your response to two decima 1; aces). b) What is the value of E? ~va ampuw wal‘z‘tia 491‘ ital/$3 -' F“ \ _ 9x18 1' ‘4) *fl/ R = D m (round your response to two HECimal places). c) What are the UCL; and LCL; using 3-sz'gma? ,~\ Wm pie 55:28, new tit/ice M i“? 5 5,1 Upper Control Limit (U CL;) = 1:] mm (round your response to gym gecimal places). z. u, y —- next a, die/i311; 354.73 +(?>0?5§(43w’1~\) =1” 1% {Q} Liam-g “7 Dag Lower Control Limit (LCL;) = D m (round your refponse to twp, decline] places). a a -1 {a t '54 ~75 "@059 “17335). ’65'9133 :L... -~!"~"" Int/Vt .. ; :1 was What are the and 1151Ilg . Upper Control Limit (UCLR) = [:1 mm (found yoursrejponsextp two decimal places). 1. /77(4,3éj {:75 Lower Control Limit (LCLR) = 1:] mm (round your response to two decimal places). “yrfis 1 R51?) (4, 35) 2;. 6273 n e) If the true diameter mean should be 15 mm and you wantthis as your center (nominal) line, what are the new UCL; and LCLjH‘ it“ ’37 1’5 2521:»; IM‘ ,,,,,,,,,,,, M W.ww,.WWWw~,.MWWWWWWWNWM, "Mamas,mW.““mewmeNwmwmwwWWWW.MMWW wax. *'“ ‘ Page 2 152» + (1 ecstaflmrg x . Upper Control Limit (UCL;) = [3 mm (round your response to two decimalplaces). “~~ ...... ‘W_M,/ Student: Caroline Uffman Instructor: Peter Kelle Assignment: HW 3: Ch S6 numerical Date: 6/4/12 Course: ISDS 3115 Summer 2012 problems Time: 3:30 PM Book: Heizer/Render: Operations Management, 10e ND 2. (cont) Lower Control Limit (LCL;) = [1 mm (round your response to two dices). :55 - (2.055314% $0533.89} Definition Sample Mean Factor, Upper Range, Lower Range, Size, n A 2 D4 D 3 2 1.880 3.268 0 3 1.023 2.574 0 4 0.729 2.282 0 5 0.577 2.115 O 6 0.483 2.004 0 7 0.419 1.924 0.076 8 0.373 1.864 0.136 9 0.337 1.816 0.184 10 0.308 1.777 0.223 . 12 0.266 1.716 0.284 Rw—v (L ——-—y 4.0 comic] :11: Meek: w m 5+ 01 (“wild Pa Ael‘ecwa In {2th Page 3 lg “Qaftwfc Jim : 0-0000 '2 AP / “"3", H [A “a LLLK f} a. F? I Prme‘i‘fl “(\P‘S‘Jfltfill ‘ (tel-am in L3“ mate E: fisfimem'bt Anagram) b) If the sample size is 100, the 3 6 control limits are: C/Ammcjt ) V”"-J:l‘l‘k.\1”§iifil W t k 'i a t‘" Equipln‘llx Mia {HMS l *- VT) Student: Caroline Uffman Date: 6/4/12 Time: 3:30 PM Assignment: HW 3: Ch S6 numerical problems Instructor: Peter Kelle Course: ISDS 3115 Summer 2012 Book: Heizer/Render: Operations Management, lOe ND The defect rate for data entry of insurance claims has historically been about 2.00%. 6 P V3 nurse P chattel This exercise contains only parts a, b, c, d, and e. /\ -. a) If you wish to use a sample size of 200, the_3_-§igma countrfl‘limits are: U P ’ UCLp = D (enter your response as a fraction between 0 and I, rounded to three decimal .olrg(ooO‘7> places). : o ‘0 %\ l pat—2:: (-00%= LCLp = D (enter your response as a fraction between 0 and [jflrgvumnmde‘d to three decimal) 0 70) places). ‘ O a -33 L _ 00 z - 0 .05 ;:~ ,flo to” EC . I.) , L ' H30 V UCLp = D (enter your response as a fiaction between 0 and 1, rounded to three deClmal 3- /‘ .o'l/‘it’ / \. , .... ---=“*‘"‘;‘" w .09: + e (.014 pope-A ,‘2 LCLp = 1:] (enter your response as a fraction between 0 and 1, rounded to three de ' al “12.3 l_{l-~ all places). places). tea—bl 0143‘- — D‘CélQ.-~~~a<§@5® 9% , 0 ooqq c) If the sample size is 200, the 20' control limits are: UCLp = '1 (enter your response as a fraction between 0 and I, rounded to three decimal r, x .... A V .oa+€a\l.octw * LCLp = U (enter your response as a fraction between 0 and 1, rounded to three decimal Places) ,53 — a ( cot/lb .ooo‘él V d) If the sample size is 100, the 26 control limits are: places). UCLp = D (enter your response as a fraction between 0 and 1, rounded to three decimal "mm"; |. .c a + at .0145 = arkéfiigii LCLp = [1 (enter your response as a fraction between 0 and 1, rounded to three decimal ~03l~ Q (10/45 1 “$008 ‘ e) What happens to as when the sample size is larger? places). places). When the sample size is larger, on is p Student: Caroline Uffman Instructor: Peter Kelle Assignment: HW 3: Ch S6 numerical . Date: 6/4/12 Course: ISDS 3115 Summer 2012 problems Time: 3:30 PM Book: Heizer/Render: Operations Management, 10e ND 4. The results of inspection of DNA samples taken over the past 10 days are given below. Sample size is 100. % Day 1 2 3 4 5 6 7 8 9 10 .... ,. Defectives 6 8 8' 10 6 7 o :2. 97) first I V/ _, 2 345:: ~— L l ' ‘ a) The upper and lower 3-sz'gma control chart limits are: I O - “7 .1) ("a a 6 ,. em: _ 7 "I ’"v ‘ “« UCLP = D (enter your response as a fraction betweeflnuO and l, rounded to three decimal A I ~ places). _ 06“ + 1L > 9‘ Sofia, i). (if?) i J ( ’ M LCLP = I: (enter your response as a fraction between 0 and 1, rounded to three decimal places). t m m u _; “ ~: ly hwy}; ' I) ' E) » Om we; - 5W \l “I b) Given the limits in part a, is the process in control for the following days as shown in the table below? «39. b o 70‘? . Day Number of Defects In-Control \oD‘“ . 0.0T“ I497” 11 10 yes \00 12 9 V65 W! Q N '2 , o w“ W 13 11 Q 5, The defect rate for your product has historically been. about 1.50%. For a sample size of 300, the0 ‘07 P . . . u or and lower 3-sz control chart limit : -. . E 0 pp ...... S are *wwgl ! 0% O W N UCLP = [I (enter your response as a fraction between 0 and I, rounded to four decimal ’_ // places). .0'5 f3('0070é> _: J LCLP = E] (enter your response as a fraction between 0 and 1, rounded to four decimal e916 7 a (new) I ,, Student: Caroline Uffman Instructor: Peter Kelle Assignment: HW 3: Ch S6 numerical Date: 6/4/12 Course: ISDS 3115 Summer 2012 problems Time: 3:30 PM Book: Heizer/Render: Operations Management, 10e ND 6, Five data entry operators work at the data processing department of the Georgia Bank. Each day for 30 days, the number of defective records in a sample of 200 records typed by these operators has been noted, as follows: ‘1 n a Sample No. No. Defectives No. Defectives No. Defectives Sample No. Sample No. p—a \DOOQO’NM-bwlx) 00 u—n 0 Hi 9r! 3f& . a) Establish 36 upper and lower control limits. UCLp = [:1 (enter your response as a fiactz'on between 0 and I, rounded to three decimal places). ' 050333 4*. ’5 (.0; 64613:) —:: LCLp = [:1 (enter your response as a fraction between 0 and 1, rounded to three decimal fix-“WK Places) 0.50553 ~?: (016%?) mQMtha/i b) Why can the lower control limit not be a negative number? \1’2. PM mOx4 c o _ \1 (1 \ 0A. Since the upper control limit is positive. T/ Since the percent of defective records cannot be a negative number. I Mr“ 00 " ~llU‘ __/ {:3 C. Since the upper control limit cannot be a negative number. r90 ” C} D. Since the percent ’of defective records is always a positive number. p.0UU ’ c) The industry standards for the upper and lower control limits are 0.10 and 0.01, respectively. What does this imply about Georgia Bank's own standards? The industry standards are l those at Georgia Bankfxllg 3‘: F $23)! Student: Caroline Uffman Instructor: Peter Kelle Assignment: HW 3: Ch 86 numerical Date: 6/4/12 Course: ISDS 3115 Summer 2012 problems Time: 3:30 PM Book: Heizer/Render: Operations Management, 10c ND 7 _ An ad agency tracks the complaints, by week received, about the billboards in its city: % z : 21 a / s 2 Q; Week No. of Complaints H 2 h b \ Q (T! 1 6 e» ’ ’V’: Q {3 ‘ 2 6 UL, it Q i 5 \V V‘ I 1 6 an \0 m M g 3 7 g u 9 o (b \ M h, a I \ 4 11 ’ L0 g3 , (9 3L 0 5 5 v~—- WV; V7 , a 6 11 c - .. cfb 00 m \/ _.__—.—.__—— '3‘ ‘\ g. N 7th exerczse contaan only parts a, b, and c. g[ 8 a) The type of control chart that is best to monitor this process is \s b) Using 2 = 3, the control chart limits for this process are (assume that the historical complaints rate 1s unknown): Jr 5rd «5 a 5.1%, a g UCLC = [:l complaints per weekp(round your response to two decimal places). g 77.7 + gar/.7 a» 19.03: 8 T LCLC = Cl complaints per week (round your response to two decimal places and fiz‘f\your N answer is negative, enter this value as 0). 7. 7 ~ 7 '3 ’ 3-9 3V”? “‘“0713, c) According to the control limits, the process is . in bril- 7'7 /S [7/4 ' w W M ,W M WWW,me limit: «473! 8, Meena Chavan Corp.'s computer chip production process yields DRAM chips with an average life of l,®Q,hours and c.5480 hours. The tolerance upper and lower specification limits are 2,200 hours and 1,500 hours, respectively. Based on the given information, the process capability ratio, 0 was“? mg“? Cp = I iround your response to two decimal places). Based on the process capability ratio (GP) for the given information, one can say that the process is ,._ a 0 ._____-_ agowow-L‘igfi W, ,3, 1 ,1 . I f producing the chips to the design specifications. a (so) ‘ M ’ gift, For the given information, the process capability index (C 17k) = (round your response to two w decimal places). .435 A». Based on the process capability index (C Pk) for the glven mformatlon, one can say that the process the specification. @1104 -—‘ ._ \~ Page 7 L11 - 'X _ ‘1" my} [I , ,3 WET/n] ,1" aaooaléw . 629 hm « is t I A Student: Caroline Uffman Instructor: Peter Kelle Assignment: HW 3: Ch S6 numerical Date: 6/4/12 Course: ISDS 3115 Summer 2012 problems . Time: 3:30 PM Book: Heizer/Render: Operations Management, 106 ND 9. Blackburn, Inc., an equipment manufacturer in Nashville, has submitted a sample cutoff valve to improve your manufacturing process. Your process engineering department has conducted experiments and found that the valve has a mean (u) of 10.00 and a standard deviation (6) of 0.06. Your desired performance is u = 10.00 i- 3 standard ,deyiations, where o # 0.080. 7 ,¢+Q5(0gb-:anullo»3¢.ag):aa,n5‘ For the given information, the process capability index (Cpk) = 1:] (round your response to three decimal places). 10_ The specifications for a plastic liner for concrete highway project calls for thickness of 5.0 m i 0.12 mm. The standard deviation of the process is estimated to be 0.02 mm. The upper specification limit for this product = mm (round your response to three decimal places). The lower specification limit for this product = [1 mm (round your response to three decimal places). The process is known to operate at a mean thickness of 5.0 mm. The process capability index (Cpk) = D (round your response to three decimal places). The upper specification limit lies abo a tandard deviations from the centerline (mean thickness). Wm“ ................ A W ............. ‘WW .... “WWW __________ Mmrm.wmsw ...................... chM» W ....................................................... tttttttttttttttttttttttttt .WW ___________ r . b-Aaiii53“> ) 5— 41182 5 i 13% Twfig- Ic-é é (Nag ‘ *3/59‘, ) (034.:199tfllé‘3‘”99? L ‘0 ’3 ; superman, a, I £1 Pages I 3K’VS)) l. 3 .m 0th (Le ‘c. n HAW? . 'La Ham 0 .5? W": m C0! "1 W (J 5‘ I. design {1 ‘Chm’r 2‘; mm mu 5,, 1 Mywuf are m loww 1* 5' WWW Umi » 913 .L .g 15in one chosen ’0 b fiu s. a) n M ’Lga L:>L+20r’7?‘/m 7915+ 50+ 0 q 1 “My gLQ/32:7:"%O’2 52”"); '20"V1’2—: 2‘4?" . b Wha} 02?, Hut MM”?! W W Cw] dau Pram Huh -. u"? 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