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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 : 343543412;
. 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 3sz'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}
Liamg “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
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LLLK f} a. F? I Prme‘i‘ﬂ “(\P‘S‘Jﬂtﬁll ‘
(telam 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 popeA ,‘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 3sz'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 InControl \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 3sz 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’NMbwlx)
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 ’ 39 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 .______
agowowL‘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
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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 . bAaiii53“> ) 5— 41182 5 i 13% Twﬁg
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