This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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
(L ———y 4.0 comic] :11: Meek: w m 5+ 01 (“wild
Pa Ael‘ecwa In {2th Page 3 lg “Qaftwfc Jim : 00000 '2 AP / “"3", H [A “a
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
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 . bAaiii53“> ) 5— 41182 5 i 13% Twﬁg
Icé 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"? UC‘” ‘ $30“ 20 1 “Wk ‘1‘???) f E: if}
Lev : 50' 7% 1%3‘6 ’3 am or we?» M <
zdmj mean “y range ‘ 1.3 \ 6+, . a whahm‘hgv \U a 5‘ = ",5 >0N oL’Y'g
’1 4 . 
3’5 7 3 ' ' J . 4 
E‘1 1cm. 5‘0 : "5 ‘
me we CVCL wust :ﬁaxawosomcu—«mm
A; + ‘11:, =‘«MbbL’I%IJI€OEWW :szvo Qumxepﬁyg/Qw, 23 a! W i D t? : W? w): M03 ‘ ' Lav. . . .225 4:10" law L'J’UWUZ Tm ‘6‘;— ’ w = we azoxms‘m Janna
hLO/c v45" 302150 3 F3 “‘3 Pewm :3 cm * d 5% hex ‘1
L—~—~w WU ' 9 MW 3w .m k! g A, 9;” n 4,:
Mme—mg” +a Mme”. P‘ m t3= :
mwjm < gm”) * : 2011
__.__.__W_.____ ._.__.___._._.__ , A5930! LCL 1,011 46(0451w531ms)
__ ' 002% ma», u o h Wzocexsi (.0 «or k1beg'iowh‘
__m ‘ Ms ax g ' MW My“ 7. “ a ‘ 1* 0? 4811661’1 W'mnheol m ‘2‘: 10 as
_ ‘1 1 <0 v63
13 ‘3 gm
‘5 H4 éa I £124; *500 I.” 4 “W4? 2's, macaan
0L : 3M £61 ’5;
023 \,,OQUH £9 ,
a ‘ (Tbo J , 1‘ 1’4“,— L‘Q) wwo’ moo ~
\ La, U
, __ qlo
(:7 ﬂ i}
_ 015,10 ) c ‘
: = ,W
In 
"‘ S
94 d
6‘4 x
m I a "T; " AV ‘ Ug dvs m, i P
, ‘w ,3: he; I shgwﬁ
54:" **<;,%4_ a U . GLOW: 5 \m‘f" 2.3m: ’ m i Qvahk‘»)  Arrhfl' CNN“. / (t, a)!" ﬁx I I; Y ‘l l.‘ g /{ “ﬁrm”! {I ﬁ\'x r it.’!‘ ‘~ ‘ “y (VI—I“ _ ..
“ “ k’lkd /W" (\{I\/F\"",‘; [WYluer “UL” “My ‘ x ‘ J ”\;.
5‘ l“\i”‘,(ﬁw. > (:w I ‘Vrf ix Kin1 \H Vi)qu Egg/MU” (“‘6 VI," 134,. J3, Mi 1” (in KL 1:1,}, ’3‘ fir», . 5“: {3/ pin; UC‘WCWJHHLJ *4: HM g“
u\" LHMAOPHF‘KC: ((iﬂkii (SCI) >m((,»( HRH.“ "2 [\1 WU \R} l)€'»('1‘“.Dw~(i,"+ "6 u ~ 4”C‘( j m“ ’i 13““‘57‘V‘é "‘"""‘"'ii’~".,ivj K1 W'i’ ' H Tadprjiﬁ l ‘ 1' mun3".» ;".¢"“. «211 “IV "Vi‘uvrgi X {Emilia f‘» 6; gF’C C/ tryHiv'isvgama b 1; \) '. \thn V ,. p WW” 3’: ‘W’ylﬂik‘m "3. 5s ﬁrm" "MM" M u‘x‘xfr' M 3‘00“; “WWW/NJ] [j ' (9 UN ’3‘ 5'” W” ' WW“? I .mjm AH‘WR‘nMHh, '2 x" (“mm
r7\ )
fv V, H 3H f.._ \J‘Q‘Vigﬂx L f ‘ A ; ﬁn 0% ﬂ , vain/1‘”le
n W . ‘ (I ‘ l
6y“ "1— “‘“m 4' 17C” “ICU A’ﬁ'W S h “/ (Mr/M :3 ; io‘hth 1: s_<§\.r\mmn _ f
1 "J’wﬁni MU, "J4 ' (ﬂ ‘3':er A m v (lV'U mm L. _ l ' /
. HWWMM (“yran ' Lo (Hmé
/ (I r i’a‘ffﬂhzx. MP3 W‘f‘ﬂhag‘: 2.51/37}va '” W,” (thigh ...
View
Full Document
 Fall '08
 WOOSLEY

Click to edit the document details