ARIMA - 4 vak Modal-Bmd APPmack W Tt‘wue ENG/b Modefs...

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Unformatted text preview: 4 vak. Modal -Bmd APPmack W Tt‘wue ENG/b Modefs Rm (I‘moUHv‘Q SPL um Cavalng ckaw+ Engineer's Computation Pad 513 Omaha Cox'wz ‘— |_._.___——————1 \ 0° “Pestokuod g do'khq 2 E: ~—- MCDOL?J “(‘R‘TY‘OVM M Ouxho CtsY‘r-QIMHM @6710 {wk-o. [M 1% almkos g __ Regioqu/s W /\ (Kw—2 We! (Kw SPC Chadd: Tl‘m 9.06% M042“ ‘. Gnu-{'0 “(taro/54v: W0d2l£ (i‘vR “maoleLC _. McmhrwzS owwa Wadi/S (MA model, ._ Combinwfi‘m «t aubccgfirwsviwc amd “chafing AV—Q‘I‘O‘ZQ Woolzk — I’m blaw'lld "mow‘vta oNOxwax Week‘s (ELM A" W0 oLQ‘S " 0 " a k GD.“ ‘ _____ -L;-, . V-» _ _. __ _. -_ -__ é E : 2. 1 Q | - _ _ _ _ _ . . _ _ _ _ __ > ”. ___-L§..A, -. h- ch 3 w M ‘ +3.- 4. -. _ mA_ “fl. _ _ __ h a _ _ _ __ - M- m w _ w__._.-u_,..________.W__fl__ ___ ”_ i ii A H+?§2_T'f€l°~+i€/I\ mac-Hm” _ 4958.91). __ ,- _ _ _fi.., , _H?§H. _ ._,__...h~i",. ._ .. - _ uh“.-._m.»m éflmu w _ . _. -__,.,___. A.V..__Vv,_-.. _.. ._ .- h- _._ ~ - e. a. . .. ~ - _. “#1:”? ._ -. -_ .__...u. .7 W _.. A...~._____.._. _|_._.__‘__....__._.___.___A.:k ‘_,_- . /._L/ ._ H..-“ _ w ¥ — » __ _.‘ .. ~V,w_...., _v_ H _,._ “mm..-__r,____.-_"-mmmw m __ .. __. “z z 5 1 | f ._, ..L A____:A.A ,A...~___.-.; _ ..-W.4.,_m 4‘_4J.A.___A g4, N L—"m ugfizfiyfiwhad | nauanag.“«nipuuflmfii- A A 1 § 3. 9.:qu “934.. | ’ a - . (.43._m.x.7_mm.;.=¢w;.wm;a 2 go 73" a. . on Poiu‘iiv‘) W......._.___..‘-‘_._.,_,,-_,.4__..___._~-_.~_~__.__-__H___._,-____._-,_~_-___._..W-_ ” __ _ _ - M- w ‘ ._ A _ ._ _. _, _. - 995).“:{51‘3 _ _ -w _-_ _ i i Les-L... -fiwmfifiwfififlvll WE: mega-‘9,” fl W- ; x m cw’rv Co-«rve\m'H cm‘ 0.75 ~ 0501- O.25[’ ’ I of O 5 10 I 15 k ~025— Figure 10.1 Autocorrelation fun- “0-50 ction for the AR(1) process xt = 5 + 0.8m.1 + 6;. -O,75'- -1.00— . c-‘Hw claw KxFawmi-{aua- (St/{WM CngVA-H‘cfll/S In +042 (kw—L2); arr-a qu‘HwLQY Cow‘r—dmfim) 40F O__L____L_._.L__| L___l J_ I __I I O 10 20 30 4O 5O 60 7O 80 90 100 T Figure 10.2 Realization of the ARM) process xt=5+ 0.8xt_1+et. Cansbtwr WC’W ARC” PTOM ’Qc :5—o~8=cfgt_I+é{_ 6. :5 "HA-K Free—2A6 £+m+~€mara? 10 Figure 10.3 Autocorrelation fun- ction of the AR(1) process act: , 5—0.8xz_1+e,. 4W+fi_CBTY-§\fir+£m,., _. _K3~5‘h7”.zmeku‘laf+x RR Pew—QMMM «$26923, bu+ 044+» Cmrekafiwé arm POSF'HW fin: awn ‘1 i log‘s md._ma0~r¥-€W.«€°V 649‘. Kasgsz. WWW-M8 Enos Amish; wwch .OsdllM-e/A. ,15. E {aridgflg 1O -1o Figure 10.4. Realization of the ARM.) process x,=5-— 0.8xt_1+€t. fi_fi%%wfn7%eng%§t9a.J§E£%lmf%9¢yémgénmm _flin _ w # a. w __._ * _~__ w _ w“. ,. A Cms'iw. 4m M29.) Pram -%-2. at“: 2: HI: + 0?be Figure 19.6 Autocorrelation fun- ction of the ARCZ) process xt+ 10 + 0.9xt_ 1 _ 0.8xt _ 2 + 6:. d—émfi 01.2% an o» GiaJWPw ci>ll+4i~¢g =.- Lo jug-+41% (4'8) :- maafi <0 {i' 3‘ x. ‘v E ‘1‘] 3 ‘ 33 m P/mualo hwoouc bahcwi‘o‘f “1-50 av?c{£m*— A?” m MW“ i 2 / . a. 20 '; O .1 J_J __I l l__J | ~' 10 20 30 4O 5O 60 7O 80 90 100 1 Figure 10.5 Realization of the AR(2) process x: = 10 + 0.9xt_1_0-8xt—2+€t' _,, moflifl .7 _ _ ,7 7 _ __ 77 7 77-77.-.. 7 77 7M , Prqceéémwrww _ 91:747. 99 at?” 7°_+°‘°"__7 _. “*Y’T'C‘L' N ' 7am___7777’?’< . 77 7. 7_ 7 _ _ _ 7. _ 77:94:). 7_. _. _ 77- _ 7:57;” _ 7__ -7 _____777_7 _ -7 -_ M7777 _ ‘77 _7 “951.9%? 7 " 7 7. 731s 7.. NJ . :7; 7. .7 _ 77-“. 7 .7 Cuff-e] .7 7 . .7 7. 7. __ i l h —— — __ .._—1 A —— ’7 ~— -£~- -— . “— ...rV.../‘ w7~ w fiA .. fl _ ._ __ i ._. 7 "7 H- 7M7,“ i % i aiu - .. i .2 T7. _ E‘me?)~€ 4'. CcmSdex m MAM) Paw W H Y’: I, 2%: “Hf ‘1 + 91‘? é'éf'. 1.00 > ‘ 075 > ' ‘ 0.50 0.25 I .‘f o of 1 j} - ' ' ‘ —O‘25 Figure 10.3 Autocorrelation fun— -' g . ction of the MAM) process _ —0.50 xt=10+ €t+ 0.96t_1. ‘ —0.75 400 k Cmrv‘elmfiw_ Wcfim AS _w+€>% W8“! HI :1} RTE/6x QbSaWa‘Hm (or. law.“ Ob-ieewafi'cm) aw» W KSWHEA .‘E—amb fiallnwwi (cw law)... Obssaawwvw. ‘ 16— . \ Vflflvfif ' I 1 1O 20 30 4O 50 60 7O 8O 90 100 T Figure ‘26.? A realization of the MAG) process xt = 10 + 6,: + 0.9g _ l. i 1 - _. _ aw. _ 2 ‘i , . -7. __ _.... .‘%<_._._ .._‘ ._. a i _. v-.. _ Ada“. - ,. _ _. Ix . 7 ii i I l _. __ _ WWW. w A v _ ' E ii I ._ _ w AWL” - _ ‘ E x :5 E W -__ - A M- u __ _ ._‘__ _. .‘L .‘ v 4"; 47;: < A g u w E“ U T .__K 4 W“.V_;rv.,_m.§..wmw.?__.,_.fa :4. t ...Z ‘ 1H» _.V....r‘.‘.....n_.. .MV ,. .. _.._... M"... . . _ ‘ "w WWW--.“ _. .. _ . ., H”, r..l,_..M_....‘_ w... v 7.. .7... .. . ., :El § @hfiwfglfirfim ‘ . N n.1,..-mwanquml 4.04“, A. “A __ i I Tun i é $392091“ €598» 0:! “an 1. I) k m V K \ r, M- ..._ Y°C—9¥§, N ML). i—s — ~~-— ~~~g~--— - ’ ~ w“- -43. w M. fl w _ _ M _ “NM H “wt _ 1W _ 3939» _= 3T. , _,_ WWW , —w~———;«_§— w— w —— — - ——— —~ ~ ~ «m ~— ‘3 ____ _ ,A .fi '1 ,. 11?. . __ _- _ a fi Mm“ - __ __ _~ - fl.“ m r ‘ -, m _ - -, LMRW __F__ m, m.“ ., _ __‘.._..N. .v.“ .v .. -._‘V..__W.,...N_.,_.m...._..-.___._____-V._‘ .- “WNW- u __.._ ..-.. w. M. . ._ ._.. ..__.__.__....__- ,___ _‘ Wm“ __ _____m-.-.,. .,. “_ -w _.._._, .. 1 3 ,aka—892mm.i.an..0L_Hm.-_..g{f.__K“._ A... Tot-«v02 -¢'_~(____x_.__ .. 3‘ , , _ Cg xxgles. ._ _____,_ 5. , M, _ ._ V w w E i i -_,.~....q. wr‘.__‘_.—.__. .,._,.... A.V_.__...__.‘_¥V_ , , 0-.» -A V._._”_.-._.§_?__...._,_‘___.___. ,_,., . , . , .7 . _,. ‘.___..___-.._w. H -M. V.._.,_ V. “-HMWWHW. « _._ .5 1 W “w ‘fwwm4h .tfl: n. Efokg Ccmmm am 147 :L [Q ‘+ (3‘8 :%£5' gig: mifln 6;}; 4.6 1.00:— 0.75L 050} 025— —025— ~O.50}- ‘075r —1.OOL— ARM-A Chi) Prom + 6% +°'°‘ 51»: Figure 10.9 Autocorrelation fun- ‘ ction of the ARMA( 1.1) process H xt=10+0.6xt_1+ Et+0.96t_*1. %¢+ 0% mbm‘vxé .ng‘sigmaax‘. alfav Umhéa. 9% Q / EAR W2 ARM} whhn c5 “20% .a~:r \ssmcak m-‘ra.cu»rv-a\o~-Hanxéwc*+m dew? me 4o 35* x ?E 30L -NW"“ 20'- :3“; S .22 15_ 1o— 5r l__.1__ I | l __L__ !_I 1.. 1 I O 10 20 30 4o 50 60 f 70 80 90 100 Figure 10.10 Realization of the ARMAU, 1) process xi: 10+ 0.6xt_1+ at +0.9€t_1. (0) Time series The? is nonsm’rionery in The mean mun. V-.,._.‘;..._.‘... T (b) Time series That is nonsToTionory in mean and slope Figure 10.11 Two nonstationary time series. (a) Time series that is non- stationary in the mean. (5) Time series that is nonstationary in mean and slope. my. . .qv: ~ -.._._.s__._....‘_- LA.-.“ j MDT) $+a+iwama (JV? m Mam Cam Isa—m Tammi“ % IDkamg M I) 3—3 Vac ’6 +~1 JV 5+m4-s‘v-Amwa' (73 WW Woe Iglqu “ sc'é‘l) g. (;E%_‘ _ 2 x4: ~ag:£.g~‘+9(+-l_r 2. '2 v SQ? 1} 120 i a 100'— 6 80'- e O :1 b so} a X a 40— a 1; so '1 O A? i 1 1 1.1.1 1 I 1 1 1 _L___L__| 123456789101112131415 i 1 ‘ (a) O 1234567891011121314 1 (b) 3 2 N; Figure “18.12 Reducing a dis— D 1 a m o e e a e e 9 o e e 9 crete nonstationary signal to a stationary signal by successive O 1 1 1 1 1 1 1 1 1 I J 1 I 1 differencing. 12 34 56178901111213 ChE376K Process Evaluation and Quality Control ' Project Due on 8th Dec 2007 Evening 5 PM Instruction: Work in groups of three. Please select your group before 29th Nov, and mail the list to me. I will assign a data number that you can use to download the particular data you have to use. Please do not share your data and Excel file with your friend. Please write or type your answer and conclusion separately. Please upload the executable EXCEL files or JUMP files in blackboard. - 1. (60% Project) In class we talked in detail about EWMA control chart to detect small shifts in process mean and when we have small rational subgroup (usually when n = 1). Another method that can be used when n = 1 and to detect small shift in process mean is cumulative Sum Control Chart (CUSUM). (Read p681-687 in your text book). The data can be downloaded from blackboard. (a) Is there a serious problem with the autocorrelation in these data? Plot both scattered plot and autocorrelation function plot. Is the process stationary? (b) set up a control chart for individuals with a moving range used to estimate process variability. what conclusion can you draw from this chart? ‘ (c) Design a CUSUM control scheme for this process, assuming that the observations are uncorrelated. How does the CUSUM per— form? (d) Set up an EWMA control chart with A = 0.15 for this process. How does this chart perform? (e) Set up a moving average control chart with w = 6 (6 day average). How does this chart perform? (f) Set up a moving center-line EWMA control chart? How does this perform? (g) Select an ARIMA model to represent the data. Find the param- eters of ARIMA model. Give reasons for your choice. (h) Repeat part .(b), (c), (d), (e), and (f) using the'uncorrelated residues. . 2. (40 % Project) In this course we learned about Shewhart Control Charts for process mean. Sometimes instead of process mean, we would like to compare the process means of current batch with a standard ref- erence batch. The reference batch consists of the same product that are produced under controlled process conditions except possibly being influenced by uncontrolled factors. ” Difference Control Charts” are de— veloped to monitor the difference between the reference mean ,LLRef and the current production lot mean your (a) Assume that URef and do,“ are known. ii. iii. iv. v. 1. Calculate central line, UCL, and LCL for a Difference Control Chart. ' How would you calculate type-I error. How would you calculate type—II error. Construct an OC—curve plot. Construct an ARL curve plot. (b) Assume that variance is not known but (afief = can). Repeat the calculations as in the previous part. ChE37 6K Process Evaluation and Quality Control . Homework-9 Due on 27th Nov 2007 Evening 5 PM 1. Consider the EWMA statistic 2;» = A331- + (1 — A)zi_1 (a) What are the center line and control limits for the EWMA and Moving Average control chart for rational subgroup of size n = 1? (b) How will you modify these limits when considering a rational sub— group of size n > 1? (c) Show that if 2 (w+1) for the EWMA control chart, then this control chart is equivalent to a w-period moving average control chart in the sense that the control limits are identical in the steady state. (d) The current data mi in EWMA statistic is weighted by a factor A. Compute the weights used for sci--1, $¢_2, and $i_3. /\= 2. A shewhart 50 chart has center line at 10 with U CL = 16 and LCL = 4. Suppose you wish to supplement this chart with an EWMA control chart using A = 0.1 and the same control limit width in a—units as employed on the 53 chart. What are the values of the steady state upper and lower control limits on the EWMA Chart? 3. An EWMA control chart uses A :04. How wide will the limits be on the Shewhart control chart, expressed as a multiple of the width of the steady—state EWMA limits? 4. Consider the AR(1) and AR(2) processes xt = + 0.6$t_1 + 6: 0'; = 4 mt = + V0.8:L't_1 — 0.25375—2 'l— 613 0'? = 1 (a) Are these processes stationary? (b) Graph the theoretical autocorrelation function. (0) If $45 = 12, would you expect $46 to be greater than or less than the mean of the series? ...
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