Chng 6: M HNH HA V TI U HA THC NGHIM6.1. Cc phng php qui hoch thc nghim Trong cng ngh ha hc, chng ta nghin cu mt i tng cng ngh thng ph thuc ng thi vo nhiu yu t m bn cht qui lut ca qu trnh xy ra bn trong i tng cha c bit r. Da vo nhng hiu bit ban u v i tng, trc khi tin hnh thc nghim chng ta cn xc nh s b m hnh ton hc ca i tng, cn gii thch nhng yu t no phi thay i trong qu trnh lm th nghim, nhng yu t no gi mc c nh v mc tiu cn t c ti u. Phng php m hnh ha ton hc l phng php tnh ton v phn tch cc qu trnh k thut, l vn chn cng thc thc nghim v c lng cc tham s ca cng thc . c lng gi tr thc ca cc i lng c kho st v chnh xc ca cc c lng cn phi tin hnh x l cc s liu thc nghim. Phng php x l s liu c s dng l phng php phn tch hi qui. i tng trong cng ngh ha hc ph thuc vo cc yu t cng ngh c th iu chnh nh nng , p sut, nhit , pH, v cc yu t ca bin ngu nhin khng iu khin c. Hm mc tiu ca i tng l hiu sut, cht lng sn phm, chi ph sn xut, tm mi quan h gia hm mc tiu y v cc yu t u vo xi bng thc nghim, chng ta cho thay i tc ng cc bin u vo xi (i lng c th o v iu khin c) v o hm p ng u ra y (i lng o c nhng khng iu khin c) theo m hnh thc nghim nh hnh 6.1.x
X1 X2 Xi
y
xi bin u vo (1 i k). y thng s u ra (bin b iu khin). x - bin ngu nhin khng iu khin c. Hnh 6.1. M hnh nghin cu thc nghim cc i tng cng ngh. Ta cn thit lp quan h: y = f(x1, x2,,xk) + x Hay: y = f(X) + x (6.1) (6.2)
Khi gi thit bin ngu nhin c phng sai D(x) = 2 v c k vng ton hc E(x) = 0 th i lng ngu nhin ny tun theo mt lut phn b no (nh phn b chun) dng: x = N(0, 2) (6.3)
1
Nu loi tr c nh hng ca nhiu (x = 0) th ta thu c cc dng m hnh thng k nh sau: - Nu m hnh thng k vi cc bin u vo cha yu t thi gian th gi l m hnh ng, khng cha yu t thi gian l m hnh tnh. - Nu m hnh ch p dng cho mt min gii hn nht nh ca bin (iu kin rng buc ca bin) gi l m hnh cc b (a phng). Vi m hnh a phng, trin khai hm f(x1, x2,,xk) di dng chui Taylor: Vi 1 i j k Cc h s 0, j, jj, c xc nh t s liu thc nghim th phng trnh thu c gi l phng trnh hi qui thc nghim ca h thng, khi :
Phng trnh hi qui thc nghim (6.5) ph thuc vo b n th nghim v phng php x l s liu thc nghim. Qu trnh tin hnh th nghim sao cho s ln th nghim t nht, tnh ton n gin v thu c kt qu chnh xc nhng vn kho st c nh hng ng thi ca nhiu thng s tc ng n i tng cng ngh phn nh chnh xc bn cht ca qu trnh v gim chi ph tin hnh th nghim. Do , cn tin hnh th nghim theo k hoch nh trc sao cho c tnh trc giao, ngha l b tr thc nghim theo ma trn bin u vo c dng: (ma trn n dng, (k + 1) ct) [ ]
(6.6)
Ma trn ct c dng:
[ ]
(6.7)
Ma trn cc h s hi qui tuyn tnh c dng:
[
]
(6.8)
Ma trn chuyn v ca ma trn X c dng: (ma trn (k + 1) dng, n ct)
2
[
]
(6.9)
Nu cc kt qu thc nghim c biu din theo phng trnh hi qui tuyn tnh bng phng php bnh phng cc tiu th ta c dng ma trn ca h phng trnh chun: ( ) ( ) (6.11) (6.10)
Suy ra:
Vi (XTX)-1 l ma trn nghch ca ma trn XTX khi nh thc ca ma trn XTX khc khng hay ma trn XTX khng suy bin. Ma trn trc giao X c nhng tnh cht sau: - Tnh trc giao: tch v hng ca hai vect ct bt k ca X bng 0.n
xim xij =0i=1
vi j, m= 0, k
(6.12)
- Tnh cht i xng: tng cc phn t trong mt ct bt k u bng 0.n
xij =0i=1
vi j
0.
(6.13)
p dng tnh cht ca ma trn trc giao, chng ta c ma trn XTX tr thnh ma trn ng cho:
[
]
Khi , ma trn nghch o (XTX)-1 trong qui hoch trc giao c dng:
[ 3
]
Vi:n
i=1
vi j= 0, k
(6.16)
Ta tnh ma trn XTY:
[
][
]
[
]
Khi ma trn cc h s hi qui c tnh theo cng thc (6.11). Trng hp tng qut, nu cc s liu thc nghim c biu din bng a thc bc th s h s hi qui trong phng trnh hi qui bng . S dng mt s php bin i, chng ta bin i phng trnh hi qui a thc bc v phng trnh hi qui tuyn tnh bng cch thay cc s hng phi tuyn ca thc bng cc s hng tuyn tnh. Phng trnh ny c gi l phng trnh tuyn tnh ha theo thng s. Nu bc ca a thc cha bit trc th vic tnh ton thng phi tin hnh vi ln, tng dn bc ca a thc n khi no phng trnh hi qui nhn c tng thch vi thc nghim. Mi ln tng bc a thc, ton b vic tnh ton lin quan n phn tch hi qui u phi tnh ton li. S thay i bc ca a thc hoc loi b mt phn cc s hng trong phng trnh hi qui dn n thay i cc gi tr tnh ton cc h s cn li trong a thc. Khi xut hin s khng xc nh trong vic c lng cc h s hi qui, gy kh khn cho vic gii thch nh hng ca cc bin c lp vo i lng nghin cu. loi b cc kh khn ni trn, chng ta tin hnh thc nghim theo qui hoch trc giao v ma trn thc nghim trc giao phi tha mn cc iu kin sau: - Cng thc tnh cc h s phng trnh hi qui bj. - H s bj l c lng trng ca cc h s j. - Phng trnh hi qui l c lng trng ca y. 6.1.1. Qui hoch thc nghim yu t ton phn 2k Qui hoch thc nghim yu ton phn l thc nghim m mi t hp cc mc ca cc yu t u c thc hin nghin cu. B tr thc nghim yu t ton phn theo ma trn trc giao v phn b i xng cc bin c lp vi tm i xng. Ma trn X tha mn iu kin trc giao v c thm tnh cht chun ha: Tng bnh phng cc phn t ca mt ct bng s th nghim. 4
n
i=1
=N
vi j= 0, k
(6.18)
Vi k l cc yu t, n l s mc th s th nghim N = nk . Nu cc th nghim ch thc hin hai mc, thng l hai gi tr bin ca mi yu t kho st th N = 2k. Gi s kho st bin Zj vi hai gi tr bin l aj < bj (1 j k), chng ta thc hin mt s php bin i sau: t: Khi ta i bin s theo biu thc: Suy ra: Zj [aj, bj] th xj [-1, 1]. Khi ta p dng cng thc (6.18) vo cng thc (6.14) ta c: [ ] = N.I
Suy ra: [ ]
Ma trn h s hi qui c tnh theo cng thc (6.11):
(
)
[ Hay: Khi phng trnh hi qui c dng:
]
Trong trng hp phng trnh hi qui c m t di dng: 5
Cc h s bij c xc nh theo cng thc sau: Vi s lng h s bij c xc nh theo cng thc:
6.1.2. Qui hoch thc nghim yu t tng phn 2k - p Khi s bin k trong m hnh qui hoch thc nghim yu t ton phn 2k cng ln th lm cho s th nghim N cng ln. iu ny lm cho qui hoch tr nn cng knh, chi ph ln, km hiu qu. khc phc iu , ngi ta dng qui hoch thc nghim yu t tng phn N = 2k p, vi p l gi tr c trng cho tng phn, l s hiu ng tng tc c thay bng s hiu ng tuyn tnh. Thc cht y l qui hoch thc nghim yu ton phn bt i p ct ca k thng s c lp, s th nghim gim i 2p ln nhng vn m bo tnh trc giao ca ma trn X. Qu trnh qui hoch thc nghim yu t tng phn theo cc bc sau: Bc 1: chn ra r thng s chnh nh hng n hm mc tiu trong k thng s u vo: r = k p. Lp qui hoch thc nghim yu ton phn 2r vi s th nghim l N = 2r. Tuy nhin, khi la chn gi tr p phi m bo iu kin sau: k + 1 N = 2r = 2k-p 2k (6.24)
Bc 2: tin hnh thit lp cc biu thc tng quan sinh biu din cc mi tng quan gia mi thng s p vi mt tch cc thng s trong r thng s chnh. Cc biu thc tng quan sinh c th l tch ca cc thng s trong r thng s chnh mang du dng hay m. Bc 3: kim tra tnh tin li ca m hnh c lp: nu ma trn X khng c cc ct ging nhau hoc ngc du nhau th vn m bo tnh trc giao, qui hoch thc nghim t yu cu. Bc 4: tin hnh xc nh v kim tra ngha ca cc h s hi qui bj, kim tra s tng thch ca phng trnh hi qui thu c. 6.1.3. Qui hoch trc giao cp 2 Qui hoch trc giao cp 2 l qui hoch thc nghim v x l s liu thc nghim theo phng php xy dng m hnh hi qui cp 2 vi cc iu kin tng t nh qui hoch thc nghim yu t ton phn (qui hoch trc giao cp 1). Phng trnh hi qui bc 2 y c dng: 6
Xy dng ma trn trc giao X bao gm ba loi th nghim: - Phn c s gm n = 2k th nghim theo qui hoch thc nghim yu t ton phn. - Phn im * gm nk = 2k im nm trn cc trc ta ca khng gian k yu t v cch tm phng n khong cch > 0. - Phn tm gm n0 (n0 1) th nghim tm phng n dng xc nh phng sai ti hin trong cng thc kim tra ngha ca cc h s hi qui. Tng s th nghim trong phng n l N = 2k + 2k + n0 Tuy nhin, nhng phng n cu trc c tm khng trc giao doN
> 0 nn:
i=1 N
0
vi j= 1, k
(6.26)
i=1
0
vi j= , j 1, k
u
(6.27)
V vy, khi xy dng ma trn trc giao X cn chn sao cho trc giao ha cng thc (6.26), (6.27). Gi s xt qui hoch thc nghim k = 2 yu t v n0 = 1 th ma trn X c dng:
(6.28)
[ t (i = 1, 2) th ta c:
]
(6.29)
[ S dng cc iu kin trc giao ca ma trn X:
]
Tch v hng ca vect ct 1 v ct 6 bng 0, tc l ta c ng thc sau: 22(1 - ) + 22 2.2 =0 - n0. = 0
Hay tng qut vi k yu t: 2k(1 - ) + 22 2k 7
Suy ra:
(6.30)
Tch v hng ca vect ct 5 v ct 6 bng 0, tc l ta c ng thc sau: 22(1 - )2 - 2.2 ( 2- ) +2
=02
Hay tng qut vi k yu t: 2k(1 - )2 - 2k ( 2- ) + n0 . Suy ra:
=0 (6.31)
Th cng thc (6.30), (6.31) vo tnh c ma trn X. Khi cc h s hi qui c tnh theo cng thc (6.11): ( ) 6.1.4. Qui hoch thc nghim tm cc tr Cc phng trnh hi qui thu c l biu din gn ng i tng cng ngh m ta ang kho st. Vn t ra l tm , ,, sao cho: (6.32) Khi hm hi qui t chnh xc cn thit, ta dng qui hoch ton hc tm cc tr: - Nu hm tuyn tnh ta dng phng php qui hoch tuyn tnh; - Nu hm phi tuyn ta p dng phng php qui hoch phi tuyn. Nu chnh xc cha t yu cu, kt qu cn th, km tin cy th trc khi s dng phng php qui hoch ton hc cn thu hp vng cha im cc tr, tc l c th tm vng cc tr bng qui hoch thc nghim. Phng php qui hoch thc nghim tm cc tr c chia lm hai giai on: - Tm vng cha im cc tr bng qui hoch trc giao cp 1. - Tm phng trnh hi qui cp 2 bng qui hoch trc giao cp 2 v cui cng l dng phng php qui hoch phi tuyn tm cc tr: qui hoch li hoc qui hoch ton phng. Vect gradient ca hm y = f(x) ti x* (k hiu l grad[f(x*)]) l vect c chiu biu th s bin thin nhanh nht ca hm y ti x*, gi tr ca grad[f(x*)] thay i t im ny sang im khc trong khng gian yu t. Vi m hnh tuyn tnh k yu t: [ Vi: + ] : o hm ring ca hm f(x) theo bin xj ti x*. (6.33)
+ ij (j = 1, 2, ,k) : vect n v theo cc trc ta . Xt mt min con D0 c tm Z0 ng vi x0, gi s phng trnh hi qui c dng:
8
Kim nh s tng hp ca , nu tng hp th c ngha l mt cong c xp x bng mt phng th D0 khng cha im cc tr. Chuyn sang vng D1 theo hng gradient [f(x0)] cho n khi y khng tng c na. Lp li qu trnh ny n khi hm khng tng hp th chuyn sang vng cha im cc tr, cn chuyn sang bc 2. Theo thut ton leo Box Wilson nh sau: Tnh cc thnh phn ca gradient theo trin khai Taylor:
Nu ta ly gn ng n s hng bc nht v t: Lng m hm f(x) tng l:
Bc tin nhanh nht tng ng vi s hng lm (6.37) tng nhanh nht j* l: | | j = (6.38)
di cc bc hj ca cc yu t c tnh theo j*:
Ch : di cc bc khng nn qu ln hoc qu nh. Chuyn ng theo grad phi bt u t im 0. Mc c s ca mi yu t v ngng li nu tm c im ti u hoc nu nhng hn ch t vo cc yu t lm cho chuyn ng tip tc theo grad khng hp l na. Cch tin hnh: gi Z0 l tm min D0 trong ta bin tht k hiu l M0. im M1 c . Ti y ta lm th nghim xc nh c cc ta xc nh theo: y2. Lp li ta s thu c dy: y0, y1, , yn. Khi thc hin: y0 < y1 tip tc lm y2. y2 < y3 tip tc lm y3. y3 < y4 tip tc lm y4. 9
yp-1 > yp mt cong bt u. Dng li Mp-1 vi tm mi Mp-1 = Zp-1. Ti Xp-1 li tip tc qu trnh xp x bng mt phng. Kim nh s ph hp ca m hnh bc nht, nu tha mn th tip tc tm di bc lm th nghim. Tin hnh cho n khi m hnh bc nht khng ph hp th l vng cc tr. 6.2. Xc nh cc h s ca phng trnh hi qui Chn phng n th nghim vi k bin c lp, thc hin n th nghim, khng c th nghim lp li v b tr thc nghim sao cho ma trn X c tnh trc giao. 6.2.1. Tnh h s hi qui bj ca phng trnh hi qui tuyn tnh p dng cng thc (6.11), (6.15), (6.16), (6.17) ta c:
(
)
[ Suy ra:
] [
]
[
]
6.2.2. Tnh h s hi qui bj ca phng trnh hi qui bc 2 Tin hnh qui hoch thc nghim theo m hnh qui hoch trc giao cp 2 th cc h s hi qui c xc nh theo cng thc (6.11) sau khi lp c ma trn trc giao X, hay xc nh theo cc cng thc sau: bj = 10 N xij y i=1 N x2 i=1 iji
(6.42)
Khi phng sai
c xc nh theo cng thc:
6.3. Kim tra s tng hp ca phng trnh hi qui Sau khi tnh ton c cc h s ca phng trnh hi qui, chng ta cn tin hnh kim nh s tng hp ca phng trnh hi qui vi s liu thc nghim. Qu trnh kim nh c tin hnh theo hai bc sau: Bc 1: kim tra s c ngha ca cc h s phng trnh hi qui bng tiu chun thng k (chun s) Student t (vi : mc ngha). Bn cht ca phng php l kim tra cc h s bj = 0 hay khng, hoc kim nh xem thc cht c bao nhiu yu t nh hng n hm mc tiu. Chn thng k:| |
(6.46)
sbj: lch qun phng ca h s th i. Phn sai c xc nh theo cng thc: Vi phng sai ti hin c tnh theo s th nghim lp tm n0 theo cng thc: Trong : - N: s th nghim. - n0: s th nghim lp tm. - fth = n0 1 : bc t do ti hin. Thng thng chn mc ngha = 0,05. Tra theo chun Student ta c c gi tr t (fth). - Nu tbi > t(fth) th h s bi c gi li trong phng trnh hi qui, nh hng ca yu t xi c ngha i vi vic thay i thng s ti u y. - Ngc li, nu tbi < t(fth) th h s bi b loi khi phng trnh hi qui. Bc 2: kim tra s tng thch ca phng trnh hi qui theo tiu chun Fisher:
Phn sai
c xc nh theo cng thc: 11
Trong :
- L: s h s c ngha trong phng trnh hi qui. Tra theo chun Fisher ta c c gi tr F(, ftt, fth), vi ftt = N L , fth = n0 1. - Nu F < F th m hnh thng k ph hp vi s liu thc nghim. - Nu F F th m hnh thng k khng ph hp vi s liu thc nghim. 6.4. ng dng Excel ti u ha thc nghim gii bi ton ti u ha thc nghim chng ta cn tin hnh cc bc sau: - Bc 1: chn phng n tin hnh th nghim. - Bc 2: lp ma trn thc nghim X. - Bc 3: tin hnh th nghim xc nh gi tr bin u ra Y. - Bc 4: xc nh cc h s trong phng trnh hi qui. - Bc 5: nh gi phng trnh hi qui thu c. - Bc 6: xc nh ch thc nghim ti u. Mi bc c th thc hin php lp nu cha t yu cu v mc tiu ca cng ngh, sai sCc bc 1, 2 v 4, 5 c trnh by phn trn v phng php thc hin. Bc 3 c thc hin phng th nghim, ph thuc vo trang thit b, dng c th nghim v tay ngh ca ngi lm th nghim. Bc 6 c ti u ha theo m hnh tm ra. Kho st phng php thc hin bc 4 v 5 bng phn mm Microsoft Excel s dng trong tnh ton bng cc hm: - Nhn hai ma trn: MMULT(array1, array2). - Tnh nh thc ca ma trn: MDETERM(array). - Tnh ma trn nghch o: MINVERSE(array). - Tnh ma trn chuyn v: TRANPOSE(array). - Tnh gi tr trung bnh cc s hng: AVERAGE(number1, number2,). - Tnh tng bnh phng cc s hng - Tnh tng bnh phng lch - Tnh lch chun ca mu
: SUMSQ(number1, number2,). : SUMXMY2(array_x, array_y). : STDEV(number1, number2,).
- Tra chun s Student: TINV(p1, p2). 12
- Tra chun s Fisher: FINV(, p1, p2). Lu : thc hin cc php ton ma trn trong Excel, chng ta phi chn cc phn t ca ma trn cn tnh (qut khi kch thc ma trn) v n gi ng thi ba phm Ctrl + Shift + Enter. V d 1: Tin hnh qui hoch thc nghim nghin cu nh hng ca 3 yu t Z1, Z2, Z3 ln hm mc tiu y vi s liu thu c nh sau:Bin thc Z1 Z2 Z3 y n=2 1 150 30 15 3,0 2 300 30 15 6,0 3 150 90 15 10,0 4 300 90 15 12,0k
n0 = 3 5 6 300 30 45 23,0 7 150 90 45 12,0 8 300 90 45 18,0 9 225 60 30 12,0 10 225 60 30 13,8 11 225 60 30 13,2
150 30 45 15,0
Hy tm mi quan h gia y v cc bin Z1, Z2, Z3 theo m hnh trc giao cp 1? Gii: Chn phng trnh hi qui biu din mi quan h gia y v cc bin Z 1, Z2, Z3 c dng theo (6.21) th qui trnh ngh gii theo cc bc sau: Bc 1: lp 1 bng tnh excel vi cc thng tin sau:
Trong : - Gi tr mc thp, mc cao ca cc bin thc ly t s liu cho. - Zj0 : mc c s, l trung bnh cng ca mc thp v mc cao nn ta nhp cng thc cell D2 = AVERAGE(B2:C2)
Sau dng draf ko chut xung cc hng trong ct Zj0. 13
Khong bin thin Zj c tnh theo cng thc trong cell E2 = (C2-B2)/2
Sau dng draf ko chut xung cc hng trong ct Zj. Bc 2: lp bng tnh chuyn i cc bin thc sang bin m ha:
Nhp gi tr cc trong ct x0 u bng 1. nn cell G8 ta nhp: =(C8-$D$2)/$E$2, sau dng draf ko chut xung cc hng trong ct x1.
nn cell H8 ta nhp: =(D8-$D$3)/$E$3, sau dng draf ko chut xung cc hng trong ct x2. 14
nn cell I8 ta nhp: =(E8-$D$4)/$E$4, sau dng draf ko chut xung cc hng trong ct x3.
Bc 3: xc nh cc h s phng trnh hi qui bng phng php ma trn: ( ) , ta lp bng excel nh sau:
Ma trn h s X: do c 3 th nghim c thc hin tm nn ma trn X l ma trn 8 dng, 4 ct. xc nh gi tr cc phn t trong ma trn X ta copy gi tr cc F8:I15 vo cc A21:D28. Ma trn cc bin u ra y: tng t nh ma trn X, ta copy gi tr cc J8:J15 vo cc F21:F28. Ma trn chuyn v XT: l ma trn c kch thc 4 dng, 8 ct. xc nh gi tr cc phn t trong ma trn XT ta lm nh sau: sau khi qut khi cc H21:O24, ta nhp cng thc: =TRANSPOSE(A21:D28) v nhn ng thi ba phm Ctrl + Shift + Enter.
15
Ma trn XTX: l ma trn c kch thc 4 dng, 4 ct. Nhn hai ma trn XT vi ma trn X: ta qut khi cc H27:K30 v nhp cng thc: =MMULT(H21:O24,A21:D28) v nhn ng thi ba phm Ctrl + Shift + Enter.
Kim tra nh thc det (XTX): ta nhp cng thc =MDETERM(H27:K30). Ma trn nghch o (XTX)-1: qut khi cc M27:P30, nhp cng thc:
=MINVERSE(H27:K30) v nhn ng thi ba phm Ctrl + Shift + Enter.
Ma trn XTY: l ma trn c kch thc 4 dng, 1 ct. Ta nhp cng thc: =MMULT(H21:O24,F21:F28). Ma trn h s hi qui B: ta qut khi cc D31:D34, nhp cng thc = MMULT(M27:P30,A31:A34) nhn ng thi ba phm Ctrl + Shift + Enter.
16
Ta thu c cc h s hi qui: b0= 12,375; b1= 2,375; b2= 0,625; b3= 4,625. Bc 4: tin hnh lp bng tnh kim tra ngha ca cc h s hi qui:
Phng sai ti hin
: c tnh theo s th nghim lp tm n0=3, theo cng thc (6.48)
nn ti O8 ta nhp cng thc: =STDEV(J16:J18)^2. lch qun phng Sbj: c tnh theo cng thc (6.47) nn ti P8 ta nhp cng thc: =SQRT(O8/B15). Cc h s tbj: tnh theo cng thc| |
nn ta nhp cng thc ti Q8: =Abs(N8)/$P$8
v sau dng draf ko chut xung cc hng trong ct tbj.
Chun s student t(fth): chn mc ngha = 0,05, bc t do ti hin f th = n0 1 = 2 nn ta nhp cng thc ti O12: =TINV(0.05,2). Kt qu chn h s hi qui: ti R8 ta nhp cng thc: =IF(Q8
