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8/8/2019 Bai Giang Qui Hoach Thuc Nghiem 22-3-2010
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1
I HC NNGTRNG I HC S PHM
--- ---
BI GING MN
QUY HOCH THC NGHIMUY HOCH THC NGHIM(CC PHNG PHP THNG K X L S LIUCC PHNG PHP THNG K X L S LIU
THC NGHIM)HC NGHIM)
Ngi son: Giang Th Kim Lin
Nng, 2009
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Chng 1. CC KHI NIM CHUNG
1.1. Qui hoch thc nghim - bc pht trin ca khoa hc thc nghim
Nhiu cng trnh nghin cu khoa hc cng ngh thng a n gii bi
ton cc, tm iu kin ti u tin hnh cc qu trnh hoc la chn thnh
phn ti u tin hnh cc qu trnh hoc la chn thnh phn ti u ca h
nhiu phn t. Chng hn, khi xem xt cc qu trnh CN ha hc mi, nhim v
nghin cu thng l thay i nhit , p sut v t l cc cht phn ng tm
hiu sut phn ng cao nht, tnh ton, la chn gi tr thch hp nht ca cc
thng s cu trc v ng hc, nhm t n cht lng lm vic v hiu qu
kinh t cao nht ca qu trnh. Nhng bi ton ny thng gii quyt cc mc
nghin cu cc yu t nh hng n h, lp m hnh biu din mi phthuc gia cc phn t ca h, iu khin h theo mc ch cho trc, hoc a
v trng thi ti u theo nhng ch tiu nh gi chn. Thng thng cc h
cn iu khin v ti u rt phc tp, i tng nghin cu ngy cng a dng
hn, tr thnh nhng h thng cng knh vi tp hp ln cc yu t nh hng
v ch tiu nh gi. Mi quan h gia cc thnh phn trong h thng cng
khng th m t bng cc hm l thuyt. V vy, a s cc bi ton cc tr c
gii quyt bng thc nghim.Ngy nay ngi ta thng cp ti phng php kt hp gia l thuyt
v thc nghim. Ty theo mc hiu bit v c ch ca qu trnh, ngha ca
nghin cu l thuyt thng c gii hn tc dng nh hng ban u, h
tr gim bt khi lng cng vic, rt ngn thi gian cho nghin cu thc
nghim. Bn cnh , thc nghim c tc dng tr li, b sung cho kt qu
nghin cu l thuyt, xc nh r hn c ch ca hin tng.
Vai tr ca thc nghim cng ln th mc tiu ra cho chng cng cao,
v vy thc nghim cng c nhu cu pht trin v tr thnh i tng nghin
cu, mt ngnh khoa hc.
C th ni, l thuyt qui hoch thc nghim t khi ra i thu ht s
quan tm v nhn c nhiu ng gp hon thin ca cc nh khoa hc.
Nhng u im r rt ca phng php ny so vi cc thc nghim c in l:
- Gim ng k s lng th nghim cn thit.
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- Hm lng thng tin nhiu hn r rt, nh nh gi c vai tr qua li
gia cc yu t v nh hng ca chng n hm mc tiu. Nhn c m hnh
ton hc thng k thc nghim theo cc tiu chun thng k, nh gi c sai
s ca qu trnh thc nghim theo cc tiu chun thng k cho php xt nh
hng ca cc yu t vi mc tin cy cn thit.- Cho php xc nh c iu kin ti u a yu t ca i tng nghin
cu mt cch kh chnh xc bng cc cng c ton hc, thay cho cch gii gn
ng, tm ti u cc b nh cc thc nghim th ng.
1.2. Nhng khi nim c bn ca qui hoch thc nghim
Qui hoch thc nghim l c s phng php lun ca nghin cu thc
nghim hin i. l phng php nghin cu mi, trong cng c ton hc
gia vai tr tch cc. C s ton hc nn tng ca l thuyt qui hoch thc
nghim l ton hc xc sut thng k vi hai lnh vc quan trng l phn tch
phng sai v phn tch hi qui.
* nh ngha qui hoch thc nghim: qui hoch thc nghim l tp hp
cc tc ng nhm a ra chin thut lm thc nghim t giai on u n giai
on kt thc ca qu trnh nghin cu i tng (t nhn thng tin m phng
n vic to ra m hnh ton, xc nh cc iu kin ti u), trong iu kin
hoc cha hiu bit y v c ch ca i tng.
* i tng ca qui hoch thc nghim trong cc ngnh cng ngh: Lmt qu trnh hoc hin tng no c nhng tnh cht, c im cha bit
cn nghin cu. Ngi nghin cu c th cha hiu bit u v i tng,
nhng c mt s thng tin tin nghim d ch l s lit k s lc nhng
thng tin bin i, nh hng n tnh cht i tng. C th hnh dung chng
nh mt hp en trong h thng iu khin gm cc tn hiu u vo v u
ra, nh hnh 1.
3
HP EN(QU TRNH
LM VIC CAH THNG)
Z
E
T
Y
I TNGNGHIN CUZ
e
T
Y
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Hnh 1.S i tng nghin cu Hnh 2.S i tng nghin
cu vi nhiu e c tnh cng
- Cc tn hiu u vo c chia thnh ba nhm:
1) Cc bin kim tra c v iu khin c, m ngi nghin cu c th
iu chnh theo d nh, biu din bng vect:
Z = [Z1
, Z2
, ..., Zk
]
2) Cc bin kim tra c nhng khng iu khin c, biu din bng
vect:
T = [T1
, T2
, ..., Th
]
3) Cc bin khng kim tra c v khng iu khin c, biu din
bng vect:
E = [E1
, E2
, ..., Ef
]
- Cc tn hiu u ra dng nh gi i tng l vect Y = (y1
, y2
,...,
yq
). Chng thng c gi l cc hm mc tiu. Biu din hnh hc ca hm
mc tiu c gi l mt p ng (b mt biu din).
Phng php tan hc trong x l s liu t k hoch thc nghim l
phng php thng k. V vy cc m hnh biu din hm mc tiu chnh l cc
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m hnh thng k thc nghim. Cc m hnh ny nhn c khi c cng tnh
nhiu ngu nhin. Cu trc m hnh thng k thc nghim c dng nh hnh 2.
Trong tp hp cc m hnh thng k khc nhau, m hnh c quan tm
nhiu nht trong thc t l m hnh ca phn tch hi qui. M hnh hi qui c
biu din bng quan h tng qut:
Y = (Z1
, Z2
, ..., Zk
; T1
, T2
, ..., Th
; 1
, 2
,..., k
) + e = [(Z, T) ; ] + e
Trong = (1
, 2
,..., k
) l vect tham s ca m hnh.
Dng hm c n nh trc, cn cc h s l cha bit, cn xc nh
t thc nghim
xc nh cc tham s ca m t thng k thc nghim ta phi lm cc
thc nghim theo k hoch thc nghim. i tng nghin cu chnh ca l
thuyt qui hoch thc nghim l cc thc nghim tch cc. l cc thc
nghim ch bao gm cc yu t u vo thuc nhm Z, ngi thc nghim ch
ng thay i chng theo k hoch thc nghim vch sn.
* Cc phng php qui hoc thc nghim :
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- Thc nghim sng lc : l thc nghim m nhim v ca n l tch
nhng yu t nh hng ng k ra khi nhng yu t u vo tip tc
nghin cu chng trong cc thc nghim cn thit.
- Thc nghim m phng : l thc nghim lin quan ti vic m phng
hin tng cn nghin cu. C nhiu dng m phng, y ch quan tm n
dng thc nghim c hon tt bng m hnh hi qui a thc.
- Thc nghim cc tr : l thc nghim c pht trin t thc nghim m
phng. Nhim v ca n l xy dng m hnh ton thc nghim, theo xc
nh gi tr ti u ca hm mc tiu v cc ta ti u ca hm. Ni cch
khc l xc nh b kt hp gi tr cc yu t m ti hm mc tiu t cc
tr.
* K hoch thc nghim :
i vi cc thc nghim tch cc, min tc ng l min cc gi tr c th
c ca cc yu t Z trong thc nghim. Trong min tc ng c min qui hoch
- min gi tr ca cc yu t vo Z - trong cha va cc im th nghim
ca thc nghim. Ni cch khc, l min to bi phm v thay i cc yu t
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Z theo k hoch thc nghim xc nh. K hoch thc nghim bao gm cc
im th nghim gi l im ca k hoch. l mt b (cn gi l phng n)
kt hp cc gi tr c th ca cc yu t vo Z, ng vi iu kin tin hnh mt
th nghim trong tp hp cc th nghim ca thc nghim. Ti im th i ca k
hoch, b kt hp cc gi tr Zji
bao gm gi tr c th ca k yu t u vo :
Zji
= [Z1i
, Z2i
, ..., Zkj
]
Trong : i = 1, 2, ..., N l im th nghim th i ca k hoch th
N l s im th nghim ca k hoch.
j = 1, 2, ..., k l yu t th j ; k l s yu t u vo.
* Cc mc yu t :
Cc gi tr c th ca yu t vo Z c n nh ti cc im k hoch
gi l cc mc yu t. Khi nim mc yu t dc s dng khi m t cc im
c trng trong min qui hoch: mc trn, mc di, mc c s, mc sao *.
Mc c s Z
0
jca cc yu t l iu kin th nghim c qun tm c
bit. Thng thng vect cc yu t u vo ti mc c s Z
0
= [Z
0j
, Z
0j
, ...,
Z
0j
] ch ra trong khng gian yu t mt im c bit no gi l tm k
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hoch, m trong vng quanh n phn b ton b cc im k hoch. Cc ta
Z
0j
ca vect Z
0
c chn theo cng thc:
j
jj
j
Z
ZZX
=
0
; j = 1, ..., k
2
minmax
jjZZ
Zj
= ; j = 1, ..., k
* Gi tr m ha: tin tnh cc h s thc nghim ca m hnh hi qui
ton hc v tin hnh cc bc x l s liu khc, trong k hoch thc nghim
ngi ta s dng cc mc yu t theo gi tr m ha. Gi tr m ha ca yu t
l i lng khng th nguyn, qui i chun ha t cc mc gi tr thc ca
yu t nh quan h :
minmax
00 )(2
jj
jj
j
jj
jZZ
ZZ
Z
ZZx
=
=
Trong ti liu ny chng ta gi nguyn cc k hiu: Zj
l gi tr thc ca
yu t (gi l bin thc) ; xj
l gi tr m ha ca yu t (gi l bin m).
Nh vy, theo t l qui chun, mc c s m ha ca yu t u vo l : x
0j
= 0.
Gc ta ca cc xj
trng vi tm thc nghim, bc thay i ca cc
bin m xj
ng vi cc bc xj
chnh l 1 n v.
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1
2
minmax=
=
j
jj
jZ
ZZx
*Ma trn k hoch thc nghim: l dng m t chun cc iu kin tin
hnh th nghim (cc im th nghim) theo bng ch nht, mi hng l mt th
nghim (cn gi l phng n kt hp cc yu t u vo), cc ct ng vi cc
yu t u vo.
Trong ma trn k hoch Z c th c mt s hng m mi thng s vo
u ging nhau, v d, c mt s hng m mi thng s vo u mc c s,
mi Z
0j
.
Ma trn k hoch thc nghim X l ma trn ch gm ton cc bin m xj
.
Cc ct bin m hon ton khc nhau.
1.3. Cc nguyn tc c bn ca qui hoch thc nghim
1.3.1. Nguyn tc khng ly ton b trng thi u vo
c thng tin ton din v tnh cht hm mc tiu v nguyn tc cn
tin hnh v s cc thc nghim trong min qui hoch.
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V d, trong trng hp c hai yu t, nu cho mi yu t bin i lin
tc t -1 n +1 th min thc nghim s l hnh vung cha v s im M(x1
,
x2
) c trng cho trng thi u vo.
V l thuyt nu khng tin hnh tt c cc thc nghim th c th b
st c im no ca hm mc tiu, tuy nhin thc t khng th thc hin
c iu . Do vy ngi nghin cu ch c th ly nhng gi tr ri rc,
chn mc bin i no cho cc yu t. S la chn ny cn c c s khoa
hc, n gn lin vi s la chn dng hm, tc l dng m phng ca b mt
p ng. Dng hm thng thng l bc mt hoc bc 2 v s mc bin i
thng l hai hoc ba.
1.3.2. Nguyn tc phc tp dn m hnh ton hc
10
O
* M(x1, x
2)
+1
-1
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Khi cha c thng tin ban u v cc tnh cht ca hm mc tiu, th
khng nn xy dng m hnh phc tp ca i tng trnh chi ph v ch v
thi gian, phng tin vt cht nu khng dng n m hnh . V th l
thuyt qui hoch thc nghim hng dn nn bt u t nhng m hnh n
gin nht, ng vi nhng thng tin ban u c v i tng.
Logic tin hnh thc nghim l nn lm t th nghim c m hnh n
gin (v d m hnh tuyn tnh), kim tra tnh tng hp ca m hnh :
- Nu m hnh tng hp, t yu cu th dng li, hoc ci tin ;
- Nu m hnh khng th tin hnh giai on tip theo ca thc nghim :
lm nhng th nghim mi, b sung ri nhn c m hnh phc tp hn (v
d m hnh phi tuyn), kim tra m hnh mi cho n khi t c m hnh hu
dng.
1.3.3. Nguyn tc i chng vi nhiu
chnh xc ca m hnh phi tng xng vi cng nhiu ngu
nhin m chng tc ng ln kt qu o hm mc tiu. Trong cng iu kin
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nh nhau, nhiu cng nh th m hnh cng phi chnh xc, phi phc tp
hn.
Bng cc cng c tnh ton thng k, ngi ta xy dng hon chnh
cc qui trnh chun theo cc tiu chun thng k gii quyt cc nhim v xc
nh tnh tng hp ca m hnh tm c, hiu chnh dng m hnh, kim tra
tnh ng n ca cc gi thit, cc tin m da vo tm ra cc m hnh.
1.4. Cc bc qui hoch thc nghim cc tr
1.4.1. Chn thng s nghin cu
Phn loi cc yu t nh hng ln i tng thnh cc nhm Z, T v E.
Mt mt a ra nhng bin php tch cc hn ch tc ng ca cc nhm
yu t T v E, mt khc phi phn tch chn t Z cc yu t nh hng
chnh, loi bt nhng yu t khng cn thit, nhm m bo tnh kh thi v hiu
qu ca thc nghim
La chn ch tiu (mc tiu) nh gi i tng, sao cho cc ch tiu ny
va p ng cc yu cu ca phng php qui hoch thc nghim, va i din
nht cho cc iu kin ti u ca i tng nghin cu.
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Cn c vo s yu t nh hng chnh, ch tiu nh gi, mc ch, nhim
v thc nghim, ngi nghin cu phi bit nhm cc yu t vo theo k hoch
thc nghim, v tnh hiu qu v kh nng lm vic ca cc m hnh hi qui ph
thuc nhiu vo kt qu xc nh yu t vo ca chng.
Trong giai on ny, min qui hoch v s mc thay i ca cc yu t
nh hng phi c xc nh s b.
1.4.2. Lp k hoch thc nghim
Chn c dng k hoch th nghim ph hp vi iu kin tin hnh th
nghim v vi c im cc yu t ca i tng.
Mi dng k hoch c trng bi cc chun ti u v tnh cht khc nhau.
Nn quan tm nhiu n iu kin th nghim v c im o c, nhn gi tr
ca mc tiu.
1.4.3. Tin hnh th nghim nhn thng tin
S dng cc phng php ring cho tng i tng
S dng mt s phng php x l s liu, kim t mt s gi thit thng
k. Vic x l nhanh cc thng tin ngay trong qu trnh nhn chng c tc dng
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tch cc, gip xc minh kp thi nhng th nghim cn b sung khi iu kin th
nghim cn ang cho php vi cc php kim tra ng nht phng sai, tnh
lin thuc ca s liu b nghi ng, mc nh hng ca cc yu t...
1.4.4. Xy dng v kim tra m hnh thc nghim
S dng phng php bnh phng nh nht v cc ni dung phn tch hi
qui, phn tch phng sai xc nh gi tr ca cc h s trong m hnh hi
qui a thc, kim tra m hnh theo tng thch v kh nng lm vic. Ty
theo loi thc nghim m m hnh l tuyn tnh hay phi tuyn. V d cc dng
phng trnh hi qui:
- M hnh bc hai tuyn tnh:
= =
+++==k
j
k
uj ujjujjk
uj
xxbxbbxxxy1 1,021
. .), . . . ,,(
- M hnh bc hai phi tuyn:
++++== =
k
jjj
k
j
k
uj
ujjujj xbxxbxbby
uj
1
2
1 1,
0 ...
Cc h s hi qui B = [b0
, b1
, b2
..., bk
, b11
, b12
, ..., bjj
] c xc nh theo
cng thc tng qut di dng ma trn :
B = [X*X]
-1
X*Y
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Trong X* - ma trn chuyn v ca ma trn k hoch
M hnh thng k thc nghim ch c th s dng sau khi tha mn
cc tiu chun thng k (Student v Fisher).
1.5. ng dng ca qui hoch thc nghim trong ha hc, cng ngh ha
hc, cng ngh vt liu v cng ngh mi trng
1.5.1. Thit lp cc m t thng k
1) Xc nh cc yu t nh hng v cu trc h
S yu t c lp nh hng ln qu trnh ha l bng s bc t do ca
h, c xc nh theo cng thc :
F = Fk
+ Fh
trong : Fk
l bc t do iu khin
Fh
l bc t do hnh hc
Ty theo yu cu ca ngi nghin cu m ch cn chn ra k yu t
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Cu trc h thc hin qu trnh ha l : l mt hp en khng bit r bn
cht bn trong m ch c mi lin h bn ngoi gia hm mc tiu v cc yu t
nh hng.
2) Xc nh cc hm ton m t h
Hm m t h l hm nhiu bin y = (x1
, x2
, ..., xk
) c phn tch
thnh dy Taylor - hm hi qui l thuyt :
== =
++++=k
j
jjj
k
j
k
uj
ujjujjq xxxxy1
2
1 1,
0 ...
Mun xc nh c cc h s hi qui l thuyt phi cn v s th
nghim. Trong thc t s th nghim N l hu hn, v vy m hnh thng k
thc nghim c dng :
++++== =
k
jjj
k
j
k
uj
ujjujjqxbxxbxbby
uj
1
2
1 1,
0 ...
Cc h s b l cc tham s ca m t thng k.
3) Xc nh cc tham s m t thng k
Cc tham s ca m t thng k c xc nh t N thc nghim nh
cc k hoch thc nghim theo phng php bnh phng cc tiu. Sau khi tnh
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c cc h s b phi kim tra tnh c ngha ca chng theo tiu chun
Student.
4) Kim tra s tng hp ca m t
S tng hp ca m t thng k vi bc tranh thc nghim c kim
chng theo tiu chun Fisher.
1.5.2. Cc phng php k hoch ha thc nghim cc tr ch yu
1) K hoch bc mt hai mc ti u
Nu khng c thng tin tin nghim cho bit h ang vng dng (vng
phi tuyn, vng cc tr) th m t qu trnh nn dng hm tuyn tnh v
khng c cc s hng bnh phng. xc nh cc tham s ca n, nn dng
k hoch bc mt hai mc ti u ca Box-Wilson l k hoch ton phn (2
k
)
hoc trong trng hp cn tit kim thi gian dng k hoch bn phn (2
k-i
).
2) K hoch bc hai
Khi m hnh tuyn tnh bc mt khng tng hp th chng t l vng
thc nghim vng phi tuyn, ta phi dng hm phi tuyn, c cc s hng
bnh phng m t.
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C cc dng k hoch bc hai c bn :
- K hoch trc giao ca Box-Wilson
- K hoch bc hai tm xoay ca Box - Hunter
- K hoch bc hai ti u ca Kiefer
1.6. Khi nim h thng v cch tip cn h thng cng ngh
H thng: l tp hp ca nhiu phn t c:
+ Cu trc bn trong nht nh.
+ Tng tc vi mi trng bn ngoi.
: - Tm c cu trc cn phn tch h thnh nhng phn t
- Nm c hnh vi ca h phi m t tp hp bn cht ca h
Vy nguyn tc tip cn h thng: phn tch v tng hp m t bn cht
ca h.
tm c bn cht ca h phi nh m hnh ho v tm ra c iu
kin cng ngh ti u nh ti u ho cc hm ton m t bn cht ca h
( thng a n gii bi ton cc tr, tc l tm iu kin ti u thc hinmt qu trnh nhm t n cht lng lm vic v hiu qu kinh t cao nht).
1.7 M hnh ho
1.7.1. M hnh
L mt i tng c mt ch th no trn c s ca s ng
dng v cu trc v chc nng dung thay th cho mt nguyn bn tng ng
c th gii quyt mt nhim v nht nh.Mt nguyn bn c th c nhiu m hnh tu thuc vo ch th cn gii
quyt.
1.7.2. M hnh ton
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Mt m hnh ton l biu din ton hc nhng mt ch yu ca 1
nguyn bn theo mt nhim v no , trong phm vi gii hn vi 1 chnh
xc va v trong 1 dng thch hp cho s vn dng.
Mt m hnh ton ca mt nguyn bn phi c 4 iu kin
+ Ch m t nhng mt chnh m ch th quan tm.+ M t trong phm vi gii hn.
+ chnh xc va .
+ Kh nng vn dng m hnh c lp trong iu kin c th.
1.7.3. Cc dng m hnh ton ca i tng cng ngh ho hc
Xt m hmh thng k thc nghim trong ho hc, CNHH ngi ta
xy dng quan h gia cc i lng trn c s thit lp cc quan h trn vic
x l thng k nhng gi tr thc nghim.
xc lp m t thng k ca i tng CNHH cn thc hin nhng bc
sau:
+ Xc nh s cc yu t c lp nh hng ln h, tc l s yu t nh
hng (k) ln 1 hay nhiu hm mc tiu.
+ Xc nh cu trc ca h s c m hnh ho.
+ Xc nh cc hm ton m t cc qu trnh xy ra trong h, v
thng l hm nhiu bin v c biu din : y = f( x1, x2,,xk).
+ Xc nh cc thng s m hnh theo s liu thc nghim.+ Kim tra s tng thch ca m hnh.
1.8. Ti u ho
1.8.1. Khi nim
L qu trnh tm kim iu kin tt nht (iu kin ti u) ca hm s
c nghin cu.
L qu trnh xc nh cc tr ca hm hay tm iu kin ti u tng
ng thc hin 1 qu trnh cho trc. nh gi im ti u cn chn chun ti u (l cc tiu chun cng
ngh).
1.8.2. Cch biu din bi ton ti u
Ga s mt h thng cng ngh c biu din di dng sau:
Y = F(x1,x2,...xk)
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x1,x2,xk : k thnh phn ca vecto thng s u vo.
Hm mc tiu : I = I (x1,x2,xk)
Bi ton c biu din I opt = opt I (x1,x2,xk) =I (x1opt,x2opt,xk )
hoc I opt = max I ( x1,x2,xk) : i vi bi ton max.
I
opt
= min I (x1,x2,xk) : i vi bi ton min.Iopt : hiu qu ti u.
x1opt,x2opt,xk nghim ti u hoc phng n ti u.
1.8.3. Thnh phn c bn ca bi ton ti u
1.8.3.1. Hm mc tiu
- L hm ph thuc.
- c lp ra trn c s tiu chun ti u c la chn.
Hm mc tiu l hm th hin kt qu m ngi thc hin phi t
c
l tiu chun ti u dng hm, ph thuc vo yu t u vo, gi tr ca
n cho php nh gi cht lng ca 1 nghin cu.
1.8.3.2. Quan h gia cc i lng
Cc biu thc ton hc m t cc mi quan h gia tiu chun ti u
ho (hm mc tiu) v cc thng s nh hng (thng s cn ti u) n gi tr
tiu chun ti u ho ny.
Cc quan h ny thng c biu din bng phng trnh c bn hocm hnh thng k thc nghim (phng trnh hi qui).
Quan h gia cc yu t nh hng vi nhau c biu din bng ng
thc hoc bt ng thc.
1.8.3.3. Cc iu kin rng buc
bi ton cng ngh c ngha thc t ,cc biu thc m t iu kin
rng buc bao gm: - iu kin bin.
- iu kin ban uCc bc gii bi ton ti u:
1. t vn cng ngh : xem xt cng ngh cn c gii quyt l
cng ngh g v chn ra nhng yu t nh hng chnh
Ch ra c hm mc tiu Y : YMAX, hoc YMIN
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2. Xy dng mi quan h gia cc yu t nh hng v hm mc tiu
theo qui lut bit trc hoc m hnh thng k thc nghim
3. Tm thut gii: l phng php tm nghim ti u ca cc bi ton
cng ngh trn c s cc m t ton hc tng thch c thit lp. a s
dn n tm cc tr ca cc hm mc tiu4. Phn tch v nh gi kt qu thu c
- Nu ph hp kim chng bng thc nghim
- Nu khng ph hp xem li tng bc hoc lm li t vic t vn
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Chng 2. CC PHNG PHP PHN TCH HI QUI TNG QUAN
2.1. Cc thng s thc nghim
2.1.1. i lng ngu nhin
- nh ngha:i lng ngu nhin (X) l tp hp tt c cc i lng m gi tr ca
n mang li mt cch ngu nhin. Tc l s xut hin l khng bit trc.
- i lng ngu nhin X c gi l ri rc khi n nhn hu hn hoc
v hn cc gi tr m c khc nhau.
- i lng ngu nhin X c gi l lin tc nu n nhn gi tr bt k
trong mt khong ca trc s.
2.1.2. Sai s oTrong thc nghim, nhng gi tr nhn c l gi tr gn ng ca
mt gi tr thc. x = x a gi l sai s o.
Vi : a l gi tr thc ca mt vt.
x l kt qu quan st c.
x l lch gia a v x.
2.1.2.1. Sai s th
- L sai s phm phi do ph v nhng iu kin cn bn ca php o,
dn n cc ln o c kt qu khc nhau nhiu.
- Cch kh sai s th :
+ Kim tra cc iu kin c bn c b vi phm hay khng.
+ S dng mt phng php nh gi, loi b hoc gi li
nhng kt qu khng bnh thng.
2.1.2.2. Sai s h thng
- L sai s khng lm thay i trong mt lot php o, m thay i
theo mt quy lut nht nh.- Nguyn nhn gy sai s: do khng iu chnh chnh xc dng c o,
hoc mt i lng lun thay i theo mt quy lut no , nh nhit
- khc phc ngi ta t mt h s hiu chnh ng vi mi nguyn
nhn.
2.1.2.3. Sai s ngu nhin
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- Sai s ngu nhin ca php o l i lng ngu nhin c trng
bng lut phn phi th hin mi quan h gia cc gi tr c th c ca sai s v
xc sut sai s ngu nhin nhn cc gi tr y.
- L sai s cn li sau khi kh sai s th v sai s h thng.
- Sai s ngu nhin do nhiu yu t gy ra, tc dng rt nh, khng thtch ring ra, v th khng loi tr c.
2.1.3. Cc c trng s ca i lng ngu nhin
2.1.3.1. K vng
1. K vng ton ca bin ngu nhin
- nh ngha:
K vng ton ca bin ngu nhin X l s c trng cho gi tr trung
bnh tnh theo xc sut ca tt c gi tr ca X.
Cho X l bin ngu nhin, k vng ton ca bin ngu nhin X c
k hiu l E(X) v xc nh nh sau:
- Nu X l bin ngu nhin ri rc v gi tr xi c th nhn cc xc sut pi
(i = 1, 2, ) th: E(X) = =
n
i 1
pixi (2.1)
- Nu X l bin ngu nhin lin tc c hm mt xc sut l f(x) th:
E(X) = +
xf(x)dx (2.2)
2. K vng mu thc nghimK vng mu thc nghim c xc nh bng gi tr trung bnh ca
cc s liu quan st ca mi php o.
X =m
1=
m
i 1xi (2.3)
Trong : xi l s o ca i lng x ln o th i.
m l s ln o.
3. Mod ca bin ngu nhin
Mod ca bin ngu nhin ri rc X l im x0 sao cho:
P(X = x0) = max P (X = xi)
i = 1, 2,, tc l ti xc sut xi l ln nht.
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2.1.3.2. Phng sai iu chnh mu thc nghim
Phng sai l c trng quan trng phn nh phn tn gi tr
bin ngu nhin xung quanh k vng v c k hiu l S2.
1. Phng sai mu thc nghim
Gi s x1, x2,xm l mu thc nghim ca X, khi S
2
gi l phngsai mu thc nghim ca X, v c xc nh nh sau:
S2 =m
1
=
m
i
ix1
( - )x 2 (2.4)
Trong : S2 l phng sai mu thc nghim.
m l s ln o hay s ln quan st c.
xi l s o ca i lng x ln o th i.
x l trung bnh mu thc nghim.
2. Phng sai iu chnh mu thc nghimGi s S2 l phng sai mu thc nghim, khi s thc S12 c gi
l phng sai mu hiu chnh ca X v c xc nh nh sau:
S12 = f1
=
m
i
xx1
)( 2 (2.5)
f = m 1 l bc t do c trng cho mu thc nghim.
2.1.3.3. lch chun (SD)
- L tham s dng xc nh phn tn ca bin ngu nhin ccng n v vi n.
- Gi s S2 v S12 l phng sai v phng sai iu chnh mu ngu
nhin ca X, khi S v S1 c gi l lch tiu chun iu chnh mu thc
nghim ca X v xc nh nh sau:
S = 2S (2.6)
S1 = 21S (2.7)
2.1.3.4. Sai s chun (SE)
- L t l gia lch chun trung bnh mu vi cn bc hai ca dung
lng mu: SE = =N
S1 (2.8)
- L thng s thng k quan trng nh gi mc phn tn ca
mu chnh n biu th sai s ca s trung bnh. Sai s y do s chnh lch c
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hc c h thng ca s liu m phng thc chn mu l mt trong nhng
nguyn nhn chnh gy nn.
- Mc ch chnh SE l xc nh mc phn tn ca gi tr trung
bnh mu v gii hn tin cy ca mu thc nghim.
2.1.3.5. ngha ca phng sai, lch chun, sai s chunPhng sai, lch chun, sai s chun gip cho ta nhn bit c
mc ng u ca gi tr thc nghim.
Nu phng sai, lch chun, sai s chun nh th cc gi tr thc
nghim tng i ng u v tp trung xung quanh gi tr trung bnh.
2.1.4. chnh xc v tin cy ca php o
- Gi s mt php o vi sai s tin cy nh sau:
XX
= X =
tin cy l xc sut kt qu cc ln o ri vo khong tin cy
( X - < X < X + ), tc l P( X - < X < X + ) = v tin cy
thng cho trc 0,95; 0,99; 0,999;...
2.2. Phn tch thng k cc kt qu thc nghim (phn tch quy hi)
Gm cc bc sau:
- Kim tra gi tr ca tt c cc h s hi qui bng cch so snh vi sai
s lp li (Sbj) hay cn gi l sai s chun.- S ph hp gia m t ton hc vi kt qu thc nghim.
2.2.1. Phng sai ti hin
Xc nh phng sai ti hin xc nh sai s ti hin.
2.2.1.1. Phng sai ti hin ca mt th nghim
Gi s mt th nghim c lp i lp li m ln vi gi tr tng ng
thu c l y1, y2,...,ym . Phng sai ti hin ca mt mu thc nghim c xc
nh nh sau: Sth2 = 1f =
m
i
iy1
( - y )2 (2.9)
hay Sth2 =1
1
m
=
m
i
iy1
( - y )2 (2.10)
Trong : f = m 1 l t do c trng cho kh nng bin i m
khng lm thay i h.
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m l s ln lp.
2.2.1.2. Phng sai ti hin ca mt cuc th nghim
Sth2 =N
1
=
N
u
uS1
2 (2.11)
M Su2
= 11
m =
m
i
uiy1
( - y u)2
(2.12)
Sth2 = )1(1
mN =
N
u 1
=
m
i
uiy1
( - y u)2 (2.13)
Trong : u = 1,2,3,...
i = 1,2,3,...
N l s th nghin khc nhau.
m l s ln lp li.
Cng thc dng tnh phng sai ti hin ca mt cuc th nghim,thng s dng cho phngn th nghin song song (phng n m mi mt
im th nghim phi tin hnh lp li).
Phng sai phn phi trung bnh cho tng th nghim c xc nh
nh sau : Sth2 ( y ) =m
1Sth2 (2.14)
V d 1: Tnh phng sai ti hin ca mt cuc th nghim tng ng
vi nhng s liu thc nghim thu c bng sau :
Bng 1:Kt qu th nghim
S.T.N
(u
)
S ln
lp (m)
Kt quyu1 yu2 yu3 yu4
1
2
34
5
6
7
8
3
3
33
3
3
3
3
73
58
5484
100
98
77
105
69
58
5994
106
90
85
95
68
64
5292
109
97
78
100
70
60
5590
105
95
80
100T bng s liu ta thy i = 1,2,3; u = 1,2,3,8; m = 3; N = 8.
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tnh phng sai ti hin ca mt cuc th nghim ta lp bng sau:
Bng 2: Phng sai ti hin ca tng th nghim
u
(yu1- y u)2
(1)
(yu2- y u)2
(2)
(yu3- y u)2
(3)
(1+2+3) Su2
1
2
3
4
5
6
7
8
9
4
1
36
25
9
9
25
1
4
16
16
1
25
25
25
4
16
9
4
16
4
4
0
14
24
26
56
42
38
38
50
7
12
13
28
21
19
19
25
=
=8
1
144u
T kt qu bng 2, ta tnh phng sai ti hin ca cuc th nghim:
Sth2 =N
1
=
N
u
uS1
2 =8
144= 8
Phng sai phn phi trung bnh cho mt th nghim:
Sth2 ( y ) =m
1Sth2 =
3
18= 6
2.2.2. Phng sai d- d l hiu gia gi tr thc nghim thu c vi gi tr tnh c theo
phng trnh hi qui ca cc thng s ti u.
- Phng sai tm c trn c s tng bnh phng cc d gi l
phng sai d, c k hiu v xc nh nh sau:
Sd2 =duf
1m u
N
u
yy =1
~( )2 (2.15)
Sd2 = duf1
u
N
uyy =1 ~( )
2 (2.16)
Trong : fd = N L t do d.
N l s th nghi m trong mt cuc th nghim.
L s h s c ngha trong phng trnh hi qui.
uy~ gi tr c tnh theo phng trnh hi qui ng vi iu kin nghim
th u.
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uy l gi tr trung bnh thc nghim ti th nghim th u
(trong iu kin mi im thc nghim c tin hnh lp li).
yu l gi tr thc nghim trong iu kin khng lm th nghim lp.
2.2.3. Kim nh thng k
2.2.3.1. Kim tra s ng nht ca cc phng sai- Kim tra s ng nht ca cc phng sai l kim tra hi t ca
cc gi tr thc nghim. Phng php kim tra ny ch c p dng trong
phng n th nghim song song.
- kim tra ngi ta ch s dng chun Cochoren.
Cc im phn v ca phn phi chun Cochoren vi P = 0,05
Trong
: N l s th nghim trong mt cuc th nghim.
f l t do ng vi th nghim c phng sai ti hin ln nht.m l s ln lp ca th nghim c phng sai ti hin ln nht.
Gb c tm thy bng vi mc ngha chn, l im gp nhau
gia hng biu th s th nghim N v ct biu th bc t do f.
* Cc bc tin hnh kim tra
- Xc nh i lng trung bnh t cc kt qu ca cc th nghim song
song.
- Xc nh cc phng sai thc nghim (Su2
) ti mi im th nghimtheo cng thc (2.9).
- Tnh tng cc phng sai =
N
u
uS1
2
- Tnh Gtn theo cng thc sau: Gtn ==
N
u
u
u
S
S
1
2
2max
; u = 1,2,3,...,N (2.17)
S T.N t do (f = m 1)1 2 3
2 0,9985
0,9750
0,9392
3 0,966
9
0,870
9
0,7977
0,000 0,000 0,000
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max Su2 l gi tr cc i ca phng sai thc nghim th u.
N l s th nghim trong mt cuc th nghim.
- Tra bng Gb vi mc ngha P chn, s th nghim N v t do f
ca im thc nghim c phng sai ti hin ln nht.
- So snh Gm v Gb.+ Nu Gtn < Gb : gi thit c chp nhn.
+ Nu Gtn < Gb : gi thit khng c chp nhn.
2.2.3.2. Kim tra ngha ca cc h s trong phng trnh hi qui
- Mc ch ca kim tra ny l xem cc h s bj trong phng trnh
hi qui c khc 0 vi mt tin cy no hay khng.
- kim tra ngha ca cc h s trong phng trnh hi qui ta phi
s dng chun Student (t).
* Cc bc tin hnh kim tra:
- Tnh chun ttn theo cng thc: ttn = tj =bj
j
S
b(2.18)
bj l h s ng vi yu t th j trong PTHQ; j = 0,1,2,
Sbj lch qun phng ca h s bj
- Tra bng tb (P,f) ng vi mc ngha P chn trc v f; f l bc t do
ng vi phng sai ti hin ca tng phng n m ngi nghin cu chn.
- So snh tj v tb+ Nu tj > tb h s bj c ngha v c gi li trong PTHQ.
+ Nu tj < tb h s bj khng c ngha v loi khi PTHQ. Cc h s
cn li c tnh li theo phng phpbnh phng ti thiu cho ti khi tt c
chng u c ngha.
2.2.3.3. Kim tra s tng thch ca PTHQ vi thc nghim
- Dng PTHQ l do ngi nghin cu t chn v cc h s trong PTHQ
c xc nh da trn cc s kiu thc nghim. V vy cn phi xem xt m t
ton hc c ph hp vi thc nghim hay khng, v ngi ta dng phn phi
Fisher (F) vi mt mc ngha no .
* Cc bc tin hnh kim tra:
- Vit PTHQ vi cc h s c ngha.
- Tnh Ftn theo cng thc: Ftn = 22
th
tt
S
S(2.19)
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Trong : Stt2 l phng sai tng thch v c tnh theo cng thc (2.15),
(2.16).
Sth2 l phng sai ti hin c tnh theo cng thc (2.8) vi
phng n th nghim ti tm hoc tnh theo cng thc (2.14) ng vi phng n
th nghim song song.- Fb tra bng fb (P, f1,f2) tc l ng vi mc ngha P chn v bc t do f1, f2
- Tiu chun kim nh (so snh Ftn v Fb)
+ Nu Ftn < Fb th PTHQ va lp ph hp vi thc nghim.
+ Nu Ftn > Fb th PTHQ va lp khng ph hp vi thc nghim v
lm tip cc cng vic sau:
* Kim tra li cng vic tnh ton.
* Xem li m hnh nghin cu ng cha.
* Chn m t ton hc (PTHQ) mc cao hn.
2.3. Cc phng php phn tch hi quy
2.3.1. Phng php bnh phng nh nht (BPNN)
L phng n c bn c hiu lc khi x l cc s liu thc nghim v
xy dng m hnh thng k cho nhiu i tng nghin cu thuc cc lnh vc
khc nhau.
Phng php ny cho php xc nh cc h s ca phng trnh hiqui chn sao cho lch ca s ph thuc cho so vi s liu thc nghim
l nh nht. min)~(
1
2=
=
N
u
uu YY (2.20)
Trong : Yu l gi tr thc nghim ng vi k thng s ti u th
nghim th u.
uY~
l gi tr theo phng trnh hi qui s ti u th nghim
th u.
2.3.2. Hi quy tuyn tnh mt bin
Phng trnh hi quy tuyn tnh mt bin s c dng:
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xbby 10 +=(2.21)
Cc h s ca phng trnh hi quy c xc nh bng phng php
bnh phng nh nht (BPNN), vi s th nghim l N.
H phng trnh chun c dng :
=+
=+
0)(
0)(
10
10
iiii
ii
xxbbxy
xbby
2.3.3. Hi quy parabol
Phng trnh hi quy parabol - bc hai mt bin c dng:
2
1110 xbxbby ++=
Cc h s ca phng trnh hi quy cng c xc nh bng phng
php bnh phng nh nht (BPNN), vi s th nghim l N.
Trong trng hp ny :
1)(
0
=
b
xf
;x
b
xf=
1
)(
;
2
11
)(x
b
xf=
H phng trnh chun c dng :
=++
=++
=++
iiiii
iiiii
iii
yxxbxbxb
yxxbxbxb
yxbxbNb
24
11
3
1
2
0
3
11
2
10
2
1110
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2.3.4. Hi quy hm s m
Khi s thc nghim N b, nu tng bc ca a thc c th dn n vic
tng phng sai d. Lc ny gim s cc h s khng xc nh, ta dng hi
quy hm s m. Vic xc nh cc h s ca phng trnh hi quy c th rt
kh khn do phi gii h phng trnh phi tuyn. Vic tnh ton s tr nn n
gin hn nu tin hnh thay th cc bin s v h bc a thc.
V d cc quan h kiu hm s m nh sau :
xbby 10 =1
0 bxby =
c logarit ha :
10 lglglg bxby += xbby lglglg 10 +=
Sau khi t :
txababzy ==== lg;lg;lg;lg 0011
Ta s nhn c phng trnh tuyn tnh vi cc bin s :
xaaz 10 += tbaz 10 +=
Cc h s a0
, a1
, b1
c xc nh theo phng php BPNN. T a0
v a1
c th tnh c b0
v b1
.
2.3.5. Hi quy nhiu bin
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Nu cn nghin cu lin kt tng quan gia nhiu i lng ngi ta
dng phng trnh hi quy nhiu bin :
kkxbxbxbby ++++= ... 22110
y, chng ta gp khng phi ng hi quy, m l mt phng hi quy
khi k=2 v mt hyper khi k>2. Trong trng hp chung, b mt ny gi l b
mt mc hoc b mt p tr. Khi xy dng b mt mc trn trc ta ca
khng gian yu t cn phi t cc gi tr bng s ca cc yu t ln h ta .
Phi chuyn t quy m t nhin sang quy m chun. Ngha l phi tin hnh
chun ha tt c cc gi tr ca cc i lng ngu nhin theo cc cng thc
thng k v chuyn t bin thc sang bin c m ha khng c th nguyn.
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Chng 3. PHNG PHP CHN LA CC YU T NH HNG
3.1. La chn cc yu t u vo
Yu cu i vi cc bin c la chn l cc yu t u vo ca nghin cu
thc nghim :
- L cc bin c lp, iu chnh c, s thay i gi tr ca chng theo cc
mc quy hoch l hon ton c lp, khng ph thuc v ko theo s thay i ca
cc yu t khc. Cc vc t ca chng trong ma trn k hoch phi c lp tuyn
tnh.
- L cc yu t nh lng, v vy cc yu t nh tnh khng c tr s xc
nh c th nh : phng php to mu, mu sc ca i tng, hnh dng ca b
phn lm vic khng th a vo lm yu t nghin cu ca quy hoch thc
nghim.
- C hiu ng nh hng r nt n hm mc tiu nh gi hnh vi i
tng nghin cu.
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Cc cn c la chn cc yu t u vo : thng tin tin nghim, kt qu
nghin cu l thuyt, kin chuyn gia, cc thc nghim thm d v thc
nghim sng lc.
3.1.1. Thng tin tin nghim
Thng tin c c nh kt qu quan st trc tip lm vic ca i tng
nghin cu v kt qu tm hiu ti liu tham kho. Phn ln cc i tng nghin
cu c nghin cu bng l thuyt hoc thc nghim. l nhng qu trnh
tng t din ra trong mi trng khc, nhng c cng bn cht vt l, cng quy
lut tc ng y l nhng thng tin s b, nh hng.
3.1.2. Kt qu nghin cu l thuyt
Trong nhiu trng hp, ngi nghin cu tuy cha th hiu bit v xy dng
nhng m hnh l thuyt c bn v ton din v i tng, nhng t nhng l
thuyt ca khoa hc c s, hoc t cc cng trnh l thuyt tng t, c th m t
bng cng thc gii tch mt s tnh cht hoc hnh vi no ca i tng nghin
cu.
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3.1.3. kin chuyn gia
Thng thng, thng tin t cc ti liu rt t v khng ton din v i tng
nghin cu. Do vy c th s dng phng php xin kin chuyn gia nh gi
mc quan trng ca cc yu t nh hng. Phng hasp ny c hiu qu tt
nu s yu t cn nh gi ln v s chuyn gia ng. y l phng php
c chun ha, c th p dng cho nhiu i tng nghin cu khoa hc khc
nhau.
3.1.4. Cc thc nghim thm d, thc nghim sng lc
i khi, sau cc bc ni trn vn cn li vi yu t nh hng ng nghi ng
m vic loi b hay gi li lm yu t nghin cu cn nh n kt qu kim chng
thc nghim. Hoc khi i tng nghin cu qu mi m, thng tin ban u t v
cha tin cy, vic sng lc cc yu t cn tin hnh ht sc cn thn. Nu b st
yu t quan trng xj no , th cc kt qu nghin cu s ch l 1 thit din ca
mt mc tiu to bi mt phng xj = const. Nhng trng hp ny i hi phi tin
hnh cc thc nghim thm d.
a. Thc nghim thm d n yu t
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- Thc hin th nghim vi mt yu t thay di, cc yu t cn li c n
nh cc gi tr c nh.
- X l s li trong c kim tra gi thit v tnh ng nht phng sai v
nh gi mc nh hng ca yu t theo kt qu phn tch phng sai.
- Xc nh m hnh ton thc nghim n yu t tin hnh cc pjaan tch
v d bo cn thit. Bc ny thc hin theo phng php bnh phng b nht.
b. Thc nghim sng lc a yu t
Thc nghim sng lc a yu t cn p ng cc yu cu:
- S th nghim so vi s yu t cn kho st l ti thiu, cho php a vo k
hoch t a cc yu t thay , m s th nghim l chp nhn c, tn t cng
sc.
- Cho php phn tch v so snh i chng hiu ng tc ng ca cc yu t
ring r, hoc cc cp yu t theo iu kin t ra ban u.
3.2. Phng php chuyn gia
Cc chuyn gia thuc nhiu trng phi khc nhau s c ngh sp
seexp cc yu t nh hng n i tng theo trnh t gim dn v mc nh
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hng n cc mc tiu ti u. Mi chuyn gia khi c hi phi in vo phiu
iu tra, ghi sn cc yu t, th nguyn v khong bin thin ca chng.
Cc chuyn gia cn phi ghi v tr th t ca mi yu t cng quan trng cng c
th t hng trng s cng ln, nu cn thit c th b sung vo phiu nhng yu t
mi hoc b bt yu t c hoc nu kin v min bin thin ca chng. m
bo nh gi khch quan th s chuyn gia c hi cng nhiu cng tt.
3.3. Cc thc nghim sng lc theo phng n bo ha
Sau tt c cc bc: nghin cu ti liu tha kho, ly kin chuyn gia,
phn tch l thuyt nu s yu t nh hng cn li kh ln th thc nghim sng
lc ng vai tr sng lc quyt nh. Ty theo gi thit ban u, ngi ta phn
thnh phng n bo ha, siu bo ha (cn i ngu nhin) v loi k tip.
Thc nghim c gi l bo ha khi ton b s bc t do ca thc nghim
c dng c lng cc h s ca m hnh ton thc nghim. Gi s s h s
trong phng trnh hi quy thc nghim l L, s th nghim ca thc nghim l N,
th thc nghim bo ha l thc nghim m:
L = N
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3.4. Nhm cc yu t vo v chn mc tiu nh gi
Mc d qua cc bc sng lc, nhng nhiu thc nghim s yu t cn
nghin cu cn li kh ln. Xt theo quan im h thng, a cng nhiu yu t c
nh hng thc s vo mt k hoch thc nghim, ngi nghin cu cng c iu
kin tm c ti u c cht lng cao ca i tng. Tuy nhin li c mt s mt
nhc im.
Trc ht, khc vi thc nghim sng lc, cc thc nghim tm ti u giai
on sau phi p ng cc tiu chun ti u nghim ngt ca k hoch thc
nghim. V th, khi s yu t vo kh ln (ch cn khi k 7 ) th s th nghim
trong k hoch tng ln rt nhiu. mi im li phi tin hnh mt s th
nghim song song (lp li). Ton b cc th nghim cn tin hnh theo trnh t
ngu nhin ha. Cc yu cu ny cng lm tng khi lng v thi gian thc
nghim.
Ngi nghin cu ng trc s la chn: xy dng v tin hnh mt k
hoch thc nghim vi ton b s yu t nh hng chn, v d k = 7, hay tch
ra thnh 2 k hoch song song vi k1
= 3; k2
= 4.
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3.5. nh hng ca cc tin ca phn tch hi qui n s la chn cc yu
t c lp
Phn tch hi quy c xy dng vi nhng tin m chng c lin quan
n mi trng v iu kin thc nghim. iu kin v mi trng thc nghim
li b rng buc bi im cc thng s nghin cu v ch tiu nh gi. Mc
tha mn cc tin vcuar phn tch hi quy ph thuc nhiu vo cch chn nhm
v xc nh mc, khong bin thin ca cc yu t nh hng, vo nhy v
chnh xc ca cc gi tr quan st ch tiu u ra. Do vy, ngi nghin cu cn
bit r cc yu cu ny c nhng quyt nh ti u ngay bc xc nh cc
yu t nghin cu.
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Chng 4. CC PHNG PHP HOCH NH THC NGHIM
4.1. QUY HOCH TRC GIAO CP I
CC BC QUY HOCH TRC GIAO CP I
1. XC NH MIN BIN THIN
Zjmin
< Zj < Zjmax
v TM QUY HOCH : Zjo = 0.5(Zjmin + Zjmax)
2. CHN DNG PHNG TRNH HI QUY
sau khi m ha : xj = 2( Zj - Zjo ) / ( Zjmax - Zjmin )
+ chn dng tuyn tnh :
y1 = b0 + b1x1 + ..+ bkxk
( + hoc dng :
y1 = b0 + b1x1 + + bkxk + b12x1x2+ +bk-1,kxk-1xk)
3. THC HIN N TH NGHIM N = 2k
TNH TON XC NH CC H S HI QUY bj
bng phng php Bnh phng cc tiu
4. KIM NH S C NGHA CA CC H S HI QUY bj
vi chun StudentThc hin cc th nghim ti tm quy hoch hoc s dng cc th
nghim song song, lp li.Loi b ccbj khng c ngha, tnh ton li cc
bj v kim nh li cho ti khi ch cn cc bj c ngha
5. KIM NH S C NGHA CA PHNG TRNH HI
QUY vi chun Fisher
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4.1.1. Thc nghim yu t ton phn TYT 2k
Tu thuc thng tin ban u m ngi nghin cu t chc cc th
nghim nhn c m hnh thng k thc nghim dng tuyn tnh hoc phi
tuyn
Chn qui hoch thc nghim yu t ton phn v tng phnNhng thc nghm m mi t hp cc mc ca yu t u c thc
hin nghin cu gi l thc nghim yu t ton phn (TYT nk)
N = nk (4.1)
Trong : N : lng th nghm
n : s lng mc ca cc yu t
k : s yu t nh hng
Xt thc nghim yu t ton phn 2 mc k yu t nh hng
4.1.1.1. Cch t chc th nghim trc nghim trc giao cp I
1 S th nghim cn thc hin
N = 2k
2 Mc c bn Zj0 =2
minmax
jj ZZ + (4.2)
Trong : Zj0 l mc c bn ( tm phng n).
Zjmax l mc trn (mc cao).
Zjmin
l mc di (mc thp).Vect vo ti mc c bn Zj0 (j = 1,2,...k) ch ra khng gian cc yu t
ca mt im c bit gi l tm thc nghim.
3 Khong bin thin
=I2
minmax
jj ZZ (j = 1,2,3...k) (4.3)
I l khong bin thin theo trc Zj.
V d: Xem trang 25 (Gio trnh Quy hoch thc nghim nghin cu
v ng dng ca Nguyn Th Lan).
4 Bin khng th nguyn : k hiu xj
M ho c thc hin d dng nh vic chn tm Zj0 ca min nghin
cu lm gc to h trc .
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=
=
=
j
jj
j
j
jj
j
j
jj
j
ZZx
ZZx
ZZx
00
0
0m in
m in
0m a x
m a x
j = 1,2,3...k (4.4)
Ta th nguyn mc trn (xjmax) lun bng +1; mc di (xjmin)
lun bng 1 v ta ca tm phng n (xj) lun bng 0 v trng vi gc ta
.
=
+=
jjj
jjj
ZZ
ZZ
0m i n
0m a x
(4.5)
5 Lp ma trn thc nghim
Ma trn thc nghim vi bin thc nghim l mt dng m t chun
cc iu kin tin hnh th nghim theo bng ch nht. Mi hng l mt th
nghim,trong ma trn c mt s hng ging nhau m thng s u mc c s
Zj0
Ma trn thc nghim vi bin o l ma trn ch bao gm cc bin o
xj .Khi xy dng ma trn thc nghim a thm bin x0 = 1 v b tr cc th
nghim sao cho khng c th nghim no trng nhau. Theo kinh nghm lm nh
sau :
- Xc nh s th nghim cn thc hin theo cng thc N = 2 k, ct x0 lun
bng 1
- Lp cho tng yu t nh hng v ln lt t x1 n xk.
Ch : Ngi nghin cu nn a cc th nghim tm vo ma trnTnh Y0 (gi tr tm thc nghim); b0 c th d on c
vng nghin cu thuc vng tuyn tnh hoc phi tuyn.
6 Tnh cht ma trn trc giao cp I
Ma trn trc giao cp I c nhng tnh cht sau:
- Tnh i xng qua tm thc nghim.
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01
==
N
u
iux ; i = 1,2,3,...k (4.6)
u = 1,2,3,...N
- Tnh trc giao gia 2 ct trong ma trn thc nghim.
01 ==
N
u
iuiuxx ; i j = 1,2,3...k (4.7)
- Tnh bt bin khi quay h trc quanh tm thc nghim.
NxN
u
iu ==1
2
; i = 1,2,3,...k (4.8)
* u im ca ma trn trc giao cp I:
- Khi loi b nhng h s khng c ngha s khng phi tnh li cc h
s c ngha.
- Phng sai cc h s b (Sbj2
) trong phng trnh hi qui c gi tr tithiu, xc nh theo kt qu ca N th nghim v nh hn phng sai ti hin
Sth2.
- Tm phng n thng tin nhiu nht ch ln thc nghim lp
tm thc nghim l .
4.1.1.2. Mt s dng ca phng trnh hi qui cp I
Trc tin,phi bit c s ph thuc gia cc thng s u vo v
thng s u ra Y=f(x) chn phng trnh hi qui sao cho hp l.
i vi qui hoch thc nghim, nhng phng trnh hi qui cp I
thng chn cc khai trin ca a thc c dng tng qut sau :
Y~ = b0 + b1x1 +...+ bkxk +...+ bijxixj +...+ bijkxixjxk ; vi ij k = 1,2,3...k (4.9)
Trong : b0 l h s hi qui.
bj l h s tuyn tnh.
bij ; bijk l h s tng tc cp v tng tc ba.
n gin th chn dng phng trnh hi qui dng tuyn tnh.
Mun xy dng phng trnh hi qui y a thm vo phng
trnh tuyn tnh cc h s tng tc
Vi k = 2 (2 yu t nh hng) ta c:
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+++=
++=
211 222110
22110
~
~
xxbxbxbbY
xbxbbY(4.10)
Vi k = 3 ta c:
+++++++=
+++=
3211 2 3311 3212 3211 23322110
3322110
bY~
bY~
xxxbxxbxxbxxbxbxbxb
xbxbxb
(4.11)
4.1.1.3.Lp cng thc tnh h s b trong phng trnh hi qui1 Phng n bnh phng nh nht (BPNN)
L phng n c bn c hiu lc khi x l cc s liu thc nghim v
xy dng m hnh thng k cho nhiu i tng nghin cu thuc cc lnh vc
khc nhau.
Phng php ny cho php xc nh cc h s ca phng trnh hi
qui chn sao cho lch ca s ph thuc cho so vi s liu thc nghim
l nh nht. min)~(1
2=
=
N
u
uuYY (4.12)
Trong : Yu l gi tr thc nghim ng vi k thng s ti u th
nghim th u.
uY~
l gi tr theo phng trnh hi qui s ti u th nghim
th u.
2 H phng trnh chun tcXt k = 2, dng PTHQ nh sau:
Y~ = b0x0u + b1x1u + b2x2u + b12x1ux2u (4.13)
Thay (4.13) vo (4.12):
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( )[ ]2
1
2u1u122u21u10u0 xxbxbxbxb=
+++=N
u
uY min
(4.14)
cc tiu khi tha mn cc iu kin sau:
00 =
b ; 01 =
b ; 02 =
b ; 012 =
b (4.15)
C th vit di dng sau:
( )[ ]
( )[ ]
( )[ ]
( )[ ]
=+++=
=+++=
=+++=
=+++=
=
=
=
=
4
1
212112221100
12
4
1
22112221100
2
4
112112221100
1
4
1
02112221100
0
0
0
0
0
u
uuu
u
uu
uuu
u
uu
xxYuxxbxbxbxbb
xYuxxbxbxbxbb
xYuxxbxbxbxbb
xYuxxbxbxbxbb
(4.16)
( )
=+++
=+++
=+++
=+++
= = = = =
= = = = =
= = = = =
= = = = =
4
1
4
1
4
1
4
1
4
1
21
2
2112
2
2122
2
11210
4
1
4
1
4
1
4
1
4
1
2
2
2112
2
2221120
4
1
4
1
4
1
4
1
4
1
12
2
112212
2
1110
4
1
4
1
4
1
4
1
4
1
02112221100
u u u u u
uuuuuuuuuuu
u u u u u
uuuuuuuu
u u u u u
uuuuuuuu
u u u u u
uuuuuuu
xxYxxbxxbxxbxxb
xYxxbxbxxbxb
xYxxbxxbxbxb
xYxxbxbxbxb
(4.17)
Phng trnh (4.17) gi l h phng trnh chun tc.
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3 Cng thc tnh h s b ca phng trnh hi qui
Cc h s b trong phng trnh hi qui c lp nhau v xc nh nh
sau:
=
=
=
=
=
=
=
=
4
1
2112
4
1
22
4
1
11
4
1
00
4
1
4
1
4
14
1
u
uuu
u
uu
u
uu
u
uu
Yxxb
Yxb
Yxb
Yxb
(4.18) Cng thc tng qut tnh cc h s b trong phng trnh hi qui
ca qui hoch trc giao cp I tng ng vi k yu t nh hng nh sau:
=
=
=
=
=
=
=
=
4
1
4
1
4
1
4
1
00
1
1
1
1
u
ulujuiuij l
u
ujuiuij
u
ujuj
u
uu
YxxxN
b
YxxN
b
YxN
b
YxN
b
i j l = 1,2,3...k
(4.19)
4 ngha ca h s b trong phng trnh hi qui
Ga tr ca h s bj trong phng trnh hi qui c trng cho s ng
gp ca yu t th j vo i lng Y.
H s no c gi tr tuyt i ln nht th yu t tng ng s nh
hng n qu trnh l nhiu nht.
4.1.1.4. Kim tra ngha ca cc h b trong phng trnh hi qui
kim tra ngha ca cc h s b trong phng trnh hi qui phi
tnh phng sai ti hin (lm th nghim song song mi im thc nghim).
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H s b trong phng trnh hi qui c lp nhau v xc nh vi mt
chnh xc (Sbj).N
SS thbj = (4.20)
N: s th nghim ng mi phng n.
Tnh ngha ca cc h s b c kim nh theo chun Student (t)
xc nh nh sau :bj
j
jS
bt = (4.21)
Trong : bj l h s th j trong phng trnh hi qui tnh theo (4.19).
Sbj: lch qun phng ca h s j c xc nh theo cng thc
(4.20)
Cc bc kim tra c tin hnh nh mc kim nh thng k
(chng 2)
Cng thc (4.21) xc nh c Sbj ng vi mi phng n thc
nghim.
1 Phng n thc nghim ti tm
Khi hon tt 2k th nghm nhn phng n, ngi nghin cu phi
lm thm m (m t nht bng 3) th nghim tm phng n vi cc gi tr ng
vi th nghim tm l: 01Y ,0
2Y ,0
3Y ...
Phng sai ti hin c xc nh:
Sth2 = ( )1
1
200
=
m
YYm
i
i i = 1,2,3...m (4.22)
Sth = 2thS (4.23)
Trong : Yi0 l gi tr o c ln lp th i0
Y l gi tr trung bnh ca m ln o
m : s ln lp
Thay (4.23) vo (4.20) tm c gi tr Sbj.
2 Phng n th nghim song song
Ti mi im th nghim c lp li m ln.Trc khi tnh ton h s b
v kim nh cc thng s thng k phi kim tra s ng nht ca cc phng
sai theo chun Cohoren (G), ch c php c lng cc sai s khi phng sai
ng nht.
Phng sai ti hin ca mt cuc th nghim:
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Sth2 =( )
)1(
1
2
1
==
mN
YYm
i
ii
N
u (4.24)
Phng sai phn phi trung bnh ca mt cuc th nghim
Sth2
(Y
) = mSth
2
(4.25)
Phng sai ca h s bj
Sbj2 =N
YSth )(2
(4.26)
Sai s chun ( lch qun phng) ca h s bj
Sbj =N
YSth )( (4.27)
Sau khi kim tra ngha ca cc h s bj, vit PTHQ vi cc h s c
ngha v kim tra tng thch ca phng trnh hi qui vi thc nghim.
4.1.1.5. Kim tra s tng thch ca PTHQ vi thc nghim
S tng thch ca PTHQ vi thc nghim c kim nh theo chun
Fisher (E).Cc bc kim tra c trnh by mc kim nh thng k (chng
2).
2
2
th
tt
S
SF = (4.30)
i vi phng n th nghim ti tm
tt
N
U
uu
ttf
YY
S
=
= 12
)~
((4.31)
Phng n th nghim song song,vi ln lp mi im thc nghim l
m
tt
N
uuu
ttf
YYmS = = 1
2
2 )
~
(.
(4.32)
uu YY , : l gi tr thc nghim.
Yu : gi tr tnh theo PTHQ.
ftt : t do ng vi phng sai tng thch (Stt2).
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ftt=N-L
N : s th nghim trong phng n.
L : s h s c ngha c kim tra mc (4.1.1.4).
Sau khi kim tra nu PTQH tng thch vi thc nghim s c s
dng tm kim ti u. Nu khng ph hp s phi xem xt li tng bc cabi qui hoch v chn m t ton hc mc cao hn.
4.1.2. Thc nghim yu t tng phn TYP 2k-p
m t qu trnh thc nghim th qui hoch thc nghim yu t ton
phn TYT 2k khng hiu qu khi s yu t k kh ln. S yu t k tng chm m
s th nghim tng qu nhanh (N=2k) v s c rt nhiu bc t do kim tra s
tng thch ca PTHQ vi thc nghim.
Tht kh khn v kinh t khi phi thc hin 1 cuc th nghim TYT 2k
m yu t k>4.
V vy s gim ng k s th nghim nu ta dng thc nghim yu t
tng phn (li gii tng phn) m ngi nghin cu vn thu c m hnh th
nghim m t tng thch qu trnh th nghim.
K hiu: TYP 2k-p
Trong :
2: l 2 mc ca mi yu t nh hngk: s yu t nh hng
p: c trng cho mc tng phn
4.1.2.1. Xy dng m hnh thng k thc nghim
4.1.2.2 Cch t chc th nghim trong phng n thc nghim tng phn
N=2k-p
S th nghim trong phng n tng phn bng p21
bng TYT 2k
4.1.2.3. Cng thc tnh h s b trong PTHQ ca qui hoch phn bng TYP
2k-p
cho li gii tng phn l mt phng n trc giao ta cn chn
phng n thc nghim yu t ton phn c s yu t nh hng nh hn lm
mc c s.
c p dng ging nh trong qui hoch TYT 2k
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b0 =u
N
u
u XYN
0
1
1
=
bj =ju
N
u
u XYN=1
1
u = 1,2,.,Nj = 1,2,..,k
4.1.2.4.Cc bc thc hin qui hoch phn bng
1/ Trng hp k=3, p=1
- Lp qui hoch v xy dng ma trn TYT 22.
- Thay ct c hiu ng tng tc bng hiu ng tuyn tnh (x1x2=x3).
- Lm 4 th nghim v dng kt qu ca 4 th nghim tnh h s b0, b1,
b2, b3.
- Sau khi lm 4 th nghim u, v mt l do no ngi nghin cu cho
rng tng tc cp c ngha th lm 4 th nghim ca na bng cn li, nhng
ln ny thay yu t b sung x3 = -x1x2. Nh vy qui hoch s cn na bng.
x3 = x1x2
x3 = -x1x2
2/ Trng hp k=4, p=1
- Lp qui hoch TYT 23
- Thay x3 = x1x2x3
- Lm 8 th nghim v s dng kt qu ca 8 th nghim tnh h s b 0,
b1, b2, b3, b4.
Nh vy qui hoch phn bng vi 2 na bng khi thay x4 = x1x2x3
3/ Trng hp k=5, p=2
Lp qui hoch TYT 23.
- Thay x4 = x1x2, (b qua tng tc x1x2), x5 = x1x2x3 (b qua tng tc
x1x2x3).- Lm 8 th nghim v dng kt qu ca th nghim xc nh h s
b0 v 5 h s cn li.
- Qui hoch phn bng vi 4 phn bng nh sau:
1 { 214 xxx = , }3215 xxxx =
2 { 214 xxx = , }3215 xxxx =
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3 { 214 xxx = , }3215 xxxx =
4 { 214 xxx = , }3215 xxxx =
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4.2. QUY HOCH TRC GIAO CP II
CC BC QUY HOCH TRC GIAO CP II
6. XC NH MIN BIN THIN Zjmin < Zj < Zjmax
v TM QUY HOCH : Zjo = 0.5(Zjmin + Zjmax)
7. CHN DNG PHNG TRNH HI QUY
sau khi m ha : xj = 2( Zj - Zjo ) / ( Zjmax - Zjmin )
y1 = b0 + b1x1 + ..+ bkxk + b12 x1x2 + + b11x12 + + bkkxk2
8. THC HIN N TH NGHIM N = 2k+ 2k +no
Trong : - 2k th nghim ca QHTG cp I
vi cc Zj = Zjmin hoc Zj = Zjmax
- 2k th nghim ti cc im sao : xj = TG
hoc xj = - TG
- no th nghim ti tm Zj = Zjo
9. TNH TON XC NH CC H S HI QUY bj
bng phng php Bnh phng cc tiu
10. KIM NH S C NGHA CA CC H S HI QUY bj
vi chun Student
Thc hin cc th nghim ti tm quy hoch hoc s dng cc th
nghim song song. Loi b ccbj khng c ngha, tnh ton li cc bj v
kim nh li cho ti khi ch cn cc bj c ngha
11. KIM NH S C NGHA CA PHNG TRNH HI
QUY vi chun Fisher
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4.3. Ti u ha qui hoch thc nghim
Bc 1
- Xc nh mt im xut pht nm trong min gii hn tng th ca
cc bin u vo. Chn im lm mc c bn, chn khong bin thin ca
tng bin xc nh min gii hn ca quy hoch thc nghim trc giaocp mt.
Bc 2
- Lm cc th nghim theo quy hoch trc giao cp mt
- Xy dng phng trnh hi quy bc nht .
Nu phng trnh hi quy bc nht khng tng thch th chuyn ti thc
hin bc 4.
Nu phng trnh hi quy bc nht tng thch th thc hin bc 3.
Bc 3
- Xc nh vect gradient ca hm mc tiu ti mc c bn v xut pht
t mc c bn xc nh ta cc im thc nghim nm cch u nhau trn
hng ca vect gradient vi khong cch t chn ph hp vi i tng
nghin cu. Lm thc nghim xc nh mt im c gi tr hm mc tiu tt
nht trn hng gradient. Chn im tm c lm im xut pht mi v quayv bc 2 .
Bc 4
- Lm cc th nghim theo quy hoch cp hai (trc giao hoc quay).
Bc 5
- Xy dng phng trnh hi quy bc hai.
- Nu phng trnh hi quy bc hai khng tng thch th chuyn ti thc
hin bc 6.
- Nu phng trnh hi quy bc hai tng thch th thc hin bc 7.
Bc 6
- Thu hp khong bin thin ca cc bin u vo ri quay v bc 5.
Bc 7
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- Tm cc tr ca hm mc tiu thu c dng phng trnh hi quy bc
hai thu c bc 5 v lm li thc nghim kim chng v nh gi kt
qu.
4.3.1. Ti u ha theo phng php leo dc
Bc 1: Chn im xut pht X(0) (x1(0), , xn(0))
Chn cc gi tr y> 0 v x> 0
Xc nh y(X(0))
Bc 2: Xc nh vect gradient ti im X(0)
Bc 3 : Chn s dng;T im X(0) xc nh X(1) :
01
)0(
1
)1(
1
XXx
yxx
=
=
02
)0(
2
)1(
2
XXx
y
xx=
=
. . . . . . . . . . . . . .
0
)0()1(
XXn
nnx
yxx
=
=
( du + khi tm max , du - khi tm min )
Xc nh y(X(1)
)Bc 4: So snh y(X(1)) vi y(X(0)).
Nu y(X(1)) tt hn y(X(0)) tip tc lp li bc 3 leo dc ti X(2),
X(3), , X(k)
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Nu y(X(k)) xu hn y(X(k-1)) Thc hin php gn X(1) = X(k-1) v y(1)
= y(X(k-1)), sau chuyn sang bc 5
Bc 5: Kim tra iu kin dng:
yyy )0()1(
hoc / vxnn xxxx ++
2)0()1(2)0(
1
)1(
1 )(...)( (*)
- Nu (*) khng tha mn:
+ Chn X(1) lm im xut pht mi ( ni cch khc : thc hin php gn
X(0) = X(1) v y(0) = y(1) )
+ Quay li bc 2
- Nu (*) tha mnkt lun : yt gi tr ti u ti X(1)
4.3.2. Phng php lun phin tng bin gii bi ton ti u phng nh
*Bc 1: Chn im xut pht X(0) (x1(0), , xn(0)),
Chn cc gi tr y> 0 v x> 0
Lm thc nghim xc nh gi tr y(0)
*Bc 2: Thc hin n phin gii bi ton ti u ln lt vi tng bin x i t im xut pht X(0) (x1(0), , xn(0) ) tm ra im X(1) (x1(1), x2(1), , xn(1)) tt
hn.
- Phin 1: C nh (n-1) bin, gii bi ton ti u vi bin cn li (gi s
x1) khi cho x1 chy trong min gi tr ca n. Gi s y tt nht ti X(*1) = (x1(1),
x2(0)
, x3(0)
,, xn(0)
)- Phin 2: Tin hnh tng t vi bin x2 (c nh cc bin cn li trong
x1 = x1(1) ). Tm c gi tr y tt nht ti im X(*2) = (x1(1), x2(1), x3(0),, xn(0)) .
- Phin th k : Gii bi ton ti u vi bin xk (c nh cc bin cn li
trong x1 = x1(1),, xk-1 = xk-1(1),xk+1 = xk+1(0),xn = xn(0), ). Tm c gi tr y tt
nht ti im X(*k) = (x1(1), ,xk(1), xk+1(0),, xn(0)) .
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- Phin th n : Gii bi ton ti u vi bin xn (c nh cc bin cn li
trong x1 = x1(1), , xk-1 = xk-1(1), xk+1 = xk+1(0), ,xn = xn(0), ). Tm c gi tr
y tt nht ti im X(*n) = (x1(1), ,xk(1), xk+1(1),, xn(1)) .
t X(1) = X(*n) ; y(1) = y(X(1))
*Bc 3: Kim tra iu kin dng:
yyy )0()1(
hoc/vxnn xxxx ++
2)0()1(2)0(
1
)1(
1 )(...)( (*)
trong y(1) = y(X(1)) = y(x1(1), , xn(1))
- Nu (*) khng tha mn:
+ Chn X(1) lm im xut pht mi ( ni cch khc : thc hin php gnX(0) = X(1) v y(0) = y(1) )
+ Quay li bc 2
- Nu (*) tha mn: kt lun yt gi tr ti u ti X(1)
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Chng 5. NG DNG QUY HOCH THC NGHIM TRONG
CC QU TRNH CNG NGH HA HC
5.1. Bi ton 1.
Mc ch: Nghin cu ti u ho quy trnh c nh t bo nm men
bng Alginat ln men ru. Quy trnh cng ngh c m t theo s (trang 1).
Nghin cu cc yu t nh hng n mng li gel: nng alginat;
nng glucose; nng t bo:
Sau qu trnh ln men, vt cc ht gel ra v xc nh t l (%) ht gel b
nt. T l ht gel b nt cng thp cng tt ngha l ht gel cng chc cng tt.
Hm mc tiu: Y = Y(Z1,Z2,Z3)
Bi ton ti u: Xc nh nng alginat; nng glucose; nng t
bo nm men ht gel bn nht trong qu trnh ln men ru bng t bo nm
men, c nh bng alginat.
Ymin = min Y(Z1,Z2,Z3)
Sau khi tin hnh cc th nghim thm d, tc gi chn vng kho stnh sau:
Z1 = 1 4%
Z2 = 10 18%
Z3 = 10 20%
y l bi ton ti u phng nh, gii bi ton theo phng php leo dc.
Phng n qui hoch thc nghim: phng php trc giao cp 1.
S th nghim phi lm: 2k =23 =8
Vi Z1min =1 Z1 4=Z1maxZ2min =10 Z2 18=Z2maxZ3min =10 Z3 20=Z3max
im xut pht tm phng n:
Z0 = (2.5; 14; 15)
58
Z1
Nng alginat
Z2
Nng glucose T l ht gel b nt Y(%)
Z3
Nng t bo
Ln men trong dungdch ng glucosebng t bo nm
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Gi tr ca hm mc tiu ti im Z0 c xc nh bng thc nghim:
Y(Z0) = 7.5
Ma trn thc nghim c b tr nh sau:
S
TN Z1 Z2 Z3 Y1 4 18 20 12.352 4 18 10 8.873 4 10 20 12.084 4 10 10 6.925 1 18 20 42.136 1 18 10 13.517 1 10 20 22.198 1 10 10 4.57
Phng trnh hi qui c dng:
Y = B0 + B1Z1 + B2Z2 + B3Z3
Trong h m ho khng th nguyn ta c c:
Mc trn: - k hiu +1
Mc c s: - k hiu 0
Mc di: - k hiu 1
Cng thc chuyn t h n v thc qua n v m ho khng th nguyn:
j
0
jj
jZ
ZZX
= ; j = 1, ..., k
2
ZZZj
min
j
max
j = ; j = 1, ..., k
Thu c ma trn thc nghim vi cc bin m nh sau:
S
TNX0 X1 X2 X3 Y
1 1 1 1 1 12.352 1 1 1 -1 8.873 1 1 -1 1 12.084 1 1 -1 -1 6.925 1 -1 1 1 42.136 1 -1 1 -1 13.517 1 -1 -1 1 22.198 1 -1 -1 -1 4.57
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T kt qu thc nghim, tnh ton cc h s Bj:
N
Y iB
n
i
== 1
0
N
YXB
n
i iij
i
== 1
N
XXB
n
i ij
j
== 1 i1
1
.Y).(
T s liu thc nghim trn, p dng cc cng thc trn ta xc nh c
gi tr B0 , B1 , B2 v B3 thu c kt qu:
B1 B2 B3 B0-5.2725 3.8875 6.86 15.3275
tnh phng sai ti hin, tc gi lm thm 3 th nghim tm.
Kt qu cc th nghim tm:
N0 Yu0 0Y Yu0- 0Y
(Yu0-0
Y )2 (Yu0
0Y )2
1 5.65
7.5033
3
-1.8533
3.4347
2
8.22742 7.19 -0.3133
0.0981
8
3 9.67 2.1667
4.6945
9Phng sai ti hin c tnh theo cng thc:
2
1
002
)(1
1= =
m
i ith YYmS
1
)(1
200
=
=
m
YYS
m
i i
th
trong m l s th nghim tm phng n.
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- S c ngha ca h s hi quy c kim nh theo tiu chun Student:
ibS
ib
ti =
bi: l h s th i trong phng trnh hi quy.
Sbi: lch qun phng ca h s th i.
N
SS th
bi=
Phng sai ti hin: S2th = 4.11
kim nh ngha cc h s, tc gi tnh cc h s tj (theo cng thc
trang 5), thu c kt qu sau:
t0 t1 t2 t321.3746 7.35263 5.42122 9.56644
Tra bng phn phi phn v Student vi mc ngha p = 0.05, f = N0-1 = 2
ta c t0.05(2) = 4.3. Vy cc h s tj u ln hn t0.05(2) nn cc h s ca phng
trnh hi qui u c ngha.
Phng trnh hi qui c dng sau:
L =15.3275-5.2725X1+3.8875X2+6.86X3
Kim nh s tng thch ca phng trnh hi qui vi thc nghim:
STT L YiYi-
L(Yi-
L)2
1
20.8
1
12.3
5 -8.46 71.57162 7.09 8.87 1.78 3.1684
3
13.0
3
12.0
8 -0.95 0.90254 -0.69 6.92 7.61 57.9121
531.3
542.1
310.7
8 116.208
6
17.6
3
13.5
1 -4.12 16.9744
7
23.5
7
22.1
9 -1.38 1.90448 9.85 4.57 -5.28 27.8784
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Phng sai d (theo cng thc trang 5):
LN
YYS
N
i ii
d u
=
= 1
2
^
2)(
(N l s th nghim, L l h s ngha)
Ta c: S2d = 74.13Tiu chun Fisher:
F= S2d / S2th = 74.13/4.1 = 18.08
Tra bng phn v phn b Fisher vi p = 0.05; f1 = N-L = 4; f2 = N0-1 = 2;
ta c:
F1-p = F0.095(4,2) = 19.3. Vy F < F0.95(4;2). Phng trnh hi qui tng
thch vi thc nghim.
*Ti u ho thc nghim bng phng php leo dc tm gi tr Ymin.
2. NHN XT
- Tc gi xc nh hm mc tiu, bi ton ti u v phng n qui hoch
trc giao cp 1 l ph hp. Cc s liu c tc gi tnh ton hu nh khng sai
lch so vi cc s liu c tnh ton li.
- Tuy nhin, cc s liu thc nghim (Yi) bin thin bt hp l (khng theo
qui lut tuyn tnh). ng thi, ba gi tr Y0 ca th nghim ti tm sai lch
nhau qu nhiu v khc rt xa so vi h s B0. (V nu cc s liu thc nghim
ng tin cy v tnh ton chnh xc th Y0 phi xp x B0) v gi tr trung bnh
ca chng l 7.503. Cc cng thc v php tnh c kim tra li l ng, v
vy c th ni cc s liu thc nghim Yi cha c chnh xc.
- C th thc hin bi tan ti u vi h s tng tc.
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0.6
0.7
0.8
0.9
1
1 2 3 4 5 6
%anthocyani
1
1.5
2
2.5
3
3.5
4
mu
% anth
m
5.2.Bi ton 2.
1. Nghin cu nh hng ca mt s yu t cng ngh n qu trnh chit
tch anthocyanin
1.1. nh hng ca t l dung miNghin cu nh hng ca t l dung mi n hm lng v mu
anthocyanin thu c. Ngi N/C lm 6 th nghim trong cc iu kin nh
sau:
-Nhit chit: 300C
-Thi gian chit: 45 pht
-Chit trong h dung mi c t l dung mi nc: ethanol thay i nh
bng 1.
Bng1. Cc thng s ban u v kt qu th nghim
S
TT
%
Vnc
%
Vethanol
T l
nc / ethanol
Hm lng %
anthocyanin, mu
18
020 4/1 0,827 3,53
27
030 7/3 0.890 3,50
36
0 40 3/2 0.870 3,46
45
050 1/1 0,857 3,40
54
060 2/3 0,845 3,37
63
070 3/7 0,840 3.34
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0. 6
0. 7
0. 8
0. 9
1
30
pht
45
pht
60
pht
75
pht
90
pht
%a
nthocyanin
1
1. 5
2
2. 5
3
3. 5
4
mu
% anthocyani
mu
Hnh 2: Biu biu din nh hng ca thi gian chit n hmlng v mu anthocyanin thu c
1.2. nh hng ca thi gian chit
Tin hnh 5 th nghim trong cc iu kin sau:
- Nhit chit: 300C
- Chit trong dung mi c t l nc: ethanol l 7/3
- Thi gian thay i t 30- 90 pht.
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Hnh 3: Biu biu din nh hng ca nhit n hmlng v mu anthocyanin
0.5
0.6
0.7
0.8
0.9
1
30oC
40oC
50oC
60oC
70oC
%
anthocyanin
0
0.5
1
1.5
2
2.5
3
3.5
4
mu
% anthocyani
mu
1.3. nh hng ca nhit
Nghin cu nh hng ca nhit , tin hnh 5 th nghim trong cng
iu kin:
- Chit trong h dung mi c t l nc: ethanol l 7/3
- Thi gian chit: 45 pht.- Nhit chit thay i t 300C 700C. Cc thng s c th v kt qu
c th hin trn biu hnh (3).
Qua nghin cu nh hng ca mt s yu t cng ngh n kh nng thu
nhn anthocyanin chng ti nhn thy: Dung mi, t l dung mi, thi gian
chit, nhit chit u nh hng n kh nng chit tch anthocyanin t bp
ci tm. ng vi mi iu kin khc nhau chng ti thu c anthocyanin c
hm lng v mu khc nhau. T cc kt qu nghin cu chng ti chn
c min kho st thch hp ca cc yu t cng ngh cho cc nghin cu tip
theo nh sau:- Chit trong h dung mi c t l nc: ethanol dao ng t 7/3 1/1
- Nhit chit t 30400C
- Thi gian chit trong khong 4575 pht
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2. Ti u ho iu kin chit tch anthocyanin c mu cao t bp ci
tm
Vi mc ch ca ti l thu nhn v s dng cht mu anthocyanin,
chng ti tin hnh ti u ho iu kin chit tch trong khun kh bi ton
ti u a mc tiu thu nhn cht mu anthocyanin c hm lng v mucao nht.
2.1 Chn cc yu t nh hng
Trong qu trnh chit tch anthocyanin phi chu tc ng ca nhiu yu
t cng ngh, song y chng ti chn 3 yu t c thm d phn trn:
- Z1: Nhit chit, 0C
- Z2: Thi gian chit, pht
- Z3: T l nc trong h dung mi,%
- Y1: Hm lng anthocyanin, %
- Y2: mu
Phng trnh biu din mi quan h c dng:
Y1 = f ( Z1 , Z2, Z3 ) Y2 = (Z1, Z2, Z3)
Y1 Max Y2 Max
Y1 l hm mc tiu hm lng Y2 Hm mc tiu mu.
2.2 Cc bc thc hin bi ton quy hoch
2.2.1 Chn phng n quy hoch xc nh hng i ca ti v nhanh chng tin ti min ti u chng
ti chn phng n quy hoch trc giao cp I (TYT 2k) thc nghim yu t ton
phn 2 mc, k yu t nh hng.
Phng trnh hi qui c dng:
Y = b0 + b1x1 + b2x2 + b3x3 + b12x1x2 + b13x1x3 + b23x2x3 + b123 x1x2x3 (1 )
Trong :
b0: H s hi qui.b1, b2, b3 : H s tuyn tnh
b12, b23, b13: H s tng tc i
b123: H s tng tc ba
Mi h s b c trng cho nh hng ca cc yu t n qu trnh chit
tch.
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2.2.2. T chc th nghim trc giao cp I
Theo [1] s th nghim trong phng n l 2k= 8, (k = 3) v iu kin th
nghim c ghi bng (2).
Bng 2 : iu kin th nghim c chn:
Cc mc Cc yu t nh hngZ1, 0C Z2, pht Z3, %VMc trn (+1) 40 75 70
Mc c s (0) 35 60 60
Mc di (-1) 30 45 50Khong bin
thin5 15 10
T cch chn phng n v iu kin th nghim, chng ti xy dng ma
trn thc nghim theo bin m v tin hnh th nghim theo ma trn.Kt qu c ghi bng (3)
Bng 3. Ma trn thc nghim trc giao cp I, k=3, v kt quSTT Bin m
Y1x1 x2 x3 x1x2 x1x3 x2x3 x1x2x3
1 + + + + + + +0,97
5
2 _ + + _ _ + _ 1,10
2
3 + _ + _ + _ _ 0,849
4 _ _ + + _ _ +1,10
9
5 + + _ + _ _ _ 0,85
4
6 _ + _ _ + _ +0,71
7
7 + _ _ _ _ + +0,94
4
8 _ _ _ + + + _
0,81
3
T1 0 0 0 0 0 0 00,91
5
T2 0 0 0 0 0 0 00,93
5
T3 0 0 0 0 0 0 00,95
5
Trong :
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- x1: Nhit chit, 0C
- x2: Thi gian chit, pht
- x3: T l nc trong h dung mi,%
- Y1: Hm mc tiu hm lng anthocyanin,%
- Y2: Hm mc tiu mu.2.2.3 Xy dng m t ton hc cho hm mc tiu hm lng anthocyanin
a) Chn phng trnh hi qui:
Phng trnh hi qui c chn theo phng trnh (1) mc 2.2.1.
Cc h s b1, b2, b3...b123 c tnh theo s liu thc nghim hm mc
tiu hm lng anthocyanin (Y1).
b) Tnh h s b:
V phng n c chn l quy hoch trc giao, theo [1] cc h s bj trong
phng trnh hi qui (1) c xc nh theo cng thc sau:
bj = )(1
1
u
N
u
ju yxN
=
vi : j = (1, k )
bij = ujuN
u
iu yxxN
)(1
1
i j = (1, k ) (2)
bijk = ukujuN
u
iu yxxxN
)(1
1
=
i j k = (1, k )
T s liu thc nghim bng (3), p dng cc cng thc (2) ta tnh c
cc h s b:
b0 = 0,9208 b12 = 0,017
b1 = -0.07 b13 = -0,0255
b2 = 0,0488 b23 = -0,019
b3 = 0,088 b123 = 0,0155
c) Kim nh mc ngha ca cc h s b trong phng trnh 3.1
Cc h s c kim nh theo tiu chun Student (t)
jb
j
jS
bt = (3)
So snh tj vi tp(f) . Trong : - tp(f) l chun student tra bng ng
vi xc sut tin cy p v bc t do f, f = n0 1.
bj : l h s trong phng trnh hi quy chn.
Sbj l lch ca cc h s bj
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Nu tj > tp(f) th h s bj c ngha.
Nu tj < tp(f) th h s bj b loi khi phng trnh.
kim nh theo chun Student (t) ta thay h s bj, Sbj vo cng thc
(3.3) ta c cc gi tr tj:
t0 = 130,21 t12 = 2,510t1 = 10,01 t13 = 3,606
t2 = 6,894 t23 = 2,687
t3 = 12,445 t123 = 2,192
Tra bng tiu chun Student ta c tp(fth) = t0,05(2) = 4,3
Do t12 < tp(fth), t13 < tp(fth), t23 < tp(fth), t123< tp(fth) nn cc h s b12, b13, b23,
b123 loi ra khi phng trnh. Phng trnh ng hc c dng:
y 1 = 0,9208 - 0,07x1 + 0,04875x2 + 0,088x3 (4)
d) Kim nh s ph hp ca phng trnh hi qui vi thc nghim
S tng thch ca phng trnh vi thc nghim c kim nh
theo tiu chun Fisher (F). 2th
2
d
S
SF = = 7,8406
Tra bng tiu chun Fisher ta c F1-p( f1, f2) = F0,95 (4,2) = 19,3
Ta c F < F 1-p. Vy m hnh ton hc chn ph hp vi thc nghim.
2.2.4 Ti u ho thc nghim thu c hm lng anthocyanin cao nht
a) Tnh cc bc chuyn ng j:
T mc c s Z0j, v phng trnh hi qui tuyn tnh i vi hm mc
tiu hm lng chng ti tnh bc chuyn ng j(j = 1, 2, 3) cho mi yu
t. Kt qu c ghi bng 4.
Bng 4. Kt qu tnh bc chuyn ng j ca cc yu t
Cc mcCc yu t nh hng
Z1,0CZ2,
phtZ3, %
Mc c s 35 60 60Khong bin thin ( j
)5 15 10
H s bj -0,070 0,048 0.088bj j -0,350 0,731 0,88
Bc chuyn ng ( j -1,980 4,150 5
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)Lm trn -2 4 5
Theo bng s liu (4) ta c : max33b = 0,88, theo ti liu [1]
Chn bc chuyn ng =3 0,5 . 3 = 0,5.10 =5.
Cc bc chuyn ng ca yu t x1, x2 c tnh:
33
1131
=
b
b = -1,98
33
22
32
=
b
b = 4,1
b) T chc th nghim leo dc:
T kt qu cc bc chuyn ng j bng (4), chng ti t chc th
nghim leo dc v im xut pht t tm thc nghim.
Th nghim theo hng chn, kt qu c biu din bng 5.
Bng 5: Kt qu th nghim theo hng leo dc
Yu t
TN
Z1,0CZ2,
pht
Z3,
%y1,% y2
1( Tn titm ) 35 60 60 0,927
2 33 64 650.96
2
3 31 68 700,98
5
4 29 72 751,11
34,720
5 27 76 800,99
7Nhn vo bng 5, kt qu th nghim tt nht th nghim th t. Ti nhit
chit 290C, thi gian chit 72 pht, t l nc trong h dung mi l 75%
chng ti thu c hm lng anthocyanin cao nht l 1,113 %. Ti th nghim
ny, chng ti xc nh mu ca anthocyanin l: 4,720. y cha phi l
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mu thu c cao nht. V th, chng ti tin hnh tm iu kin chit tch ti
u thu c anthocyanin c mu cao.
2.2.5 Xy dng m t ton hc vi hm mc tiu mu.
a) Chn phng trnh hi qui:
Phng trnh hi qui c chn theo phng trnh (1) mc 2.2.1. Cc hs b1, b2, b3...b123 c tnh theo s liu thc nghim hm mc tiu mu
(Y2).
b) Kim tra mc ngha ca h s b trong phng trnh hi qui:
Sau khi kim tra mc ngha ca cc h s b ta c: t13 < tp(fth), t23 < tp(fth),
t123< tp (fth) nn cc h s b13, b23, b123 b loi ra khi phng trnh.
Phng trnh hi qui c dng:
=
y 3,7709 - 0,55x1 0,2826x2 + 0,5291x3 0,2224x1x2 (5)
c) Kim nh s ph hp ca phng trnh hi qui vi thc nghim:
Cc bc kim tra c trnh by ph lc 6.
Sau khi kim tra ta c phng trnh hi qui (5) ph hp vi thc nghim.
2.2.6: Ti u ho thc nghim thu c anthocyanin c mu cao.
Sau khi kim tra phng trnh hi quy ph hp vi thc nghim,
chng ti tin hnh ti u ho thc nghim bng phng php leo dc thu
c anthocyanin c mu cao.
* Tnh bc chuyn ng ca cc yu t
Cng t mc c s Zj v phng trnh hi qui i vi hm mc tiu
mu. Chng ti tnh bc chuyn ng j (j = 1, 2, 3) tng t nh mc 2.2.3.
Kt qu c th hin bng 6.
Bng 6: Tnh bc chuyn ng j ca cc yu t
Cc mcCc yu t
Z1,0CZ2,
pht Z3, %
Mc c s 35 60 60
Khong bin thin ( j
)5 15 10
H s bj -0,55-
0,28260,5291
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2.3. Ti u ho hm a mc tiu bng phng php chp tuyn tnh.
Qu trnh chit tch cht mu anthocyanin c mu cao t bp ci tm
c c trng bi hai phng trnh (4), (5). Hai phng trnh ny th hin s
tc ng ca cc yu t cng ngh n hm lng v mu anthocyanin thu
c.Khi c s thay i ca b s liu (x1, x2, x3) trong bng ma trn thc
nghim (56) th cho cc gi tr thc nghim y1, y2 khc nhau v y1max, y2max cng
khc nhau.
Vi mc ch thu nhn cht mu anthocyanin c hm lng v mu
cao, nhim v ca chng ti phi ti u ho hm a mc tiu tm gii php
cng ngh thc tin tt cho c hai hm mc tiu, ng thi nng cao tnh ton
din v tnh thuyt phc cho kt qu thu c.
Thc t khng th c mt nghim chung cho c hai qu trnh t c
y1max, y2max m ch tm c nghim tho hip (x1, x2, x3) cc gi tr y1, y2 nm
gn y1max, y2max. tm c nghim tho hip chng ti s dng phng php
chp tuyn tnh :
yL = 1y1 + 2y2
Trong :
- 1 l h s quan trng ng vi hm mc tiu hm lng (y1)
- 2 l h s quan trng ng vi hm mc tiu mu (y2)Vi mc ch thu nhn cht mu anthocyanin ng dng lm cht ch th
trong ho phn tch. chng ti u tin cho hm mc tiu hm lng.
Chn: - 1 = 0,6, 2 = 0,4
Ta c phng trnh hm a mc tiu : yL = 0,6y1 + 0,4y2
Cc h s ca phng trnh hi quy c tnh theo bng 8
Bng 8. Tnh h s ca phng trnh hi quy
H s b y1 y2 yLb0 0,928 3,7709 2,06b1 -0,07 -0,55 -0,262b2 0,04875 -0,2826 -0,0837b3 0,088 0,5291 0,2644b12 -0,2223 -0,22
Ta c phng trnh hi quy:73
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yL = 2,06 0,262x1- 0,08379x2 + 0,2644x3 0,222x1x2 ( 6 )
Tin hnh ti u ha hm a mc tiu tm gii php cng ngh thc tin
ph hp.
*Tnh cc bc chuyn ng j: cho HMT yL
Cng tng t nh mc 2.2.3, chng ti tnh bc chuyn ngj
(321 ,, ) cho cc yu t nh hng. Kt qu c biu din bng 9.
Bng 9. Tnh bc chuyn ng ca cc mc yu t
Cc mcCc yu t
Z1,0CZ2,
phtZ3, %
Mc c s 35 60 60
Khong bin thin (
j )5 15 10
H s bj-
0,262
-
0,083970,2644
bj j -1,31-
1,4273
Bc chuyn ng (
j)-1,48
-
1,4273
Lm trn -1,5 -1,5 3
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* T chc th nghim leo dc cho hm mc tiu YL:
Bng 10: Kt qu th nghim theo hng leo dc ca hm chp YL
K
TN
Cc yu t nh hngY1,% Y2 YLZ1,0C
Z2,
phtZ3,%
1 ( TN
ti tm)35 60 60
0,92
7
3,99
12,149
2 33,558,
5 630,94
3
4,40
3 2,287
3 32 57 660,97
2
4,91
12,548
4 30,555,
569
0,98
3
4,95
22,553
5 29 54 721,11
0
4,96
72,656
6 28,5
52,
5 75
0,91
5
4,50
0 2,397
Nhn vo bng 10, ti th nghim th 5 hm chp yL t gi tr ln nht
yLmax= 2,656. So snh vi cc th nghim leo dc vi tng hm mc tiu ta c
PA.TN Z1, 0C Z2,
pht
Z3, % y1,% y2 yL
Theo
% Anth
2
9
75 7
5
1,11
3
4,72 2,556
Theo
mu
2
9
51 7
2
0,97
5
5,00
1
2,585
Theo
hm chp
2
9
54 7
2
1,11
0
4,96
7
2,656
T bng trn chng ti tm c iu kin tt chit cht mu
anthocyanin t bp ci tm trong mi trng trung tnh l:
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Nhit chit 29 0C
Thi gian chit l 54 pht
H dung mi nc -ethanol l 72-28
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5.3. Bi ton 3.
KT QU NGHIN CU NH HNG CA NHIT V
NNG KIM N QU TRNH TCH TP CHT RA KHI X
SI XENLULO
(Trch mt phn ti lun vn Thc s ca CN Nguyn B Trung khoa Ha, trng i hc S phm HN)
I. KT QU NGHIN CU T TI
1. Nghin cu nh hng ca nhit v nng kim n qu trnh tch
tp cht ra khi x si xenlulo.
1.1. nh hng ca thi gian v nng NaOH n lng tp cht tch
ra.
Gai b sau khi c tch s b, dng tay tc gai b kh ra thnh
nhng si mng, ct li thnh b, cc b c khi lng xp x nhau khong 15
gam. Ngm cc b si vo dung dch NaOH cc nng nghin cu trong
thi gian tng ng. Kt qu thu c bng sau:
Bng 1.1: Kt qu kho st nh hng ca thi gian v nng
n lng tp cht tch ra
Thi gian
C%
5 gi 10 gi 15 gi 20 gi
1% NaOH 10.098 10.978 12.46 13.013% NaOH 12.92 14.1 15.4 14.815% NaOH 14.01 14.67 15.598 14.46
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6
8
10
12
14
16
18
0 5 10 15 20 25
thoi gian (gio)
%t
achduoc
1%
3%
5%
Hnh 1.1. nh hng thi gian v nng NaOH n lng tp cht tch
ra.
Nhn xt: T 3 ng biu din nng 1%, 3% v 5% ta nhn
thy rng nng kim 1% th hiu qu tch l khng cao. Khi tng nng
kim ln 3% v 5% th hiu qu tch tng ln nhiu. Tuy nhin hiu qu tch
trong khong nng t 3 5% l khng khc nhau lm trong khong thi
gian t 10 15h.iu ny c th c gii thch nh sau: nng qu long 1%, thi
gian ngn ban u cha ho tan cc tp cht bao bc bn ngoi nn hiu
qu ca qu trnh tch khng cao. Sau thi gian t 10 15 gi, cc lp bn
ngoi b ho tan nn to iu kin thun li cho NaOH thm nhp vo bn
trong ho tan hemixenlulo, lignin v cc cht c phn t lng thp khc c
trong cc b si.
Nh vy qua th cho ta thy rng c thi gian v nng u c nh
hng r nt n qu trnh tch tp cht ra khi x si gai.
1.1.2. Ti u ho thc nghim qu trnh tch tp cht ra khi si gai.
tin ti min ti u, chng ti chn phng n thc nghim yu t ton
phn. Hai yu t nh hng n qu trnh l nng (Z1) v thi gian ngm (Z2-
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). Hm mc tiu cn t c l lng tp cht tch ra khi si l ln nht hay
ni cch khc hiu qu tch l cao nht.
quy hoch thc nghim ton phn, chng ti tin hnh b tr th
nghim thay i ng thi cc yu t, mi yu t c tin hnh 3 mc: mc
trn, mc di v mc c s th nghim tm phng nMc trn, mc di, khong bin thin c trnh by bng 1.2, ma trn
quy hoch thc nghim c trnh by bng 1.3.
Bng 1.2. Cc mc th nghim.
Mc di Mc c s Mc trn Khong bin
thin ( )Z1 (C% NaOH) 3 4 5 1Z2 (thi gian ngm) 10 12.5 15 2.5
Lp ma trn quy hoch:
Vi 2 yu t nhit v nng (k = 2), mi yu t c hai mc l mc
trn v mc di. Vy s th nghim c tin hnh l N = 22 = 4 th nghim.
Phng n tin hnh trnh by bng sau:
Bng 1.3. Ma trn quy hoch thc nghim.
NCc yu t theo t
l thcCc yu t theo t l
m hoGi tr oc
Z1 (C%) Z2 (t) X0 X1 X2 X1X2 Y1 5 15 1 1 1 1 15.5982 3 10 1 -1 -1 1 14.13 5 10 1 1 -1 -1 14.674 3 15 1 -1 1 -1 15.45 4 12.5 1 0 0 0 14.826 4 12.5 1 0 0 0 14.8
7 4 12.5 1 0 0 0 14.75
Thit lp phng trnh hi quy:
Tnh h s hi quy: Cc h s hi quy c tnh theo cng thc ton hc
(3), (4), (5). T s liu thc nghim ta xc nh c cc gi tr b0, b1, b2 nh
sau:
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b0 = 14,982 ; b1 = 0,192 ; b2 = 0,557 ; b12 = - 0,093
Vi kt qu trn ta c phng trnh hi quy theo ton hc:
Y = 14,982 + 0,192 X1 + 0,557 X2 0,093 X1X2
kim nh ngha ca h s hi quy v s tng thch ca phng
trnh hi quy vi thc nghim, ta phi tm phng sai ti hin S2th. Do vy
chng ta phi lm thm 3 th nghim tm phng n v thu c gi tr 0Y .
T cng thc tnh phng sai ti hin, ta c: S2 th = 0,0013 ; Sth =
0.03605551
Kim nh cc h s c ngha ca phng trnh hi quy:
S c ngha ca h s hi quy c theo tiu chun Student:ib
i
S
bti =
Bng cch tnh nh trn ta thu c cc gi tr ti nh sau:
t0 = 828,833 t2 = 30,896
t1 = 10,650 t12 = 5,158
Tra bng tiu chun Student ta c t0,05 (2) = 4,303.
Qua bng s liu trn ta thy cc c h s b0, b1, b2 , b12 l c ngha vi
tin cy P
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- Trong khong kho st mun tng hiu qa tch tp cht th phi tng
nng kim v thi gian ngm. Tuy nhin trong 2 yu t trn th thi gian
ngm ng vai tr quan trng hn do gi tr h s b2 l ln hn so vi b1.
Ti u ho thc nghim.
Chng ti tin hnh ti u ho qu trnh tch tp cht ra khi s si gaibng phng php thc nghim leo dc nht. Vi cc h s ca phng trnh
hi quy nh trn, tng nng kim v thi gian th hiu qu tch tp cht s
tng, n mt gii hn no th gi tr phn trm tp cht tch ra l cc i v
qu gii hn th gim xung.
Ti u ho c thc hin nh sau:
Chn bc nhy ca yu t x1 l 1 = 0,05. Da vo 1 ta tnh c gi
tr 2 theo cng thc sau :33
1131
=
bb
33
22
32
=
b
b
c: 36,01.192,0
5,2.557,0.05,02 == T cc thng s tnh c, chng ti tin
hnh thc nghim ti u ho cc mc c s v bc nhy nh sau:
Bng 1.4. Kt qu ti u phng trnh hi quy Y1 m t nh hng
ca nng v thi gian ln hm lng cc cht tch ra.Cc yu t X 1 (nng
kim C%)
X2 (thi gian ngm)
Mc c s 4 % 12,5 gi = 750
phtBc nhy 1 = 0,05 2 = 0,36 gi
20 phtNng
NaOH
Thi gian ngm Y (lng cht
tch ra)4,05 770 pht 14,34,10 790 pht 14,814,15
(TNTT)
810 pht -
4,20
(TNTT)
830 pht -
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4,25 850 pht 15,344,30 870 pht 15,624,35 890 pht 15,2
Nh vy hiu qu tch tp cht ra khi x si gai th phi tin hnh
cc thng s sau nhit phng.Cc thng s ti u Gi tr
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