Chuong 6 _ Toi Uu Hoa Bang Excel

03/03/2013

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CHNG 6: M HNH HA &

TI U HA

TIN HC NG DNG TRONG

HA HC

TRNG I HC CNG NGHIP TP.HCM

KHOA CNG NGH HO HCNI DUNG

1. GII H PHNG TRNH TUYN TNH

2. BI TON TRUYN NHIT

3. BI TON CHNG CT

4. TI U HA THC NGHIM


HPT tuyn tnh n phng trnh, n n s:

a11.x1 + a12.x2 + + a1n.xn = b1

a21.x1 + a22.x2 + + a2n.xn = b2

.

an1.x1 + an2.x2 + + ann.xn = bn

Hay vit di dng ma trn: A. X = B

HPT c nghim khi det A 0, khi nghim ca h

xc nh theo phng php ma trn : X = A-1.B


Mt s hm trong Excel:

Tnh nh thc ma trn A: MDETERM(A)

Tm ma trn nghch o A-1: MINVERSE(A)

Nhn 2 ma trn A-1 v B: MMULT(A-1,B)

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Gii h phng trnh:

2,75X1 + 1,78X2 + 1,11X3 = 13,62

3,28X1 + 0,71X2 + 1,15X3 = 17,98

1,15X1 + 2,70X2 + 3,58X3 = 39,72

gii bng phng php ma trn


Bc 1: lp bng s liu

Phng php ma trn


Bc 2: tnh det (A)

Phng php ma trn


Bc 3: Tnh ma trn A-1Phng php ma trn

n ba phm ng thi

Shift + Ctrl + Enter

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Bc 4: Tnh nghim X

Phng php ma trn

n ba phm ng thi

Shift + Ctrl + Enter


Bc 5: Nhp gi tr XT v tnh BT

Phng php ma trn


Bc 6: tnh BT

Phng php ma trn


Bc 7: tnh X nhn kt qu nghim.

Phng php ma trn

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Kt qu gii bng Solver:

BI TP

1. Gii cc phng trnh sau vi chnh xc 10-5:

a. ln(8x) x 0,5 =0

b. ln (7x) 3x + 1 = 0

c. ln(6x) x 0,4 = 0

2. Gii phng trnh sau:

a. x6 + 4x4 - 3x - 5 = 0 vi x thuc on [1;2]

b. x5 + 5x 2 = 0 vi x thuc on [0;1]


Thnh l t c 3 lp:

Gch chu nhit dy 120 mm, h s dn nhit 0,81 W/ m.K

Gch cch nhit dy 65 mm, h s dn nhit 0,23 W/ m.K

Thp chu lc dy 10 mm, h s dn nhit 45 W/ m.K

Nhit l: 8000C, h s cp nhit trong l: 69,6 W/ m2.K

Nhit khng kh: 350C, h s cp nhit khng kh: 13,9 W/

m2.K

Yu cu: Xc nh cc nhit b mt cc lp.


1= 0,12m, 1= 0,81 W/ m.K

2= 0,065 m, 1= 0,23 W/ m.K

3= 0,01 m, 3= 45 W/ m.K

t1= 8000C, 1=69,6 W/ m

2.K

t2= 350C, 2=13,9 W/ m

2.K

1

2

3Gch chu

nhit

Gch cch

nhit

Thp chu

lc

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1

2

3

1

2

q1=1(t1-tT1) q2=1

(tT1-tT2)

q3=2

(tT2-tT3)q4=

3

(tT3-tT4)

q5=2(tT4-t2)

Lu : q1 = q2 = q3 = q4 = q5 = q

2. BI TON TRUYN NHITa. S dng hm Goalseek

Bc 1: Lp bng tnh nh sau:


Bc 2: Nhp gi tr tT1 v tnh q1


Bc 3: tnh gi tr tT2, tT3, tT4 thng qua q1

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Bc 4: tnh gi tr q5 thng qua tT4 v t2


Bc 5: Lp biu thc so snh q5 vi q1


Bc 6: Thay i tT1 biu thc so snh 2%


Kt qu:

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2. BI TON TRUYN NHITb. S dng hm SolverBc 1: chuyn iu kin v dng h phng

trnh tuyn tnh:

q1=1(t1-tT1)

q2=1

(tT1-tT2)

q3=2

(tT2-tT3)

q4=3

(tT3-tT4)

q5=2(tT4-t2)

q +1.tT1 =1.t1

1 .q- .tT1 + .tT2 =0

2 .q - .tT2 + .tT3 =0

3 .q - .tT3 + .tT4=0

q -2.tT4 =-2.t2

2. BI TON TRUYN NHITb. S dng hm SolverBc 2: lp bng s liu:

2. BI TON TRUYN NHITb. S dng hm SolverBc 3: nhp gi tr bin v tnh VT

2. BI TON TRUYN NHITb. S dng hm SolverBc 4: gi hm Solver

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2. BI TON TRUYN NHITb. S dng hm SolverBc 5: ci t VT = VP

2. BI TON TRUYN NHITb. S dng hm SolverKt qu nghim:

3. BI TON CHNG CT

Chng ct hn hp nc Acetic vi nng nhp liu 20%

(KL), sn phm nh 95%(kl), sn phm y 0,5% (kl), ch s

hi lu bng 4.

Yu cu: Xc nh s a l thuyt.

3. BI TON CHNG CT

Cc cng thc s dng:

Phng trnh on ct:

yL =

+1 +

+1

= 0,8x + 19,698

Phng trnh on chng:

yc =+

+1

1

+1xw = 1,2419x 0,3986

Phng trnh ng cn bng (xy dng t s liu thc

nghim): y*=4.10-5x3-0,0106x2+1,6853x-0,5523

f = F/ D

Chuyn nng phn mol:

xA =

+

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NI DUNG

4.1 PP QUI HOCH THC NGHIM

4.2 KIM TRA S TNG HP CA PTHQ

4.3 TI U HA THC NGHIM BNG

EXCEL

4. TI U HA THC NGHIM 4. TI U HA THC NGHIM

M hnh thc nghim ca i tng nghin cu:

Bin u vo: nng , p sut, nhit , pH,

Hm mc tiu: hiu sut, cht lng sn phm, CPSX,



Ta cn thit lp quan h: y = f(x1, x2,,xk) +

hay y = f(X) +

Trin khai hm di dng chui Taylor:

f(x1, x2, . . . , xk) = 0 + j=1

k

j xj + j=1

k

jj xj2 +

ji

k

ij xixj+. . .

Vi 1 i j k

PTHQ thc nghim: y = f x1, x2, . . . , xk



B tr thc nghim theo ma trn bin u vo - ra:

X =

1 x111 x21

x12 x1kx22 x2k

1 xn1 xn2 xnk

Y =

y1y2yn

(n dng, k+1 ct)

Ma trn cc h s hi qui tuyn tnh c: B =

b0b1bk

Theo phng php bnh phng cc tiu, ta c:

XTX . B = XTY B = XTX1. (XTY)

Vi: XT l ma trn chuyn v ca ma trn X

(XTX)-1 l ma trn nghch ca ma trn XTX


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Ma trn trc giao X c nhng tnh cht sau:

+ Tnh trc giao: tch v hng ca hai vect ct bt k caX bng 0.

i=1

n

ximxij=0 vi j, m= 0, k

+ Tnh cht i xng: tng cc phn t trong mt ct bt ku bng 0.

i=1

n

xij=0 vi j 0.



Mt vi phng php qui hoch thc nghim:

+ Qui hoch thc nghim yu t ton phn 2k ( k l cc yu

t, n l s mc th s th nghim N = nk)

Ma trn X c thm tnh cht chun ha:

i=1

n

xij2=N vi j= 0, k

+ Qui hoch thc nghim yu t tng phn 2k p (p l gi tr

c trng cho tng phn)



+ Qui hoch trc giao cp 2: xy dng ma trn trc giao X bao gm ba loi th nghim:

- Phn c s gm n = 2k th nghim theo qui hoch thcnghim 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 khongcch > 0.

- Phn tm gm n0 (n0 1) th nghim tm phng ndng xc nh phng sai ti hin trong cng thc kimtra ngha ca cc h s hi qui.

Tng s th nghim trong phng n l N = 2k + 2k + n0



Mt vi dng phng trnh hi qui:

+ hi qui bc 1: y = b0 + j=1k bj xj

+ hi qui bc 1 y : y = b0 + j=1k bj xj + ji

k bij xixj

( vi Ck2 h s bij)

+ hi qui bc 2 y :

y = b0 + j=1k bj xj + ji

k bij xixj + j=1k bjj xj

2


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Bc 1: kim tra ngha ca cc h s phng trnh hi qui

bng tiu chun Student t (vi : mc ngha, = 0,05)

Chn thng k: tbj =bj

sbj

Vi sbj: lch qun phng ca h s th i.

Nu tbi > t(fth) th h s bi c gi li trong phng trnh

hi qui.(fth = n0 1 : bc t do ti hin)

Nu tbi < t(fth) th h s bi b loi khi phng trnh hi qui.



Phng sai sbj2 c xc nh theo cng thc:

sbj2 =

sth2

i=1n xi

2 =sth2

Phng sai ti hin sth2 :

sth2 =

1

n0 1 i=1

n0

(yi0 yo)2



Bc 2: kim tra s tng thch ca phng trnh hi qui

theo tiu chun Fisher:

F =sd2

sth2

Phng sai sd2 c xc nh theo cng thc:

sd2 =

1

N L i=1

n

(yi y)2

Vi L: s h s c ngha trong phng trnh hi qui.

Nu F < F(, ftt, fth) th m hnh thng k ph hp vi s liu

thc nghim.(ftt = N L , fth = n0 1)



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.

4.3 TI U HA THC NGHIM BNG EXCEL

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Mt s hm thng dng trong ti u ha:

+ 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 x2: SUMSQ(number1, number2,)



Mt s hm thng dng trong ti u ha:

+ Tnh tng bnh phng lch (y )2:

SUMXMY2(array_x, array_y).

+ Tnh lch chun ca mu i=1n (xi x)2

n1:

STDEV(number1, number2,).

+ Tra chun s Student: TINV(p1, p2).

+ Tra chun s Fisher: FINV(, p1, p2).



Hy tm mi quan h gia hm mc tiu y v cc bin Z1, Z2,

Z3 theo m hnh trc giao cp 1 vi s liu thu c nh sau:

Bin

thc

n = 2kn0 = 3

1 2 3 4 5 6 78 9

10 11

Z1 150 300 150 300 150 300 150 300225

225 225

Z2 30 30 90 90 30 30 90 9060 60 60

Z3 15 15 15 15 45 45 4545 30 30 30

y 3,0 6,0 10,0 12,0 15,0 23,0 12,0 18,012,0 13,8 13,2



Bc 1: lp 1 bng tnh excel vi cc thng tin sau:

Zj0 : mc c s.

Khong bin thin Zj.

Zj0 : D2 = AVERAGE(B2:C2).

Zj : E2 = (C2-B2)/2


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Bc 2: chuyn i cc bin thc sang bin m ha:

x0= 1

x1=

G8=(C8-$D$2)/$E$2

4.3 TI U HA THC NGHIM BNG EXCEL4. TI U HA THC NGHIM

Bc 3: xc nh cc h s B = XTX1. (XTY)



Bc 4: kim tra ngha ca cc h s hi qui:

sbj2 =

sth2

i=1n xi

2 =sth2

; sth

2 =1

n0 1 i=1

n0

(yi0 yo)2tbj =

bjsbj

STDEV(J16:J18)^2

TINV(0.05,2)

4.3 TI U HA THC NGHIM BNG EXCEL4. TI U HA THC NGHIM

PTHQ c dng: y = 12,375 + 2,375 x1 + 4,625 x3


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Bc 5: kim tra s tng hp ca phng trnh hi qui:

sd2 =

1

N L i=1

n

(yi y)2


F =sd2

sth2

F(, ftt, fth) = F(0.05,5,2): FINV(0.05,5,2)

PTHQ c dng: y = 12,375 + 2,375 x1 + 4,625 x3 Add your company slogan

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Chuong 6 _ Toi Uu Hoa Bang Excel