4.1. Hàm hồi quy bậc 2
Gia cong vat lieu SKD tren may phay CNC Thoi gian chay chuong trinh:24-Jan-2009
***************************************************************Ket qua hoi qui voi do tin cay 95 phan tram la: Ket qua hoi qui voi do tin cay 95 phan tram la:
Phuong trinh Hoi qui tuyen tinh cap II co dang:
y=b0+b1*x1+b2*x2+b3*x3+b12*x1*x2+b13*x1*x3+b23*x2*x3+b11*x1^2+b22*x2^2+b33*x3^2 CAC HE SO HOI QUI CUA NANG SUAT Q:
b0=157.057599 b1=-2.650311 b2=-4.934117 b3=-56.147729 b12=-0.046275 b13=0.448740 b23=0.512351 b11=-0.672684 b22=-1.211865 b33=5.130406 He so xac dinh R^2= 0.944028
Gia tri ham so thong ke de kiem dinh gia thiet F0=9.369973 va gia tri p-value=0.011976
Kiem dinh ket qua CAC SO THONG KE
Phan bo Student voi muc y nghia 0.05 bac tu do 2 t_anpha=2.919986
Phan bo F voi muc y nghia 0.01 bac tu do 5 va 2 F_anpha=99.415852
--- Phuong sai du (binh phuong):0.028185 Phuong sai du (binh phuong):0.028185 phuong sai tai sinh (binh phuong)=0.000015 --- KIEM DINH TETAj=0?
t(1)=157024.928927 > t_anpha=2.919986, He so beta(1) co nghia t(2)=2139.164021 > t_anpha=2.919986, He so beta(2) co nghia t(3)=6329.584292 > t_anpha=2.919986, He so beta(3) co nghia t(4)=321744.442412 > t_anpha=2.919986, He so beta(4) co nghia t(5)=47.870004 > t_anpha=2.919986, He so beta(5) co nghia t(6)=2075.945656 > t_anpha=2.919986, He so beta(6) co nghia t(7)=3767.069519 > t_anpha=2.919986, He so beta(7) co nghia t(8)=624.230866 > t_anpha=2.919986, He so beta(8) co nghia t(9)=2203.192549 > t_anpha=2.919986, He so beta(9) co nghia t(10)=168626.171675 > t_anpha=2.919986, He so beta(10) co nghia ---KIEM TRA y^:----
F^=S^2du/S^2ts =9.369973<F(anpha) = 99.415852, Mo hinh phu hop ---
Q= e^(157.057599-2.650311*lnt-4.934117lnS-56.147729lnV-
0.046275lnt*lnS+0.448740lnt*lnV+0.512351lnS*lnV-0.672684lnt^2-1.211865lnS^2+5.130406lnV^2) ---
***************************************************************CAC HE SO HOI QUI CUA NHAN BE MAT Rz: CAC HE SO HOI QUI CUA NHAN BE MAT Rz:
b0=28.977480 b1=8.491366 b2=-9.752964 b3=-10.234771 b12=0.217090 b13=-1.449336 b23=1.647837 b11=-0.160584 b22=-0.284649 b33=0.936091 He so xac dinh R^2= 0.775675
Gia tri ham so thong ke de kiem dinh gia thiet F0=1.921012 va gia tri p-value=0.244398
Kiem dinh ket qua CAC SO THONG KE
Phan bo Student voi muc y nghia 0.05 bac tu do 2 t_anpha=2.919986
Phan bo F voi muc y nghia 0.01 bac tu do 5 va 2 F_anpha=99.415852
Phuong sai du:0.009668
phuong sai tai sinh (binh phuong)=0.000148
t(1)=9232.840119 > t_anpha=2.919986, He so beta(1) co nghia t(2)=2184.187362 > t_anpha=2.919986, He so beta(2) co nghia t(3)=3987.194584 > t_anpha=2.919986, He so beta(3) co nghia t(4)=18690.547651 > t_anpha=2.919986, He so beta(4) co nghia t(5)=71.568653 > t_anpha=2.919986, He so beta(5) co nghia t(6)=2136.757828 > t_anpha=2.919986, He so beta(6) co nghia t(7)=3861.135494 > t_anpha=2.919986, He so beta(7) co nghia t(8)=47.489946 > t_anpha=2.919986, He so beta(8) co nghia t(9)=164.919739 > t_anpha=2.919986, He so beta(9) co nghia t(10)=9805.197496 > t_anpha=2.919986, He so beta(10) co nghia ---
---KIEM TRA y^:----
F^=S^2du/S^2ts =1.921012<F(anpha) = 99.415852, Mo hinh phu hop ---
SAU KHI MU HOA- Nham be mat co quan he theo dang:
Rz= e^(28.977480+8.491366*lnt-9.752964lnS-10.234771lnV+0.217090lnt*lnS- 1.449336lnt*lnV+1.647837lnS*lnV-0.160584lnt^2-0.284649lnS^2+0.936091lnV^2) ---
CAC HE SO HOI QUI CUA MON hs: b0=-170.914172
b1=-0.142169 b2=29.140555 b3=57.368954 b12=1.410713
b23=-3.766014 b11=0.936300 b22=2.272186 b33=-4.733311 (adsbygoogle = window.adsbygoogle || []).push({});
He so xac dinh R^2= 0.916914
Gia tri ham so thong ke de kiem dinh gia thiet F0=6.130994 va gia tri p-value=0.029964
Phuong sai du (binh phuong):0.133978 phuong sai tai sinh (binh phuong)=0.077456 --- KIEM DINH TETAj=0?
t(1)=2378.465950 > t_anpha=2.919986, He so beta(1) co nghia t(2)=1.597203 < t_anpha=2.919986, He so beta(2) bi loai t(3)=520.322715 > t_anpha=2.919986, He so beta(3) co nghia t(4)=4575.778578 > t_anpha=2.919986, He so beta(4) co nghia t(5)=20.312649 > t_anpha=2.919986, He so beta(5) co nghia t(6)=51.268097 > t_anpha=2.919986, He so beta(6) co nghia t(7)=385.413421 > t_anpha=2.919986, He so beta(7) co nghia t(8)=12.093674 > t_anpha=2.919986, He so beta(8) co nghia t(9)=57.497828 > t_anpha=2.919986, He so beta(9) co nghia t(10)=2165.446665 > t_anpha=2.919986, He so beta(10) co nghia ---KIEM TRA y^:----
F^=S^2du/S^2ts =6.130994<F(anpha) = 99.415852, Mo hinh phu hop ---
SAU KHI MU HOA- Luong mon co quan he theo dang:
hs= e^(-170.914172-0.142169*lnt+29.140555lnS+57.368954lnV+1.410713lnt*lnS+0.796190lnt*lnV- 3.766014lnS*lnV+0.936300lnt^2+2.272186lnS^2-4.733311lnV^2)
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4.2. Tối ưu hoỏ quỏ trỡnh gia cụng vật liệu SKD61
Thoi gian chay chuong trinh:24-Jan-2009
*********************************************************NANG SUAT Q: NANG SUAT Q:
---
SAU KHI MU HOA- Nang suat co quan he theo dang: Q= e^(157.057599-2.650311*lnt-4.934117lnS-56.147729lnV-
0.046275lnt*lnS+0.448740lnt*lnV+0.512351lnS*lnV-0.672684lnt^2-1.211865lnS^2+5.130406lnV^2) ---
Ham muc tieu hoi tu Nang suat Q dat max tai: t= 0.8540
S= 0.4210V= 358.600 V= 358.600 Qmax=153.224292 So buoc lap :5
Thuat toan :medium-scale: SQP, Quasi-Newton, line-search So lan tinh toan gia tri ham muc tieu:24
Final step size:1 Fisrt order optimality:0
Gia tri GRADIENT cua ham muc tieu :-43.353851 Gia tri GRADIENT cua ham muc tieu :-66.881903 Gia tri GRADIENT cua ham muc tieu :-1.578574 ********************************************************* KET QUA TOI UU HOA
************************************************************* NHAM BE MAT Rz: NHAM BE MAT Rz:
---
SAU KHI MU HOA- Nham be mat co quan he theo dang:
Rz= e^(28.977480+8.491366*lnt-9.752964lnS-10.234771lnV+0.217090lnt*lnS- 1.449336lnt*lnV+1.647837lnS*lnV-0.160584lnt^2-0.284649lnS^2+0.936091lnV^2) ---
Ham muc tieu hoi tu
Nham be mat Rz dat min tai: t= 0.2460
S= 0.1780V= 358.600 V= 358.600 Rz min=1.966820 So buoc lap :5
Thuat toan :medium-scale: SQP, Quasi-Newton, line-search So lan tinh toan gia tri ham muc tieu:24
Final step size:1 Fisrt order optimality:0
Gia tri GRADIENT cua ham muc tieu :0.334163 Gia tri GRADIENT cua ham muc tieu :6.829685 Gia tri GRADIENT cua ham muc tieu :-0.000185 LUONG MON hs:
--- (adsbygoogle = window.adsbygoogle || []).push({});
SAU KHI MU HOA- Luong mon co quan he theo dang:
hs= e^(-170.914172-0.142169*lnt+29.140555lnS+57.368954lnV+1.410713lnt*lnS+0.796190lnt*lnV- 3.766014lnS*lnV+0.936300lnt^2+2.272186lnS^2-4.733311lnV^2)
--- Ham muc tieu hoi tu Ham muc tieu hoi tu
Luong mon hs dat mmin tai: t= 0.2460
S= 0.2944V= 309.996 V= 309.996 hs_min=0.004788 So buoc lap :8
Final step size:1
Fisrt order optimality:1.011943e-004
Gia tri GRADIENT cua ham muc tieu :0.001436 Gia tri GRADIENT cua ham muc tieu :0.000009 Gia tri GRADIENT cua ham muc tieu :0.000101 *********************************************************
4.3. Đồ thị xỏc định miền tối ưu hoỏ quỏ trỡnh gia cụng