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Finding minimum value of function of several variables?

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Le Dung
Le Dung el 15 de Feb. de 2017
Comentada: Jan el 15 de Feb. de 2017
Hi everybody. I have a code to find minimum value of function of several variable (Mr) as below:
% GIOI THIEU
% Chuong trinh tinh toan toi uu hoa mat cat dam ban BTCT
% Tac gia: Le Tuan Dung
% Bo mon Cong trinh - Co so Thai Nguyen - DHCNGTVT
%-------------------*************--------------------------
% PHAN1: THAM SO DAU VAO CUA BAI TOAN
% 1.1. Cac gia tri trong luong rieng:
% Trong luong rieng be tong dam va ban
gamabt=25; % Don vi: kN/m3
% Trong luong rieng lop phu
gamalp=22.5; % Don vi: kN/m3
% 1.2. Cuong do cua vat lieu
% Cuong do be tong dam (f'c)
fcdam=30; % Don vi: Mpa
% Cuong do be tong ban (f'c)
fcban=30; % Don vi: Mpa
% Cuong do cot thep DUL (dieu 5.9.3)
fpu=1860; % Cuong do keo dut (Mpa)
fpy=0.9*fpu; % Cuong do chay deo (Mpa)
fpe=0.8*fpy; % Cuong do huu hieu (sau khi tru toan bo mat mat US) (Mpa)
% Modun dan hoi cua thep DUL (dieu 5.4.4.2)
Eps=197000; % (Mpa)
% 1.3. Tham so mat cat ngang cau:
Bcau=12; % % Be rong MCN cau (m)
hbbien=0.1; % Chieu day ban lien ket ben ngoai (m)
hbtrong=0.21; % Chieu day ban len ket ben trong (m)
Sodam=12; % Cau co 12 dam chu
daylp=0.08; % Chieu day lop phu mat cau (m)
Aclc=0.231; % Dien tich chan lan can
mlc=0.04; % Khoi luong/chieu dai cua lan can (kg/m)
gtt=9.81; % Gia toc trong truong (m/s2)
Ldam=24; % Chieu dai dam (m)
agoi=0.35; % Dau dam cach goi cau (m)
% 1.4. Cac he so tinh toan mo men:
neta=0.95; % He so dieu chinh
gamatt1=1.25; % He so tai trong tinh tai giai doan 1
gamatt2=1.5; % He so tai trong tinh tai giai doan 2
gamall=1.75; % He so tai trong cua hoat tai
IM=0.25; % He so xung kich
phif=0.9; % He so suc khang uon
% He so quy quy khoi ung suat chu nhat:
if fcdam <= 28
beta1=0.85;
elseif 28<fcdam<56
beta1=0.85-0.05*(fcdam-28)/7;
else
beta1=0.65;
end
% 1.5. Tai trong Hoat tai:
Ptruck=[145 145 35]; % (kN)
Kctruck=4.3; % (m)
Ptandem=[110 110]; % (kN)
Kctandem=1.2; % (m)
qlan=9.3; % (kN/m)
% 1.6. Tham so kich thuoc cua mat cat ngang:
disp('MOI CAC BAN NHAP CAC THAM SO DAU BAI');
B1=input('Gia tri can duoi be rong B1(cm): ');
B2=input('Gia tri can tren be rong B2(cm): ');
bs1=input('Gia tri can duoi be rong suon bien bs1(cm): ');
bs2=input('Gia tri can tren be rong suon bien bs2(cm): ');
bsg1=input('Gia tri can duoi be rong suon giua bsg1(cm): ');
bsg2=input('Gia tri can tren be rong suon giua bsg2(cm): ');
H1=input('Gia tri can duoi chieu cao H1(cm): ');
H2=input('Gia tri can tren chieu cao H2(cm): ');
hnt1=input('Gia tri can duoi chieu cao nap tren hnt1(cm): ');
hnt2=input('Gia tri can tren chieu cao nap tren hnt2(cm): ');
hnd1=input('Gia tri can duoi chieu cao nap duoi hnd1(cm): ');
hnd2=input('Gia tri can tren chieu cao nap duoi hnd2(cm): ');
Bslk=input('Gia tri chieu rong ban be tong lien ket Bslk(cm): '); % Bslk= 1 (m)
hslk=input('Gia tri chieu day ban be tong lien ket hslk(cm): '); % hslk=0.1 (m)
ntao1=input('Gia tri can duoi cua so tao cap ntao1: ');
ntao2=input('Gia tri can tren cua so tao cap ntao2: ');
dtao=input('Nhap duong kinh cua mot tao thep DUL dtao(mm): ');
B=(B1:B2); % Be rong MCN (0,8 - 1,0 met)
bs=(bs1:bs2); % Be rong suon bien (0,1 - 0,2 met)
bsg=(bsg1:bsg2); % Be rong suon giua (0,1 - 0,2 met)
H=(H1:H2); % Chieu cao MCN (0,8 - 1,0 met)
hnt=(hnt1:hnt2); % Chieu day ban tren (0,1 - 0,2 met)
hnd=(hnd1:hnd2); % Chieu day ban duoi (0,15 - 0,25 met)
ntao=(ntao1:2:ntao2); % So tao cap DUL la so chan(30 - 40 tao cap)
% Kiem tra tham so nhap vao:
Ktra=[B1 B2 bs1 bs2 bsg1 bsg2 H1 H2 hnt1 hnt2 hnd1 hnd2 Bslk hslk ntao dtao];
V={'B1','B2','bs1','bs2','bsg1','bsg2','H1','H2','hnt1','hnt2','hnd1','hnd2','Bs','hs','ntao','dtao'};
if find(Ktra<0)>0
disp('GIA TRI NHAP PHAI LA SO DUONG');
end
for i=1:numel(Ktra)
if Ktra(i)<0
%Cach bieu dien ten bien so
fprintf('Nhap lai gia tri cua: %s \n',V{i});
end
end
% PHAN 2: TOI UU HOA MAT CAT
% 2.1. Xac dinh tai trong tinh tai:
% Trong luong phan ban lien ket:
qlk=(2*(hbbien+hbtrong)*Bcau/2)/(2*Sodam); % Don vi: kN/m
% Tai trong do trong luong lop phu mat cau
qlp=(Bcau*daylp*gamalp)/Sodam;
% Tai trong do trong luong phan lan can
qlc=Aclc*gamabt+mlc*gtt;
% 2.2. Xac dinh cac tai trong do hoat tai:
% Chieu dai nhip tinh toan
Ldamtt=Ldam-2*agoi;
% Gia tri tung do duong anh huong:
y1=Ldamtt/4;
y2=((Ldamtt/2-4.3)*y1)/(Ldamtt/2);
y3=((Ldamtt/2-8.6)*y1)/(Ldamtt/2);
y4=((Ldamtt/2-1.2)*y1)/(Ldamtt/2);
Mrmin=Inf;
imin=1;
jmin=1;
kmin=1;
zmin=1;
lmin=1;
mmin=1;
nmin=1;
for i=1:numel(B)
for j=1:numel(bs)
for k=1:numel(bsg)
for z=1:numel(H)
for l=1:numel(hnt)
for m=1:numel(hnd)
for n=1:numel(ntao)
% Be rong cua lo rong:
br=((B(i)-2*bs(j)-bsg(k))/2)*0.01; % (m)
% Chieu cao phan lo rong
hr=(H(z)-hnt(l)-hnd(m))*0.01; % (m)
% TINH TOÁN MAT CAT QUY DOI
% Chieu cao canh tren:
ht=hnt(l)*B(i)/B(i)*0.01; % (m)
% Chieu cao canh duoi:
hd=hnd(m)*B(i)/B(i)*0.01; % (m)
% Chieu cao phan suon dam:
hsd=H(z)*0.01-ht-hd; % (m)
% Be rong phan suon dam:
bsd=(2*bs(j)*0.01*hr+bsg(k)*0.01*hr)/hsd; % (m)
% Tai trong do trong luong ban than dam chu:
qdc=(B(i)*0.01*hd+bsd*hsd+B(i)*001*ht)*gamabt; % (kN/m)
% TINH TOAN GIA TRI MOMEN UON
% Momen uon do tinh tai:
Mttai=neta*(gamatt1*qdc*Ldamtt*Ldamtt/8+gamatt1*qlk*Ldamtt*Ldamtt/8+...
gamatt1*qlc*Ldamtt*Ldamtt/8+gamatt2*qlp*Ldamtt*Ldamtt/8);
% Tinh he so phan bo ngang dam giua:
K=2.5*power(Sodam,-0.2);
de=0;
e=1.04+de/7600;
gg1lan=K*power((B(i)*0.01)/(2.8*Ldamtt),0.5)*power(1,0.25); % He so pbn 1 lan dam giua
gg2lan=K*power(B(i)*10/7600,0.6)*power((B(i)*0.01)/Ldamtt,0.2)*power(1,006); % He so pbn 2 lan dam giua
% Tinh he so phan bo ngang dam bien:
gb1lan=0.2; % Tinh theo PP don bay
gb2lan=e*gg2lan;
glay=[gg1lan gg2lan gb1lan gb2lan];
% He so phan bo ngang de tinh toan:
gpbn=max(glay);
% Momen uon do xe tai thiet ke:
Mtruck=neta*gpbn*(1+IM)*gamall*(Ptruck(1)*y1+Ptruck(2)*y2+Ptruck(3)*y3); % kN.m
% Momen uon do xe 2 truc:
M2truc=neta*gpbn*(1+IM)*gamall*(Ptandem(1)*y1+Ptandem(2)*y4); % kN.m
Mxetk=[Mtruck M2truc]; % kN.m
% Momen uon do tai trong lan xe
Mlanxe=neta*gpbn*gamall*9.3*0.5*(1/4)*Ldamtt*Ldamtt; % kN.m
% Momen uon do hoat tai
Mhtai=max(Mxetk)+Mlanxe; % kN.m
% Momen uon su dung tinh toan
Mutt=Mttai+Mhtai;
% TINH TOAN DAC TRUNG HINH HOC MAT CAT QUY DOI
% Dien tich mat cat ngang:
Aqd=B(i)*0.01*hd+bsd*hsd+B(i)*0.01*ht; %(m2)
% Momen tinh cua mat cat so voi day mat cat:
Sqd=(1/2)*B(i)*0.01*hd*hd+(0.5*hsd+hd)*hsd*bsd+(H(z)*0.01-0.5*ht)*ht*B(i)*0.01; % (m3)
% K/cach tu trong tam cua tiet dien quy doi toi day
% mat cat:
yd=Sqd/Aqd; %(m)
% K/cach tu tho tren cung toi truc trung hoa:
yt=H(z)*0.01-yd; %(m)
% Momen quan tinh cua mat cat doi voi truc trung
% hoa:
Jqd=(B(i)*0.01*ht*ht*ht)/12+power((yt-0.5*ht),2)*ht*(B(i)*0.01)+bsd*power(hsd,3)/12+...
power((yt-ht-hsd/2),2)*hsd*bsd+(B(i)*0.01)*power(hd,3)/12+power((yd-0.5*hd),2)*hd*(B(i)*0.01);
% TINH TOAN DAC TRUNG HINH HOC MAT CAT LIEN HOP
% Modun dan hoi dam:
Edam=0.043*power(gamabt*100,1.5)*sqrt(fcdam); % Mpa
% Modun dan hoi dam:
Eban=0.043*power(gamabt*100,1.5)*sqrt(fcban); % Mpa
% He so quy doi tu be tong ban sang be tong dam:
n1=Eban/Edam;
% Dien tich mat cat ngang:
Alh=Aqd+n1*Bslk*0.01*hslk*0.01; %(m2)
% Mome tinh cua mat cat so voi day mat cat:
Slh=Sqd+n1*Bslk*0.01*hslk*0.01*(H(z)*0.01+0.5*hslk*0.01); % (m3)
% K/cach tu trong tam cua tiet dien lien hop toi
% day ban:
ydlh=Slh/Alh; % (m)
% K/cach tu tho tren cung toi truc trung hoa:
ytlh=H(z)*0.01+hslk*0.01-ydlh; % (m)
% Momen quan tinh cua mat cat lien hop:
Jlh=(B(i)*0.01)*power(ht,3)/12+power((ytlh-hslk*0.01-0.5*ht),2)*ht*(B(i)*0.01)...
+bsd*power(hsd,3)/12+power((ytlh-hslk*0.01-ht-0.5*hsd),2)*hsd*bsd+...
(B(i)*0.01)*power(hd,3)/12+power((ydlh-0.5*hd),2)*hd*(B(i)*0.01)+...
n1*(Bslk*0.01*power(hslk*0.01,3)/12+Bslk*0.01*hslk*0.01*power((ytlh-0.5*hslk*0.01),2)); % (m4)
% TINH TOAN DAC TRUNG HINH HOC MAT CAT CO COT THEP
% He so quy doi tu cot thep DUL sang be tong dam:
n2=Eps/Edam;
% Dien tich mat cat ngang:
Act=Alh+(n2-1)*pi*dtao*0.001*dtao*0.001*ntao(n)*(1/4); %(m2)
% K/cach tu tim hang cot thep ngoai cung toi tho
% duoi:
ctduoi=0.06; % (m)
% K/cach tim toi tim giua cac hang cot thep:
timct=0.045; % (m)
% K/cach tu trong tam cot thep toi day mat cat:
dpsduoi=ctduoi+timct/2; %(m)
% K/cach tu trong tam dien tich cot thep toi tho tren cung cua MC:
dps=H(z)*0.01+hslk*0.01-dpsduoi;
% Momen tinh cua mat cat so voi day mat cat:
Sct=Slh+(n2-1)*pi*dtao*0.001*dtao*0.001*ntao(n)*(1/4)*(H(z)*0.01+hslk*0.01-dps); % (m3)
% K/cach tu trong tam tiet dien lien hop toi day MC
ydct=Sct/Act; %(m)
% K/cach tu tho tren cung toi TTH:
ytct=H(z)*0.01+hslk*0.01-ydct; % (m)
% Momen quan tinh cua mat cat co cot thep:
Jct=(B(i)*0.01)*power(ht,3)/12+power((ytct-hslk*0.01-ht/2),2)*ht*(B(i)*0.01)+...
bsd*power(hsd,3)/12+power((ytct-hslk*0.01-ht-hsd/2),2)*hsd*bsd+...
(B(i)*0.01)*power(hd,3)/12+power((ydct-hd/2),2)*hd*B(i)*0.01+...
n1*(Bslk*0.01*power(hslk*0.01,3)/12+Bslk*0.01*hslk*0.01*power((ytct-(hslk*0.01)/2),2))+...
(n2-1)*pi*dtao*0.001*dtao*0.001*(1/4)*ntao(n)*power((ydct-dpsduoi),2); % (m4)
% XAC DINH SUC KHANG UON CUA MAT CAT NGANG
% Chieu cao khoi ung suat hinh chu nhat:
acn=beta1*ytct; %(m)
if ytct>(ht+hslk*0.01)
% Suc khang uon danh dinh:
Mn=pi*dtao*0.001*dtao*0.001*(1/4)*ntao(n)*fpe*1000*(dps-acn/2)+...
0.85*fcdam*1000*(B(i)*0.01-bsd)*ht*(acn/2-ht/2)+0.85*fcdam*1000*(Bslk*0.01-bsd)*hslk*0.01*(acn/2-(hslk*0.01)/2); % kN.m
else
Mn=pi*dtao*0.001*dtao*0.001/4*ntao(n)*fpe*1000*(dps-acn/2)+...
0.85*fcdam*1000*(Bslk*0.01-B(i)*0.01)*hslk*0.01*(acn/2-(hslk*0.01)/2); % kN.m
end
% Suc khang uon tinh toan:
Mr=phif*Mn; %kN.m
if (Mr<=Mrmin)&&(Mr>=Mutt)
Mrmin = Mr;
imin = i;
jmin = j;
kmin = k;
zmin = z;
lmin = l;
mmin = m;
nmin = n;
end
end
end
end
end
end
end
end
disp('DA CHAY XONG CHUONG TRINH');
if ytct>(ht+hslk*0.01)
disp('TRUC TRUNG HOA DI QUA SUON DAM');
else
disp('TRUC TRUNG HOA DI QUA BAN CANH TREN');
end
% CAC GIA TRI KICH THUOC MAT CAT NGANG VA SO TAO THEP TOI UU
disp(Mrmin);
disp(B(imin));
disp(bs(jmin));
disp(bsg(kmin));
disp(H(zmin));
disp(hnd(lmin));
disp(hnt(mmin));
disp(ntao(nmin));
And when program finished, it gives me:
Mrmin=Inf;
80
10
10
80
10
15
30
of cause they (above values) are lower bonds of values that i must enter in paragraph:
B1=input('Gia tri can duoi be rong B1(cm): ');
B2=input('Gia tri can tren be rong B2(cm): ');
bs1=input('Gia tri can duoi be rong suon bien bs1(cm): ');
bs2=input('Gia tri can tren be rong suon bien bs2(cm): ');
bsg1=input('Gia tri can duoi be rong suon giua bsg1(cm): ');
bsg2=input('Gia tri can tren be rong suon giua bsg2(cm): ');
H1=input('Gia tri can duoi chieu cao H1(cm): ');
H2=input('Gia tri can tren chieu cao H2(cm): ');
hnt1=input('Gia tri can duoi chieu cao nap tren hnt1(cm): ');
hnt2=input('Gia tri can tren chieu cao nap tren hnt2(cm): ');
hnd1=input('Gia tri can duoi chieu cao nap duoi hnd1(cm): ');
hnd2=input('Gia tri can tren chieu cao nap duoi hnd2(cm): ');
Bslk=input('Gia tri chieu rong ban be tong lien ket Bslk(cm): '); % Bslk= 1 (m)
hslk=input('Gia tri chieu day ban be tong lien ket hslk(cm): '); % hslk=0.1 (m)
ntao1=input('Gia tri can duoi cua so tao cap ntao1: ');
ntao2=input('Gia tri can tren cua so tao cap ntao2: ');
dtao=input('Nhap duong kinh cua mot tao thep DUL dtao(mm): ');
So, who can explanin to me why?
  1 comentario
Jan
Jan el 15 de Feb. de 2017
Why what? Why you have to enter the values or why they are the lower bounds? I do not understand the comments and have no idea, what's going on inside the code.

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