I have to optimise the cost of a liquid pipeline with single variable( i.e. diameter of pipeline) genetic algorithm and I have written the algorithm but my program is not giving me the results. Please correct the error in my algorithm.

{ function [TC]=costopti(x)

D=x(1);

%Diameter to be optimized

L=[11.298,21.394,25.0410,25.743,20.717,24.203]; TC=0;

%distance between sv stations along the pipeline route in (km).Also includes the highest point

e=.045;

%internal roughness of the pipeline in (nm)

Qo=640;

MAOP=1000;

%Design factor of offshore liquid pipelines

SMYS=60000;

T=6.35;

DI=D-(2*T);

%thickness of the pipelines in (mm)

V=4.7290;

%Kinematic Viscosity at 40?C, V (cst)

S=0.8359;

%Specific Gravity at 40?C, (no units)

F=15;

%Transmission factor (no units)

H=[15,-10,-91,-301,-36,-25];

%Elevation along the pipeline route in (meter)

P1(1,1)=x(2);

%Inlet pressure to the pipeline in (kg/m2) considering the possible diameters to find the pressure provide along the pipeline

Re=(353678*Qo)/(V*DI);

%Reynolds number for all possible diameters (no units)

F=-4*log10(e/(3.7*DI)+(1.255*(F/Re)));

%Finding the transmission factor by trial and error method for each diameters taken(no units)

Pkm=24.99*(10^10*((Qo/F)^2)*(S/DI^5));

%Pressure drop per km in kPa due to friction for the possible diameter for each diameter pressure loss due to friction and pressure loss due to elevation changes is to be found

for i=1:6

Pf(i,1)=Pkm*L(i)*0.0102;

%Pressure drop due to friction along the pipelines route

(kg/cm2)

Pe(1,i)=(H(i)*S/.102)*0.0102;

%Pressure drop due to elevation along the pipeline route

P1(i+1,1)=P1(i,1)-Pe(1,i)-Pf(i,1); PL(i,1)=(P1(i+1,1)-P1(i,1));

%pressure loss between stations in (kg/cm2)

totalPL1=sum(PL); Prloss=-[totalPL1];

%Pressure profile along the pipelines route for 12 inch diameter pipeline end

Le=128.396;

%Total length of pipelines in Km

Cpt=600;

%Cost of carbon steel per tonne in USD

PMC=0.02463*Le*(D-T)*T*(Cpt);

%Pipe Material Cost for the whole length of pipeline in USD

PC=PMC*60;

%Pipe material cost in rupees

V1=7.855*(10^-4)*(DI^2);

VL=V1*128.396;

LFC=VL*1000*55;

b=3.8620;

if x(2)-b>=0;

%by calculations it is known that if Hassan pumps are commissioned the product will be delivered at a pressure of 3.8620 kg/cm2 at Bangalore

Pr=x(2);

%operating cost is calculated by considering Bangalore outlet pressure

PPr=1.2*x(2);

%Mostly pumps run at 80 percent of its maximum capacity

Psig1=Pr*14.2233;

%kg/cm2 to psig conversion

Head1=2.31*Psig1/S;

%head necessary for operation in (ft)

Q=16924.45;

%flow rate in (lb/min)

HP1=(Q*Head1)/33000;

%Horse power required while operation

Psig=PPr*14.2233;

Head=2.31*Psig/S;

HP=(Q*Head)/33000;

%Horse Power required for structuring pump

PSC=1500*HP*60;

%Pump station cost by considering a multiplication factor

BHP=HP1/(.80*.90*.95);

%Brake Horse of Bangalore pumps power by considering pump efficiency ,motor or driver efficiency ,mechanical efficiency

OpC=BHP*24*350*0.08*0.746*60*30;

%Operating cost of Bangalore pumps by considering 30 years of operation

TC=PC+PSC+OpC+LFC;

%total cost by considering pipeline cost, pumps station cost, operation cost of Bangalore pumps else

Pr=x(2)-5.0438+26.9421;

%Outlet pressure from Bangalore if Hassan pumps are commissioned by considering pressure loss/gain due to elevation and friction

Psig1=Pr*14.2233; Head1=2.31*Psig1/S; Q=16924.45;

HP1=(Q*Head1)/33000;

BHP=HP1/(.80*.90*.95); OpC=BHP*24*350*0.08*0.746*60*30;

%Operating cost of Hassan Pumps by considering 30 years of operation

TC=PC+OpC+LFC;

disp('Total Cost') disp(TC) %total cost by considering pipe cost & operation cost Of Hassan pumps

end }

function [gineq geq]=progconstraint(x) D=x(1);

P1=x(2);

P2=x(3);

P3=x(4);

P4=x(5);

P5=x(6);

P6=x(7);

P7=x(8);

L=[11.298,21.394,25.0410,25.743,20.717,24.203];

e=.045;

%internal roughness in nm.

Qo=640;

%Maximum flow rate

T=6.35;

DI=D-(2*T);

V=4.7290;

S=0.8359;

F=15;

H=[15,-10,-91,-301,-36,-25];

Re=(353678*Qo)/(V*DI);

F=-4*log10((e/(3.7*DI)+1.255*(F/Re)));

Pkm=24.99*(10^10*((Qo/F^2)*(S/DI^5)));

for i=1:6;

Pf(1,i)=Pkm*L(i)*0.0102;

Pe(1,i)=(H(i)*S/.102)*0.0102;

end

geq(1)=P2-(P1-Pe(1,1)-Pf(1,1));

geq(2)=P3-(P2-Pe(1,2)-Pf(1,2));

geq(3)=P4-(P3-Pe(1,3)-Pf(1,3));

geq(4)=P5-(P4-Pe(1,4)-Pf(1,4));

geq(5)=P6-(P5-Pe(1,5)-Pf(1,5));

geq(6)=P7-(P6-Pe(1,6)-Pf(1,6));

%equality constraints

gineq(1)=3-P1;

gineq(2)=3-P2;

gineq(3)=3-P3;

gineq(4)=3-P4;

gineq(5)=3-P5;

gineq(6)=3-P6;

gineq(7)=3-P7;

%In equality constraints .Pressure at any points along the pipeline should be more than 3kg/cm2 to avoid loss due to interface mix end

[Alan Weiss]

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Alan Weiss

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