Investment Decision on the Number of Producing Units

Problems with modeling
Posts: 3
Joined: 2 months ago

Investment Decision on the Number of Producing Units

Postby unyelf » 1 month ago

I am trying to satisfy a demand by different production units which have different capacity constraints, different efficiencies, and different initial purchasing prices. I want to obtain the minimum total cost of the system in a year, and also the number of the most optimal units that I should purchase.

My code as below which gives integer infeasible error:

Code: Select all

Option optcr=0.001
i pumps                 /i1*i8/

pump(i)               /i1*i4/
vscond_recip(i)        /i1/
vscond_screw(i)        /i2/
fx_recip(i)           /i3/
fx_screw(i)             /i4/

air(i)                    /i5*i8/
dvscond_recip(i)        /i5/
dvscond_screw(i)        /i6/
dfx_recip(i)            /i7/
dfx_screw(i)            /i8/

t time                     /t1*t8760/


pmin(i)   minimum generation level (kW)            /i1*i4 500, i5*i8  300/
pmax(i)   maximum generation level (kW)            /i1*i4 1500, i5*i8 1000/
c(i)      investment costs             ($)         /i1 15.000, i2 10.000, i3 13.000, i4 15.000, i5 15.000, i6 15.000, i7 12.000, i8 12.000/
u_init(i) initial online status                    /i1*i8 0/

*demand is for hourly for a year, 8760 inputs.
d(t) demand /
$include Demand.csv

parameter hour(t)  "hour of the day from 1 to 24";
parameter p_el(t)  "electricity price";
hour(t) = mod(ord(t), 24);
p_el(t) =0.03 ;
p_el(t)$(hour(t) >= 6 and hour(t) <= 18) = 0.08;
p_el(t)$(hour(t) > 18 and hour(t) < 24)  = 0.05 ;

*COP is for hourly for a year, 8760 inputs.
table COP(t,i)

$include COP.csv

display d,COP;


plantlife   life time of the plant                /20/
irate       interest rate                         /0.05/  ;

u(i,t) online status
v(i)   investment     "number of investments for the respective unit(s)"
p(i,t) production level

z total costs

free variable

Positive variable
binary variables
u ;


costs  total costs
Mingen(i,t) minimum generation
Maxgen(i,t) maximum generation
Demand(t)   demand
investment(i,t) investment

Mingen(i,t) .. p(i,t)=g=pmin(i)*u(i,t);
Maxgen(i,t) .. p(i,t)=l=pmax(i)*u(i,t);

costs.. z=e=sum( (i,t), p_el(t)*(d(t)/COP(t,i)) )+sum(i,v(i)*(c(i)/(1-((1+irate)**(plantlife/irate)))));

Demand(t).. d(t)=e=sum(i,p(i,t));

Model UC /all/
Solve UC using mip minimizing z ;

display z.l;

There are two files attached for you to execute the program and hopefully solve my problem:)
(115.63 KiB) Downloaded 4 times
(872.48 KiB) Downloaded 5 times

Return to “Modeling”

Who is online

Users browsing this forum: No registered users and 1 guest