Question

Hi I am working with the brute force method to examine possible combinations of “panels” and “turbines”

My code is

 for number_panels = 0:5

for number_turbines = 0:10


    for n = 1:24 % number of hours per day



deficit(n) = Demand(n) - (PV_supply(n)*number_panels) -...
 (WT_supply(n)*number_turbines);% hourly power deficit

        if deficit(n)<0

            deficit(n) = 0;

        end

The problem I have above is that I haven’t yet figured the correct indexing of this code.

What I am trying to do is find the “deficit” for the “number_panels” , “number_turbines” and “n”. As it stands I can only find the “deficit” for the last for loop.
How can I code so that I can have the option to access the nth row (or sets of “n” i.e 1-24) and also for the “number_panels” “number_turbines” option?

---

thanks – in order to find the sum of each deficit(n) value and thus have the respective total deficit of the 24 hour period I have done the following which seems to me to be able to do what I am asking but I am getting incorrect answers:

 daily_deficit(number_panels + 1, number_turbines + 1) =...
 sum(deficit(number_panels + 1, number_turbines + 1,n)) –

Answer 1

function calcDeficit.m:

File Edit Options Buffers Tools ObjC Help                                                                              
function deficit = calcDeficit (Demand, PV_supply, WT_supply)

  % initialize the size (good practice)
  deficit = zeros(6,11,24);


  for number_panels = 0:5

    for number_turbines = 0:10

      for n = 1:24 % number of hours per day
        deficit(number_panels+1,number_turbines+1,:) = Demand(n) - (PV_supply(n)*number_panels) -...
          (WT_supply(n)*number_turbines);% hourly power deficit

        if deficit(n)<0 
          deficit(n) = 0;
        end
      end
    end
  end
return

example call:

Demand=randn(24,1); PV_supply=randn(24,1); WT_supply=randn(24,1); test(Demand,PV_supply,WT_supply)

You access Demand by

Demand(panels+1,number_turbines+1,hour)