Code to calculate Bi-Hourly energy tariff

Hi!

I'm working on a project on energy production and consumption. I have a table (called "d") that includes a row with the datetime information of a year in 15min intervals (called "date", 35040 lines), and other with the electric consumption of that year. And I need to calculate the cost of the electricity according to the bi-hourly energy tariff.

In this code I'm just trying to create a vector (pe) that has the correct cost of electricity (€/kWh) in that specific moment, throughout the year, to later multiply by the consumption.

It doesn't work and takes over 10min to run. I'm out of ideas, I would appreciate if someone could help me :)

Bi-hourly tariff:

night time: 0.0969 €/kWh (00:00 - 08:00 / 22:00 - 24:00)

day time: 0.2028 €/kWh (08:00 - 22:00)

 peFV = 0.2028;
 peV = 0.0969;

 pe = zeros(35040,1);

 for i=1:length(pe)
    %particular case for the days in which the hour changes

    if day(d.date(i)) == 25 && month(d.date(i)) == 3    %25 march
        for k=1:92
            if k>=29 && k<=84
                pe(i)=0.2028;

            else
                pe(i)=0.0969;
            end
            i=i+1;
        end
        i=i-1;

    elseif day(d.date(i)) == 28 && month(d.date(i)) == 10     %28 October
        for k=1:100
            if k>=37 && k<=92
                pe(i)=0.2028;

            else
                pe(i)=0.0969;
            end
            i=i+1;
        end
        i=i-1;

    else     %all other days

        for k=1:96
            if k>=33 && k<=88
                pe(i)=peFV;

            else
                pe(i)=peV;
            end
            i=i+1;
        end
        i=1-1;

    end
end

1 day has 96 lines, so i'm using the line number to distinguish between hours.

Sorry for the rudimentary code, I'm a beginner.

Answer 1

[Akira Agata]

I think you can do this type of task much easier by using timetable and retime function. The following is an example.

% Sample data
rng('default'); % for reproducibility
date = (datetime(2017,1,1,0,0,0):minutes(15):datetime(2017,12,31,23,45,0))';
consumption = 10*rand(size(date));
d = timetable(date,consumption);

% Bi-hour tarrif ( pe(1): day time / pe(2): night time)
pe = [0.2028; 0.0969];

% Aggregate electric consumption [kWh] hourly 
d2 = retime(d,'hourly',@(x) sum(x)*0.25);

% Assign tarrif type (1: day time / 2: night time)
d2.group = zeros(height(d2),1);
idx = d2.date.Hour >= 8 & d2.date.Hour < 22;
d2.group(idx) = 1;
d2.group(~idx) = 2;

% Calculate hourly cost [euro]
d2.hourlyCost = d2.consumption .* pe(d2.group);

Then, the total cost in euro for this year can be obtained by:

>> sum(d2.hourlyCost)
ans =
 6.9600e+03

[Ammar Saif]

pe = [0.2028; 0.0969; 0.888; 0.6545; 0.877; 0.7646];

how can we expand the pe to 6 elements?

because " index exceeds matrix dimensions" appears as the code runs

[Akira Agata]

OK. Then, the program needs to be slightly modified, like:

% Sample data
rng('default'); % for reproducibility
date = (datetime(2017,1,1,0,0,0):minutes(15):datetime(2017,12,31,23,45,0))';
consumption = 10*rand(size(date));
d = timetable(date,consumption);

% Hour tarrif (pe(k) : 4*(k-1) o'clock to 4*k o'clock)
pe = [0.2028; 0.0969; 0.888; 0.6545; 0.877; 0.7646]; % <- modified 

% Aggregate electric consumption [kWh] hourly 
d2 = retime(d,'hourly',@(x) sum(x)*0.25);

% Assign tarrif type (group = k for 4*(k-1) o'clock to 4*k o'clock)
d2.group = floor(d2.date.Hour/4)+1; % <- modified 

% Calculate hourly cost [euro]
d2.hourlyCost = d2.consumption .* pe(d2.group);