# use previous element in 5 dimensional matrix for equation in current cell

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Andrew Alkiviades on 27 Jul 2012
Hi I have a 5 dimensional matrix as shown in the code below. What I am trying to do is find the "battery_capacity" shown in the "if function" what uses amongst other variables, the "battery_capacity" value of the previous cell in the loop (i.e h-1). I know the "battery_capacity" at the first cell which is "battery_capacity(:,:,:,1,1).
The relevant part of the code is below and I get the error "Subscript indices must either be real positive integers or logicals."
I believe its due to the (h-1) format which I can't really find a solution to. Also I need to be able to separate the days and thats why I have the hours and days as two separate dimensions.
% number_of_days = 3;
number_of_hours = 24*number_of_days;
number_panels = 1:5;
for idx_number_panels = 1:length(number_panels) % range of PV panel units examined
for number_turbines = 0:1 % range of wind turbine units examined
for number_batteries = 1:2 % range of battery units examined
for h=1:24 %# hours
battery_capacity(:,:,1,1) = max_battery_capacity*number_batteries;
for d = 1:number_of_days %# which day
n = h + 24*(d-1);
% hourly_deficit_1(...,..., h, d)= Demand(n)-(PV_supply(n)... %
battery_capacity(idx_number_panels , number_turbines + 1,number_batteries, h,d) = battery_capacity(idx_number_panels , number_turbines + 1,number_batteries, h-1,d)+...
(hourly_total_RES(idx_number_panels, number_turbines + 1,number_batteries, h,d)-hourly_annual_demand(n))*charging_efficiency;code
end
Thank you

Kevin Claytor on 27 Jul 2012
"The relevant part of the code is below and I get the error "Subscript indices must either be real positive integers or logicals.""
Yes, you have h ranging from 1:24, and call h-1 (ranges from 0->23) Matlab matrix indices start at 1 (not 0 as in C), so trying to access an element indexed by 0 is not allowed. To avoid this, a quick solution is to start at 2 instead of 1; (Example loops through the values of a vector (1D matrix if you will) x, and adds the previous value to the current one.)
x = [45, 34, 14, 65, 12]
for ii = 2:5
fprintf('ii = %d \t x(ii) + x(ii - 1) = %d', ii, x(ii)-x(ii-1))
end