Hello Tim,
Yes. batt is the is the State of Charge of the battery.
P(n,2) (kN) is the power taken rom battery, engine or both. I will
calculate the battery in As, so P(n,2) is only an in alternate value -
but thats fare to complicated for the example code.
"Proving that it's
correct will be
Depending on how often you switch between battery and generator, and
how icky-picky you're willing to be, there may be a way to reduce
computation.
It looks like the term P(n,2) * (P(n+1,1) - P(n,1)) is always there,
and you're either adding it to 'gen' (is it energy production?) or
subtracting it
Thanks for all the answers.
I feared that there is no way around a loop. During the process batt
(Battery) is charged and discharged. In my example, it is only
discharged. I will code the entire problem with a loop, maybe somebody
knows something to speed up the process with the full problem.
Le 10/05/2017 à 17:25, Frieder Nikolaisen a écrit :
There was a piece of code missing:
P = [
1, 0;
2, 50;
5, 110;
10, 80;
11, 200
15, 0];
batt = 1000;
gen = 0;
n = 1
for n=1:5
if P(n,2) > 100 then
if batt > 800 then batt
What it's the equation you need to compute ?Perhaps I can help.
I think it's possible to compute with something in this way:
map = find (P(:,2) > 100 );
if batt > 800 then
batt = batt - P(map,2) * (P(map+1,1) - P(map,1));
end
On Wed, 2017-05-10 at 17:23 +0200, Frieder Nikolaisen wrote:
> Hello
I had a cool solution for you, but then I saw that you're implementing
a difference equation, with current values of batt (for instance)
depending on prior values.
I think that no, you can't eliminate the loop. If speed is a concern
and if you're using the code heavily, you can put it in a C func
There was a piece of code missing:
P = [
1, 0;
2, 50;
5,
110;
10, 80;
11, 200
15, 0];
batt = 1000;
gen = 0;
n = 1
for n=1:5
if P(n,2) > 100 then
if batt > 800 then batt = batt - P(n,2) *
(P(n+1,1) - P(n,1))
else
gen = gen + P(n,2) * (P(n+1,1) - P(n,1))
end
else
batt = batt - P(n,2
Hello,
I did write an example code, but I do not like the time
consuming way I solved the problem. With 50 000 lines in the matrix, it
wouldn't be fun.
How can I avoid using the for-loop?
10, 80;
11,
200
15, 0];
batt = 1000;
gen = 0;
n = 1
for n=1:5
if P(n,2) > 100
then
if batt > 800
