How could I code for loop with this values?

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Mohmmad Abu Yousuf
Mohmmad Abu Yousuf am 21 Feb. 2020
Hi guys! I need a simple help from you.
I want to create a for loop with nested if-else condition. Afterthat I want to plot them. Could you guys help me please! I have written the pseudocode below. I need the coding based on this. Please help me out.
Thank you in advance.
%'A' consists some values (e.g. 40, 60, 30, 80, 70, 50, 90, 30);
for A % it will take first 2 values
% 'B' is the maximum value consisting of 2 values from 'A'
% e.g. max (40,60) = 60 then this 60 will be first value of A and last
% value will be 30 so max(60,30) is 60. Afterthat first value is 60 and 2nd value is 80
% so, max(60,80) is 80 likewise.
B = max (A);
% then 'C' is the last value from 'A' which is taken to get the 'B', will be assigned to 'C'.
C = 60; % for 2nd itaration C = 30, for 3rd C = 80, for 4th C = 70, for 5th C = 50 and so on..
% If C reduces 30% of B then,
x = C/10
fprintf ('This is your value %d',x );
else
fprintf ('This is not your value %d',x );
end
end
plot ('x' vs simulation time)
  3 Kommentare
madhan ravi
madhan ravi am 21 Feb. 2020
Illustration of expected result would help.
Mohmmad Abu Yousuf
Mohmmad Abu Yousuf am 22 Feb. 2020
Thank you,
Result would be like this. Could you please help me Sir.
for 1st iteration:
A is 40 and 60,
so, B is 60,
C is 60 then
if C reduces 30% of B
x = C/10
else
x = C/5
end
for 2nd iteration:
A is 60 and 30,
so, B is 60,
C is 30 then
if C reduces 30% of B
x = C/10
else
x = C/5
end
for 3rd iteration:
A is 60 and 80,
so, B is 80,
C is 80 then
if C reduces 30% of B
x = C/10
else
x = C/5
end
for 4th iteration:
A is 80 and 70,
so, B is 80,
C is 70 then
if C reduces 30% of B
x = C/10
else
x = C/5
end
this will go on till the last value of A

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Akzeptierte Antwort

darova
darova am 22 Feb. 2020
Bearbeitet: darova am 23 Feb. 2020
This script should give you success you are looking for
A = [40, 60, 30, 80, 70, 50, 90, 30];
B = A(1);
for i = 2:length(A)
B = max(B,A(i));
C = A(i);
if C/B < 0.7
x = C/10;
else
x = C/5;
end
end
  3 Kommentare
darova
darova am 23 Feb. 2020
Oh, understand. My bad. I corrected the code. See now
Mohmmad Abu Yousuf
Mohmmad Abu Yousuf am 23 Feb. 2020
thank you again Sir. :D

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