Create a for cycle over an array

In my case I have X different job with Y different duration

D = [2 3 4 4 5 6 7 3 ];

so job X=1 will last Y=2.

I need to write a code that give me the following result:

tin1= 0 , tfin1=2, tin2= 2 , tfin2=2+3=5, tin3= 5 , tfin3=5+4=9, tin4=9 , tfin4=9+4=13........ So "tinx" means the inital time of job x, while "tfinx" means the final time of job x. if job one finish at 2, the job two will finish at 2 plus the duration of the job 2 (3) obtaining 5, and so on.

I want to write a code that is able to generate all this value

Does someone help me? the length of D should be variable so I would like to obtain a code that is independent from its length

[Jan]

I've cleaned the tags. It is not useful to use a pile of tags, which have no relation to your problem.

[Steven Lord]

I'm trimming more of the tags.

Answer 1

[Jan]

Do not create a bunch of variables with an index hidden in the names: See TUTORIAL: Why and how to avoid Eval

I assume all you need is:

D = [2 3 4 4 5 6 7 3 ]
T = cumsum([0, D])

Now e.g. T(1) and T(2) contain everything you want. This is much better than tin(1) and tfin(1). If you really need 2 variables:

Tini = T(1:end-1);
Tfin = T(2:end);

[luca]

I would like to obtain this values with a for loop, instead of CUMSUM

k = size(M,2)
for k=1:k
    
    tin1(1)=0 
    tfin1(1)=M(1)
    ....
    tink(k)=tfink-1(k-1)  % of the last element
    tfin(k)=M(k)
    
end

but I don't know how to solve it.

[Fabio Freschi]

why?

[Jan]

Len = size(D, 2);
s   = 0;
for k = 1:Len
    tini(k) = s;
    tfin(k) = D(k) + s;
    s       = tfin(k);
end

Again: Do not hide indices in the names of the variables, but use indices for this job.

Or:

Len  = numel(D);
tini = zeros(1, Len);  % Pre-allocation
tfin = zeros(1, Len);
s    = 0;
for k = 1:Len
    tini(k) = s;
    s       = s + D(k);
    tfin(k) = s;
end

[luca]

Cause this is a general case, then I have to implement the for loop that I've asked for in a more complicated case

[luca]

thanks !

Answer 2

[darova]

It is a job for cumsum() function:

D    =   [2 3 4  4  5  6  7 3 ];
tin1 = [0 2 5 9 13 18 24 31];
tin2 =   [2 5 9 13 18 24 31 34];]

Answer 3

[Fabio Freschi]

Try this

>> tfin = cumsum(D)

tfin =

     2     5     9    13    18    24    31    34

>> tin = [0 tfin(1:end-1)]

tin =

     0     2     5     9    13    18    24    31