Can someone help me in finding a way in arriving at the solution for the below problem.

I am trying to optimize the work flows in a manufacturing environment

Let A, B and C are continuous tasks, implies C is dependent of B and B on A.A,B and C combined is a single unit

let A could be finished in 2 or 3 or 4 days.denoted as an array below.

A=[1 2 3]
B=[4 5 6]
C=[7 8 9]

I want the complete list of POSSIBLE ways that I can take to finish the three tasks.

Output should look something like..

Column 1: Values of A matrix Column 2: Values of B matrix Column 3: Values of C matrix Column 4: Sum of the values in that particular row.

Possible scenarios

     1     4     7   12(sum of the values)
     1     4     8   13
     1     4     9   14
     1     5     7   13
     1     5     8   14
     1     5     9   15

Like we continue further the output will be

1 6 7 14

1 6 8 15

1 6 9 16

2 4 7 13

2 4 8 14

2 4 9 15

3 4 7 14

3 4 8 15

3 4 9 16

[Tien Tran]

it means that A is column 1, B is column 2 and C is column 3? So, you want to calculate B = a+ (...); and then C = B + ..., isn't it? I still don't understand your question

[Walter Roberson]

Where did the 3 come from in that table? Don't you need one of A, one of B, one of C?

[Suneel Kumar]

Yes, Tran. A is column 1, B col 2 and C col 3. I want the list of possible cases with the matrices A, B and C.

Like we continue further the output will be

1 6 7 14

1 6 8 15

1 6 9 16

2 4 7 13

2 4 8 14

2 4 9 15

3 4 7 14

3 4 8 15

3 4 9 16

[Suneel Kumar]

Sorry @Walter for the confusion. I corrected it.

Answer 1

[Guillaume]

You're asking for the cartesian product of A, B and C. There are entries for it in the file exchange, e.g. ALLCOMB, or you can do it easily with ndgrid:

A=[1 2 3];
B=[4 5 6];
C=[7 8 9];
[ca, cb, cc] = ndgrid(A, B, C);
allcombs = [ca(:), cb(:), cc(:)]

The sum is then trivial to obtain:

allcombswithsum = [allcombs, sum(allcombs, 2)]

[Suneel Kumar]

Thank you so much Guillaume. Appreciate your help!