Ok Thank you I got it working after it caused errors. The source of the error was the string nature of the imported names. ToExpression seems to have fixed it.

Even then it wasn't working on the real problem because the names had special characters like "-" which treated it as a negative sign, and ' sign which just threw some error.

I eventually found out by using your code and copying and pasting the names, which pasted as string, (bingo), then looking over the list and finding things being out of order due to the hyphens and apostrophes (double bingo)

When I fixed these issues code ran no problems.

if I can I ask you guys a couple of more questions,

How would I select the cases where the mathematica has put 1 next to other than to scroll through a long list.

I've tried

max= NMaximize[{ans.as + bns.bs, constraints}, vars]
Select[max[[2]], #-> 1&]

Apparently that's not the correct syntax.

The goal is to make a list of the money spent (condition in the constraint ans.as+bns.bs <= 400), players chosen, points added up which is given by NMaximize in a long list.

Also from what I understand, you can seed NMaximize randomly as to change the outcome? Or does NMaximize always converge to the global optimum?

If you would like to see your code in action using the fixed excel sheet and my nb I've attached them both.

Attachments:

So it becomes an integer linear program. I suspect the linear relaxation is always optimal since it looks like the type of problem for which that will hold, but I am not certain after a brief look. Anyway, can be done as below. Note that all I changed from my last version is the constraints.

as = {3, 1, 4, 5, 5, 8};
an = {100, 50, 100, 150, 50, 100};
ans = {a1, a2, a3, a4, a5, a6};
bs = {3, 5, 1, 6, 8, 5, 5, 10};
bn = {100, 200, 150, 50, 150, 200, 50, 150};
bns = {b1, b2, b3, b4, b5, b6, b7, b8};
vars = Join[ans, bns];
constraints = 
  Flatten[{Total[ans] == 1, Total[bns] == 2, ans.an + bns.bn <= 400, 
    Map[0 <= # <= 1 &, vars]}];
NMaximize[{ans.as + bns.bs, constraints}, vars]

(* Out[69]= {26., {a1 -> 0., a2 -> 0., a3 -> 0., a4 -> 0., a5 -> 0., 
  a6 -> 1., b1 -> 0., b2 -> 0., b3 -> 0., b4 -> 0., b5 -> 1., 
  b6 -> 0., b7 -> 0., b8 -> 1.}} *)

Ok Will watch where I post the questions to. I figured those groups would be relevant since it deals with converting data from excel into mathematica and using the data to maximize which a lot of those disciplines have experiences in. But I'll keep it to those groups you mentioned.

Oh yes, that is the better idea! I was wondering what bored dude might have had in mind with these ans and bns.

Hi bored dude,

in your example there is no variable according to which anything could be maximized. Instead you have a small set of items from which you can simply select the right one. I would do it like this:

a = Permutations[{0, 1, 0, 0, 0, 0}];
as = {3, 1, 4, 5, 5, 8};
an = {100, 50, 100, 150, 50, 100};

b = Permutations[{0, 1, 1, 0, 0, 0, 0, 0}];
bs = {3, 5, 1, 6, 8, 5, 5, 10};
bn = {100, 200, 150, 50, 150, 200, 50, 150};

(* all combinations from permutations a and b: *)
ab = Flatten[Outer[List, a, b, 1], 1];

(* members of 'ab' which fullfill the constrain: *)
constrOK = Select[ab, First[#].an + Last[#].bn <= 400 &];

(* finding the best combination simply by sorting: *)
{optA, optB} = Last@SortBy[constrOK, First[#].as + Last[#].bs &]
(* Out:  {{0,0,0,0,0,1},{0,0,0,0,1,0,0,1}} *)

Regards -- Henrik