Need help regarding optimization

Hi, here's the operation I want to do

1. the following operations:

A=rand(10,2)*100;

B=rand(10,2)*100;

c=[ones(10,1)*80.5 ones(10,1)*89.27];

result=A-B;

result(~(A-c>=0)) = 0

2. the following optimization (wich i don't know how to do exept with a looop but I want something faster and more straight forward)

I want to do the sum of the variable"results" and get the maximum.In this case the result would be in the form of: 1x2

So clearly I want to find the "c" that maximise "result2":

result2=sum(result)

(in the previous example the "c" were 80.5 and 89.27

So the point is when we do

result=A-B;

result(~(A-c>=0)) = 0

Some "A-B" yield a negative results so I want to find the c that

optimize positively. Let's set the "c" as c1 and c2

the final optization is going to be in the following dimesnion:

c=[ones(4,1)*c1 ones(4,1)*c2];

Thank you in advance

Gimpy

[Gimpy]

Let's take the following example (4x1, but I'am going to have something like 22000x18)

If IF A >= C then A-B

example: A=[45;37;32;50]

B=[17;100;200;10]

The alwser would be 45.00 or something between 45.0=<C>37

45.0 would be a good (I don't need an interval)

[Gimpy]

In this case the maximun of A-B is 68 if 45.0=<C>37

[Matt Fig]

Please take the time to give just one good example that captures all of the salient features of your real problem. If you have 22000-by-18, make a specific example with 5(or 6,7 or 8)-by-3. Show exactly (give the numeric arrays) the inputs and the exact expected output(s) for your specific example.

Thanks.

[Gimpy]

ok here's my example again:

A=[45;37;32;50]

B=[17;100;200;10]

c=??(the value we look for)

The condition to test:

A=[45;37;32;50]

B=[17;100;200;10]

c=? (the value we are looking for)

the condition to test:

A >= C then A-B, else A-B=0

in my example the C we look for is something between 37.01 and 50 because.

In excel for example to be more specific:

cells A2= 45,A3=37,A4=32,A5=50

B2=17, B3=100, B4=200, B5=10

C2:C5=37.01 (in this case i know the results by trial and error).

The condition to test is:

=IF(A2>=C2,(A2-B2),0)= 28
=IF(A3>=C3,(A3-B3),0)=0
=IF(A4>=C4,(A4-B4),0)=0
=IF(A5>=C5,(A5-B5),0)=40

Total: 68

Any value between 37.01 and 50 would give the save result. A way to find the lower and uper bound would be appreciate of only one...

Thanks again for your patience and help.

[Matt Fig]

Where did the values of 37.1 and 50 come from?

Inputs --- A=[45;37;32;50], B=[17;100;200;10]
Outputs --- C = [37.01 50]

Is this correct? If so, please show the MATLAB for loop you say you have that gets this result.

[Gimpy]

the value the values of 37.1 and 50 come from trial and error in Excel by testing:

=IF(A2>=C2,(A2-B2),0)= 28

=IF(A3>=C3,(A3-B3),0)=0

=IF(A4>=C4,(A4-B4),0)=0

=IF(A5>=C5,(A5-B5),0)=40

In this case it's the result we are looking for.

[Gimpy]

Here’s my problem with more explanation again:

Matrix A,B,C :22 500 x18

Matrix c : each colum are scalar

Exemple :[50;50;50;50 72.25;72.25;72.25;72.25]

… Here’s what I want to do :

For each elements in A, lets call them ai, if ai >ci then

ai - bi

else

ai - bi =0

Here’s what I do in matlab(for simplification I do a 4x1)

a1=[88;57;42;100];

a2=[98;87;32;80];

A=[a1 a2];

b1=[78;5;32;106];

b2=[88;87;35;88];

B=[b1 b2];

c=[ones(4,1)*80.5 ones(4,1)*89.27];

result=A-B;

result(~(A-c>=0)) = 0

result2=sum(result)

****Now here’s the point : I want to find the matrix c that’s going to maximise :

result2=sum(result).

The results should be an interval

In the revious example I’am expecting MATLAB to give something like:

resultat2= 68

C_lower= 37

C_upper=45

Or at least one one of the 2 results (37 or 45).

I found this result by calculating result2=sum(result) by trial and error until

I get the maximum value(result2=68).

Please keep I mind that I want to work with 22500x18