how to find the minimum value based on two independent value

Hi, guys, i have a problem and dont know how to solve. i have two independent variale:x1, x2, y is dependent output value.i want to find when y will be the minimum value. I heard about surface response method is useful to solve this problem, but it is a little complicated. could you give me an advice to solve problem using matlab, thank you! here list the results

x1 x2 y

0.15 1.0 0.863

0.15 1.5 0.050

0.15 2.0 1.267

0.15 2.5 3.082

0.20 1.0 0.635

0.20 1.5 0.010

0.20 2.0 0.690

0.20 2.5 1.900

0.25 1.0 0.629

0.25 1.5 0.009

0.25 2.0 0.617

0.25 2.5 1.522

0.30 1.0 0.701

0.30 1.5 0.016

0.30 2.0 0.593

0.30 2.5 1.452

[Andrew Newell]

Are you wanting to find where y is a minimum in a table like the one above, or are you trying to minimize a function?

[suby]

thank you for your comments, i am trying to find where y is a minimum in a table.it means i can find x1 and x2 to make y be a minimum

[Andrew Newell]

First, do you know how to assign the values in the table to x1, x2 and y?

[suby]

you mean the range of both x1 and x2? Actually, there is a basement that x1=0.02 and x2=0.15,then y=0.010. Based on that, in order to find a better result, I changed both two inputs x1 and x2 a little bit. so I did not confine the range of both inputs, I hope i make me clear, thank you!

[suby]

in addition to that,I am clear that the minimum value for y is close to x1=0.02 and x2=0.15

[Andrew Newell]

I think you're having trouble articulating what you really want. You talk about finding the minimum in a table, which is pretty trivial in this case because you can just eyeball the third column. The smallest value is 0.009, for which x1 = 0.25 and x2 = 1.5. However, you're talking about changing inputs and the minimum being "close to" x1=0.2 and x2=1.5. That implies there is a function that generates y. That is quite a different problem, involving optimization.

Answer 1

[John D'Errico]

Ok. So since you now tell us that you have ONLY a table of data, with no function.

Then you need to recognize that without some way to predict a value between the tabular points, there is no way to find a minimum value. You need some scheme to interpolate, or some approximate function that fits the data. The simple approach is to interpolate, here using interp2 as the logical choice.

The problems with interp2 in this approach are:

 1. You will need to use the 'spline' method, as a lower order scheme will be a bit boring.
 2. Interp2 does not extrapolate well, and extrapolation of a spline is just silly. So you will want to use a bounded solver.

We can see how this works on your data.

M = [0.15 1.0 0.863;
0.15 1.5 0.050;
0.15 2.0 1.267;
0.15 2.5 3.082;
0.20 1.0 0.635;
0.20 1.5 0.010;
0.20 2.0 0.690;
0.20 2.5 1.900;
0.25 1.0 0.629;
0.25 1.5 0.009;
0.25 2.0 0.617;
0.25 2.5 1.522;
0.30 1.0 0.701;
0.30 1.5 0.016;
0.30 2.0 0.593;
0.30 2.5 1.452];

Simple observation shows a minimum near (x,y) = (0.25,1.5), but we can find that programmatically easily enough using the min function.

x = M(:,1);
y = M(:,2);
z = M(:,3);

[minz,ind] = min(z)
minz =
        0.009
ind =
    10

M(10,:)
ans =
         0.25          1.5        0.009

Now lets define a function that can be sued to interpolate this tabular set of data.

xi = unique(x);
yi = unique(y);
fun = @(xy) interp2(xi,yi,reshape(z,4,4),xy(1),xy(2),'spline');

So for example...

fun([.25 ,1.5])
ans =
        0.009

fun([.2,1.75])
ans =
        0.23531

It is always a good idea to plot EVERYTHING.

contour(xi,yi,reshape(z,4,4),[.1 .2 .3 .4 .5 .75 1 2 3])

Now lets use a solver on fun.

lb = [.15 ,1];
ub = [.3, 2.5];

[xymin,zmin] = fminsearchbnd(fun,[.25,1.5],lb,ub)

xymin =
      0.22732       1.4453

zmin =
    -0.001039

See that I used my own fminsearchbnd, found on the file exchange. It allows me to add bounds to the search, so I need not worry about the solver going outside of the bounds.

I could also have used a model of this data. A polynomial model should be sufficient, and nothing of too high an order, as the function is a simple one.

p = polyfitn([x,y],z,3)
p = 
      ModelTerms: [10x2 double]
    Coefficients: [-375.33 58.84 200.68 -7.124 -8.072 -43.525 -1.3727 10.692 -18.922 14.639]
    ParameterVar: [87301 314.28 40837 3.1428 102.77 2120.7 0.087301 2.5751 9.3821 14.673]
    ParameterStd: [295.47 17.728 202.08 1.7728 10.138 46.052 0.29547 1.6047 3.063 3.8305]
             DoF: 6
               p: [0.25101 0.016023 0.35902 0.0069707 0.45623 0.38108 0.0035193 0.00055274 0.00082733 0.0087443]
              R2: 0.99415
      AdjustedR2: 0.98537
            RMSE: 0.060688
        VarNames: {'X1'  'X2'}

psym = vpa(polyn2sym(p),16)
psym =
- 375.3333333333113*X1^3 + 58.83999999999785*X1^2*X2 + 200.6799999999904*X1^2 - 7.123999999999711*X1*X2^2 - 8.072*X1*X2 - 43.52466666666535*X1 - 1.372666666666667*X2^3 + 10.6924*X2^2 - 18.92153333333317*X2 + 14.63875

pgrad = gradient(psym);
groots = solve(pgrad);

groots.X1
ans =
                                        0.29927874279651926226688855470807
                                        0.20749483993099085441217447597139
 0.32278193339635663611070377153234 - 0.040069824732610684465833401918639i
 0.32278193339635663611070377153234 + 0.040069824732610684465833401918639i

groots.X2
ans =
                                        1.431025448839533109541965980622
                                       1.4500035963111515399851171239296
 2.660439130764919498188972022711 + 0.062485973176614710041936383492915i
 2.660439130764919498188972022711 - 0.062485973176614710041936383492915i

Which of those roots is the minimum?

syms X1 X2
subs(psym,{X1,X2},{groots.X1,groots.X2})
ans =
                                      0.041173813812552132138095037065494
                                      -0.10874928276576993381362733327689
 1.4706001978222628873166798300997 + 0.061660508055862572262913258951546i
 1.4706001978222628873166798300997 - 0.061660508055862572262913258951546i

So the model solution lives at:

[groots.X1(2),groots.X2(2)]
ans =
[ 0.20749483993099085441217447597139, 1.4500035963111515399851171239296]

Don't forget to plot the modeled surface...

ezsurf(psym,[.15,.3],[1,2.5])

[suby]

Thank you very much for your solution! I think it works!

Answer 2

[Andrew Newell]

I'm going to assume you're really trying to minimize a function of two variables and show how you can explore it to get some insight. First, I'll read your table in as a matrix:

M = [0.15 1.0 0.863;
0.15 1.5 0.050;
0.15 2.0 1.267;
0.15 2.5 3.082;
0.20 1.0 0.635;
0.20 1.5 0.010;
0.20 2.0 0.690;
0.20 2.5 1.900;
0.25 1.0 0.629;
0.25 1.5 0.009;
0.25 2.0 0.617;
0.25 2.5 1.522;
0.30 1.0 0.701;
0.30 1.5 0.016;
0.30 2.0 0.593;
0.30 2.5 1.452];

Now, a contour plot is a good idea, but that requires rearranging the points:

x1 = 0.15:.05:.3;
x2 = 1:.5:2.5;
[x1,x2] = meshgrid(x1,x2);
y = reshape(M(:,3),size(x1));

You'll find that [x1(:) x2(:) y(:)] gives you back your table. Now we'll plot the data:

contourf(x1,x2,y);
xlabel('x1')
ylabel('x2')
colorbar

This appears to be a fairly smooth function with a minimum somewhere near where you have guessed it is. You should be able to find an accurate minimum using fminsearch, but I can't tell you the details because you haven't told me what your function is.

[suby]

sorry for late, thank you very much for your concern, actually there is no function. but this is an optimization.I want to optimize an configuration in terms of x1 and x2 to minimize y. all the data is from simulation results

Answer 3

[Alan Weiss]

As with the other comments you have gotten, I am going to assume that there is a function, say myfun(x1, x2), that gives you your y variables. You are looking for the minimum of y, and want to know how to find the x1 and x2 that give the minimum.

This is a job for fminsearch. The only issue is getting your function of two variables to be a function of one vector variable x = [x1,x2]. Do it like this.

 fun = @(x)myfun(x(1),x(2)); % x is a 2-component vector
 solution = fminsearch(fun,[0.2,1.5]) % Put in a good start guess

For more information, see the documentation.

Alan Weiss

MATLAB mathematical toolbox documentation

[suby]

thank you! Actually there is no function, I want to optimize an configuration in terms of x1 and x2 to minimize y. all the data is from simulation results

[Image Analyst]

Explain why you want an interpolated value that does not exist in your table. What are you going to do with that information? Why is the value in the table not sufficient for what you want to do? In other words, describe your "use case".