Why does the the f vector in a linear equation always consist of negative values?

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Kevin Juan 2020-9-12

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评论： Kevin Juan 2020-9-12

A simultaneous linear equation usually will consist of two types of equation(I will be using a fundraiser event as an example), which are the parameters(budget, deadlines, location) and a value to be maximized(i.e. the money raised). Normally we would solve these types of equations in MATLAB using the command simlp. From what I understand, simlp requires three different vectors to use. A and b, which represents the parameters for the problem, and f, represents the function of the maximized value. Now my question is, why are the values in vector f always negative, while some values of vector A and b still can remain positive? Attached below is a sample question.

We let w, x, y, and z, denote the number of residences canvassed in the four cities Gotham, Metropolis, Oz, and River City, respectively.

Then the linear inequalities specified by the given data are as follows:

Nonnegative data: w ≥ 0, x ≥ 0, y ≥ 0, z ≥ 0;

Pamphlets: w + x + y + z ≤ 50,000; T

ravel cost: 0.5w + 0.5x + y + 2z ≤ 40,000;

Time available: 2w + 3x + y + 4z ≤ 18,000;

Preferences: w ≤ x, x + y ≤ z; 282

Contributions: w + 0.25x + 0.5y + 3z ≥ 10,000.

The quantity to be maximized is: Voter support: 0.6w + 0.6x + 0.5y + 0.3z.

The vector form is as follows:

f = [-0.6 -0.6 -0.5 -0.3];
A = [1 1 1 1; 0.5 0.5 1 2; 2 3 1 4; 1 -1 0 0; 0 1 1 -1; -1...
    -0.25 -0.5 -3; -1 0 0 0; 0 -1 0 0; 0 0 -1 0; 0 0 0 -1];
b = [50000; 40000; 18000; 0; 0; -10000; 0; 0; 0; 0];

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Answer 1

Matt J 2020-9-12

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编辑：Matt J 2020-9-12

在 MATLAB Online 中打开

Well, I've never heard of simlp, but many solvers, such as linprog() will only solve minimization problems, so if you have an objective function that you want to maximize, you need to multiply f by -1 in order to recast the maximization as a minimization.

Note however, that if you have R2017 or higher, you can use the problem-based optimization framwork, which will let you specify explicitly that the objective is to be maximized. Your example problem above could therefore be implemented without a sign flip as follows:

  w=optimvar('w',1,'LowerBound',0);
  x=optimvar('x',1,'LowerBound',0);
  y=optimvar('y',1,'LowerBound',0);
  z=optimvar('z',1,'LowerBound',0);
  
  prob=optimproblem('ObjectiveSense','maximize', 'Objective', 0.6*w+0.6*x+0.5*y*0.3*z);
  prob.Constraints.Pamphlets= w + x + y + z <=50,000;
  prob.Constraints.TravelCost= 0.5*w + 0.5*x + y + 2*z<= 40,000; 
  prob.Constraints.TimeAvailable= 2*w + 3*x + y + 4*z <= 18,000; 
  prob.Constraints.Pref1= w <= x;
  prob.Constraints.Pref2= x + y <=z;
  prob.Constraints.Contributions= w + 0.25*x + 0.5*y + 3*z >= 10,000; 
  sol=solve(prob);  

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Kevin Juan 2020-9-12

"you need to multiply f by -1 in order to recast the maximization as a minimization." Exactly what I need. Just noticed that the parametric functions are searching for maximum values, while the objective function f searches for the minimum. Thanks!

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