minimising function of multivariable constraints Matlab - matlab

Question

How to minimise this function please teach me the code in MATLAB
Minimize Ra = (0.237 − 0.00175v + 8.693f − 0.00159z)
subjected to 124.53 ≤ v ≤ 167.03
0.025 ≤ f ≤ 0.083
6.2 ≤ z ≤ 14.8

Answer 1

You are looking at a (very degenerate) linear program.
Your unknown vector x = [v; f; z], the objective coeff vector is c = [-0.00175; 8.693; -0.00159].
In your particular case, you have no equality or inequality constraints, only lower and upper bounds lb = [124.53; 0.025; 6.2] and ub = [167.03; 0.083; 14.8] respectively.
You are trying to minimize
argmin c^T x
s.t. lb <= x <= ub
Use linprog
x = linprog( c, [], [], [], [], lb, ub );
Note that the constant coefficient 0.237 is not participating in the optimization since it has no effect on the argmin.

Answer 2

You could use fmincon. Let's say v = x(1), f = x(2), and z = x(3) so that tehy are all in a vector. First you define a handle to your function as:
h = #(x)(0.237 - − 0.00175*x(1) + 8.693*x(2) − 0.00159*x(3))
You need to define your constraints in the form A*x ≤ b. In your example, we rewrite the constraints as both being less than some function:
v ≤ 167.03
-v ≤ -124.53
f ≤ 0.083
-f ≤ -0.025
z ≤ 14.8
-z ≤ -6.2
Now you can convert the above into matrix form A*x ≤ b:
A = [1, 0, 0;
-1, 0, 0;
0, 1, 0;
0, -1, 0;
0, 0, 1;
0, 0, -1]
b = [167.03; -124.53; 0.083; -0.025; 14.8; -6.2]
Finally, you define an initial guess x0 for your variable x, and optimize by calling fmincon as:
xOpt = fmincon(h, x0, A, b)
The initial guess can be any values that lie within your constraints, so x0 = [167; 0.08; 14] could work.
EDIT: You could also define your lower and upper bound vectors as lb = [124.53; 0.025; 6.2] and ub = [167.03; 0.083; 14.8] and call fmincon using:
xOpt = fmincon(h, x0, [], [], [], [], lb, ub)
Note: if you want to randomize your initial guess, you can use your lb and ub vectors to do so with x0 = lb + (ub - lb)*rand().
This is a bit more powerful of a tool than you need for this problem, but hey, teach a man to fish.

Answer 3

Well, it's very simple:
>> Min_Ra = 0.237 − 0.00175*167.03 + 8.693*0.025 − 0.00159*14.8
ans =
0.1384905
:)
(just subtract the maximum permissible amounts while adding the minimum permissible amounts)