Still be confused in how to add a slack variable

1

Now i am trying to add a slack variable to this optimal problem: Why does the optimal value become NaN sometimes?

And here is my reference about adding slack variable section 6 in this website https://yalmip.github.io/debugginginfeasible/

And here is the slack code of the website,but there are somethings i don’t understand in it

slack1 = sdpvar(N,1);
slack2 = sdpvar(N,1);
Constraints = [slack1>=0]
for i = 1:N
 Constraints = [Constraints, something1 <= slack1(i)];
 Constraints = [Constraints, something2 == slack2(i)];
end

My questions about the example are
1.so if i have three constraints,i need to produce three slack variables?
2.do i need to build Constraints = [slack2>=0] also?
3.what does the "something " mean in his example code?value?vector?matrix? or a formula
4.How do i define the N ? because in one constraint,there are not only one vector,but also the other vector,some vectors may be 5 by 1,may be 4 by 1,so i don’t know what value should i assume to the N

Here is my code below,i don’t think this is right,unless i know the explanation of the question i ask ,the N i assume is 4 ,because i only have one kind of vector,4 by 1 vector

slack_for_C3 = sdpvar(4,1);
slack_for_C5  = sdpvar(4,1);
slack_for_C10  = sdpvar(4,1);
Constraints = [ slack_for_C3 >=0]
for i = 1:4
 Constraints = [Constraints, something1 <= slack_for_C3(i)];
 Constraints = [Constraints, something2 == slack_for_C5(i)];
 Constraints = [Constraints, something3 == slack_for_C10(i)];
end
2

Use a slack variable, or element of a slack variable, for each constraint which you wish to diagnose by slack.variables.

variable X(N)
if A*x <= b is infeasible, then introduce a slack vector
variable slack_vector(M) nonnegative
Change the constraint A*x <= b to
A*X <= b + slack_vector
Then instead of your original objective, use
minimize(sum(slack_vector))

The optimal slack_vector will have positive elements for rows of A*X <= b which are not feasible.

For equality constraints, you can change
A*x == b
to

variable slack_vector_equality(M)
A*x == b + slack_vector_equality
minimize(slack_vector_equality'*slack_vector_equality)

The optimal slack_vector_equality will have significantly non-zero elements for rows of A*X == b which are not feasible.

If you have both inequalities and equalities, you can add the objective terms for both into one objective.

3

Can you take my code for example? Why does the optimal value become NaN sometimes? Because it seems that you just said something information about slack variable,but i still don’t know how to use it in my code

4

Do not use the original objective function of the optimization problem, or any part of that objective function.

For each <= constraint, add a nonnegative slack variable to the RHS, and include a term in the objective to minimize the sum of all those slacks.

For a semidefinite constraint
LHS == semidefinite(n) (or LHS >= 0 in sdp mode),
change it to
LHS + slack*eye(n) == semidefinite(n) (or LHS + slack*eye(n) >= 0 in sdp mode) , where slack is a nonnegative variable, and include a term in the objective to minimize the sum of all those slacks.

For each == constraint, add an unconstrained slack variable to the RHS (or LHS, it doesn’t matter), and include a term in the objective to minimize the sum of squares of all those slacks,

That’s all there is to it.