I have an optimization problem with a minimization cost function and two constraints to meet. Without considering one of the constraints, I can reduce the optimization problem to an NP-Complete problem. But with both constraints, I don't have any idea to reduce the problem to a known NP-Complete or NP-hard problem.
Suppose in a graph, I want to select the minimum number of nodes that address two different conditions. These conditions are independent. For example one of them is addressing the minimum dominating set problem and another is ensuring that nodes with some structural features are selected. So, I should find the minimum number of nodes that both dominant all nodes of the network, and also ensure nodes with specific structural features are selected. Without the second constraint, I can prove that the optimization model is an NP-hard problem. But with both of them, I can't find any known problem to reduce.
Therefore, I want to know that, is it possible to say that as the optimization problem with one of the constraints is NP-hard or NP-Complete, so with both constraints has also the same complexity class?
If you are asking if this is true in general, the answer is no.
Example 1: Satisfiability problem (3-SAT) is NP-Complete. Add a constraint that the clauses have atmost one positive literal (Horn clause), we get Horn-satisfiability (HORN-SAT) problem which is solvable in polynomial time. Here, adding a constraint made the problem less complex and easier to solve.
Example 2: Minimum Spanning Tree (MSP) is of complexity O(E log(V)). If we add the constraints that there can be no branches in the tree, we get the Traveling Salesman Problem (TSP) which is NP-Complete. Here, adding a constraint made the problem more complex and difficult to solve.
So, adding a constraint can either increase or decrease complexity. Therefore, the answer to your question is - No.