Final answer:
To show P3‖Cₘₐₓ and Pm‖Cₘₐₓ are NP-hard, we demonstrate their decision versions are NP-complete by proving that they are in NP and that an NP-complete problem can be reduced to them. This classification informs us of their computational complexity.
Step-by-step explanation:
To establish that the P3‖Cₘₐₓ problem is NP-hard, we start by considering its decision version, asking whether there is a schedule for the jobs such that the maximum completion time is less than or equal to a given threshold. To prove that it is NP-complete, we must show two things: first, that P3‖Cₘₐₓ is in NP, and second, that it is NP-hard by reducing a known NP-complete problem to it. If a problem is NP-complete, it means that it is at least as hard as any other problem in NP, and if a polynomial-time algorithm were found for it, then all problems in NP could be solved in polynomial time.
Similarly, the decision version of Pm‖Cₘₐₓ for fixed m>3 can be shown to be NP-complete, thus establishing that it is NP-hard. We again need to demonstrate that this problem is in NP, and that any instance of an NP-complete problem can be polynomially transformed into an instance of Pm‖Cₘₐₓ.
The significance of proving a problem to be NP-complete is that it implies the problem is as difficult as the hardest problems in NP, which are widely believed not to have polynomial-time solutions. Therefore, identifying a problem as NP-hard is important for understanding its computational complexity and for setting practical expectations for finding exact solutions.