*3.3. NP-Hardness Proof*

According to Garey and Johnson [55], any decision problem that can be reduced from an NP-complete problem, whether it is a member of NP or not, is not solvable in polynomial time unless P = NP since it is as hard as the NP-complete problem. In order to prove the NP-hardness of FPP, its computational complexity is analyzed. Therefore, the decision counterpart of the fleet placement problem (FPP)–FPP–D is introduced. The decision counterpart FPP–D inherits all parameters from FPP.

In this section, the NP-hardness of FPP, through proving the NP-completeness of FPP–D, is demonstrated. For FPP–D, the question is to determine whether there exists a solution with *f* station(s) such that all buildings are covered.

#### **Proposition 1.** *The FPP is NP-hard in the strong sense even if there is only one user in each building.*

**Proof.** We introduce a polynomial-time transformation to the FPP–D from the strongly NP-complete problem "Set Cover Problem (Minimum Cover Problem)" [55,56].

Set Cover Problem or SCP can be defined as follows: given a universe U of R elements, a collection of subsets of *U*, *G* = {*g*1, *g*2, *g*3, ..., *gL*} and a positive integer *K* ≤ | *<sup>G</sup>*|, the question is "Does G contain a cover for U of size K or less, i.e., a subset *G* ⊂ *G* with |*G*|≤ *K* such that every element of U belongs to at least one member of *G*?"

Given an instance of SCP, we introduce the following instance of FPP–D. Firstly, let all buildings in *B* be the equivalence of universe *U* in SCP and |*B*|= *R*. Then, let S be the direct transformation of collection G where *S* = {*<sup>s</sup>*1,*s*2,*s*3, ...,*sL*}, such that *sl* = *gl*, *l* = 1, 2, 3,..., *L*. In addition, we let *w*, *z* = 1 and *r* = 0 so that the building is covered if it is connected to the street node (*sl*). With the prior assumption, the distances in matrix *D* are assumed to be one if the street node is a 1-hop neighbour of the building and zero, otherwise. Therefore matrix *D* reflects the membership of *S* and is used to constitute the membership of collection *S* in FPP–D. We also assume that there are L stations, hence *n* = *L*. Next, we assume *pj* = 1; *j* ∈ {1, 2, 3,..., |*B*|} which means there is only user in building *j*. Let *cj* be one if a collection *si* contains a building *bj* where *i* ∈ {1, 2, 3,..., *n*} and *j* ∈ {1, 2, 3,..., |*B*|}. With the aforementioned assumptions, ∑*<sup>R</sup> j*=1(*cj* × *pj*) = |*B*|. Finally, we let the threshold value *f* = *K*.

Let X be a solution to SCP. A solution for FPP–D is constructed in which the buildings in *B* (*U*) are covered by *f* stations where *s l* = *gl* ∈ *X*, such that *xl* = *s l*, if *s l* ∈ *X* and *xl* = ∅ if *s l* ∈/ *X*. Since X is a cover of *U* (in SCP), all buildings in *B* are covered and the number of stations in the corresponding solution (for FPP–D) is *f* = |*X*|.

Now assume that there exists a solution Y in FPP–D with |*Y*|≤ *K* and |*Y*| should not exceed *K*, otherwise, |*Y*|> *K* and the condition will not hold. Therefore, there are at most *K* station(s) with *yi* = ∅. Since all buildings form *B* and all buildings in *B* belong to at least one member of *Y*, the selected stations with *yi* = ∅ represents a solution to SCP, given a polynomial transformation from SCP to FPP-D. Since all input numbers in the FPP–D instance have a size most polynomial in the size of the input, FPP is strongly NP-hard.

SCP (as an optimization problem) was proved to be polynomially non-approximable within the ratio *c* · ln |*G*|, for some constant *c* > 0 [57]. Therefore, we propose the following statement.

**Statement 1.** There exists no polynomial (*c* · ln *n*)- approximation algorithm for the FPP where *n* is the input size, unless *P* = *NP*.

## **4. Optimization Methods**

In this section, the state-of-the-art algorithms used in our experiments, PolySCIP [58], heuristics [5], and the Non-dominated Sorting Genetic Algorithm-II (NSGA-II) [45] are described. Each of them represents a different category of problem solver.
