2.2 Minimum Power BS Activation Problem for Uplink Coverage-Limited
2.2.1 Complexity Analysis of Minimum Power BS Activation Problem 29
The minimum power BS activation problem for minimizing total power by activating a set of BSs while preserving coverage can be closely related to the minimum-weight disk cover problem in which the goal is to minimize the total cost to cover a set of points in the plane by selecting a subset of disks [37–39]. It is known to be NP-hard [39]. Recently, a polynomial time approximation scheme (PTAS) is proposed in [37] for a minimum-weight disk cover problem. However, the authors in [38] show that PTAS for set cover does not exist (unless P = NP ) for circles of roughly the same size. To cover a set of points by unit disks, an algorithm with constant approximation ratio is proposed for minimizing the summation of the weight of disks in [39]. Since the design objective is to cover a set of points instead of covering the entire region, the authors in [39] indicate that one hole or many holes will be present with the proposed algorithm. From the above, we can find that the results in the minimum-weight disk cover problem cannot be directly applicable in our green network coverage problem in which coverage holes shall be prevented.
We first examine the complexity of the problem and derive key properties for preserving network coverage which inspire us to design the BS activation algorithm.
To analyze the complexity of this problem, we look at a network in which 1) the entire network can be covered when all BSs are activated, 2) the entire network is a continuous region within which an MS can move from any position to another position, and 3) no BSs will be placed at the same position.
Given a target covering region A and a set of BSs B, we have the following definitions for the coverage problem.
Definition 2.1. A point i∈ A is said to be covered by an active BS Bm if it is within the coverage range of Bm, D(i, Bm) < Rm in which D(·, ·) is the Euclidean distance between two positions. Otherwise, the point i is said to be not covered by Bm.
Definition 2.2. A boundary intersection is a crossing point between the boundary of two sets. The set can be the coverage of a BS or the target coverage region. Formally,
Region III
Figure 2-2: The relative positions between an uncovered point and the active BS closest to the uncovered point. (a)The regional classification of an uncovered point. (b)The position of an uncovered point in region I. (c)The position of an uncovered point in the region II.
the set of all boundary intersections in the target area A, denoted by IA, can be represented as
where the boundary of coverage Cm is denoted by ∂Cm and the boundary of the target region is denoted by ∂A.
We first derive the sufficient conditions for an activated BS set, denoted by Bon, to maintain network coverage and then show the minimum BS activation problem is NP-hard.
Lemma 2.3. If all boundary intersections of an activated BS set Bon in the target area are covered by Bon itself, all positions in the target region will be covered.
Proof. We prove this lemma by contradiction. Specifically, we will prove that if there exists a point in the network not covered by the set of active BSs, there must exist a boundary intersection which is not covered by the set of active BSs. Suppose that point u ∈ A in the network is not covered by any active BSs. That is, {u ∈ A|u /∈ S
Bn∈BonCn}. If Bm ∈ Bon is the activated BS closest to the uncovered (not covered) point u ∈ A, we will prove that the boundary intersection closest to u at the boundary of Cm will not be covered by any active BSs. Denote i∗m,u ∈ A as the boundary intersection between the coverage of Bm and the coverage of other BS in Bon closest to u. We will prove that i∗m,u will not be covered by any BSs in Bon if u
is an uncovered point in the network. It will be a contradiction to our assumption.
Suppose that the boundary intersection i∗m,u is covered by an active BS Bn ∈ Bon. Since Bn covers i∗m,u, there will be two intersections between boundary intersection
∂Cm and ∂Cn. See Figure 2-2 (a) as an example. In the figure, v and w are the intersections between ∂Cm and ∂Cn. Because Bm is the active BS closest to the uncovered point, the uncovered point will be at the upper side of vw. Otherwise, Bn
will be closer to the uncovered point than Bm. Here we classify the possible positions of the uncovered point u into three regions and prove that if i∗m,u is covered byCn, we will prove that i∗m,u will not be the intersection on the boundary of Cm closest to u.
We first consider the uncovered position u located in Region I, see Figure 2-2 (b) for example. In this region, because ∠wi∗m,uv is an inscribed angle centred at Bm
outside chord vw, ∠wi∗m,uv > π2 and ∠xvi∗m,u = ∠vwi∗m,u+∠vi∗m,uw will be larger than
π
2. Moreover, because D(u, Bn) ≥ D(u, Bm), ∠uvx ≥ 0. The equality holds when u is at −wv. Thus, ∠uvi→ ∗m,u = ∠xvi∗m,u+ ∠uvx > π2. By law of sines in △uvi∗m,u, we can obtain ui∗m,u > uv. Similarly, if i∗m,u is in Region III, we can prove ui∗m,u > uw. On the other hand, if the uncovered position u is at i∗m,uv or i∗m,uw, ui∗m,u ≥ uv, because u is either outside ∂Cm or on ∂Cm. A contradiction to i∗m,u is a boundary intersection on ∂Cm closest to u.
Finally, consider that an uncovered point u in Region II, see Figure 2-2 (c) as an example. In the figure, because ∠xvi∗m,u and ∠xwi∗m,u are two opposite angles of a cyclic quadrilateral on ∂Cm, ∠xvi∗m,u + ∠xwi∗m,u = π. Because u is on −−−→i∗m,ux either outside the circle or on the circle, x is an interior point either in △uwv or x is at the same position of u such that ∠uvi∗m,u+ ∠uwi∗m,u = ∠xvi∗m,u+ ∠xwi∗m,u+ ∠uwx +
∠uvx ≥ π. Equality holds when u is on the circle. Thus, we can find that at least
∠uvi∗m,u or ∠uwi∗m,u will be larger than π2. By law of sines, at least ui∗m,u ≥ uv or ui∗m,u ≥ uw. As a result, i∗m,u is not the boundary intersection on Bm closest to u.
This leads to a contradiction.
From the above, we have proven that if there exists a point in the network not covered by the set of active BSs, there must exist a boundary intersection between two active BSs which is not covered by the set of active BSs. It is a contradiction to
our assumption.
Lemma 2.3 can be easily realized by observations as follows. If a boundary inter-section between two active BSs can be covered by another active BS, there will be no coverage hole in the interior triangle region of the three active BSs. By recursively examining the intersections between two active BSs, we can know whether coverage holes exist among the active cells.
Lemma 2.4. If all boundary intersections in the target region IA are covered by an activated BS set Bon, the entire target region A is covered by Bon.
Proof. In Lemma 2.3, we prove that the sufficient condition to cover the network is that all boundary intersections of the activated BSs in the target region are covered.
The set of all boundary intersections of BSs in the target region is a super set of the set of boundary intersections of activated BSs in the target region. If the super set is covered, we can obtain the network will be covered by the activated BS setBon from Lemma 2.3.
From Lemma 2.4, if the target coverage region contains boundary intersections which are not covered by active BSs, the target coverage region is not covered. One can easily examine whether coverage holes exist in the target coverage region by checking only the boundary intersections of all BSs instead of checking every position in the target region.
Theorem 2.5. To find the minimum activated BS set to cover the network is NP-hard.
Proof. From Lemma 2.4, we can reduce the minimum BS activation problem from a continuous region covering problem to a discrete point covering problem. By this transformation, this problem is equivalent to the minimum disk cover problem [37,40].
Specifically, the minimum disk cover problem asks to cover a set of points P in the plane with a subset of disks D with minimum cardinality. This minimum disk cover
problem is known to be NP-hard by a reduction from the planar 3SAT problem [41].
In our green cellular coverage problem, the set of points P to cover is all boundary intersections in the target area region IA. We try to find a subset of BS coverages D = {C1, . . . ,CM} to cover IA with minimum cardinality. Henceforth, the minimum BS activation problem corresponds to the minimum disk cover problem and is thus an NP-hard problem.
2.2.2 Cell Activation Algorithm– Cell Overlap Minimization with Intersection Covered (COMIC)
Here our goal is to activate a specific set of BSs which can minimize the network power consumption with network-wide coverage and can also support the network traffic loads. Based on the current standards, traffic volume information is usually collected in a group unit (e.g., RNC in UMTS). Here we propose a centralized algorithm which may be more fitting with the current standardization directions. Decentralized mechanisms may need the development of extra signalling mechanisms and may easily cause ping-pong effects (a BS is switching on and off repeatedly) [42] with only local information.
We propose a load-aware algorithm called Cell Overlap Minimization with Inter-section Covered (COMIC) for activating an energy-efficient BS set while maintaining network coverage. With the obtained optimal coverages for minimizing area power consumption, we try to find an optimal set which minimizes the overlaps while avoid-ing coverage holes in the target area. Here our solution is designed for the network in which the location of BSs is available. As shown in [20], the optimal position of a BS to minimize the overlap of the coverage of two other activated BSs shall be located such that the distances from the joint boundary intersection point to the other two boundary intersections are exactly the same. Specifically, if i∈ ∂Cm∩∂Cn, the optimal coverage to cover i (denoted by Ci∗) shall satisfy 1) i ∈ ∂Ci∗ and 2) D(i, j) = D(i, k) for j ∈ ∂Cm ∩ ∂Ci∗, j 6= i and k ∈ ∂Cn∩ ∂Ci∗, k6= i.
The steps of the COMIC algorithm are as follows.
1. Choose a BS which has the maximal coverage range from B (say B1) as the initial active BS, and add it as the first element of Bon (i.e., Bon ={B1}).
2. Find Bk which has the maximal coverage range fromB \ {B1} and Ck∩ C1 6= ∅.
Set Bk as the new active BS. Add the new active BS Bk toBon. Denote the set of uncovered boundary intersections between activated BSs as I and initiate I = ∅.
3. For each Bm ∈ Bon, find boundary intersection i between Bk and Bm which is not covered by any BS in the set of Bon\ {Bm}. Add boundary intersection i toI. Repeat until no such intersections exist for Bm, i.e.,
I = I[{i|i ∈ \
Bj∈{Bon\{Bm}}
Cj′
\∂Cm\
∂Ck\ A},
where Cj′ represents the complement of Cj, ∂Cm is the boundary of Cm, and A is the target covering region.
4. For each new i ∈ I, find the optimal BS for i. The optimal BS for i is the BS which is closest to the optimal BS position of i and can cover the boundary intersection i. Set these BSs as candidate BSs.
5. Choose Bk which minimizes the distance from its position to the optimal posi-tion of the corresponding boundary intersecposi-tion from the candidate BSs as the new active BS. Add the new active BS Bk to Bon.
6. For each i ∈ I, if i is covered by Bk, remove it fromI.
7. Go to Step 3 until I = ∅.
8. Find uncovered boundary intersections at network boundaries, i.e.,
I∂A = [
Bm∈Bon
∂Cm ∩ Con′ ∩ ∂A, (2.23)
where I∂A is the set of uncovered boundary intersections between active cells and the target area A, Con =S
Bm∈BonCm, and Con′ is the complement of Con.
9. If I∂A 6= ∅, choose Bk such thatCk minimize the overlap withCon and can cover a boundary intersection j ∈ I∂A. For each j ∈ I∂A, if j is covered by Bk, remove it from I∂A. Add the new active BS Bk to Bon.
10. Go to Step 3 until I∂A =∅.
The rationale behind COMIC is to activate a BS which can cover a boundary intersection of any two active BSs with minimum overlap until no uncovered boundary intersection is detected in the target region. Other BSs which are not activated by COMIC can be switched off to save power. Practically, more BSs around a hot spot area may need to power on to satisfy throughput requirements. Alternatively, low-power picocells or femtocells can be deployed to support the capacity in hot spot areas.