5.3 Energy Restricted Sweep Coverage Problem
5.3.1 Algorithm and Analysis
Let G = (U, E) be the complete weighted graph with each PoI as a vertex and the line segment joining between two PoIs on the plane as an edge. We define two weight functionsw1 :E −→R+andw2 :E∪U −→R+as follows. For any edgee= (ui, uj)∈E, w1(e) = the Euclidean distance between ui and uj. For the edge e = (ui, uj) ∈ E, w2(e) = w1v(e) ·µ and for each vertex ui ∈ U, w2(ui) = µi. For a subgraph H of G, w1(H) = P
e∈E(H)w1(e) and w2(H) = P
v∈V(H)w2(v) +P
e∈E(H)w2(e). Basically, for any path P in G, w1(P) gives the total distance a mobile sensor must travel to cover each point on P and w2(P) gives the total energy loss by a mobile sensor for moving along P. The Algorithm 5 (ERSweepCoverage) is proposed to solve the ERSweep coverage problem which returns the number of mobile sensors needed to guarantee t- sweep coverage of the given set of PoIs by maintaining the energy restriction of the mobile sensors.
Algorithm 5 ERSweepCoverage
1: for k = 1 to n do
2: Find the minimum spanning forest Fk with k components on G with respect to the weight functionw1. Let C1, C2,· · ·Ck be the connected components of Fk.
3: Find toursT1,T2,· · ·, Tk by doubling the edges ofC1, C2,· · ·Ck, respectively and short cutting.
4: for i= 1 to k do
5: if Γ≥t then
6: Set xi=PartitionTour1(Ti, t,Γ,Z)
7: else
8: Set xi=PartitionTour2(Ti, t,Γ,Z)
9: end if
10: end for
11: Nk=Pk j=1xj
12: end for
13: N = min{N1, N2,· · · , Nn}, and let J be the index such that NJ =N
14: Repeat step 2 to step 11 for k =J.
15: Deploy one mobile sensor at each of the partitioning points on each tour. The mobile sensors start moving at the same time along the respective tours in same direction.
To find the best possible solution, from step 2 to step 11 of the Algorithm 5 is executed in n iterations. In kth iteration (1 ≤ k ≤ n), the minimum spanning forest Fk with k connected components with respect to w1 is computed. After that k disjoint
tours T1, T2,· · · , Tk are found by doubling all edges of the components C1, C2,· · · , Ck, respectively and short cutting. Depending on the values of Γ and t, each tour Ti is partitioned using one of the following two partitioning strategies.
PartitionTour1(Ti, t,Γ,Z): This partitioning strategy is applied when Γ ≥ t.
For each tour Ti, we partition Ti intol
w2(Ti) Z
m
parts with weight at most Z with respect to w2. Let P r1, P r2,· · ·, P rlw2(Ti)
Z
m be the partitions. Now, partitionP rj into lw
1(P rj) vt
m parts with weight at most vt with respect to w1. Finally, this partitioning strategy returns total number of partitioning points P
lw
2(Ti) Z
m
j=1
lw
1(P rj) vt
m
for all partitions.
PartitionTour2(Ti, t,Γ,Z): This partitioning strategy is applied when Γ < t.
Let α be the positive integer such that α · Γ ≤ t < (α + 1)· Γ. Partition Ti into lw1(Ti)
vt
m
parts with weight at most vt with respect to the weight function w1. Let P r1, P r2,· · · , P rlw1(Ti)
vt
m be the partitions. Now, partitionP rj into lw
2(P rj) α·Z
m parts with weight at most α· Z with respect tow2. Finally, this partitioning strategy returns total number of partitioning points P
lw
1(Ti) vt
m
j=1
lw
1(P rj) α·Z
m for all partitions.
Minimum over the number of partitions of all iterations is chosen as the output of the Algorithm 5.
Theorem 5.3.2. The Algorithm 5 ensures t-sweep coverage of each PoI and energy constraint of each mobile sensor.
Proof. Let ui ∈U be visited by a mobile sensor mj at timet0. According to Algorithm 5, mobile sensors are initially deployed within vt distance apart, after that they start moving with speedv in the same direction. Soui will be again visited by the next mobile sensor within time t+t0. Also the partitioning strategies guarantee each mobile sensor to consume at most Z energy in any time interval [jΓ + Γ0,(j+ 1)Γ + Γ0], where j is a positive integer.
Letoptbe the number of mobile sensor in the optimal solution. The following lemmas give lower bounds on opt.
Lemma 5.3.3. opt ≥ w1(Fvtopt), where Fopt is the minimum spanning forest of G with respect to w1 having opt number of components.
Proof. Consider movements of the mobile sensors in optimal solution during the time interval [t0, t+t0]. Let P1, P2,· · · , Popt be the movement paths of the mobile sensors.
Then w1(Pi) ≤ vt. P1, P2,· · · , Popt form a forest with opt number of components that spans all the vertices of G. Since Fopt is the minimum spanning forest withopt number of components, opt·vt≥w1(Fopt), i.e., opt≥ w1(Fvtopt).
Lemma 5.3.4. If Γ≥t then opt≥ w2(FZopt).
Proof. LetP1, P2,· · · , Popt be the movement paths of the mobile sensors during the time interval [t0, t+t0]. Then, w2(Pi)≤ Z for i= 1 to opt. Now,
w2(P1) +w2(P2) +· · ·+w2(Popt) =
opt
X
i=1
X
u∈V(Pi)
w2(u) + X
e∈E(Pi)
w2(e)
= X
u∈V(G)
w2(u) + X
e∈E(Sopt i=1Pi)
w2(e)
= X
u∈V(G)
w2(u) + µ
v · X
e∈E(Sopt i=1Pi)
w1(e)
≥ X
u∈V(G)
w2(u) + µ
v · X
e∈E(Fopt)
w1(e)
= w2(Fopt).
Therefore, Z ·opt≥w2(Fopt), i.e., opt≥ w2(FZopt).
Lemma 5.3.5. If Γ< t then opt≥ w(α+1)2(Fopt·Z), whereα is the integer such that α·Γ≤t <
(α+ 1)·Γ.
Proof. Let P1, P2,· · · , Popt be the movement paths of the mobile sensors in the optimal solution during the time interval [t0, t+t0]. Since α·Γ≤t <(α+ 1)·Γ, the maximum energy consumption by a mobile sensor for its activities in the time interval [t0, t+t0] is less than (α+ 1)· Z. Therefore, w2(Pi)≤(α+ 1)· Z, fori= 1 to opt. Again, from the proof of Lemma 5.3.4, we have w2(P1) +w2(P2) +· · ·+w2(Popt)≥w2(Fopt). Therefore, (α+ 1)· Z ·opt≥w2(Fopt), i.e., opt≥ w(α+1)2(Fopt·Z).
Lemma 5.3.6. The number of partition xi on the tour Ti is given by
xi ≤ w2(Ti)
Z + w1(Ti)
vt + 1 if Γ≥t
≤ w2(Ti)
α· Z + w1(Ti)
vt + 1 if Γ< t
Proof. For Γ ≥ t each tour Ti is partitioned into parts with weight ≤ Z with respect to w2. Let P r1, P r2,· · · , P rlw2 (Ti)
Z
m be the partitions with w2(P ri) ≤ Z. According to PartitionTour1(Ti, t,Γ,Z), total number of partitions on Ti is
xi = P
lw
2(Ti) Z
m
j=1
lw
1(P rj) vt
m ≤ P
lw
2(Ti) Z
m
j=1
w1(P rj)
vt +l
w2(Ti) Z
m ≤ w1vt(Ti) +l
w2(Ti) Z
m ≤ w1vt(Ti) +
w2(Ti)
Z + 1.
For Γ < t each tour Ti is partitioned into parts with weight ≤ vt with respect to w1. Let P r1, P r2,· · · , P rlw1(Ti)
vt
m be the partitions with w1(P ri) ≤ vt. According to PartitionTour2(Ti, t,Γ,Z), total number of partitions on Ti is
xi = P
lw
1(Ti) vt
m
j=1
lw
2(P rj) α·Z
m
≤ P
lw
1(Ti) vt
m
j=1
w2(P rj) α·Z +l
w1(Ti) vt
m
≤ w2α(TZi) +l
w1(Ti) vt
m
≤ w1vt(Ti) +
w2(Ti) α·Z + 1.
Theorem 5.3.7. The Algorithm 5 is a 5-approximation algorithm for Γ≥t.
Proof. Algorithm 5 chooses the minimum over allNk fork= 1 to n. Let us consider the iteration of the algorithm when k =opt. For Γ≥t, the total number of mobile sensors required is given by
k
X
i=1
xi ≤
k
X
i=1
w2(Ti)
Z +
k
X
i=1
w1(Ti)
vt +k (Lemma 5.3.6)
≤ 2
k
X
i=1
w2(Ci)
Z + 2
k
X
i=1
w1(Ci) vt +k
= 2w2(Fk)
Z + 2w1(Fk) vt +k
≤ 2k+ 2k+k (Lemma 5.3.4 andLemma 5.3.3)
= 5opt.
Therefore if Γ≥t, the Algorithm 5 is a 5-approximation algorithm.
Theorem 5.3.8. Algorithm 5 is a (5 + α2)-approximation algorithm for Γ< t, where α is the integer such that α·Γ≤t <(α+ 1)·Γ.
Proof. The Algorithm 5 chooses the minimum over all Nk for k = 1 to n. Let us consider the iteration of the algorithm when k = opt. For Γ < t, the total number of mobile sensors required is given by
k
X
i=1
xi ≤
k
X
i=1
w2(Ti) αZ +
k
X
i=1
w1(Ti)
vt +k (Lemma 5.3.6)
≤ 2
k
X
i=1
w2(Ci) αZ + 2
k
X
i=1
w1(Ci) vt +k
= 2w2(Fk)
αZ + 2w1(Fk) vt +k
≤ 2k
α+ 1 α
+ 2k+k (Lemma 5.3.5 and Lemma 5.3.3)
=
5 + 2 α
k
=
5 + 2 α
opt.
Therefore if Γ< t, the Algorithm 5 is a (5 +α2)-approximation algorithm.