Energy-Optimal Online Algorithms for Broadcasting in Wireless Network Shay Kutten Hirotaka ono David Peleg Kunihiko Sadakane Masafumi Yamashita.

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Presentation transcript:

Energy-Optimal Online Algorithms for Broadcasting in Wireless Network Shay Kutten Hirotaka ono David Peleg Kunihiko Sadakane Masafumi Yamashita

Outline Background Model Problems and Results Algorithms and Analyses –Single Receiver Case –Multiple Receivers Case Conclusion

Background Design energy-efficient online message broadcasting protocols in ad-hoc wireless networks –energy-efficiency : save the battery-resource –online property: ・・・ ad-hoc network non-static network

Problem sender The sender does not know the distances to receivers Broadcast a message to nearest n-1 hosts Receiving ack from the n-1 hosts ack Problem: Design receiver a good online algorithm

Problem Communication requires energy consumption. The energy consumption depends on the distance between the sender and receivers. ( The distance is longer, the energy must be larger. ) The sender/receiver have no distance information. s sends some message (e.g., beacon) to r. with some energy consumption. If r receives the beacon, he needs to send “ack” to s with the same amount of energy consumption.

Problem Problem: Design Energy: 20 (broadcast) Energy: 20 (ack) The total cost (energy) is Broadcastack Energy: 20 (ack) a good online algorithm (1×20) + (20×6 ) = 140

Problem Problem: Design Energy: 5 (broadcast) Energy: 20 (broadcast) Energy: 5 (ack) Energy: 20 (ack) The total cost (energy) is Broadcastack Energy: 20 (ack) a good online algorithm (5+20)+(5×1 +20×5 )= 130 an energy-efficient online algorithm < 140

Model The attenuation of signal power P s is where d(s,r) is the distance between s and r, and δ ≧ 1 is the distance-power gradient. γ is the minimum power to decode a message. –The maximum distance to which a message can derivered from s is Only a direct broadcast is allowed. (No multi-hop delivery is allowed.) Synchronous Communication ・・・ (We can utilize a global clock and unique IDs of nodes ) Collision-free and Failure-free

Related Works Range Assignment Problem – offline : The distances between any pair of hosts are given. – Minimizing the total energy consumption to broadcast a message to a set of recipients –Constructing energy-efficient multicast tree with several properties: connectivity from a source, strong connectivity, small radius, and so on “ On the Complexity of Computing Minimum Energy Consumption Broadcast Subgraphs”, [CCPRV 2001] “Power Consumption in Packet Radio Networks”, [KKKP 2000]

Problems and Results

Algorithms and its Performance Minimize the total energy consumption Our model is “online”, i.e., no a-priori information. Use competitive analysis: The performance of algorithm A (competitive ratio) is cost* : the minimum value of the total energy consumption with complete information

Problems and Results (1) Problem BA2 (Broadcast+Ack-2) –one sender s and one receiver r –s sends a message to r. –r sends an ack to s after receiving the message. – Theorem s r message ack The optimal competitive ratio of problem BA2 is (No online algorithm whose competitive ratio is smaller than )

Problems and Results (2) Problem BAn (Broadcast+Ack-n) –one sender s and n-1 receivers, r 1, r 2, …,r n-1 –s sends a message to r 1, r 2, …,r n-1. –Each r sends an ack to s after receiving the message. s d1d1 d2d2 d3d3 n 1 receiversn 2 receiversn 3 receivers

Problems and Results (3) Theorem The optimal competitive ratio of problem BAn is.

Algorithms and Analyses

Generic Protocol (Algorithm) Procedure SendMessage(t,msg) 1.i := 1, f := true 2.while f 3. do Transmit(msg,pi) with power pi. 4. wait. 5. if received ack from t 6. f := false; 7. i := i+1;

Procedure SendMessage(t,msg) 1.i := 1, f := true 2.while f 3. do Transmit(msg,pi) with power pi. 4. wait. 5. if received ack from t 6. f := false; 7. i := i+1; i =1 i =2 i =3 ack The total cost (energy consumption) of the procedure is s t

Single Receiver Case (1) Algorithm: DA[β] (Doubling Algorithm) –In SendMessage, s β = 2 γ =1 1 2 4

Single Receiver Case (2) Proposition: DA[β] algorithm achieves the competitive ratio is for problem BA2. The value is minimized when, and it is.

Single Receiver Case (3) Theorem The optimal competitive ratio of problem BA2 is Sketch of Proof: 1. Assume an optimal online algorithm with competitive ratio c ≦ 3/2+√2 and its output, say x 1,x 2,x 3 …. 2. From the competitive ratio property, we have the following inequality,

Sketch of Proof: (continued) 3. The parameter sequences is a Cauchy sequence, so converges to a real value, and we have the following quadratic equation: 4.From the condition of the existence of is 5.The algorithm achieves the competitive ratio; i.e., it gives the upper bound. Q.E.D

Multiple Receivers Case(0) Problem BAn (Broadcast+Ack-n) –one sender s and n-1 receivers, r 1, r 2, …,r n-1 –s sends a message to r 1, r 2, …,r n-1. –Each r sends an ack to s after receiving the message. Offline case: A simple Dynamic Programming can solve this in liner time. s d1d1 d2d2 d3d3 n 1 receiversn 2 receiversn 3 receivers

Multiple Receivers Case (1) Consider a simple special case : Problem UBAn (Broadcast+Ack-n) –one sender s and n-1 receivers, r 1, r 2, …,r n-1 all at the same distance d from s. –s sends a message to r 1, r 2, …,r n-1. –Each r sends an ack to s after receiving the message. s n-1 receivers d

Multiple Receivers Case (2) Proposition Theorem The optimal competitive ratio of problem UBAn is The previous proofs can be extended to this case. The competitive ratio of DA[β] algorithm fixing, for UBAn is at most

Procedure DDA(n,msg) 1.p:= γ 2.while n>1 3. do Transmit(msg,p) with power p. 4. wait. 5. n:=n-#(received ack) 6. p:= p ・ β n Dynamic Doubling Algorithm To solve the general case, we propose the following online algorithm:

Results of Multiple Receivers (1) Theorem The optimal competitive ratio of problem BAn is. Sketch of Proof: Lower bound : Consider the following situation (instance): s 1 receiver n-2 receivers This dominates the total energy consumption.

Sketch of Proof: Upper bound : DDA algorithm achieves the competitive ratio. 1. The problem instance can be considered the union of and. 2.In the part, DDA algorithm achieves competitive ratio 3.By applying this discussion repeatedly, the competitive ratio of each part is at most, so in total the competitive ratio is. Q.E.D

Conclusion Direct broadcast on online setting Single receiver and multiple receivers Energy-optimal online algorithms –doubling algorithm and dynamic doubling algorithm –The optimal competitive ratios are both

Future Work Not only energy-efficient but also time-efficient online algorithm Considering failure, collision, and so on