On the displacement for covering a unit interval with randomly placed sensors

• We prove that total movement to power a   of the sensors with range r=12n is (a2)!2a2(1+a)1na2−1+O(1na2), when a is an even natural number.
• We give the Algorithm 1 that uses expected O(1na2−1(ln⁡nn)a2), total movement to power a  , where a>0a>0 and r≥62n.Algorithm 1. Displacement to the power a when a > 0, p=94(2+a), q=34(2+a), x0 is the real solution of the equation x94(2+a)ln⁡x=3 such that x0 ≥ 3.Figure optionsDownload as PowerPoint slide

Consider n   mobile sensors placed independently at random with the uniform distribution on a barrier represented as the unit line segment [0,1][0,1]. The sensors have identical sensing radius, say r. When a sensor is displaced on the line a distance equal to d   it consumes energy (in movement) which is proportional to some (fixed) power a>0a>0 of the distance d traveled. The energy consumption of a system of n sensors thus displaced is defined as the sum of the energy consumptions for the displacement of the individual sensors.We focus on the problem of energy efficient displacement of the sensors so that in their final placement the sensor system ensures coverage of the barrier and the energy consumed for the displacement of the sensors to these final positions is minimized in expectation. In particular, we analyze the problem of displacing the sensors from their initial positions so as to attain coverage of the unit interval and derive trade-offs for this displacement as a function of the sensor range. We obtain several tight bounds in this setting thus generalizing several of the results of [10] to any power a>0a>0.