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simulated annealing - sensor positioning in sensor networks

开发者 https://www.devze.com 2023-02-01 01:25 出处:网络
Hi I have a litte problem with understanding problem of localization sensors in wireless sensor networks. Based on that article

Hi I have a litte problem with understanding problem of localization sensors in wireless sensor networks. Based on that article http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.110.2833&rep=rep1&type=pdf I'm about to write a little simulation program which would solve the problem of localization sensor in sensor network.

The optimalization problem looks like that

Assume we have a set of m sensors (anchor ndoes), each with known position and n sensors (non-anchor ndoes) with unknown location. Every node have an ability to measure the distance between itself and neigbour nodes (the measurments are corrupted with noise).

My task is:

Having noisy distance measurements and position of anchor nodes estimate the location of all nodes withy unknown positions.

In the article(which I mentioned at the begining of my questions) is also a cost function which I don't understand. I just don't know how the position of anchor nodes helps me with estimation of location of all nodes.

I hope somebody will understand what I'm writting about开发者_如何学编程 :) Sorry for my English


You start out by guessing positions for the non-anchor nodes. When you have a position for that node, you can calculate the distance from it to each other node using the pythagorean theorem, which the paper calls estimated distance. You also have a noisy measured distance, using radio strength or whatever.

You subtract those two distances, square it, and sum that value for each measured distance available in your network. That's your cost function, which you're trying to get as small as possible.

Where simulated annealing comes in is obviously you're not going to guess the right position for the non-anchor nodes at first. You move one of your guesses slightly in a random direction and see if the cost function gets better or not. If it does, you keep the new guess. Otherwise, you try again. Do this long enough, and eventually your guesses are pretty close to the actual position.

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