0:00:13 | and |
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0:00:14 | and i'm the last one i'll try to make it a fish and |

0:00:18 | a is but yeah yeah that time and now we can spend a whole leaving here no i don't think |

0:00:23 | um |

0:00:25 | this article of a |

0:00:26 | this paper or uh is a knapsack problem uh formulation for really selection in secure corporate people that wireless was |

0:00:33 | communication |

0:00:34 | i i this work yeah yeah um is a collaboration with a one Q will uh |

0:00:39 | a professor for trouble profess of a and myself |

0:00:43 | and will start by a short introduction talking about the uh um |

0:00:48 | uh a cooperative jamming uh the system will introduce the system the model that we are considering for this |

0:00:54 | um and then we will uh formulate their uh an set |

0:00:58 | uh |

0:00:59 | problem uh for the selection of um minimal set |

0:01:03 | of really is uh to cooperate um |

0:01:06 | a a in the jamming |

0:01:08 | uh as uh this result that you know uh optimization problem that requires exponential complexity we will um |

0:01:15 | all uh all four three uh alternative a solution that um |

0:01:20 | oh four as um significant the |

0:01:22 | saving in complexity |

0:01:24 | and we will introduce a it would show some assimilation analysis of but these algorithms |

0:01:29 | and finally some concluding remarks |

0:01:33 | a so we know |

0:01:35 | a a cooperative the wireless communication system |

0:01:37 | the additional uh um |

0:01:39 | and degrees of freedom that are offered by those cooperating uh really is uh maybe use them to degrade great |

0:01:45 | uh channel |

0:01:47 | to uh potentially drop or |

0:01:49 | uh and they are uh |

0:01:52 | they can use the uh uh jointly |

0:01:54 | some way a |

0:01:56 | B to send a a collaborative them |

0:01:59 | a jamming signal |

0:02:00 | uh |

0:02:01 | to that you've drop |

0:02:03 | uh this type of uh a scheme was uh considered uh a would for protection meaning that if we have |

0:02:09 | a a an available really we use all in available relays |

0:02:14 | a for the purpose of jamming |

0:02:16 | uh and they a send a way uh version of the same jamming signal uh |

0:02:23 | for the two D pro |

0:02:25 | and this work was been done in two thousand nine by |

0:02:28 | don't hand problem |

0:02:30 | uh since no |

0:02:32 | it has been shown that |

0:02:33 | at some point |

0:02:34 | and as the that lot number of uh |

0:02:38 | grows |

0:02:39 | at some point |

0:02:40 | the that we get a introducing the relay for the jamming |

0:02:44 | is is small |

0:02:47 | uh and this has been a shown by um |

0:02:50 | one and uh |

0:02:52 | swindle to her in two thousand nine |

0:02:54 | uh considering that that and the fact that |

0:02:57 | the larger the system is |

0:02:59 | the larger uh the communication that a it is and uh |

0:03:03 | the synchronization requirement |

0:03:05 | that bring just to to the full question uh when you have a and collaborating really is and we previously |

0:03:11 | used all of them |

0:03:12 | uh can now the sec you rate requirement be a cheap tool close to to the maximum be achieved |

0:03:17 | we do a smaller or you're |

0:03:20 | uh uh uh activity |

0:03:23 | and the objective that |

0:03:24 | we we put forth and this is to identify i mean a more set of of uh relays |

0:03:30 | such that the predetermined secrecy receive rate the objective is a G |

0:03:35 | uh |

0:03:36 | first let's in the system that are considering |

0:03:39 | assuming we have a a a a a source transmitting a signal to the destination and |

0:03:45 | that you drop or is hearing the same message |

0:03:48 | oh we know the channel a between this uh a source and the destination by a just zero and between |

0:03:54 | the source and the proper or by G zero |

0:03:57 | also seem we have and |

0:03:59 | uh cooperating relays |

0:04:01 | and i and for each one of the really a uh with assume we have |

0:04:05 | we know the channel uh from |

0:04:07 | for example a really really high |

0:04:09 | to the destination marked by a giant again from the really |

0:04:13 | uh to that you drop or might but by G I and you seem we know we have knowledge of |

0:04:17 | the channel and bob |

0:04:20 | uh we assume um white gaussian noise of be zero mean and variance sigma squared |

0:04:25 | assume the energy for the |

0:04:28 | transmitted symbol X is uh one |

0:04:31 | and at the sort transmission power for the |

0:04:34 | the |

0:04:35 | so the symbol is P |

0:04:37 | uh all S |

0:04:39 | uh the really transmit meet a common jamming signal uh Z and we assume that each one transmit a weighted |

0:04:46 | version of signal |

0:04:50 | a for this uh scenario we basically have here |

0:04:53 | for the the signal received at the destination |

0:04:57 | is composed of the first element that is the message that was sending |

0:05:01 | uh in the second part is uh the version of the jamming signal that we have transmitted |

0:05:08 | multiply force by the channel to the destination |

0:05:11 | um |

0:05:12 | the signal received at the E dropper again has the same structure where we have |

0:05:16 | a version of the |

0:05:18 | symbol is received at the dropper |

0:05:20 | uh once the jamming signal that we have |

0:05:23 | france |

0:05:25 | uh we have here the notation for the various vectors |

0:05:29 | and basically in this case the secrecy rate may be written in this form where the first |

0:05:33 | element here represents the rate |

0:05:36 | yeah to the destination |

0:05:37 | and the second to rake at T |

0:05:41 | um |

0:05:42 | as i said the the first uh |

0:05:44 | what that has been done on um |

0:05:46 | and this uh uh and is assumed that |

0:05:49 | or and |

0:05:51 | a really are called right |

0:05:53 | the G |

0:05:54 | and for this and really is |

0:05:57 | uh a are the power |

0:05:59 | uh the power P S and the vector or all the weights vector |

0:06:03 | uh only got |

0:06:04 | maybe be uh such that |

0:06:06 | yeah we get a maximum |

0:06:08 | sec receive and this is basically the optimization problem that |

0:06:12 | design |

0:06:12 | a P S O |

0:06:14 | oh make the |

0:06:15 | uh oh yeah O optimal |

0:06:17 | um |

0:06:18 | such that |

0:06:20 | that set was rate is |

0:06:21 | max |

0:06:23 | as we said were looking for more |

0:06:25 | uh energy efficient the uh weighting and we trying to minimize the number of really at |

0:06:31 | so we are basically um defining a racial and the threshold these is with here is the are sub S |

0:06:36 | a requirement |

0:06:38 | and we we uh define it as a fraction of the maximum a C rate that |

0:06:44 | so |

0:06:44 | if we are all K be eighty percent of a |

0:06:46 | that would be a zero point eight as we set this threshold |

0:06:50 | uh and therefore we |

0:06:52 | next we want to define an optimization problem that will help us identify |

0:06:57 | the best realise to use to get this accuracy rate which is the minimal one as well |

0:07:04 | uh to do so we first started um |

0:07:07 | um by introducing |

0:07:09 | a binary variable |

0:07:12 | and we just a vector Q where each element of this queue Q why |

0:07:16 | can be wanting to really |

0:07:18 | i exactly or zero if it's not |

0:07:22 | uh our objective eventually is to write it and that combinatorial uh a framework |

0:07:26 | and uh |

0:07:28 | uh yeah |

0:07:29 | and specifically in a knapsack problem yeah therefore we take |

0:07:33 | the original a sec receiver rate expression that we had before and we rewrite it so we can use it |

0:07:39 | in this frame |

0:07:41 | and we define um |

0:07:43 | are uh uh that the sec C rate |

0:07:46 | uh we |

0:07:47 | and E here |

0:07:48 | a out which is a uh a and X exponent um |

0:07:53 | function of the do what we can secrecy rate |

0:07:55 | and it a to be an expression and basically is a sum and high and J work Q one Q |

0:08:00 | J |

0:08:01 | uh |

0:08:01 | as you you know you recall from here basically represent present a really is T one and |

0:08:08 | uh in the um |

0:08:10 | uh jamming |

0:08:11 | um multiplies applies |

0:08:12 | a a function here are are i J |

0:08:15 | but are i J is you can see is a function of Q as well |

0:08:19 | uh so this uh F F of to here is is given here |

0:08:23 | and you can see to F age of Q and F G of Q |

0:08:27 | where the dependencies still and Q one Q J |

0:08:30 | which is given a general form here |

0:08:33 | uh so it's a nonlinear your function basically of Q i and Q |

0:08:37 | but this type of formulation now enables us to right |

0:08:42 | uh the search |

0:08:43 | uh |

0:08:44 | problem for the minimal set E as an that |

0:08:48 | and that six probably it means that we are trying to minimize |

0:08:51 | the number of element |

0:08:53 | or the cost of that element there are putting in and a |

0:08:56 | well at this is a are capacity this is what we want to go |

0:09:00 | and a and as i said it's a combinatorial optimisation problem Q why can be zero |

0:09:06 | um |

0:09:07 | as you can see when we use in here that can a sec receive a to are using it read |

0:09:11 | or my got a star and P star |

0:09:14 | to mean that we are using it with a |

0:09:17 | vector or a mean god that was to my |

0:09:20 | for the sec rate so |

0:09:22 | you can see here |

0:09:23 | but to basically uh do for a given set of Q let's say that we follow that we we |

0:09:29 | we are looking for Q when where |

0:09:30 | oh forwarding a possible solution for Q |

0:09:33 | for this Q we are going and optimising in and then P S |

0:09:37 | such that it maximises the set or C |

0:09:41 | uh so this is this is the do know um |

0:09:44 | and knapsack optimization problem |

0:09:47 | and |

0:09:47 | one of the well known problems all of this is that a complexity it it has an exponential complexity |

0:09:53 | uh it's uh |

0:09:55 | you for larger number of of of and it's |

0:09:58 | impossible possible to solve |

0:10:00 | so of course |

0:10:01 | to make it feasible and and usable they are looking for |

0:10:04 | a a fast approximation algorithm that was still be a fish |

0:10:09 | and um |

0:10:10 | in this work we have proposed a three different |

0:10:13 | uh |

0:10:13 | a approximation algorithms |

0:10:16 | each one has its advantages or disadvantages as we will discuss later |

0:10:20 | and each one has different the a level of complexity of course |

0:10:25 | uh the first one is uh |

0:10:27 | yeah individual sectors you rate than a base |

0:10:30 | really selection |

0:10:31 | uh |

0:10:32 | the second is a weighted norm really selection and the last one |

0:10:37 | is a successive a a really uh selection algorithm that is based on |

0:10:41 | what power uh start local search uh approach |

0:10:46 | um we start for the first one |

0:10:47 | the first one is is the |

0:10:49 | think five one |

0:10:51 | and what it it does it it relaxes the original position problem |

0:10:56 | and says we not going to optimize it's for each vector |

0:10:59 | what what are going to do we are going to |

0:11:01 | first take each individual really |

0:11:05 | are going to find a a |

0:11:07 | the optimal value for a i and P S time |

0:11:11 | that maximise the sectors C |

0:11:12 | for this uh a specific one |

0:11:15 | and |

0:11:15 | once we generate a set of uh of value |

0:11:19 | we are so like think the the we are successively selecting them by |

0:11:23 | they're value so we're selecting the highest one first and keep on adding |

0:11:27 | uh by there |

0:11:29 | and feel we we basically have some |

0:11:31 | uh um threshold uh pass |

0:11:36 | um |

0:11:38 | i don't note and how what what what is basically down we use this a figure to explain how |

0:11:44 | we basically find uh the the palm maximum um |

0:11:48 | oh values for P S and all my got to this process |

0:11:52 | and you have your plot for to to really |

0:11:55 | call really and really be |

0:11:56 | and you can see them |

0:11:58 | sec to see wait for each one of them as a function of the |

0:12:01 | of P S |

0:12:02 | and basically it it the the algorithm that to show identifies stills maximal point and the appropriate P S for |

0:12:08 | them |

0:12:09 | which are then used |

0:12:11 | a selection out |

0:12:14 | uh the second approach |

0:12:16 | yeah |

0:12:17 | is going if |

0:12:18 | first we went for the individual one here we going on the full system |

0:12:22 | so we assume we take all and really |

0:12:24 | and we doing what has been that in previous work which is optimising |

0:12:28 | the weight the vector or my guy and P S |

0:12:30 | such that the total |

0:12:32 | step C rate is all is met |

0:12:35 | no when not |

0:12:36 | repeating that that actually taking the values that we got |

0:12:39 | are this optimal value |

0:12:41 | and we are using them |

0:12:43 | in our uh expression for the set to see so yeah we not |

0:12:47 | calculate that again and again |

0:12:49 | so we're saving uh on this process |

0:12:52 | and uh we are selecting uh |

0:12:55 | uh the the the larger one of course which use a basically equivalent to |

0:12:58 | selecting the |

0:13:00 | uh i |

0:13:01 | a really is with their a larger storm and this is like like weighted norm uh |

0:13:06 | selection |

0:13:09 | the third one |

0:13:10 | does not assume any uh |

0:13:13 | relaxation on the original problem |

0:13:15 | that's on basically takes the we problem |

0:13:18 | and |

0:13:20 | is um |

0:13:22 | in a sense |

0:13:23 | in a greedy way |

0:13:25 | ah |

0:13:25 | and it's a a really is to the group |

0:13:28 | so we have a starting point |

0:13:30 | we we we start the let's say we do really one |

0:13:34 | and we keep and adding really |

0:13:37 | and each time we adding relays |

0:13:40 | we are we |

0:13:41 | going and maximizing uh the vectors |

0:13:45 | uh W and P again |

0:13:47 | and we are choosing the relay that |

0:13:49 | maximise the sec receiver in this case |

0:13:53 | now once we add a a a a a a ad |

0:13:55 | the the one that maximises with keep on adding and till we cross trash |

0:14:00 | uh |

0:14:01 | for a |

0:14:02 | a better performance in terms of of uh reading the optimum we we be |

0:14:06 | this search for every possible forced really |

0:14:09 | and this is one call |

0:14:11 | a most people start |

0:14:12 | yeah i am with people start a local search |

0:14:16 | uh |

0:14:17 | we eventually end up with them |

0:14:20 | and optimal solutions |

0:14:22 | and we choose the the one with the least number or |

0:14:27 | um |

0:14:30 | sure an example of that |

0:14:32 | uh |

0:14:33 | by the way the the here though optimization used done using um |

0:14:38 | an algorithm that was suppose to the proposed by um the at a bit trouble in web |

0:14:44 | and two thousand ten at each that |

0:14:48 | um um |

0:14:49 | to show uh an example of that |

0:14:52 | a you see C or scenario all of a source destination |

0:14:56 | and and you drop or or and we assume we have a set of relays |

0:14:59 | a a spreading a given area |

0:15:02 | ah |

0:15:03 | we assume fifteen uh really is are a given here |

0:15:06 | um and |

0:15:09 | uh we we normalize the sec to C rate compare sent to the maximum achievable one |

0:15:14 | and we we used to threshold one of them is ninety percent of zero point nine five and seventy five |

0:15:19 | percent |

0:15:19 | zero point somebody five |

0:15:22 | i you see for an example |

0:15:24 | of of the results that we get for this scenario |

0:15:27 | and |

0:15:29 | in in in black though it's not seen because the green is over that |

0:15:33 | and i and like to use a a is a uh the results of an exhaustive search |

0:15:37 | so basically if i would take for example for |

0:15:40 | really is |

0:15:41 | this points says |

0:15:43 | if i would do an exhaustive search for the best for really is this would be the the value the |

0:15:47 | to see way that that would get and so |

0:15:50 | and |

0:15:51 | the third on go with him a very nice we'd we'd the optimal one |

0:15:56 | uh uh we it for different scenario in in general it fold was very close to the optimal |

0:16:01 | uh it also more we'll bossed uh there there was information in paper it also more boss to the location |

0:16:07 | of the the sensors |

0:16:09 | respect to |

0:16:10 | if dropped or and |

0:16:12 | nation |

0:16:13 | uh in terms that it falls again candle |

0:16:16 | uh a the on a two |

0:16:17 | uh uh um |

0:16:19 | there are behaviour is more uh reliant on location |

0:16:23 | uh in many cases uh |

0:16:25 | as expected algorithm uh be performs better than algorithm a a a board to many have the stalls |

0:16:31 | slow reaction as you can see from here so |

0:16:34 | um uh for for low uh |

0:16:37 | low threshold if forms a we well but when the threshold is high as you can see here |

0:16:42 | it will give fifteen sensors or something like that |

0:16:44 | C right there |

0:16:46 | uh it would give sort twelve |

0:16:48 | a a sensor as the number of sensor that that we need for that |

0:16:52 | um |

0:16:53 | we also checked or robust most of that too small error in the estimation of the channel |

0:16:58 | uh between the source and you drop or |

0:17:02 | and uh again the the algorithm C is falling that before forming quite well in in uh in this sense |

0:17:08 | as well and most of them |

0:17:10 | a stay within a reasonable before |

0:17:13 | so to wrap things up uh |

0:17:16 | i the problem of selecting mean set of really seeing um |

0:17:20 | well wire score a cooperative system uh for the purpose of jamming |

0:17:24 | uh has been formulated as a knapsack problem |

0:17:28 | uh the first two algorithm of four or complexity all and uh |

0:17:32 | order of L |

0:17:34 | a while the um |

0:17:35 | the third one is |

0:17:37 | it's it's here |

0:17:38 | it sort of a a L |

0:17:39 | and time L and L is the number of a is we eventually end up |

0:17:44 | having in the subset |

0:17:46 | um |

0:17:48 | um |

0:17:53 | uh in terms of complexity what that explains is a little what this so uh in the sense that in |

0:17:58 | small or for smaller or uh on |

0:18:00 | a threshold |

0:18:02 | uh i'll go "'em" one free form relatively good but in large a threshold it you can perform as well |

0:18:08 | well do for like special algorithm be free phone |

0:18:10 | or not converge conversion to to the optimal one and you know the case as well |

0:18:15 | so there is a trade of your between performance of course and um |

0:18:19 | and complexity not to large trade off but there is a tradeoff |

0:18:22 | and |

0:18:23 | gives a few option of uh |

0:18:25 | uh |

0:18:26 | implementing that |

0:18:27 | in real |

0:18:29 | thank you |