0:00:14and i'm the last one i'll try to make it a fish and
0:00:18a is but yeah yeah that time and now we can spend a whole leaving here no i don't think
0:00:25this article of a
0:00:26this paper or uh is a knapsack problem uh formulation for really selection in secure corporate people that wireless was
0:00:34i i this work yeah yeah um is a collaboration with a one Q will uh
0:00:39a professor for trouble profess of a and myself
0:00:43and will start by a short introduction talking about the uh um
0:00:48uh a cooperative jamming uh the system will introduce the system the model that we are considering for this
0:00:54um and then we will uh formulate their uh an set
0:00:59problem uh for the selection of um minimal set
0:01:03of really is uh to cooperate um
0:01:06a a in the jamming
0:01:08uh as uh this result that you know uh optimization problem that requires exponential complexity we will um
0:01:15all uh all four three uh alternative a solution that um
0:01:20oh four as um significant the
0:01:22saving in complexity
0:01:24and we will introduce a it would show some assimilation analysis of but these algorithms
0:01:29and finally some concluding remarks
0:01:33a so we know
0:01:35a a cooperative the wireless communication system
0:01:37the additional uh um
0:01:39and degrees of freedom that are offered by those cooperating uh really is uh maybe use them to degrade great
0:01:45uh channel
0:01:47to uh potentially drop or
0:01:49uh and they are uh
0:01:52they can use the uh uh jointly
0:01:54some way a
0:01:56B to send a a collaborative them
0:01:59a jamming signal
0:02:01to that you've drop
0:02:03uh this type of uh a scheme was uh considered uh a would for protection meaning that if we have
0:02:09a a an available really we use all in available relays
0:02:14a for the purpose of jamming
0:02:16uh and they a send a way uh version of the same jamming signal uh
0:02:23for the two D pro
0:02:25and this work was been done in two thousand nine by
0:02:28don't hand problem
0:02:30uh since no
0:02:32it has been shown that
0:02:33at some point
0:02:34and as the that lot number of uh
0:02:39at some point
0:02:40the that we get a introducing the relay for the jamming
0:02:44is is small
0:02:47uh and this has been a shown by um
0:02:50one and uh
0:02:52swindle to her in two thousand nine
0:02:54uh considering that that and the fact that
0:02:57the larger the system is
0:02:59the larger uh the communication that a it is and uh
0:03:03the synchronization requirement
0:03:05that bring just to to the full question uh when you have a and collaborating really is and we previously
0:03:11used all of them
0:03:12uh can now the sec you rate requirement be a cheap tool close to to the maximum be achieved
0:03:17we do a smaller or you're
0:03:20uh uh uh activity
0:03:23and the objective that
0:03:24we we put forth and this is to identify i mean a more set of of uh relays
0:03:30such that the predetermined secrecy receive rate the objective is a G
0:03:36first let's in the system that are considering
0:03:39assuming we have a a a a a source transmitting a signal to the destination and
0:03:45that you drop or is hearing the same message
0:03:48oh we know the channel a between this uh a source and the destination by a just zero and between
0:03:54the source and the proper or by G zero
0:03:57also seem we have and
0:03:59uh cooperating relays
0:04:01and i and for each one of the really a uh with assume we have
0:04:05we know the channel uh from
0:04:07for example a really really high
0:04:09to the destination marked by a giant again from the really
0:04:13uh to that you drop or might but by G I and you seem we know we have knowledge of
0:04:17the channel and bob
0:04:20uh we assume um white gaussian noise of be zero mean and variance sigma squared
0:04:25assume the energy for the
0:04:28transmitted symbol X is uh one
0:04:31and at the sort transmission power for the
0:04:35so the symbol is P
0:04:37uh all S
0:04:39uh the really transmit meet a common jamming signal uh Z and we assume that each one transmit a weighted
0:04:46version of signal
0:04:50a for this uh scenario we basically have here
0:04:53for the the signal received at the destination
0:04:57is composed of the first element that is the message that was sending
0:05:01uh in the second part is uh the version of the jamming signal that we have transmitted
0:05:08multiply force by the channel to the destination
0:05:12the signal received at the E dropper again has the same structure where we have
0:05:16a version of the
0:05:18symbol is received at the dropper
0:05:20uh once the jamming signal that we have
0:05:25uh we have here the notation for the various vectors
0:05:29and basically in this case the secrecy rate may be written in this form where the first
0:05:33element here represents the rate
0:05:36yeah to the destination
0:05:37and the second to rake at T
0:05:42as i said the the first uh
0:05:44what that has been done on um
0:05:46and this uh uh and is assumed that
0:05:49or and
0:05:51a really are called right
0:05:53the G
0:05:54and for this and really is
0:05:57uh a are the power
0:05:59uh the power P S and the vector or all the weights vector
0:06:03uh only got
0:06:04maybe be uh such that
0:06:06yeah we get a maximum
0:06:08sec receive and this is basically the optimization problem that
0:06:12a P S O
0:06:14oh make the
0:06:15uh oh yeah O optimal
0:06:18such that
0:06:20that set was rate is
0:06:23as we said were looking for more
0:06:25uh energy efficient the uh weighting and we trying to minimize the number of really at
0:06:31so we are basically um defining a racial and the threshold these is with here is the are sub S
0:06:36a requirement
0:06:38and we we uh define it as a fraction of the maximum a C rate that
0:06:44if we are all K be eighty percent of a
0:06:46that would be a zero point eight as we set this threshold
0:06:50uh and therefore we
0:06:52next we want to define an optimization problem that will help us identify
0:06:57the best realise to use to get this accuracy rate which is the minimal one as well
0:07:04uh to do so we first started um
0:07:07um by introducing
0:07:09a binary variable
0:07:12and we just a vector Q where each element of this queue Q why
0:07:16can be wanting to really
0:07:18i exactly or zero if it's not
0:07:22uh our objective eventually is to write it and that combinatorial uh a framework
0:07:26and uh
0:07:28uh yeah
0:07:29and specifically in a knapsack problem yeah therefore we take
0:07:33the original a sec receiver rate expression that we had before and we rewrite it so we can use it
0:07:39in this frame
0:07:41and we define um
0:07:43are uh uh that the sec C rate
0:07:46uh we
0:07:47and E here
0:07:48a out which is a uh a and X exponent um
0:07:53function of the do what we can secrecy rate
0:07:55and it a to be an expression and basically is a sum and high and J work Q one Q
0:08:01as you you know you recall from here basically represent present a really is T one and
0:08:08uh in the um
0:08:10uh jamming
0:08:11um multiplies applies
0:08:12a a function here are are i J
0:08:15but are i J is you can see is a function of Q as well
0:08:19uh so this uh F F of to here is is given here
0:08:23and you can see to F age of Q and F G of Q
0:08:27where the dependencies still and Q one Q J
0:08:30which is given a general form here
0:08:33uh so it's a nonlinear your function basically of Q i and Q
0:08:37but this type of formulation now enables us to right
0:08:42uh the search
0:08:44problem for the minimal set E as an that
0:08:48and that six probably it means that we are trying to minimize
0:08:51the number of element
0:08:53or the cost of that element there are putting in and a
0:08:56well at this is a are capacity this is what we want to go
0:09:00and a and as i said it's a combinatorial optimisation problem Q why can be zero
0:09:07as you can see when we use in here that can a sec receive a to are using it read
0:09:11or my got a star and P star
0:09:14to mean that we are using it with a
0:09:17vector or a mean god that was to my
0:09:20for the sec rate so
0:09:22you can see here
0:09:23but to basically uh do for a given set of Q let's say that we follow that we we
0:09:29we are looking for Q when where
0:09:30oh forwarding a possible solution for Q
0:09:33for this Q we are going and optimising in and then P S
0:09:37such that it maximises the set or C
0:09:41uh so this is this is the do know um
0:09:44and knapsack optimization problem
0:09:47one of the well known problems all of this is that a complexity it it has an exponential complexity
0:09:53uh it's uh
0:09:55you for larger number of of of and it's
0:09:58impossible possible to solve
0:10:00so of course
0:10:01to make it feasible and and usable they are looking for
0:10:04a a fast approximation algorithm that was still be a fish
0:10:09and um
0:10:10in this work we have proposed a three different
0:10:13a approximation algorithms
0:10:16each one has its advantages or disadvantages as we will discuss later
0:10:20and each one has different the a level of complexity of course
0:10:25uh the first one is uh
0:10:27yeah individual sectors you rate than a base
0:10:30really selection
0:10:32the second is a weighted norm really selection and the last one
0:10:37is a successive a a really uh selection algorithm that is based on
0:10:41what power uh start local search uh approach
0:10:46um we start for the first one
0:10:47the first one is is the
0:10:49think five one
0:10:51and what it it does it it relaxes the original position problem
0:10:56and says we not going to optimize it's for each vector
0:10:59what what are going to do we are going to
0:11:01first take each individual really
0:11:05are going to find a a
0:11:07the optimal value for a i and P S time
0:11:11that maximise the sectors C
0:11:12for this uh a specific one
0:11:15once we generate a set of uh of value
0:11:19we are so like think the the we are successively selecting them by
0:11:23they're value so we're selecting the highest one first and keep on adding
0:11:27uh by there
0:11:29and feel we we basically have some
0:11:31uh um threshold uh pass
0:11:38i don't note and how what what what is basically down we use this a figure to explain how
0:11:44we basically find uh the the palm maximum um
0:11:48oh values for P S and all my got to this process
0:11:52and you have your plot for to to really
0:11:55call really and really be
0:11:56and you can see them
0:11:58sec to see wait for each one of them as a function of the
0:12:01of P S
0:12:02and basically it it the the algorithm that to show identifies stills maximal point and the appropriate P S for
0:12:09which are then used
0:12:11a selection out
0:12:14uh the second approach
0:12:17is going if
0:12:18first we went for the individual one here we going on the full system
0:12:22so we assume we take all and really
0:12:24and we doing what has been that in previous work which is optimising
0:12:28the weight the vector or my guy and P S
0:12:30such that the total
0:12:32step C rate is all is met
0:12:35no when not
0:12:36repeating that that actually taking the values that we got
0:12:39are this optimal value
0:12:41and we are using them
0:12:43in our uh expression for the set to see so yeah we not
0:12:47calculate that again and again
0:12:49so we're saving uh on this process
0:12:52and uh we are selecting uh
0:12:55uh the the the larger one of course which use a basically equivalent to
0:12:58selecting the
0:13:00uh i
0:13:01a really is with their a larger storm and this is like like weighted norm uh
0:13:09the third one
0:13:10does not assume any uh
0:13:13relaxation on the original problem
0:13:15that's on basically takes the we problem
0:13:20is um
0:13:22in a sense
0:13:23in a greedy way
0:13:25and it's a a really is to the group
0:13:28so we have a starting point
0:13:30we we we start the let's say we do really one
0:13:34and we keep and adding really
0:13:37and each time we adding relays
0:13:40we are we
0:13:41going and maximizing uh the vectors
0:13:45uh W and P again
0:13:47and we are choosing the relay that
0:13:49maximise the sec receiver in this case
0:13:53now once we add a a a a a a ad
0:13:55the the one that maximises with keep on adding and till we cross trash
0:14:01for a
0:14:02a better performance in terms of of uh reading the optimum we we be
0:14:06this search for every possible forced really
0:14:09and this is one call
0:14:11a most people start
0:14:12yeah i am with people start a local search
0:14:17we eventually end up with them
0:14:20and optimal solutions
0:14:22and we choose the the one with the least number or
0:14:30sure an example of that
0:14:33by the way the the here though optimization used done using um
0:14:38an algorithm that was suppose to the proposed by um the at a bit trouble in web
0:14:44and two thousand ten at each that
0:14:48um um
0:14:49to show uh an example of that
0:14:52a you see C or scenario all of a source destination
0:14:56and and you drop or or and we assume we have a set of relays
0:14:59a a spreading a given area
0:15:03we assume fifteen uh really is are a given here
0:15:06um and
0:15:09uh we we normalize the sec to C rate compare sent to the maximum achievable one
0:15:14and we we used to threshold one of them is ninety percent of zero point nine five and seventy five
0:15:19zero point somebody five
0:15:22i you see for an example
0:15:24of of the results that we get for this scenario
0:15:29in in in black though it's not seen because the green is over that
0:15:33and i and like to use a a is a uh the results of an exhaustive search
0:15:37so basically if i would take for example for
0:15:40really is
0:15:41this points says
0:15:43if i would do an exhaustive search for the best for really is this would be the the value the
0:15:47to see way that that would get and so
0:15:51the third on go with him a very nice we'd we'd the optimal one
0:15:56uh uh we it for different scenario in in general it fold was very close to the optimal
0:16:01uh it also more we'll bossed uh there there was information in paper it also more boss to the location
0:16:07of the the sensors
0:16:09respect to
0:16:10if dropped or and
0:16:13uh in terms that it falls again candle
0:16:16uh a the on a two
0:16:17uh uh um
0:16:19there are behaviour is more uh reliant on location
0:16:23uh in many cases uh
0:16:25as expected algorithm uh be performs better than algorithm a a a board to many have the stalls
0:16:31slow reaction as you can see from here so
0:16:34um uh for for low uh
0:16:37low threshold if forms a we well but when the threshold is high as you can see here
0:16:42it will give fifteen sensors or something like that
0:16:44C right there
0:16:46uh it would give sort twelve
0:16:48a a sensor as the number of sensor that that we need for that
0:16:53we also checked or robust most of that too small error in the estimation of the channel
0:16:58uh between the source and you drop or
0:17:02and uh again the the algorithm C is falling that before forming quite well in in uh in this sense
0:17:08as well and most of them
0:17:10a stay within a reasonable before
0:17:13so to wrap things up uh
0:17:16i the problem of selecting mean set of really seeing um
0:17:20well wire score a cooperative system uh for the purpose of jamming
0:17:24uh has been formulated as a knapsack problem
0:17:28uh the first two algorithm of four or complexity all and uh
0:17:32order of L
0:17:34a while the um
0:17:35the third one is
0:17:37it's it's here
0:17:38it sort of a a L
0:17:39and time L and L is the number of a is we eventually end up
0:17:44having in the subset
0:17:53uh in terms of complexity what that explains is a little what this so uh in the sense that in
0:17:58small or for smaller or uh on
0:18:00a threshold
0:18:02uh i'll go "'em" one free form relatively good but in large a threshold it you can perform as well
0:18:08well do for like special algorithm be free phone
0:18:10or not converge conversion to to the optimal one and you know the case as well
0:18:15so there is a trade of your between performance of course and um
0:18:19and complexity not to large trade off but there is a tradeoff
0:18:23gives a few option of uh
0:18:26implementing that
0:18:27in real
0:18:29thank you