Time Optimal Spectrum Sensing
Garimella Rama Murthy, Rhishi Pratap Singh, Samdarshi Abhijeet, Sachin Chaudhary
aa r X i v : . [ c s . I T ] J un Time Optimal Spectrum Sensing
Garimella Rama Murthy , Rhishi Pratap Singh , Samdarshi Abhijeet , and Sachin Chaudhary Signal Processing and Communication Research CenterInternational Institute of Information Technology, Hyderabad, India Email: [email protected] Email: [email protected] Email: [email protected] Email: [email protected]
Abstract —Spectrum sensing is a fundamental operation incognitive radio environment. It gives information about spectrumavailability by scanning the bands. Usually a fixed amount of timeis given to scan individual bands. Most of the times, historicalinformation about the traffic in the spectrum bands is not used.But this information gives the idea, how busy a specific band is.Therefore, instead of scanning a band for a fixed amount of time,more time can be given to less occupied bands and less time toheavily occupied ones. In this paper we have formulated the timeassignment problem as integer linear programming and sourcecoding problems. The time assignment problem is solved usingthe associated stochastic optimization problem.
Index Terms —Spectrum Sensing, Pareto Front, Integer Pro-gramming, Source Coding, Stochastic Optimization.
I. I
NTRODUCTION
In recent years, Cognitive Radio technology [1] is proposedfor making efficient utilization of electromagnetic spectrum.At the physical layer of cognitive radio networks, varioustechniques are proposed for Spectrum Sensing [2]. One ofthe basic approaches to spectrum sensing is based on EnergyDetection. In earlier efforts of spectrum sensing, the temporalrecord/history of spectrum utilization has been completelyignored. Some researchers realized that such approach tospectrum sensing is sub-optimal[3]. The authors particularlyproposed Doubly Cognitive Network Architecture in whichIntelligent Spectrum Sensing is carried out by taking thehistorical data of spectrum utilization into account. In thisresearch paper, we make precise mathematical formulationof time optimal spectrum sensing and propose an interestingsolution.II. T
IME O PTIMAL S PECTRUM S ENSING : I
NTEGER L INEARPROGRAMMING
Consider a band of EM spectrum available for wirelesscommunication. Let this band be subdivided into sub-bandslabeled , , . . . , M . In traditional spectrum sensing basedon, say, energy detection, all the sub-bands are scanned witha fixed, constant time irrespective of the historical data aboutpacket traffic. It is logically clear that the sub bands whichare heavily occupied(based on historical traffic data) can bescanned faster(sensing time is chosen to be smaller) while theless occupied sub-bands can be scanned using larger sensingtime. The total available time for spectrum sensing of the entire band is assumed to be constant, say L seconds. Thesensing time allocated for each of the sub bands is assumedto be integer valued. Note:
The time optimal spectrum sensing problem formulatedbelow does not depend on the spectrum sensing approach.
Joint Detection-Estimation Approach to SpectrumSensing:
In the following discussion we formulate theproblem of prediction of packet traffic based on historicaldata as a Linear Mean Square Estimation problem. Alsoas in traditional spectrum sensing primary user detection isformulated as hypothesis testing based detection problem.As discussed earlier, we take the historical traffic dataon various sub-bands into account for choosing the spectrumsensing time. In this direction we model the historical trafficdata as an Auto Regressive(AR) process. In time, the unit onwhich the data is collected, could be an hour, day, month etc.Specifically, we fit a p th order AR process to the traffic data,i.e. x ( n + p ) = a x ( n ) + a x ( n + 1) + . . . + a p x ( n + p −
1) + w ( n + p ) (1)where using LMSE(Linear Mean Square Error estimation i.e.Solving Yule-Walker equations) method, the coefficients areestimated and the traffic data is predicted(on certain timeunit). Note:
The prediction tool(model) can be chosen to be moresophisticated (artificial Neural network based approach).Let the predicted data in M sub-bands be denoted by n , n , . . . , n M . We normalize the number of packets invarious sub-bands in the following manner q i = n i P Mj =1 n j for ≤ i ≤ M (2)Thus { q , q , . . . , q M } is a probability mass function,associated with packet traffic data in various sub-bands.Now, we formulate the time-optimal spectrum sensingproblem . Our goal is to allocate the total time for sensingthe entire band ( say L seconds ) into time for sensing sub- bands( i.e. T , T , . . . , T M ) such that the average sensing timei.e. ¯ T = M X i =1 T i q i with M X i =1 T i = L (3)is minimized. We reason below that if no constraints areimposed on { T i } , then we have a trivial problem. Case1:
In this case order { q i } from smallest value tolargest value, i.e. label them as { ˆ q , ˆ q , . . . , ˆ q M } Set T = L, T = 0 , . . . , T M = 0 . With such a trivial allocation, ¯ T is minimized. Case 2:
Minimum sensing time in any of thebands is lower bounded by T (i.e smallest sensingtime is at-least T ). Allocation can be as following: T , T , . . . , T , ( L − ( M − T ) with ( L − M T + T ) ≥ T Case 3:
Smallest sensing time is at-least T and othersensing times differ by at-least 1 time unit. Allocation can beas following: T , T + 1 , T + 2 , . . . , T + M − , ( L − S )) with ( L − S ) ≥ T , where S=( T ) + ( T + 1) + ( T + 2) + . . . , ( T + M − . Case 4:
Smallest sensing time is at-least T and othersensing times differ by at-least d time units. Allocation can beas following: T , T + d, T + 2 d, . . . , T + ( M − d, ( L − S )) with ( L − S ) ≥ T , where S=( T ) + ( T + 1) + ( T + 2) + . . . , ( T + M − .Thus we are naturally led to imposition of realistic (practical)constraints on the integer valued T i ’s. Case A: T i ’s are in arithmetic progression. i.e. T , T + d, . . . , T + ( M − d . These times must addup to total sensing time, L. Thus, we have M T + dM ( M − L M T + dM ( M −
1) = 2 L (4) Note:
In the above equation M, the number of sub-bandsand ’L’, the total sensing time are known. T , d areunknown variables. Since T , d are always constrained tobe integers, we have a linear Diophantine equation of theform aT + bd = 2 L , where a=2M and b=M(M-1) Thereare standard techniques for solving such an algebraic equation. Case B: T i ’s are in Geometric progression. i.e. T , ( T )( d ) , ( T )( d ) , . . . , ( T )( d M − ) . They must addup to total sensing time L. T + ( T )( d ) + ( T )( d ) + . . . + ( T )( d M − ) = LT (1 + d + d + . . . + ( d M − ) = LT ( d M − d − L (5)As discussed earlier M, L are known and T , d are unknown.Thus we need to solve the following algebraic equation T d M − Ld − ( T − L ) = 0 (6) Goal:
To solve the above algebraic equation for T , d supposewe assume that d = 2. Thus we have to decide ’ T ’ for T (2 M − − L = 0 Thus, for a given ’M’; T , (2 M − must be divisors of L. Ifnot, no solution exists. Suppose ’M’ is such that M − isa prime i.e. A Mersenne prime. If ’L’ happens to be a primenumber, no solution exists. (It should be noted that ’M’ mustnecessarily be a prime for M − to be a Mersenne prime).Thus in this case for a solution to exist ’L’ must be such thatits prime factorization contains the Mersenne prime M − .For a given M if L is divisible by M − , we have T = L (2 M − (7) Significance of this solution:
Energy detection is facilitatedby the use of FFT of certain length/size. Typically the FFTsizes are power of 2. Thus, ’d’ can be chosen to be a powerof 2, leading to explicit solution for T , i.e. T = L d − d M − (8) General Solution in Case B:
Factoring ’L’ gives all thepossibilities for T . A short computation will give the desiredsolutions, if any. Justification of AP/GP for sensing times:
As the probabilitiesdecrease, the increase in sensing times assume values in anAP i.e. the rate of increase of sensing times is linear, orSensing times increase geometrically (implemented by anFFT of suitable frequency resolution.) e.g. a, 2a, 4a, 8a, 16a, . . .
Case C: T i ’s are in Arithmetico-geometric sequence. i.e. T , ( T + d )( r ) , ( T + 2 d )( r ) , . . . , [ T + ( M − d ]( r M − ) .They must add up to total sensing time L. T + ( T + d )( r ) + ( T + 2 d )( r ) + . . . +( T + ( M − d )( r M − ) = LT [1 + r + r + . . . + r M − ] + dr [1+2 r + 3 r + . . . + ( M − r M − ] = LT (1 − r M )(1 − r ) + dr [ (1 − M r M − )(1 − r ) + ( r − r M )(1 − r ) ] = L (9)If common difference is equal to common ratio i.e d = r T (1 − d M )(1 − d ) + d [ (1 − M d M − )(1 − d ) + ( d − d M )(1 − d ) ] = L (10)Thus, the Diophantine equation whose solutions are of inter-est to us are given by above equations. Solutions must befeasible/Non-negative integer values of { T , d } . Note:
It can easily be reasoned that if d=0, the above equationreduces to (4) and if r=1, (by using L’Hospitals rule) theequation reduces to (8).
III. T
IME O PTIMAL S PECTRUM S ENSING : S
OURCE C ODING
In this section we relate the problem of Time OptimalSpectrum Sensing to the source coding problem.Let S i = n i for ≤ i ≤ M Compute ˆ p i = S i P Mj =1 S j for ≤ i ≤ M Let ˆ T = M X j =1 ˆ T j ˆ p j (11)Let X be the random variable assuming values { ˆ T , ˆ T , . . . , ˆ T M } with probabilities { ˆ p , ˆ p , . . . , ˆ p M } .Shannon Entropy of X is given by H ( X ) = − M X j =1 ˆ p j log ˆ( p j ) (12)Suppose we require the spectrum sensing times in various sub-bands i.e. { ˆ T i } Mi =1 to satisfy the Kraft inequality. i.e. M X i =1 − ˆ T i ≤ (13)Then we necessarily have the following lower bound onaverage sensing time i.e. ˆ T ≥ H ( X ) , where H(X) is theentropy of the random variable assuming values { ˆ T i } Mi =1 withthe probabilities { ˆ q i } Mi =1 . In this connection we have followinginteresting lemma. Lemma:
If the sensing times { ˆ T i } Mi =1 are increasing at-leastin an arithmetical progression with common difference 1 i.e. ˆ T = ˆ T + 1 , ˆ T = ˆ T + 2 , . . . , ˆ T M = ˆ T + ( M − then Kraftinequality is satisfied. Proof:
Refer [4]
Note:
It is immediate that if Kraft inequality is satisfied withD = 2 i.e. M X i =1 − ˆ T i ≤ then M X i =1 D − ˆ T i ≤ (14)for any D > . Also if { ˆ T i } Mi =1 increases faster thanArithmetic progression with common difference ONE (i.e. APwith common difference strictly greater than one or geometricprogression etc) then Kraft inequality is satisfied.Using Huffman coding we determine the values { ˆ T i } Mi =1 .Suppose they i.e. { ˆ T i } Mi =1 must add up to ’L’. Then the valuesof { ˆ T i } Mi =1 are scaled/ normalized such that P Mi =1 ˆ T i = 1 . IV. N
UMERICAL E XPERIMENTS
We now consider case A in section 2. We invoke thefollowing theorem on computing the solution of linearDiophantine Equation [5].
Theorem:
The linear Diophantine equation ax + by = chas a solution if and only if d | c (d divides c), where d isG.C.D.(a,b). Furthermore if ( x , y ) is a solution for thisequation, then the set of solutions of the equations consistof all integer pairs (x, y), where x = x + t ( b/d ) and y = y + t ( a/d ) ,where t= . . . ,-2,-1,0,1,2, . . . Note:
We can compute any one solution discussed inthe above theorem using Euclidean (G.C.D. Computation)algorithm. Q: How do we select the required solution? i.e. { T , d } should be non-negative. Q: What if there are multiple solutions for { T , d } ? Examples:
Case of linear Diophantine Equation case 1: aT + bd = 2 L , where a=2M, b=M(M-1) and L =Total sensing time.let M=10, L=100 T + 90 d = 200 GCD (20 ,
90) = 10 . (200 is divisible by 10.) T = 1 + t (90 / d = 2 − t (20 / (15)For t= . . . .-2,-1,0,1,2, . . . there are multiple solutions but thereis only one interesting solution is with t=0, T = 1 and d=2 case 2: if GCD(a,b) in aT + bd = 2 L is a. let M=15, T + 210 d = 1800 GCD (30 , . (1800 is divisible by 30.) T + 7 d = 60 (16)One solution can be d=8 and T =4 T = 4 + t (210 / d = 8 − t (30 / (17)For t= 0,1,2, . . . ,7 there are solutions. So there are multiplesolutions. Note:
The solution for { T , d } in case A and B is always amatching pair. Problem:
Suppose the number of solutions i.e. { T , d } in caseA, case B is strictly more than One. Goal:
We would like to arrive at solutions that minimizeboth the mean and variance of sensing time random variable.Suppose even after such optimization procedure, we arrive atmultiple solutions. Heuristically, some solutions are eliminatedon the basis of { T , d } that are too low or too high.V. T IME O PTIMAL S PECTRUM S ENSING : S
TOCHASTIC O PTIMIZATION
Case A: { q , q , q , . . . , q M } are unsorted probabilities. { p , p , p , . . . , p M } are sorted increasing probabilities. M ean = ˆ T = M X i =1 T i q i = X j ∈ R ˜ T j p j = E [ Z ] (18)where Z is spectrum sensing time random variable, ˜ T j ’s aresorted sensing time values and R is a suitable index set. E [ Z ] = X j ∈ R ˜ T j p j V ariance [ Z ] = E [ Z ] − [ E [ Z ]] (19) Note:
Minimizing E[Z] maximizes variance [Z]. Our goal isto minimize E[Z] as well as variance [Z] (Joint Optimization
Problem). We would like to arrive at a PARETO OptimalSolution.
Note:
Suppose q i ’s are all equal. Then T i = T , for ≤ i ≤ ME [ Z ] = M X j =1 ˜ T j p j = T (20) First Approach:
Suppose T i ’s are in arithmetical progression. E [ Z ] = M X j =1 [ ˜ T + ( j − d ] p j = ˜ T (1) + M X j =1 ( j − p j dE [ Z ] = ˜ T + ( µ )( d ) (21)where µ = P Mj =1 ( j − p j E [ Z ] = M X j =1 ( ˜ T j ) p j = M X j =1 [ ˜ T + ( j − d ] p j = M X j =1 [ ˜ T + ( j − d + 2( j −
1) ˜ T d ] p j = ˜ T + d M X j =1 ( j − p j + 2 ˜ T d M X j =1 ( j − p j = ˜ T + ( α ) d + (2 ˜ T d )( µ ) (22)where α = P Mj =1 ( j − p j var [ Z ] = ˜ T + ( α ) d + (2 ˜ T d ) µ − ( ˜ T + µ d + 2 µ ˜ T d )= ( α )( d ) − µ d = ( α − µ ) d (23) Note:
Optimal choice of { T , d } are decoupled. Thus, theproblem boils down to minimize E[Z] as well as var[Z]. Howcan we select the best solution? E [ Z ] = ˜ T + ( µ )( d ) var [ Z ] = ( α − µ ) d (24)where µ = P Mj =1 ( j − p j and α = P Mj =1 ( j − p j i.e. { µ, α } are determined by probabilities { ˆ p j } Mj =1 and arefixed / constants. Problem:
Determine ˜ T and ’d’ from possibly non-uniquesolutions for { ˜ T , d } (determined by Diophantine equation) Note: ˜ T does not effect var[Z] and only affects E[Z]. Sochoose minimum possible positive solution for ˜ T .Simultaneously minimize E[Z], var[Z] with respect to ’d’(treating ˜ T as constant.) E [ Z ] = f ( d ) = ˜ T + ( µ )( d ) var [ Z ] = ( α − µ ) d hence ( α − µ ) ≥ (25) Note:
If only mean needs to be minimized, choose the smallest ˜ T and matching value for d among pairs of solution of (4). Note:
It can easily be reasoned that, with ˜ T being chosen assmallest feasible value, d is chosen to be smallest matchingvalue from among all solutions of Diophantine equation 4. Lemma:
Unique optimal solution for d exists whereE[Z]=var[Z].
Proof:
For an optimal solution E [ Z ] = var [ Z ]˜ T + ( µ )( d ) = ( α − µ ) d ( α − µ ) d − µd − ˜ T = 0 ad + bd + c = 0 (26)where a = ( α − µ ) , b = − µ, c = − ˜ T b − ac > for d to be real. µ − α − µ )( − ˜ T ) > µ + 4( α − µ ) ˜ T > since ( α − µ ) > (27)The zeros are distinct, thus we are interested in the value of’d’ in the first quadrant. Thus,a unique optimal solution for’d’ is achieved. Q.E.D. Note:
We expect the optimization problem formulated in thetime-optimal spectrum sensing to arise in other applications.The above lemma provides solution.
Case B:
Suppose T i ’s are in geometrical progression. E [ Z ] = M X j =1 ˜ T j p j E [ Z ] = M X j =1 ( ˜ T d j − ) p j = ˜ T ( M X j =1 d j − p j ) E [ Z ] = M X j =1 ( ˜ T j ) p j = M X j =1 ( ˜ T d j − ) p j var [ Z ] = E [ Z ] − ( E [ Z ]) = ˜ T [ M X j =1 d j − p j ] − ˜ T [ M X j =1 d j p j ] = ˜ T [( M X j =1 d j − p j ) − ( M X j =1 d j p j ) ] (28) Note: E [ Z ] = ˜ T f ( d ) var [ z ] = ˜ T R ( d ) (29) where f ( d ) = M X j =1 d j p j and R ( d ) = ( M X j =1 d j − p j ) − ( M X j =1 d j p j ) = f ( d ) − [ f ( d )] Thus the optimal choice of minimal ˜ T will be optimalfor both E[Z] and var[Z]. But minimization of E[Z] withrespect to d will maximize var[z]. Thus we are interested inPareto Optimal Solution i.e. jointly optimal choice for ’d’ forminimizing E[Z] as well as var[z]. We now prove that if f(d)is minimized, R(d) is maximized. Claim:
If f(t) is minimized, then f ( t ) is maximized, whichleads to Pareto optimal solution.Suppose we consider the unconstrained optimiza-tion/minimization of f(t) thenLet K ( t ) = t ⇒ f ( t ) = f ( K ( t )) df ( K ( t )) dt = dfdk dkdt = ( dfdt )(2 t ) d f ( K ( t )) dt = d fdt (2 t ) + ( dfdt )(2) df ( K ( t )) dt = 0 if and only if dfdt = 0 (30)Further, the minima of f(.) are in the left half plane. Supposethey are real valued, e.g. to d fdt | t = t > with t < d f ( K ( t )) dt | t = t = d fdt (2 t ) | t = t + 0 Since t < t d f ( K ( t )) dt | t = t < (31) f ( t ) is maximized, when f(t) is minimized. Thus, we lookfor Pareto optimal solution for d i.e. denoted t here. So useclosest solution of 7 pair of T , d to Pareto optimal solution. Pareto Optimal Solution:Fixed Point Equation: E [ Z ] = ˜ T f ( d ) E [ Z ] = ˜ T g ( d ) E [ Z ] = var [ Z ] note that g(d) = f ( d )˜ T f ( d ) = ˜ T g ( d ) − ˜ T ( f ( d )) ˜ T f ( d ) − ˜ T ( f ( d )) − f ( d ) = 0 (32)Since f(d) is a polynomial in d, we have a polynomial equationwhich has multiple zeros. Q: How can we determine optimal ’d’?Choose smallest real ’d’ that is feasible.
Example:
Let M=3 f ( d ) = X j =1 d ( j − p j = p + dp + d p g ( d ) = f ( d ) = X j =1 d j − p j = p + d p + d p (33)Replace values in equation ˜ T f ( d ) − ˜ T ( f ( d )) − f ( d ) = 0 (34)Let ˜ T = 1 , p = . , p = . , p = . p + d p + d p ) − ( p + dp + d p ) − ( p + dp + d p ) = 0 After solving equations values for d = 2.43, -1.26, -0.20 ± ˜ T = 1 , p = . , p = . , p = . values for d = 2.53, -1.19, -0.06 ± ˜ T = 1 , p = . , p = . , p = . values for d = 2.69, -1.33, -0.34 ± Summary:Step1:
Based on data, predict the Probabilities related tospectrum band occupancy.
Step2:
Allocate Sensing Times in the order of probabilityvalues i.e. if a band is highly occupied (probabilistic), allocatesmaller sensing time and vice-versa.
Step3:
Assume that the sensing times are in arithmetic/Geometrical progression. Compute solution to the Integerprogramming problem (or the Diophantine equation.) If thereis more than one solution, we need to decide the solution thatmust be chosen.
Step4:
Find solution/solutions which minimize the mean,variance (assuming that the sensing time values are in AP/GP)and find unique/multiple solutions.In summary if the allocated times are in AP:1) If E[Z] and Var[Z] both require minimization, choosesmallest { T , d } pair solution to (4).2) If E[Z] is maximized and Var[Z] require minimization,it will lead to unique Pareto optimal solution.3) If Var[Z] is maximized and E[Z] require minimization,it will lead to unique Pareto optimal solution.if the allocated times are in GP:1) If E[Z] and Var[Z] both require minimization, choosePareto solution to d rounded & closest matching pair { T , d } solution to (8), with T chosen as small aspossible.2) If E[Z] is maximized and var[Z] require minimization,choose Pareto solution to d rounded & closest matchingpair { T , d } solution to (8), with T chosen as large aspossible.. Case C:
Suppose T i ’s are in Arithmetico-geometric progres-sion. E [ Z ] = M X j =1 ˜ T j p j E [ Z ] = M X j =1 [ ˜ T + ( j − d ] r j − p j = ˜ T M X j =1 r j − p j + d M X j =1 ( j − r j − p j = ˜ T f ( r ) + df ( r ) where f ( r ) = M X j =1 r j − p j and f ( r ) = M X j =1 ( j − r j − p j E [ Z ] = M X j =1 ( ˜ T j ) p j = ˜ T M X j =1 r j − p j + d M X j =1 [( j − r j − ] p j +2 T d M X j =1 ( j − r j − p j = ˜ T f ( r ) + d f ( r ) + 2 T df ( r ) where f ( r ) = M X j =1 r j − p j , f ( r ) = M X j =1 [( j − r j − ] p j and f ( r ) = M X j =1 ( j − r j − p j (35)Case: If r = d E [ Z ] = ˜ T M X j =1 d j − p j + M X j =1 ( j − d j p j = ˜ T f ( d ) + df ( d ) where f ( d ) = M X j =1 d j − p j and f ( d ) = M X j =1 ( j − d j p j E [ Z ] = ˜ T M X j =1 d j − p j + M X j =1 [( j − d j ] p j +2 T M X j =1 ( j − r j − p j = ˜ T f ( d ) + f ( r ) + 2 T f ( r ) where f ( d ) = M X j =1 d j − p j , f ( d ) = M X j =1 [( j − d j ] p j and f ( d ) = M X j =1 ( j − r j − p j Variance var [ Z ] = E [ Z ] − ( E [ Z ]) (36)For a Pareto Optimal Solution E [ Z ] = var [ Z ] E [ Z ] = E [ Z ] − E [ Z ] (37)Keep values from equation (35) and (36) and solve thefunctional equation in { T , d } .TODO: Numerical Experiments Case D: Generalization:
Sensing times form an increasing sequence (not necessarilyAP/GP). ˜ T , ˜ T , . . . , ˜ T M are such that ˜ T + ˜ T + . . . + ˜ T M = L (38)where ˜ T i > for ≤ i ≤ M We have a constrained partition problem (as in NumberTheory.) i.e. Find all possible solutions of partition problemand prune out unsuitable solutions based on some criterion. ˜ T i < ˜ T j if j > i With this constraint only, the number of possible solution needto be computed.
Case 1:
M < L (Most interesting case) L ( L − . . . ( L − ( M − L !( L − M + 2)! (39)possible solutions when there is no further constraint on valuesof . We don’t worry about other case M > L .Note: We can have a lower bound of sensing time allocatedin any of the sub-bands i.e. ˜ T i > s for ≤ i ≤ M Max number of solutions = ( L − s )( L − s − . . . ( L − s − ( M − L − s )( L − s − . . . ( L − s − M + 1)= ( L − s )!( L − s − M + 2)! (40) Effective Idea:
The most general choice of sensing times (increasing numbers)leads to the constrained partition problem. Further the sensingtimes must minimize the mean as well as variance of thesensing time random variable.The above discussion naturally leads to the following moreinteresting optimization problems (related to joint optimiza-tion of moments of a discrete random variable.) Let ’Z’ bea random variable assuming values { T , T , . . . , T M } withprobabilities { q , q , . . . , q M } respectively. E [ Z ] = M X i =1 T i q i (41)Let { T i } Mi =1 be the unknowns and { q i } Mi =1 are known con-stants. Then the mean and variance of the random variableare given by E [ Z ] = M X i =1 T i q i = f ( T , T , . . . , T M ) = f ( ¯ T ) var [ Z ] = E [ Z ] − ( E [ Z ]) = g ( T , T , . . . , T M ) = g ( ¯ T ) (42) Goal:
To see if we can optimize E[Z], var[z] jointly. Q: Do we have an interesting functional equation arising inthe joint optimization of E[Z], var[Z] ? E [ Z ] = var [ Z ] E [ Z ] = E [ Z ] − ( E [ Z ]) E [ Z ] − E [ Z ] − ( E [ Z ]) = 0 letting E [ Z ] = h ( T , T , . . . , T M )= f ( T , T , . . . , T M ) (43)The multivariate functional equation that must be solved isgiven by f ( T , T , . . . , T M ) − f ( T , T , . . . , T M ) − ( f ( T , T , . . . , T M )) = 0 (44)Is there a solution to such a functional equation? Mostly itconstitutes the Pareto Front(Non-Dominating solution set).VI. M ULTI - OBJECTIVE O PTIMIZATION : L
INEAR AND Q UADRATIC P ROGRAMMING (H YBRID P ROGRAMMING ) Objective Functions: C = [ p , p , . . . , p M ] T T = [ T , T , . . . , T M ] T D = diag { p , p , . . . , p M } E [ Z ] = C T T = T T C = C.TV ar [ Z ] = T T ¯ DT − ( C T T ) = T T ¯ DT − ( T T C ) = T T ¯ DT − T T CC T T = T T ( ¯ D − C T C ) T = T T ¯ GT ,where ¯ G = ¯ D − CC T (45)Example: ¯ G = ¯ D − CC T ¯ G = p p
00 0 p − p p p p p p p p p p p p p p p = p (1 − p ) − p p − p p − p p p (1 − p ) − p p − p p − p p p (1 − p ) (46)Inferences: • ¯ G is a laplacian like matrix. • - ¯ G is a symmetric generator matrix.Function of interest for arriving at solutions whereE[Z]=var[Z]: J ( T ) = V ar [ Z ] − E [ Z ]= T T ¯ GT − T T C = (cid:2) T T (cid:3) (cid:20) ¯ G − C T (cid:21) (cid:20) T (cid:21) (47) Note: ¯ G ¯ e = ¯0¯ G = ¯ G T Note:
We use Laplacian and Laplacian like matrix interchange-ably.
Theme:
Laplacian matrix arising in variance optimization of a discrete random variable. Q: Can (linear algebraic) properties of matrix G be capitalizedto derive new results on variance minimization?
Goal:
To study properties of laplacian like matrix ¯ G =¯ D − ¯ CC T where ¯ D = diag { p , p , . . . , p M } , ¯ C =[ p , p , . . . , p M ] T , ¯ G = ¯ G T
1) Eigen values are all real.2) ¯ G is positive semidefinite, with an eigen value at zero( ¯ e . . . all ones vector). ¯ e is in the null space of ¯ G .3) 0 is the smallest eigen value and all other eigen valueslie on real axis.4) Bounds on spectral radius of ¯ G n X j =1 G ij = n X j =1 D ij − p j X ( p + . . . + p j − + p j + . . . + p M )= p j − p j = 0 n X j =1 | G ij | = p j (1 − p j ) + | p j ( p + . . . + p j − + p j +1 + . . . + p M ) | = p j (1 − p j ) + | p j (1 − p j ) | = 2 p j (1 − p j )min j { p j (1 − p j ) } ≤ Spectral radius(G) ≤ max j { p j (1 − p j ) } (48)All eigen values of G lie in the interval [0,1). Note: ˆ G = − ¯ G is a generator matrix. ˆ G/θ + I = P , stochastic matrix. I − G/θ = Pµ . . . eigen value of ˆ G.λ . . . eigen value of P .θ . . . largest diagonal element of ˆ G.ǫ . . . eigen value of G .ǫ . . . Sp ( G ) .µ/θ + 1 = λ and µ = − ǫµ = 0 , so λ = 1 (49)Thus, by Perron Frobenius theorem, the dimension of nullspace of ¯ G is one (with ¯ e = (cid:2) . . . (cid:3) T ) i.e. all onescolumn vector using null space. Computation of determinant and trace of G: ¯ G ¯ C = ¯ D ¯ C − ¯ C ( M X j =1 p i )= p p ... p M − δ p p ... p M = p − δp p − δp ... p M − δp M ¯ G = ¯ D − ¯ C ¯ C T = ¯ D [ I − ¯ D − ¯ C ¯ C T ] Det ( ¯ G ) = Det ( ¯ D ) Det [ I − ¯ D − ¯ C ¯ C T ] Note:
Det ( ¯ G ) = 0 ¯ D − ¯ C ¯ C T = /p . . .
00 1 /p . . . ... ... ... ... . . . /p M ¯ C = p p ... p M ; ¯ D − ¯ C = p /p p /p ... p M /p M = ... ¯ D − ¯ C ¯ C T = ... (cid:2) p p . . . p M (cid:3) = F (50)It is a rank one matrix.From Kailath (”Linear Systems”), page 658, we have that ifA is a rank one matrix Det ( I + A ) = 1 + trace ( A ) Determinant:
Det (¯( I ) + ( − F )) = 1 + trace ( − F )= 1 − ( p + p + . . . + p M ) = 0 Det ( G ) = Det ( D ) Det ( I − D − ¯( C ) ¯ C T ) = N Y i =1 λ iN Y i =1 λ i = (cid:0) p p . . . p M (cid:1) (1 − p − p . . . − p M ) = 0 (51) Trace:
T race ( G ) = T race ( D ) − T race ( ¯ C ¯ C T ) N X i =1 λ i = ( p + p . . . + p M ) − ( p + p . . . + p M )= N X i =1 λ i > (52) Properties of laplacian type matrix arising in varianceexpression of a Discrete Random Variable Z:
V ar [ Z ] = ¯ T T G ¯ T ≥
1) Global minimum occurs at the eigen vector(s) of Gcorresponding to zero eigen value(null space of G).2) Let µ ≥ µ ≥ . . . ≤ µ N − ≤ µ N = 0 . Non-zerominimum value of Var[Z] is determined by the eigenvector ¯ R , corresponding to second smallest eigen value µ N − . G ¯ R = µ N − ¯ R ¯ R T ¯ R = µ N − ¯ R T ¯ R = µ N − ( || ¯ R || ) ,where || ¯ R || is the L -norm of ¯ G (53) Note:
Optimization of Var[Z] requires specification ofconstraint set on ¯ T . Example: || ¯ T || = 1 ⇒ ¯ R T ¯ G ¯ R = µ N − > (54)3) Using Rayleigh’s theorem, when constraint set is eu-clidean hyper sphere µ is the maximum value.4) Similar results are derived when the constraint set is unithypercube, lattice. Results related to Laplacian like matrix G: ¯ G = ¯ D − ¯ C ¯ C T is symmetric laplacian like matrix.1) T race ( ¯ G ) = N X i =1 ( p i − p i )= N X i =1 p i − N X i =1 p i = 1 − N X i =1 p i = 1 − Tsallis entropy of ( ¯ P = ( p , p , . . . , p N ) )Tsallis entropy and Shannon entropy are related. T race ( ¯ G ) = 1 − N X i =1 p i ≥ Note that µ i ∈ (0 , N − X i =1 µ i (55)2) Det(G) = 0 , other coefficients of characteristic polyno-mial may easily be computed.3) Computation of eigen values of G G = P N − i =1 µ i ¯ f i ¯ f iT , where f i ’s are the right eigenvectors of ¯ G . Q: How do we compute f i ’s.4) G is sub-stochastic since µ i ∈ [0 , G n n ↑ ∞−−−→ ,where G = ¯ D − ¯ C ¯ C T Using matrix binomial theorem, G n can be explicitly computed. ¯ G = ¯ D − ˆ C ,where ˆ C is rank one matrix. ˆ C = ¯ C ¯ C T ¯ C ¯ C T = ( ¯ C T ¯ C ) ¯ C ¯ C T = ( N X i =1 p i )( ¯ C ¯ C T ) = α ( ¯ C ¯ C T )ˆ C m can be computed for m ≥ C m = ( α ) m − ( ¯ C ¯ C T ) AlsoD m = diag { p m , p m , . . . , p N m } Using expression for G M , we can compute T race ( G m ) = N − X i =1 ( µ i ) m ,for m ≥ (56)5) Using Leverrier - Fadeev algorith, all the coefficientsof characteristic polynomial of G could be computedefficiently. They involve { p , p , . . . , p N } .VII. F UTURE W ORK
Consider the packet arrivals to each secondary user con-stitute a Poisson process. Let these packet streams be in-dependent. Also, let there be ’K’ channels available forcommunication. Let the service times (for transmitting thepackets from the secondary users) be exponential randomvariables. Thus, we model the associated Queuing systemto be an M/M/K queue. Using standard results in queuingtheory, various performance measures can be computed andinterpreted.
More General Stochastic Model:
By associating channelstates, a more general model based on Quasi Birth and Deathprocess is being developed and analyzed.VIII. C
ONCLUSION