Page 124 - ITU Journal Future and evolving technologies Volume 2 (2021), Issue 1
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ITU Journal on Future and Evolving Technologies, Volume 2 (2021), Issue 1




          Also, in OFDM communication, the transmitter’s strategy  only if the following condition holds:
          becomes uniform in a high SINR mode, i.e., large trans‑
                                                                                      
          mission power, [25], while in the MLCC problem the jam‑    ℒ   ,   (  )  =           +           −      = 0,     > 0,
                                                                                                             
          mer’strategytendstoauniformoneifeitherthetotaljam‑               (        + ℎ    )       ℎ    2  ≤ 0,     = 0.
                                                                                                             
                                                                                        
                                                                                  
                                                                                                 
          ming resource budget is large, or the total transmission                                          (48)
          resource budget is small.                            By (47), the NLP (13) is a convex problem. Thus, to solve
                                                               the NLP (13) we introduce Lagrangian ℒ (  ) with    is
                                                                                                     ,  
          8.  CONCLUSIONS                                      a Lagrange multiplier as follows: ℒ   ,   (  ) = −  (  ,   ) +
                                                                         
                                                                      − ∑       . Then, similarly, for a  ixed    ∈   ,    ∈   
          A problem of multi‑link communication connectivity un‑         =1    
          der jamming of a transmitter with a group of receivers  is the best response if and only if the following condition
          when the channels are affected by Rayleigh fading has  holds:
          been formulated as a zero‑sum power resource alloca‑       ℒ   ,   (  )            = 0,     > 0,
                                                                                                     
          tion game. Existence and uniqueness of the equilibrium           =             −                  (49)
                                                                                      + ℎ     ≤ 0,     = 0.
                                                                                           
                                                                                     
                                                                                                     
          in power allocation strategies have been proven. Thus,
          in contrast to Colonel Blotto games, if channels are af‑  By (48), we have that
          fected by Rayleigh fading, then the stability of commu‑
          nication connectivity in a multi‑link system can be main‑                 > 0 for any   .         (50)
                                                                                    
          tained without introducing a random factor for a decision  Then, by (48) and (49),    > 0 and    > 0 correspondingly.
          maker. Also, the problem of designing the equilibrium  By (48), only two cases arise to consider: (I)    > 0,    = 0
                                                                                                       
                                                                                                             
          power allocation strategies has been reduced to the prob‑  and (II)    > 0,    > 0.
                                                                                
                                                                         
          lem of  inding a  ixed point of a real‑valued function. An  (I) Let    > 0 and    = 0. Then, by (48),
                                                                        
                                                                                  
          algorithm based on the bisection method for  inding the
           ixed point has been developed and its convergence has                                
          been proven.                                                               =    ℎ    .            (51)
                                                                                      
                                                                                            
          ACKNOWLEDGEMENT                                      Substituting (51) into (49) implies
          This work was supported in part by the U.S. National Sci‑                     2        ≤    .     (52)
                                                                                           2
          ence Foundation under Grant ACI‑1541069, Grant CCF‑                        ℎ
                                                                                         
          1908308 and Grant CNS‑1909186.
                                                               (II) Let    > 0 and    > 0. Then, by (48) and (49), we have
                                                                         
                                                                                  
                                                               that
                                                                                                   
                                                                                       
          9.  APPENDIX                                                                    +     =           (53)
                                                                           (        + ℎ    )       ℎ    2
                                                                                   
                                                                                         
                                                                                                  
          9.1 Proof of Theorem 1                               and
                                                                                                   
                                                                                       + ℎ    =  .          (54)
          Note that   (  ,   ) is an additively separable function of                            
          (   ,    ),    ∈    and                              Thus,
                
             
                                                                                     1    ℎ   
              2
                                                                                               
                (  ,   )  = −          (        + 2ℎ    )  −  2         < 0 (46)     =      −              .  (55)
                                                                                    
                                           
                                    
                               
                                     2 2
                      2  (        + ℎ    )        ℎ    3       Substituting (55) into (53) implies
                                                   
                                 
                                       
          and                                                                 1              ℎ   
                                    2 2
                      2
                         (  ,   )                                                    +  ℎ    2  =    +             .  (56)
                                                                                        
                           2     =  (        + ℎ    ) 2  > 0.  (47)  Solving (56) on    implies that
                                            
                                       
                                                                                
          Thus,   (  ,   )isconcavein   andconvexin  , andtheresult
          follows from the Nash’s theorem [2] since sets    and    of        1 +  1 + 4              +  ℎ       
          feasible strategies of the transmitter and the jammer are                     ℎ              
          compact.                                                          =              ℎ          .     (57)
                                                                            
                                                                                   2     +        
                                                                                                 
          9.2 Proof of Proposition 1                           Thus,
          By (46), the NLP problem (12) is a concave problem. Thus,
          to solve the NLP (12) we introduce Lagrangian ℒ   ,   (  )           1 +  1 + 4               +  ℎ       
          with    is a Lagrange multiplier: ℒ   ,   (  ) =   (  ,   ) +  1   ℎ            ℎ              
                                                                       =  −                            .    (58)
                                                                       
                 − ∑   =1       . Then, for a  ixed    ∈   , following [21]                 ℎ   
                       
          and the KKT Theorem,    ∈    is the best response if and                   2     +             
          108                                © International Telecommunication Union, 2021
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