Wireless Ad Hoc and Sensor Networks
Wireless Ad Hoc and Sensor Networks Wireless Ad Hoc and Sensor Networks
Distributed Fair Scheduling in Wireless Ad Hoc and Sensor Networks 313This equation is a stable linear system (Brogan 1991), driven by abounded input uk ( ) (see Remark 2). According to linear system theory(Brogan 1991), xk ( ) converges close to its target value in a finite time.LEMMA 7.3.3If flow f is backlogged throughout the interval [ t , t ], then in an ADFS wireless nodewhere v = v( t ) and v = v( t ).1 1max.( v −v ) −l ≤W ( t , t ) , (7.30)PROOF The steps in the proof follow similar to the method of Goyalet al. (1997).maxBecause Wf ( t1, t2)≥ 0 , if φ f , l ( v2 −v1)−lf ≤0, Equation 7.30 holds trivially.maxlHence, consider the case where φ f , l ( v2 −v1)− l f > 0, i.e., v v f2 > 1+ max. Letφkf , lpacket p f be the first packet of flow f that receives service in the openinterval ( v1, v2). To observe that such a packet exists, consider the followingtwo cases:nnnCASE 1 Packet p f such that Sp ( f )< v 1 and Fp ( f )> v 1 exists.n+1Because flow f is backlogged in [ t1, t2], we conclude vAp ( ( f )) ≤ v1. FromEquation 7.7 and Equation 7.8, we get:1 2φ f , l 2 1 f f 1 22 2n+1Sp ( f )= Fp ( nf )maxn nl fnBecause Fp ( ) ≤ Sp ( ) + and Sp ( )< v 1 , we get:φfff,lf(7.31)n+1l fSp ( f )< v1+φ< v 2maxf,l(7.32)(7.33)n+1 nBecause Sp ( f ) = Fp ( f ) > v1, using Equation 7.33, we concluden+1Sp ( ) ∈( v, v).f1 2nnnCASE 2 Packet p f such that Sp ( f )= v 1 exists p f may finish service at timet< t 1 or t≥ t 1 . In either case, because the flow f is backlogged in [ t1, t2],n+1n+1 nvAp ( ( f )) ≤ v1. Hence, Sp ( f ) = Fp ( f ).nnBecause Fp ( ) ≤ Sp ( ) + φand Sp ( f )< v 1 , we get,fmaxn l ff f , ln+1l fSp ( f )≤ v1+φ< v 2maxf,l(7.34)(7.35)
314 Wireless Ad Hoc and Sensor Networksn+1 nn+1Because Sp ( f ) = Fp ( f ) > v1, using Equation 7.35, we conclude Sp ( f ) ∈( v1, v2).Because either of the two cases always holds, we conclude that packetkkp f such that Sp ( f ) ( v, v)exists. Furthermore from Equation 7.32 andEquation 7.34, we get∈ 1 2Sp vlkf( f )≤ +φ1maxf,l(7.36)k+mLet p f be the last packet to receive service in the virtual time interval( v , v ). Hence,1 2F pk +( mf )≥ v 2(7.37)From Equation 7.36 and Equation 7.37, we concludemaxl f( f )− ( f )≥ 2 − 1 −φ f,lk+mF p S p k ( v v )(7.38)But because flow f is backlogged in the interval ( v1, v2), from Equation 7.7and Equation 7.8 we known=m k+nk+mlF pfS pkf( )= ( f )+ ∑ φ f , ln=0(7.39)n=m k+nk+mlF pfS pkf( )− ( f )= ∑ φ f , ln=0(7.40)Hence, from Equation 7.38 and Equation 7.40 we getn=m k+nl fl f∑ ≥ ( v2 − v1)φ− φn=0f,lmaxf,l(7.41)n=mk+n∑lf≥ φ f lv2 − v1−, ( ) ln=0maxf(7.42)
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Distributed Fair Scheduling in <strong>Wireless</strong> <strong>Ad</strong> <strong>Hoc</strong> <strong>and</strong> <strong>Sensor</strong> <strong>Networks</strong> 313This equation is a stable linear system (Brogan 1991), driven by abounded input uk ( ) (see Remark 2). According to linear system theory(Brogan 1991), xk ( ) converges close to its target value in a finite time.LEMMA 7.3.3If flow f is backlogged throughout the interval [ t , t ], then in an ADFS wireless nodewhere v = v( t ) <strong>and</strong> v = v( t ).1 1max.( v −v ) −l ≤W ( t , t ) , (7.30)PROOF The steps in the proof follow similar to the method of Goyalet al. (1997).maxBecause Wf ( t1, t2)≥ 0 , if φ f , l ( v2 −v1)−lf ≤0, Equation 7.30 holds trivially.maxlHence, consider the case where φ f , l ( v2 −v1)− l f > 0, i.e., v v f2 > 1+ max. Letφkf , lpacket p f be the first packet of flow f that receives service in the openinterval ( v1, v2). To observe that such a packet exists, consider the followingtwo cases:nnnCASE 1 Packet p f such that Sp ( f )< v 1 <strong>and</strong> Fp ( f )> v 1 exists.n+1Because flow f is backlogged in [ t1, t2], we conclude vAp ( ( f )) ≤ v1. FromEquation 7.7 <strong>and</strong> Equation 7.8, we get:1 2φ f , l 2 1 f f 1 22 2n+1Sp ( f )= Fp ( nf )maxn nl fnBecause Fp ( ) ≤ Sp ( ) + <strong>and</strong> Sp ( )< v 1 , we get:φfff,lf(7.31)n+1l fSp ( f )< v1+φ< v 2maxf,l(7.32)(7.33)n+1 nBecause Sp ( f ) = Fp ( f ) > v1, using Equation 7.33, we concluden+1Sp ( ) ∈( v, v).f1 2nnnCASE 2 Packet p f such that Sp ( f )= v 1 exists p f may finish service at timet< t 1 or t≥ t 1 . In either case, because the flow f is backlogged in [ t1, t2],n+1n+1 nvAp ( ( f )) ≤ v1. Hence, Sp ( f ) = Fp ( f ).nnBecause Fp ( ) ≤ Sp ( ) + φ<strong>and</strong> Sp ( f )< v 1 , we get,fmaxn l ff f , ln+1l fSp ( f )≤ v1+φ< v 2maxf,l(7.34)(7.35)