Wireless Ad Hoc and Sensor Networks

Wireless Ad Hoc and Sensor Networks Wireless Ad Hoc and Sensor Networks

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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)

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)

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