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Chapter 3. Delay models in Data Networks. Section 3.2. Little`s Theorem. 3.2 Little`s Theorem. : average number of customers in system : mean arrival rate T:mean time a customer spends in system. Little`s Theorem. Proof N( t ) = number of customers in system at time t
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Chapter 3 Delay models in Data Networks
Section 3.2 Little`s Theorem
3.2 Little`s Theorem • : average number of customers in system • : mean arrival rate • T:mean time a customer spends in system
Little`s Theorem • Proof • N(t) = number of customers in system at time t • (t) = number of customers who arrived in interval [0,t] • Ti = time spent in system by the i-th customer
3.2.3 Application of Little`s Theorem • Ex3.1 • : arrival rate in a transmission line • NQ : average number of packets waiting in queue • W : average waiting time spent by a packet in queue NQ = W
Application of Little`s Theorem • If = average Tx time • = • : Average number of packets under Tx • I.e. fraction of time that s busy utilization factor
Application of Little`s Theorem • Ex3.2 • N : average number packets in network • T : average delay per packet also • Ti : average delay of packets arriving at node i
3.3 M/M/1 Queuing System • M/M/1 • First M : arrival , Poisson • Second M : service , Exponential • 1 : server number
M/M/1 Queuing System • Arrival Poisson process • A(t) : number of arrivals from 0 to time t • Number of arrivals that occur in disjoint intervals are independent • Number of arrivals in any interval of length is Poisson distributed with parameter ,
M/M/1 Queuing System • Properties of Poisson process • Inter arrival times are independent and exponentially distributed with parameter tn : time of the n-th arrival
M/M/1 Queuing System • For every t0, 0
M/M/1 Queuing System • A = A1+A2++AK is also Poisson with rate = 1+ 2++ K Poisson A1 merge A2 ……….. AK
M/M/1 Queuing System P Also Poisson with P split Poisson Poisson with (1-P) 1-P
M/M/1 Queuing System • Service time : Exponential distribution with parameter • Sn : service time of n-th customer
M/M/1 Queuing System • Properties of Exponential : memoryless
Markov chain formulation • Let's focus at the times,0,,2,…,k,… • Nk = number of customers in system at time k = N(k) • Where N(t) is continuous-time Markov Chain • Nk is discrete-time • Let Pij : transition probabilities = P{Nk+1=j|Nk=i}
Markov chain formulation • Note • During any time interval, the total number of transitions from state n to n+1 must differ from the total number of transitions from n+1 to n by at most 1 • I.e. frequency of transitions from n+1 to n = frequency of transitions from n to n+1
Markov chain formulation Take ->0 Pn=Pn+1 Pn+1=Pn, n=0, 1, …等比數列 where = / utilization Pn+1= n+1P0, n=0,1,… Since <1, and
M/G/1 System Let Ci: customer I Wi = waiting time of Ci Xi = service time of Ci Ni = # of customers found waiting in queue when Ci arrives Ri = residual service time of the customer in service when Ci arrives
M/G/1 System Xi- Ni Ri Xi-1 Ci start service Ni Ci arrives In steady-state,
M/G/1 System To calculate R, by graphical approach: Residual service time r() M(t)=# of service completion in [0, t] X1 X2 XM(t) Time X2 X1 XM(t) Ci starts service t
M/G/1 System Time avg of r() in [0, t]
M/G/1 System P-K Formula (3.53)
Ex3.15 Consider a go back n ARQ: sender 1 2 3 … n-1 n 1 time Timeout (n-1) frames 1 2 3 receiver time Prop. delay • Assume that error in the forward channel is p, return channel is error-free • Packet arrives as a Poisson process with rate packets/frame
Ex3.15 Service time X : from when a packet transmitted until it is successfully received 1 , if 1st tx is successful (1-p) X={ 1+n, if 1st tx is un- successful; 2nd is successful p(1-p) 1+kn, if 1st k is un- successful;(k+1)th successful Pk(1-p)
3.5.1 M/G/1 Queue with vacations • When the server has served all customers, it goes on vacation • If the system is still idle after a vacation interval, go on another vacation interval • If a customer arrives during a vacation, customer waits until the end of vacation. Chapter 1 section 1.3.1 page 34in Network or Transport Layer
M/G/1 Queue with vacations • Assume vacation intervals v1, v2… are iid and are independent of customers arrival & service times. →A customer must wait for the completion of the current service or vacation interval, and then the service of all customers waiting before it.
M/G/1 Queue with vacations • Where R is the mean residual time for completion of service or vacation when the customer arrives.
M/G/1 Queue with vacations Let L(t) = # of vacations completed by t M(t) = # of services completed by t
Ex3.16 : FDM, SFDM, TDM • m streams of traffic with rate /m(Poisson) • FDM system – Divide available bandwidth into m subchannels. Transmission time for a packet on each of these subchannels is m.
Slotted FDM System • Packet trans starts only at time m, 2m,…When the queue is idle, server takes a vacation of m. (if idle again after vacation, take another)
TDM System • Look at SFDM queue, ->same queue • WTDM=WSFDM
Summary Service time
Reservations & Polling Satellite Collision -> solution:polling or reservation … S1 D1 D1 S2 D2 S1 D1 D1 S2 D2 Cycle
Reservation & Polling • M Poisson traffic streams with rate /m • Gated System – only those packets which arrive prior to the user’s preceding reservation period are transmitted. • Exhaustive system – all packets are transmitted including those that arrive during this data period • Partially gated – all packets that arrive up to the beginning of the data interval.
Single-User Gated system: m=1 Di starts Di arrives Wi time S D D S D D … D Di ends tx Ri Vl(I) l(i)-th reservation interval Ni : # of packets arrive in front of Di
Single-User • A reservation(vacation)starts when the system has served all packets which arrive prior to the preceding reservation interval. • A vacation(M/G/1 queue with vacation) starts when the system has served all packets which have arrived.(corresponds to exhaustive system)