Networking Theory and Fundamentals - Lectures 9 & 10 pps

1 TCOM 501: Networking Theory & Fundamentals Lectures 9 & 10 M/G/1 Queue Prof. Yannis A. Korilis 10-2 Topics  M/G/1 Queue  Pollaczek-Khinchin (P-K) Formula  Embedded Markov Chain Observed at Departure Epochs  Pollaczek-Khinchin Transform Equation  Queues with Vacations  Priority Queueing 10-3 M/G/1 Queue  Arrival Process: Poisson with rate λ  Single server, infinite waiting room  Service times:  Independent identically distributed following a general distribution  Independent of the arrival process  Main Results: Determine the average time a customer spends in the queue waiting service (Pollaczek-Khinchin Formula) Calculation of stationary distribution for special cases only 10-4 M/G/1 Queue – Notation  i W : waiting time of customer i  i X : service time of customer i  i Q : number of customers waiting in queue (excluding the one in service) upon arrival of customer i  i R : residual service time of customer i = time until the customer found in service b y customer i completes service  i A : number of arrivals during the service time i X of customer i Service Times  X 1 , X 2 , …, independent identically distributed RVs  Independent of the inter-arrival times  Follow a general distribution characterized by its pdf () X f x , or cdf () X F x  Common mean []1/EX = µ  Common second moment 2 []EX 10-5 M/G/1 Queue State Representation:  {(): 0}Nt t≥ is not a Markov process – time spent at each state is not exponential  R(t) = the time until the customer that is in service at time t completes service  {( ( ), ( )) : 0}Nt Rt t≥ is a continuous time Markov process, but the state space is not a countable set  Finding the stationary distribution can be a rather challenging task Goals:  Calculate average number of customers and average time-delay without first calculating the stationary distribution  Pollaczek-Khinchin (P-K) Formula: 2 [] [] 2(1 [ ]) EX EW E X λ = −λ  To find the stationary distribution, we use the embedded Markov chain, defined by observing ()Nt at departure epochs only – transformation methods 10-6 A Result from Probability Theory Proposition: Sum of a Random Number of Random Variables  N: random variable taking values 0,1,2,…, with mean []EN  X 1 , X 2 , …, X N : iid random variables with common mean []EX Then: 1 [][][] N EX X EX EN++ = ⋅ Proof: Given that N=n the expected value of the sum is 111 |[][] Nnn jjj jjj EXNnEX EXnEX ===  == = =  ∑∑∑ Then: 11 11 1 |{}[]{} [] { } [][] NN jj jj nn n EX EXNnPNnnEXPNn EX nPN n EXEN ∞∞ == == ∞ =   = =× == ⋅ =   === ∑∑∑ ∑ ∑ 10-7 Pollaczek-Khinchin Formula  Assume FCFS discipline  Waiting time for customer i is: 1 12 i i i ii i i iQ i j jiQ WRX X X R X − −− − =− =+ + ++ =+ ∑   Take the expectation on both sides and let i →∞ , assuming the limits exist: 1 [] [] [] [][] [] [] [][] i i ii j i i jiQ EW ER E X ER EX EQ EW E R E X EQ − =−  = +=+⇒  =+ ∑  Averages E[Q], E[R] in the above equation are those seen by an arriving customer.  Poisson arrivals and Lack of Anticipation: averages seen by an arriving customer are equal averages seen by an outside observer – PASTA property  Little’s theorem for the waiting area only: [] [ ] E QEW = λ  [] [] [] [] [] [] [] 1 E R EW E R E X EW R EW EW=+λ⋅ =+ρ ⇒ = −ρ  [] /EXρ=λ =λ µ : utilization factor = proportion of time server is busy 0 []1 E Xp ρ =λ = − Calculate the average residual time: [ ] lim [ ] i i E RER →∞ = 10-8 Average Residual Time 2 X 1 X 1 X () D t X t ()Rt  Graphical calculation of the long-term average of the residual time  Time-average residual time over [0,t]: 1 0 () t tRsds − ∫  Consider time t, such that R(t)=0. Let D(t) be the number of departures in [0,t] and assume that R(0)=0. From the figure we see that: () 2 2 () 1 0 1 () 2 1 0 111() () 22 () 11() lim ( ) lim lim 2() Dt Dt t i ii i Dt t i i ttt X X Dt Rsds tt tDt X Dt Rsds ttDt = = = →∞ →∞ →∞ = =⋅ ⋅ ⇒ =⋅ ⋅ ∑ ∑ ∫ ∑ ∫  Ergodicity: long-term time averages = steady-state averages (with probability 1) 0 1 [ ] lim [ ] lim ( ) t i it ER ER Rsds t →∞ →∞ == ∫ 10-9 Average Residual Time (cont.)  () 2 1 0 11() lim ( ) lim lim 2() Dt t i i ttt X Dt Rsds ttDt = →∞ →∞ →∞ =⋅ ⋅ ∑ ∫  lim ( ) / t Dt t →∞ : long-term average departure rate. Should be equal to the long-term average arrival rate. Long-term averages = steady-state averages (with probability 1): () lim t Dt t →∞ = λ  Law of large numbers: () 22 2 11 lim lim [ ] () Dt n ii ii tn XX E X Dt n == →∞ →∞ == ∑ ∑ Average residual time: 2 1 [] [ ] 2 E REX=λ P-K Formula: 2 [] [ ] [] 12(1) E REX EW λ == −ρ − ρ 10-10 P-K Formula P-K Formula: 2 [] [ ] [] 12(1) E REX EW λ == − ρ−ρ  Average time a customer spends in the system 2 1[] [] [ ] [ ] 2(1 ) E X ET E X EW λ =+=+ µ −ρ  Average number of customers waiting for service: 22 [] [] [ ] 2(1 ) E X EQ EW λ =λ = − ρ  Average number of customers in the system (waiting or in service): 22 [] [] [] 2(1 ) E X EN ET λ =λ =ρ+ − ρ Averages E[W], E[T], E[Q], E[N] depend on the first two moments of the service time [...]... function M X (t ) exists and is finite in some neighborhood of t=0, it determines the distribution (pdf or pmf) of X uniquely Theorem 2: For any positive integer n: 1 dn M X (t ) = E[ X n etX ] n dt 2 dn M X (0) = E[ X n ] n dt Theorem 3: If X and Y are independent random variables: M X +Y (t ) = M X (t ) M Y (t ) 1 0-1 9 Z-Transforms of Discrete Random Variables For a discrete random variable, the moment... to set z = et and define the z-transform (or characteristic function): GX ( z ) = E[ z X ] = ∑ z j P{ X = x j } x j Let X be a discrete random variable taking values 0, 1, 2,…, and let pn = P{ X = n} The ztransform is well-defined for | z |< 1 : GX ( z ) = p0 + zp1 + z 2 p2 + z 3 p3 + ∞ = ∑ pn z n n=0 Z-transform uniquely determines the distribution of X If X and Y are independent random variables:... h ir pet 1¡(1¡p)et t e¸(e ¡1) 1 0-2 2 P-K Transform Equation We have established: L j = L j −1 − 1{L j −1 > 0} + Aj = ( L j −1 − 1) + + Aj Let πn = lim P{L j = n} be the stationary distribution and GL ( z ) = ∑ n = 0 πn z n its z-transform ∞ j →∞ Noting that ( L j −1 − 1) + and Aj are independent, we have: L E[ z j ] = E[ z ( L j −1 −1) + A ]E [ z j ] At steady-state, L j and L j −1 are statistically... factorial moments can be calculated similarly 1 0-2 0 Continuous Random Variables Distribution (parameters) Prob Density Fun fX (x) Moment Gen Fun MX (t) Mean E[X] Variance Var(X) Uniform over (a; b) 1 b¡a etb ¡eta t(b¡a) a+b 2 (b¡a)2 12 ¸ ¸¡t 1 ¸ 1 ¸ ¹ ¾2 Exponential ¸ Normal (¹; ¾ 2 ) a . 1 TCOM 501: Networking Theory & Fundamentals Lectures 9 & 10 M/G/1 Queue Prof. Yannis A. Korilis 1 0-2 Topics  M/G/1 Queue  Pollaczek-Khinchin (P-K) Formula  Embedded Markov. ] n n X n d M EX dt = Theorem 3: If X and Y are independent random variables: () () () XY X Y M tMtMt + = 1 0-1 9 Z-Transforms of Discrete Random Variables  For a discrete random variable, the moment. 1 0-6 A Result from Probability Theory Proposition: Sum of a Random Number of Random Variables  N: random variable taking values 0,1,2,…, with mean []EN  X 1 , X 2 , …, X N : iid random