M/G/1 queue (with infinite buffer)

Transcription

1 M/G/1 queue (with infinite buffer) M. Veeraraghavan, March 17, 4 1. Mean number of jobs in the system and mean response time Using Mean Value Analysis (MVA) [6], we derive the mean delay in an M/G/1 queue with an infinite buffer. Focus on the buffer occupancy at the time of departure of a job in service. n j-1 n j n j+1 # of customers in queue (j-1) j (j+1) th customer finishes service service time of j th customer Fig. 1 At the times of job departures ν j n j : number of customers arriving during the service of the customer : number of customers in the queue when the customer finishes service. j th j th ( n j 1 1) + ν j n j 1 1 n j ν j n j 1 (EQ 1) 1 x 1 n j n j 1 un ( j 1 ) + ν j where ux ( ) (EQ ) x Since the arrival process is Poisson (memoryless), the statistics found are the same at all points on the time axis (I found a derivation for this in [1], page 17 - see Section 3.). Therefore, En [ j ] En [ j 1 ] as j. (EQ 3) Taking expectations of () yields E[ ν] Eun [ ( )] (EQ 4) E[ ν] : average number of arrivals in a service interval. E[ ν] < 1 ; else the queue will become unstable. 1

2 Eun [ ( )] un ( )p n p n prob( n > ), which is the utilization. (EQ 5) Since p n is that probability that the queue is in state n, p 1. But to only find average queueing delays, we use a trick. Take squares of both sides of () and take expectations: n 1 n j n j 1 u [ + ( n j 1 ) n j 1 un ( j 1 )] + ν j + ν j ( n j 1 un ( j 1 )) (EQ 6) Let j. En [ j ] En [ j 1 ]; also Eu [ ( n) ] Eun [ ( )] E[ ν] (EQ 7) En [ j 1 un ( j 1 )] + E[ ν j ] + En [ j 1 ν j ] Eun [ ( j 1 )ν j ] (EQ 8) Enun [ ( )] En [ ] by the definition of un ( ). (EQ 9) Assume ν j and n j 1 are independent (true for Poisson arrivals), rewriting (8) En [ j 1 ] + E[ ν j ] + En [ j 1 ]E[ ν j ] Eun [ ( j 1 )] E[ ν j ] (EQ 1) As j En [ ] + E[ ν ] + En [ ] (EQ 11) ( 1 )En [ ] -- E[ ν ] (EQ 1) ( 1 )En [ ] σ v σ v ( 1 ) or En [ ] -- (Eqn. -69 of [6], page 6) (EQ 13) ( ) Given E[ ν] Variance, [ k E[ ν] ] P( ν k) (EQ 14) (see (7) above) σ v k

3 σ v ( k + ( E[ ν] ) ke[ ν] )P ( ν k ) E ν k [ ] + E[ ν ] (EQ 15) Let the service time distribution have a probability density function f τ ( τ). Then P( ν k) P( ν k τ)f τ ( τ) dτ (EQ 16) where P( ν k τ) is the conditional probability of k arrivals in τ sec. P( ν k τ) ( λτ) k e λτ k! Poisson arrival process (EQ 17) P( ν k) ( λτ) k e λτ f k! τ ( τ) dτ (EQ 18) E[ ν] kp( ν k) k Exchange integration and summation, k k ( λτ) k e λτ f k! τ ( τ) dτ (EQ 19) E[ ν] k ( λτ)k λτ e fτ( τ) dτ k! k ( λτ) k 1 ( λτ) k 1 λτ e ( k 1)! fτ( τ) dτ (EQ ) ( λτ) Since k e λτ 1, E[ ν] ( λτ)f, (EQ 1) ( k 1)! τ ( τ) dτ E[ ν] λe[ τ] k 1 The above validates the conclusion from (4) and (5) that E[ ν]. For Poisson arrivals, the average customer arrival rate is λ times the average service time E[ τ]. Similarly, derive σ ν as a function of the variance of the service time. σ ν E[ ν ] ( E[ ν] ) (EQ ) 3

4 Exchanging the integral and the summation: σ ν k ( λτ)k λτ e f (since ). (EQ 3) k! τ τ ( ) dτ E[ ν] k Since k ( λτ)k λτ e is the second moment of the Poisson distribution Ek [ ], knowing that the k! k variance σ k Ek [ Ek [ ]] Ek [ ] ( Ek [ ]) λτ Ek [ ], Ek [ ] λτ + ( λτ), σ ν ( λτ + ( λτ ) )f ( τ τ ) dτ λe[ τ] + λ E[ τ ] (EQ 4) σ ν λ + σ τ (since λ( E[ τ] ) and σ τ E[ τ ] ( E[ τ] ) ) (EQ 5) In other words, the variance of the number of arrivals within one service time is equal to the utilization plus (arrival rate) squared times the variance of the service time. Going back to (13), we get En [ ] σ ν ( ) -- λ σ τ 1 ( ) (EQ 6) The Pollaczek-Khinchine formula for the average number of jobs in queue in a M/G/1 system: En [ ]. (EQ 7) ( ) µ ( σ τ ) We show in the Appendix, Section 4.1 that (7) equals (6). Reference [] gives another version of the same formulation: where E[ τ ] (8) equals (7). Using Little s Law, En [ ] λ E τ [ ], (EQ 8) 1 ( ) is the second moment of the service time. We show in the Appendix, Section 4. that 4

5 ET [ ] En [ ] 1 µ or (EQ 9) λ ( ) µ [ σ τ ] ET [ ] -- 1 λe τ [ ] µ ( ) First term is mean service time. The mean waiting time is: EW [ ] Similarly mean number waiting in queue is λe[ τ ] 1 ( ) (EQ 3) EN [ Q ] λ E[ τ ] 1 ( ) [1] (EQ 31) See that M/G/1 equations become equal to M/M/1 equations where G is M.. Distribution of number of jobs in the system From [] - we use new notation here. Let Nt () be the number of jobs in the system (number in queue plus in service). If Nt () 1, then given the service time is generally distributed (and hence need not be memoryless), the future behavior of the system depends upon how much time was already spent on the current job in service. This means Nt () is not a Markov chain (see QT lecture on M/M/1 to see why the number of jobs in an M/ M/1 queueing system is a CTMC). But there is an embedded Markov chain in this system. Look at the system when jobs depart the server. At these departure epochs, t n, n 1,,, the number of jobs in the system X n is given by X n Nt ( n ). This stochastic process X n, n 1,,, is a DTMC. Why is this? X n 1 + Y n + 1 X n > + Y n + 1 X n X n 1 (EQ 3) Y n X 1 is the number of jobs arriving during the service time of the job. These are independent of, X,. Hence X n + 1 only depends upon X n. This is the property of a Markov chain. Hence X n is a MC. Now we can obtain the transition probabilities of this DTMC: n th 5

6 p ij PX ( n + 1 j X n i) PY n 1 Let PY ( n + 1 j) a j (and all Y n are iid), a j 1 ( + j i+ 1) if( i, j i 1) PY ( n + 1 j) if( i, j ) otherwise j Let the limiting probability of being in state j be denoted ν j (EQ 33) ν j lim PX ( n j) n (EQ 34) Using (48) from MC lesson, which is ν j ν i p ij j 1,, and ν j 1 (EQ 35) i j j + 1 ν j ν a j + ν i a j i i (EQ 36) Since j i 1, j + 1 i. If i j+, this relation is violated. Note that p ij is a j i + 1 when i 1 as per the first option in (33). Note that the DTMC underlying an M/G/1 queue is not a birth-death process because if the number of arrivals is more than one within a service time, then the state change will be from state j i + 1. Therefore we cannot use the detailed-balance equation. to some Define a moment generating function (see Appendix Section 4.4 for a recap on moment generating functions): X n i Gz ( ) ν j z j j (EQ 37) ν j z j ν a j + ν i a j i j j j + 1 i j z (using (36)) (EQ 38) Interchanging the summation order 6

7 Gz ( ) ν a j z j + ν i a j i + 1 z j ν a z j + ν j i a j i j j j + 1 i 1 Define G A ( z) a j z j, and setting k j ( i 1) j j i 1 j i z j Gz ( ) ν G A ( z) + ν i a k z k + i 1 ν G z 1 ( ) + -- ν A z i z i a k z k i 1 k i 1 k (EQ 39) (EQ 4) 1 Gz ( ) ν G A ( z) + -- [ Gz ( ) ν z ]G A ( z) (EQ 41) Gz ( ) ( z 1)ν G A ( z) z G A ( z) (EQ 4) Since G( 1) G A ( 1) 1, we can use L Hospital s rule: G( 1) 1 ν ( z 1)G' A ( z) + G A ( z) ν lim provided that G' (EQ 43) z 1 1 G' A ( z) 1 G' A ( 1) A ( 1) < 1 See Appendix Section 4.3 for a recap of L Hospital s rule. G' A ( z) ja j z j 1. (EQ 44) Therefore G' A ( 1) EY [ ], i.e., the mean number of arrivals within a service time. Let G' A ( 1). From (43), ν 1, and since ν is the probability that the server is idle, is the server utilization. Then (4) can be rewritten as: j Gz ( ) ( z 1) ( 1 )G A ( z) z G A ( z) (EQ 45) In other words, the moment generating function of the number of jobs in the system can be expressed as a function of the moment generating function of the number of arrivals within a service time. 7

8 EN [ ] lim EX [ n ] G' ( 1) because Gz ( ) p X ()z i i and G' ( z) ip X ()z i i 1 (EQ 46) Therefore to compute Gz ( ), we need G A ( z). Compute as follows (since arrival process is Poisson): n i i ( + jb t) e λt( λt)j, where B is a r.v. denoting service times (iid) (EQ 47) j! PY n 1 Therefore, a j PY ( n + 1 j) PY ( n + 1 j B t)f B () t dt (EQ 48) G A ( z) a j z j j PY ( n + 1 j B t)f B () t dt z j j (EQ 49) G A ( z) e λt( f λt)j j () t dt z j! B j ( λtz)j j! e λt j f B () t dt (EQ 5) G A ( z) e λt e λtz f () B t dt e λt( 1 z) f () t dt L B B ( λ( 1 z) ) because (EQ 51) the Laplace transform for a random variable X is L X ( s) e sx f X ( x) dx (EQ 5) dl ( G' A ( 1) B ( λ( 1 z) )) dz z 1 dl B ( λ) ds s (EQ 53) dl B ds ( λ) ( EB [ ])( λ) s -- λ µ (EQ 54) Substituting (51) into (45), we get the Pollaczek-Khinchin transform equation: 8

9 Gz ( ) ( z 1) ( 1 )L B ( λ( 1 z) ) z L B ( λ( 1 z) ) (EQ 55) The above describes the z-transform of the number of jobs in the system in terms of the Laplace transform for the service time distribution. We can obtain the mean number of jobs by: EN [ ] For an M/D/1 queue, dg λ EB [ ] dz ( ) z 1 EB [ ] 1 --, therefore µ P-K mean-value formula (derive) (EQ 56) EN [ ] ( ) (EQ 57) Athough the above was derived assuming FCFS, it holds for any scheduling discipline that assumes: 1. The server is not idle when there are jobs waiting for service (work conserving). The scheduler is not allowed to use any deterministic a priori information about job service times 3. The scheduling is nonpreemptive. Results shown in [9] are the same as above. The derivation in [1] is completely different. It uses a graphical method to derive mean waiting time. See the book for details. Finally, check Kleinrock s book to see if any more details are provided than Gz ( ), e.g. state probabilities for the M/G/1 queue. 3. Occupancy distribution upon arrival and upon departure [1], pg 171. Let the unconditional steady-state probabilities of being in different states for an M/M/1 queue be p n lim P{ N() t n} n (EQ 58) and let the steady-state occupancy probabilities upon arrival be: a n lim P{ ( N() t n) an arrival occurred just after time t} n (EQ 59) 9

10 It turns out that for an M/M/1 system, p n a n (EQ 6) Furthermore, this holds under very general conditions for queueing systems with Poisson arrivals regardless of the distribution of the service times! Applies for our M/G/1 case. The only additional assumption we need is that for every time t and increment δ >, the number of arrivals in the interval ( t, t+ δ) is independent of the number in the system at time t. This assumption holds given our assumption for queueing systems that the interarrival times and service times are independent. For a formal proof of (6): Let Att (, + δ) be the event that an arrival occurs in the interval ( t, t+ δ). Let p n () t P{ N() t n} and a n () t P{ N() t n ( an arrival occurred just after time t) } (EQ 61) Using Bayes rule: a n () t lim P{ N() t n Att (, + δ) } δ (EQ 6) a n () t lim P{ N() t natt, (, + δ) } δ P{ A( t, t + δ) } (EQ 63) a n () t P{ A( t, t+ δ) Nt () n}p{ N() t n} lim δ P{ A( t, t + δ) } (EQ 64) Given our assumptions, the event Att (, + δ) t. Therefore is independent upon the number in the system at time P{ A( t, t+ δ) Nt () n} P{ A( t, t + δ) } and therefore (64) becomes (EQ 65) a n () t P{ N() t n} p n () t (EQ 66) Great example: If the arrival process is not Poisson, say interarrival times are independent and uniformly distributed between and 4 sec, while customer service time is deterministic at 1sec. An arriving job always finds an empty queue. But the average number in the system as seen by an outside observer at a random time is 1 3 per Little s law since response time is 1sec, and λ

11 System when a job departs: d n () t PNt ( () n ( a departure occurred just after time t) ) (EQ 67) The steady-state values are denoted: d n lim d n () t, n 1,, (EQ 68) t If the system reaches steady-state with all n having positive steady-state probabilities, and changes in unit increments (true for M/M/1 or M/G/1 if < 1 ), N() t d n, n 1,, (EQ 69) a n Explanation: For every sample path and every n, the number in the system will be n an infinite number of times. This means each time the number increases from n to n+ 1, there is a corresponding decrease from n + 1 back to n. The frequency of transitions from n to n+ 1 out of transitions from any k to k+ 1 will equal the frequency of transitions from n + 1 to n out of transitions from any k + 1 to k. Therefore d n a n. In the steady-state, the system appears statistically identical to an arriving and a departing customer. If arrival process is Poisson, then the system is statistically identical as seen by an arriving or a departing customer or by an observing looking at the system at random time. All the above to justify (3)! Does it? 4. Appendix 4.1 Equivalence of (6) and (7) Eqn (7): En [ ] (EQ 7) ( ) µ ( σ τ ) En [ ] ( 1 ) 1 ( ) µ 1 ( ) σ τ (EQ 71) Since µ , and λe[ τ]. Therefore E[ τ] 11

12 En [ ] λ σ τ ( 1 ) 1 ( ) ( ) Comparing (7) with our derivation (6), we ask is λ σ τ 1 ( ) ( ) (EQ 7) ( ) --? (EQ 73) ( ) ( ) ( 1 ) ( ) ( ) QED! (EQ 74) 4. Equivalence of (7) and (8) In other words, is λ + σ τ λ E τ [ ]? (EQ 75) 1 ( ) ( ) λ E[ τ ] λ ( σ τ + ( E[ τ] ) ) λ σ τ 1 ( ) 1 ( ) + 1 ( ) ( ) (EQ 76) ( 1 ) + λ σ τ 1 ( ) ( ) λ σ τ 1 ( ) ( ) (EQ 77) Therefore En [ ] (EQ 78) ( ) µ [ σ τ ] 4.3 Recap of L Hospital s rule from Mathworld: f' ( x) If lim fx ( ) and lim gx ( ) are both zero or both at ±, and if lim is finite or the limit is x c x c x c g' ( x) ±, then fx ( ) lim x c gx ( ) f' ( x) lim x c g' ( x) (EQ 79) 1

13 4.4 Recap on moment generating functions M( θ) Ee [ xθ ] j e x j θ px ( j ) e xθ fx ( ) dx if X is discrete if X is continuous (EQ 8) for continuous rv, Laplace transform L X ( s) Ls ( ) M X ( s) e sx f X ( x) dx (EQ 81) for discrete rv, z-transform G X ( z) Gz ( ) Ez [ X ] M X ( lnz) p X ()z i i (EQ 8) i e Xθ 1 Xθ X θ Xk θ k ! k! (EQ 83) M( θ) EE [ Xθ ] 1 EX [ ]θ EXk [ ]θ k k! (EQ 84) M is called a moment generating function because moments of X can be extracted as follows: EX [ k ] k d M( θ) or EX [ k ] ( 1) k k d Ls ( ) (EQ 85) dθ k ds k θ s For a discrete rv EX [ ] dg and EX [ ] d G + dg (EQ 86) dz dz dz z 1 z 1 z 1 References [1] D. Bertsekas and R. Gallager, Data Networks, Prentice Hall, Second Edition, 199. [] K. S. Trivedi, Probability, Statistics with Reliability, Queueing and Computer Science Applications, First Edition, Prentice Hall, ISBN r. [3] A. Leon Garcia and I. Widjaja, Communication Networks, McGraw Hill,, First Edition. 13

14 [4] E. Pinsky, A. Conway and W. Liu, Blocking Formulae for the Engset Model, IEEE Transactions on Communications, vol. 4, no. 6, June 1994, pp [5] R. Syski, Introduction to Congestion Theory in Telephone Systems, Oliver and Boyd, Edinburgh, 196. [6] Mischa Schwartz, Telecommunications Networks, Protocols, Modeling and Analysis, Addison Wesley, [7] D. Gross and C. M. Harris, Fundamentals of Queueing Theory, Wiley Series in Probability and Mathematical Statistics, [8] S. M. Ross, Stochastic Processes. [9] Bob Boorstyn s notes. 14