GI/M/1 and GI/M/m queuing systems

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1 GI/M/1 and GI/M/m queuing systems Dmitri A. Moltchanov

2 OUTLINE: GI/M/1 queuing system; Methods of analysis; Imbedded Markov chain approach; Waiting time in FCFS GI/M/1; GI/M/m queuing system; Note on results for GI/M/-/-/-; Rate conservation law for GI/M/-/-. Lecture: GI/M/1 and GI/M/m queuing systems 2

3 1. GI/M/1 queuing system GI/M/1 queuing system is characterized by: generally distributed interarrival times. exponentially distributed service times; single server; infinite capacity (therefore, infinite number of waiting positions); We also assume that: FCFS service discipline when needed: mean waiting time is the same for all disciplines; PDF and pdf may be different! Note: GI/M/1 is exactly opposite to what we had for M/G/1 queuing system. Lecture: GI/M/1 and GI/M/m queuing systems 3

4 1.1. Arrival and service processes Arrival process: renewal arrivals with mean value λ; interarrival times are 1/λ; PDF, pdf and LT are: A(t), a(t), A(s). (1) Service process: service times are exponential with mean 1/µ; PDF, pdf and LT are: B(t) = 1 e µt, b(t) = µe µt, B(s) = µ s + µ. (2) Lecture: GI/M/1 and GI/M/m queuing systems 4

5 2. Methods of analysis There are a number of methods: transform approach based on imbedded Markov chain: distributions of desired performance parameters can be obtained; the idea: find points at which Markov property holds and use transforms. direct approach based on imbedded Markov chain: distributions of desired performance parameters can be obtained; the idea: find points at which Markov property holds and use convolution. method of supplementary variables: distributions of desired performance parameters can be obtained; the idea: look at arbitrary points and make them Markovian. Note: we consider only the second. Lecture: GI/M/1 and GI/M/m queuing systems 5

6 3. Direct approach based on imbedded Markov chain Recall M/M/-/-/-: let N(t) be the number of customers at time t: we know how the system evolves in time after t in t; arrival may occur with probability λ t: it is independent from previous arrival: memoryless property (A 1 = A 2 )! departure may occur with rate µ t: it is independent from previous departure: memoryless property (B 1 = B 2 )! the resulting process is actually birth-death Markov one! A 1 (t) A 2 (t) t B 1 (t)=b 2 (t) B 2 (t) t Figure 1: State of M/M/1 queuing system. Lecture: GI/M/1 and GI/M/m queuing systems 6

7 3.1. Problem with GI/M/1 queuing system Let N(t) be the number of customers at time t: do we know how the system evolves in time after t in t? service completion may occur with rate λ t: it is independent from previously received service time: memoryless property (B 1 = B 2 )! arrival may occur with probability µ t: A 2 is not the same as A 1! this process is no longer Markovian! A ( t) : A ( t) A ( t) A 2 (t) t B 2 (t) B ( t) : B ( t) B ( t) t Figure 2: State of M/G/1 queuing system. Lecture: GI/M/1 and GI/M/m queuing systems 7

8 3.2. How to choose the state of GI/M/1 system State of the system: M/G/1 queuing system number of customers in the system and time since previous arrival: (N(t), A(t + x x)): in this case we know how the system evolves in time; we know the distribution of time till the next departure: it is the same as initial one; we know the distribution of time till the next arrival: we track it: A(t + x x) = P r{(t t + x) T > x} = (A(t + x) A(x))/(1 A(x)). (3) N( t) k, k 0,1,... A( t x x) A( t x) A( x) 1 A( x) t B( t) 1 e t t Figure 3: State of GI/M/1 queuing system given by (N(t), A(t + x x)). Lecture: GI/M/1 and GI/M/m queuing systems 8

9 3.3. Steady-state distribution as seen by arrival Let ρ be offered traffic load to G/M/1 queuing system: ρ = λ µ. (4) Note: for GI/M/1 queuing system to be stable we require ρ < 1. State of the system: we consider number of customers just before arrivals; note that arriving customer is not included! time instant t i... state arrivals Figure 4: Illustration of the system state in GI/M/1 queuing system. Lecture: GI/M/1 and GI/M/m queuing systems 9

10 Define the following: n i, i = 0, 1,..., be the number of customers in the system just before arrival i; consider two successive arrivals to the system i th arrival and (i + 1) th arrival: s i+1 be the number of served customers between i th and (i + 1) th arrivals. s i+1 customers served in (i+1) th interarrival time... i th arrival (i+1) th interarrival time (i+1) th arrival t n i customers just before n i+1 customers just before Figure 5: Time diagram of GI/M/1 queuing system. Lecture: GI/M/1 and GI/M/m queuing systems 10

11 Arrivals... Server n i n i +1 s i+1 n i+1 n i+1 +1 just before just before (i+1) th interarrival time t i t i+1 Figure 6: Better view of time diagram of GI/M/1 queuing system. Lecture: GI/M/1 and GI/M/m queuing systems 11

12 We can relate n i and n i+1 using s i+1 as: note that by definition: n i+1 = n i + 1 s i+1, n i = 0, 1,.... (5) s i+1 n i + 1. (6) since there are not more than (n i + 1) customers at time point t + i. Note the following: stochastic process {n i, i = 0, 1,... } given by (5) and (7) is a continuous-time Markov chain; it defines number of customers seen by arrival. Consider this Markov chain at steady-state: t : we will be looking for steady-state probabilities as seen by arrival; to find them we have to determine transition probabilities of the imbedded Markov chain: p jk = lim t P r{n i+1 = k ni = j}, j, k {0, 1,... }, (7) p jk is the probability that (j +1 k) customers served between consecutive arrival instants. Lecture: GI/M/1 and GI/M/m queuing systems 12

13 Note the following: transition probabilities from state j to state k, k > j + 1, are not possible: p jk = 0, k > j + 1. (8) state of the system cannot increase on more that one customer between imbedded points. p n, n n-2 n-1 n n+1 p n,n-1... p n,0 p n,1 p n,n-2 Figure 7: State transition diagram of imbedded Markov chain. Lecture: GI/M/1 and GI/M/m queuing systems 13

14 Probabilities p ik can be summarized in the compact matrix form: p 00 p p 10 p 11 p P = p 20 p 21 p 22 p (9) p 30 p 31 p 32 p 33 p Let the steady-state probabilities be defined as follows: p j, j = 0, 1,.... (10) To determine p j, j = 0, 1,... we have to solve the following: p k = p j p jk, k = 0, 1,..., j=0 p k = 1. (11) k=0 Lecture: GI/M/1 and GI/M/m queuing systems 14

15 Do the following: let α n, n = 0, 1,... : probabilities that there are n departures in interarrival time. service time is exponentially distributed with rate µ; number of service completions in a fixed interval is Poisson; get α n, n = 0, 1,... by integration over all possible durations of interarrival time: α j = 0 (µx) j e µx a(x)dx, j = 0, 1,.... (12) j! this holds when the server is always busy during an interarrival time. Use (12) to write balance equations for states of the system: p j = α k+1 p k, j = 0 k=0 p j = α 0 p j 1 + α k+1 p j+k, j = 1, 2,.... (13) k=0 these have to be solved to get steady-state probabilities p j, j = 0, 1,.... Lecture: GI/M/1 and GI/M/m queuing systems 15

16 3.4. Solution Note the following: to determine steady-state probabilities we use direct approach; we will try to find the solution in the form: p n = σ n, n = 0, 1,.... (14) Do the following: substitute p n = σ n into last equation of (13) and divide by σ n 1 : σ = σ i α i. (15) we already have expression for α i, i = 0, 1,... : σ = σ i (µt) i e µt a(t)dt = i! i=0 0 i=0 0 e (µ µσ)t a(t)dt. (16) the latter is Laplace transform of interarrival times in GI/M/1 queuing system. Lecture: GI/M/1 and GI/M/m queuing systems 16

17 Therefore, we get the following equation: σ = A(µ µσ), (17) which is LT of interarrival times with parameter (µ µσ). Finding the roots: σ = 1 is the simple root: it does not satisfy normalizing condition! we have to find a root in 0 < σ < 1 if ρ < 1: testing p n = σ n in (13) shows that this is a root. normalizing p n = σ n leads to the final result: p n = (1 σ)σ n, n = 0, 1,.... (18) Conclusion: number of customers in the system as seen by arrival is the solution of: σ = A(µ µσ). (19) Lecture: GI/M/1 and GI/M/m queuing systems 17

18 3.5. Example: M/M/1 queuing system LT of exponentially distributed interarrival times are given as follows: A(s) = λ λ + s. (20) Hence the equation from which we have to determine σ reduces to: σ = λ λ + µ µσ. (21) Therefore: σ(λ + µ µσ) λ = (σ 1)(λ µσ) = 0. (22) Finally,we get σ = ρ and the arrival distribution is: p n = (1 ρ)ρ n, n = 0, 1,.... (23) due to PASTA and Kleinrock this is the same as time-averaged and seen by departure! Lecture: GI/M/1 and GI/M/m queuing systems 18

19 3.6. Example: E 2 /M/1 queuing system Assume, we are given: interarrival times are Erlang distributed with two phases and mean 2/3; Laplace transform of interarrival times is then: A(s) = (3/(3 + s)) 2. (24) µ = 4 corresponding to stable system: ρ = 3/8 < 1. Equation from which we have to determine σ reduces to: σ = (3/(7 4σ)) 2. (25) Thus: σ(7 4σ) 2 9 = (σ 1)(4σ 9)(4σ 1) = 0. (26) Therefore, the desired root is σ = 1/4: p n = (3/4)(1/4) n, n = 0, 1,.... (27) Lecture: GI/M/1 and GI/M/m queuing systems 19

20 3.7. Example: E 2 /M/1 queuing system Assume, we are given: interarrival times are generalized Erlang with two phases: µ and 2µ; Laplace transform of interarrival times are given by: service rate is µ leading to stable system. A(s) = (2µ 2 )/((µ + s)(2µ + s)). (28) Equation from which we have to determine σ reduces to: σ = It leads to following equation: 2µ 2 (2µ µσ)(3µ µσ) = 2 (2 σ)(3 σ). (29) σ 3 5σ 2 + 6σ 2 = (σ 1)(σ 2 2)(σ 2 + 2) = 0. (30) The only root σ = 2 2 is acceptable leading to: p n = ( 2 1)(2 2) n, n = 0, 1,.... (31) Lecture: GI/M/1 and GI/M/m queuing systems 20

21 4. Waiting time in FCFS GI/M/1 system What is nice about GI/M/1: we had to imbed Markov chain just before customer arrivals; steady-state distribution we found is what arrival sees; arrival has to wait until service completion of all customers it sees. Let us define the following: T : waiting time in the system with PDF F T (t), pdf f T (t) and LT F T (s); Q: waiting time prior to service with PDF F Q (t), pdf f Q (t) and LT F Q (s); X: service time with PDF B(t), pdf b(t) and LT B(s). We are looking for: distributions of T and Q. Lecture: GI/M/1 and GI/M/m queuing systems 21

22 FCFS service discipline n t i t Q: waiting time prior to service W: waiting time in the system Figure 8: Waiting time and time prior to service in GI/M/1 queuing system. To determine total delay use the following: we know distribution as seen by arrival; we know that service times are exponential; use convolution to obtain time needed to serve customers which arrival sees. Lecture: GI/M/1 and GI/M/m queuing systems 22

23 Do the following: note that it is better to work in transform domain: sum of RVs is given by convolution; property: LT of the convolution is the product of individual LTs. we get for the total delay: F T (s) = n=0 ( ) n+1 µ p n. (32) µ + s Substituting p n = (1 σ)σ 2 we get LT F T (s): ( ) n+1 µ F T (s) = (1 σ)σ n = µ + s n=0 µ(1 σ) ( ) n µσ = = µ + s µ + s = n=0 µ(1 σ) µ(1 σ) + s. (33) Lecture: GI/M/1 and GI/M/m queuing systems 23

24 Observe the latter result: F T (s) = µ(1 σ) µ(1 σ) + s. (34) one may recognize LT of exponential with with parameter µ(1 σ): F T (t) = P r{t t} = 1 e µ(1 σ)t, t 0. (35) Note the following: total waiting time is almost the same as for M/M/1 (replace ρ by σ). We may use the same approach to get F Q (t): F Q (t) = P r{q t} = 1 σe µ(1 σ)t, t 0. (36) Note: probability that customer has to wait is (1 σ)! Lecture: GI/M/1 and GI/M/m queuing systems 24

25 4.1. Mean sojourn time How to get: estimate directly from F T (t); compute using mean values approach. Note that according to Little s result we have: E[N] = λe[t ]. (37) Note the following: arrivals are not Poisson and PASTA property does not hold: E[N A ] E[N]. (38) For arbitrary arriving customer we have: E[T ] = E[N A ] 1 µ + 1 µ, (39) E[N A ] is the mean number of customers in the system as seen by arrival. Lecture: GI/M/1 and GI/M/m queuing systems 25

26 How to get N A? Express E[N A ] considering steady-state distribution at arrival time: E[N A ] = np n = n=0 n=0 n(1 σ)σ n = σ 1 σ. (40) Substitute this result into E[T ] = E[N A ](1/µ) + 1/µ to get: E[T ] = σ (1 σ)µ + 1 µ = 1 (1 σ)µ. (41) Now we can get the mean number of customers in the buffer as follows: E[N] = λ (1 σ)µ = ρ (1 σ). (42) Lecture: GI/M/1 and GI/M/m queuing systems 26

27 5. GI/M/m queuing system GI/M/m queuing system is characterized by: generally distributed interarrival times. exponentially distributed service times; multiple identical servers; infinite capacity: infinite number of waiting positions. We also assume that: FCFS service discipline when needed: mean waiting time is the same for all disciplines; PDF and pdf may be different! Note: GI/M/m is exactly opposite to what we had for M/G/m queuing system. Lecture: GI/M/1 and GI/M/m queuing systems 27

28 5.1. Arrival and service processes Arrival process: renewal arrivals with mean value λ; interarrival times are 1/λ; PDF, pdf and LT are: A(t), a(t), A(s). (43) Service process: service times are exponential with mean 1/µ; PDF, pdf and LT are: B(t) = 1 e µt, b(t) = µe µt, B(s) = µ s + µ. (44) Lecture: GI/M/1 and GI/M/m queuing systems 28

29 5.2. Direct approach based on imbedded Markov chains Let ρ be offered traffic load to G/M/m queuing system: ρ = Note: for GI/M/m queuing system to be stable we require ρ < 1. State of the system: λ mµ. (45) we consider number of customers just before arrivals at times t i, i = 0, 1,... ; note that arriving customer does not included! Arrivals m Figure 9: Graphical representation of GI/M/m queuing system. Lecture: GI/M/1 and GI/M/m queuing systems 29

30 Define the following: denote the number of customers in the system at time t i by n i, i = 0, 1,... ; process of n i changes constitutes a imbedded continuous-time Markov chain. n i n i +1 s i+1 n i+1 n i+1 +1 just before just before (i+1) th interarrival time t i t i+1 Figure 10: Interarrival time between two successive imbedded time points. Note: the only difference compared to GI/M/1 is that customers are served by m servers. Lecture: GI/M/1 and GI/M/m queuing systems 30

31 Define steady-state probabilities as seen by arrival: q j, j = 0, 1,.... (46) These probabilities should satisfy the following balance equations at t : q j = q i q ij, j = 0, 1,.... (47) i=j 1 q ij are transition probabilities from i to j between successive imbedded Markov points. qn,n n-2 n-1 n n+1 q n,n-1... q n,0 q n,1 q n,n-2 Figure 11: Transition diagram of imbedded Markov chain. Lecture: GI/M/1 and GI/M/m queuing systems 31

32 Steady-state probabilities must also satisfy normalizing condition: q j = 1. (48) j=0 What we should do to get: q j, j = 0, 1,... : we have to get transition probabilities q ij ; to get q ij we have to determine the number of service completions in interarrival time. Consider two arrival time instants t i and t i+1 : assume that i customers exist in the system just before time t i and (i + 1) just after; probability that (i j + 1) customers are served during interarrival time: integrate over all possible number of arrivals (Poisson: mµ) during interarrival time: 0 (mµx) (i j+1) (i j + 1)! e mµx a(x)dx, m j i + 1. (49) this results in exactly j customers prior to time instant t i+1. Lecture: GI/M/1 and GI/M/m queuing systems 32

33 5.3. Some performance parameters of GI/M/m queuing system Probability that arriving customer has to wait: P W = q i. (50) i=m since this event may only occur when all servers are busy; note that we use steady-state distribution as seen by arrival. PF of the number of customers ahead in the queue given that the customers waits: tag an arbitrary customer arriving to the queue; letting n be the number of customers prior to this arrival, we have: P r{n = j n m} = P r{n = j n m} P r{n m} = q m+j P W, j = 1, 2,.... (51) using previous result for probability that an arriving customer has to wait we get: P r{n = j n > m} = q m+j i=m q, j = 1, 2,.... (52) i Lecture: GI/M/1 and GI/M/m queuing systems 33

34 6. Note on results for GI/M/-/-/- Notes on results we got for GI/M/-/-/- queuing system: we got steady-state probabilities of the number of customers as seen by arrival; we got waiting time directly: thanks that imbedded points are just before arrivals. we got time-averaged parameters indirectly! Observing GI/M/s queuing system one can claim that: Klienrock principle holds: obtained distribution is the same as the distribution seen by departure. PASTA property does not hold: at this time we cannot say what is the distribution at arbitrary time t. Question: how to get distribution at arbitrary time t? Lecture: GI/M/1 and GI/M/m queuing systems 34

35 7. Rate conservation law for GI/M/-/-/- queues Consider a system with following parameters: customers arrive in a renewal arrival process with rate λ; each customer require exponentially distributed service time with mean µ 1 ; system has m, m 1 servers. Let us denote by: p j : steady-state probability that there are j customers in the system at arbitrary time; q j : steady-state probability that there are j customers in the system just before arrival. For considered system the following rate conservation law holds: λq j 1 = jµp j, j m, loss system, λq j 1 = jµp j, j m, delay system, λq j 1 = mµp j, j > m, delay system. (53) Lecture: GI/M/1 and GI/M/m queuing systems 35

36 7.1. Rate conservation law for loss system Consider loss system: example: GI/M/1/K; example: GI/M/m/m, etc. Think as follows: let S j denote j customers in the system; q j 1 is the probability that j 1 customers just prior to arrival and λ is the arrival rate: S j 1 S j. (54) at the same time RHS represents rate: S j S j 1 (55) both sides must be balanced at steady-state: λq j 1 = jµp j. (56) Lecture: GI/M/1 and GI/M/m queuing systems 36

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