M/G/1 and M/G/1/K systems

Transcription

1 M/G/1 and M/G/1/K systems Dmitri A. Moltchanov

2 OUTLINE: Description of M/G/1 system; Methods of analysis; Residual life approach; Imbedded Markov chain approach. Lecture: M/G/1 and M/G/1/K systems: part I 2

3 1. Description of M/G/1 queuing system M/G/1 queuing system stands for: single server; infinite number of waiting positions; Poisson arrival process; interarrival times are exponentially distributed. generally distributed service times: practically, any distribution. What we also assume: first come, first served (FCFS); what it may affect: PDF or pdf of waiting time! Lecture: M/G/1 and M/G/1/K systems: part I 3

4 1.1. Arrival and service processes Arrival process: Poisson with parameter (mean value) λ interarrival times are exponential with mean 1/λ; CDF, pdf and LT are: A(t) = 1 e λt, a(t) = λe λt, A(s) = 0 a(t)e st dt = 0 λe λt e st dt = λ s + λ. (1) Service process: service times are arbitrary with mean 1/µ; PDF, pdf and LT are: B(t), b(t), B(s). (2) Lecture: M/G/1 and M/G/1/K systems: part I 4

5 2. Methods of analysis There are a number of methods: residual life approach: simple to understand: only mean values can be obtained. transform approach based on imbedded Markov chain: harder to deal with; 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: harder to deal with; distributions of desired performance parameters can be obtained; the idea: find points at which Markov property holds and use direct integration. method of supplementary variables: the most complicated one: distributions of desired performance parameters can be obtained; the idea: look at arbitrary points and make them Markovian. Lecture: M/G/1 and M/G/1/K systems: part I 5

6 3. Residual life approach: mean waiting time Note the following: we want to find mean queuing time (mean time spent waiting for service); no need to assume FCFS service discipline: service order does not influence mean waiting time; we have to require that the server does not stay idle when there are waiting customers; mean values are the same as for LCFS, RANDOM, PS.... Arrivals: 1 Service: 1 Figure 1: M/G/1 queuing system. Note: we are looking for E[W Q ]: mean waiting time in the buffer. Lecture: M/G/1 and M/G/1/K systems: part I 6

7 Consider arriving customer, we have two choice: customer arrives to non-empty system or; it arrives to empty system. When the customer arrives to non-empty system it finds: n q customers waiting in the buffer; customer which is currently in service: it has R residual service time left to complete its ongoing service; residual service time is the time required to complete the ongoing service. When the customer arrives to empty system: no customers waiting n q = 0; residual service time R is zero; customer can enter service immediately. Lecture: M/G/1 and M/G/1/K systems: part I 7

8 Let us denote: RV R be the RV of residual service time: in case of exponential service times R has exponential distribution. E[R] be it mean. Note the following: PASTA property holds for M/G/1 system; distribution that arrival sees is the same as time-averaged (at arbitrary time); using Little s result we have for the mean waiting time in the buffer, E[W Q ]: E[W Q ] = E[N Q ]E[X] + E[R]. (3) E[N Q ] - mean number of packets waiting in the buffer, E[X] mean service time; E[R] - mean residual service time. Using Little s result E[N Q ] = λe[w Q ] we get: E[W Q ] = λe[w Q ]E[X] + E[R]. (4) Lecture: M/G/1 and M/G/1/K systems: part I 8

9 Denoting ρ = λe[x] and solving for E[W Q ]: E[W Q ] = ρe[w Q ] + E[R], E[W Q ] = E[R] (1 ρ). (5) to find E[W ] we, therefore, have to find mean residual service time E[R]. in what follows we take graphical approach. r(ô) ô Figure 2: Residual service times. R jumps by the amount of service time X when a new customer enters the service; decreases until the customer completes service, duration is X; R = 0 when customer completes its service. Lecture: M/G/1 and M/G/1/K systems: part I 9

10 Time averaged of r(τ) in (0, t) is: E[r[0, t)] = 1 t t 0 r(τ)dτ. (6) Approximate previous equation by the following sum: E[r[0, t)] = 1 t t 0 r(τ)dτ 1 t M(t) X 2 /2 is the square of a service triangle, M(t) arrivals in (0, t). Note the following: we have an error in the last residual service time in [0, t); i=1 1 2 X2 i. (7) approximation becomes better as t (number of components increases!). Multiplying and dividing it by M(t) we get: E[r[0, t)] = M(t) t 1 1 M(t) 2 M(t) i=1 X 2 i. (8) Lecture: M/G/1 and M/G/1/K systems: part I 10

11 Consider the last equation when t we get: lim t lim t M(t) t M(t) λ M(t) i=1 X 2 i 1 2 E[X2 ]. (9) Therefore, we have the following expression for mean residual service time: E[R] = E[r] = 1 2 λe[x2 ], (10) Substituting this result in E[W ] = E[R]/(1 ρ) we get: this result is known as Pollaczek-Khinchine formula: they simultaneously got this result. E[W Q ] = λe[x2 ] 2(1 ρ). (11) Lecture: M/G/1 and M/G/1/K systems: part I 11

12 3.1. Performance parameters Mean time spent in the buffer: E[W Q ] Mean number of customers in the buffer: E[N Q ] Mean time spent in the system: E[W ] E[W Q ] = λe[x2 ] 2(1 ρ). (12) E[N Q ] = λe[w Q ] = λ2 E[X 2 ] 2(1 ρ). (13) E[W ] = E[W Q ] + E[X] = λe[x2 ] 2(1 ρ) Mean number of customers in the system: E[N Q ] + E[X]. (14) E[N] = E[W ]λ = λ2 E[X 2 ] 2(1 ρ) + ρ. (15) Lecture: M/G/1 and M/G/1/K systems: part I 12

13 4. 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 (D 1 = D 2 )! the resulting process is actually birth-death Markov one! A 1 (t)=a 2 (t) A 2 (t) t D 1 (t)=d 2 (t) D 2 (t) t Figure 3: State of M/M/1 queuing system. Lecture: M/G/1 and M/G/1/K systems: part I 13

14 4.1. Problem with M/G/1 queuing system Let S Q (t) be the number of customers at time t: do we know how the system evolves in time after t in t? arrival may occur with rate λ t: it is independent from previous arrival: memoryless property (A 1 = A 2 )! departure may occur with probability µ t: µ µ as D 2 is not the same as D 1! this process is no longer Markovian! A 1 (t) A 2 (t) t D 2 (t) D ( t) : D ( t) D ( t) t Figure 4: State of M/G/1 queuing system. Lecture: M/G/1 and M/G/1/K systems: part I 14

15 4.2. How to choose the state of M/G/1 system State of the system: M/G/1 queuing system number of customers in the system AND time since the service started: in this case we know how the system evolves in time; we know the distribution of time till the next arrival: it is the same as initial one; we know the distribution of time till the next departure: we track it: D(t + x x) = P r{(t t + x) T > x} = (D(t + x) D(x))/(1 D(x)). (16) t A( t) 1 e t t S ( ), 0,1,... Q t k k D( t x x) D( t x) D( x) 1 D( x) Figure 5: State of M/G/1 queuing system given by (S Q (t), D(t + x x)). Lecture: M/G/1 and M/G/1/K systems: part I 15

16 Are there any problems with such description? two complicated to deal with; we better prefer to avoid it if more simple description is available! State of the system: M/G/1 queuing system number of customers in the system at special time instants d + 1, d + 2,... : choose these instants such that we should not track time since previous service started; which instants: just after service completions as D(t + x x) = D(t) when x = 0! d 1 + A( t) 1 e t d 2 + x 0 D( t) D( t x) D( x) D( t 0) D(0) D( t x x) 0 1 D( x) 1 D(0) t S ( ), 0,1,... Q t k k Figure 6: State of M/G/1 queuing system. Lecture: M/G/1 and M/G/1/K systems: part I 16

17 4.3. Steady-state distribution as seen by departure Doe the following: consider Markov chain imbedded at t i, i = 1, 2,... when i th customer departs from the system; system state: number of customers at t i, i = 1, 2,... denoted by n i, i = 1, 2,... ; departing customer is not included in the system state! Time instant t i -... n i - = k Time instant t i +... n i + = k-1 Just departed at t i Figure 7: Time instants t i and t + i in M/G/1 queuing system. Lecture: M/G/1 and M/G/1/K systems: part I 17

18 Denote: a i+1 be the number of arrivals in (i + 1) th service time. a i+1 t i t i+1 x 0 ith departure D( t) t Figure 8: Arrivals in (i + 1) th service time. Consider time diagram of the system and distinguish between: i th departure leaves non-empty system: n i > 0; i th departure leaves empty system: n i = 0. Lecture: M/G/1 and M/G/1/K systems: part I 18

19 Consider the case when n i > 0. a i+1 arrivals in (i+1) th service time... (i+1) th service time t i th departure (system is not empty; (i+1) th departure (system may be empty may be not; n i >0 n i+1 = n i a i+1 Figure 9: Time diagram of the M/G/1 queuing system when n i > 0. In this case we have the following relation between n i and n i+1 : n i+1 = n i 1 + a i+1, n i = 1, 2,.... (17) Lecture: M/G/1 and M/G/1/K systems: part I 19

20 Consider the case when n i = 0. first arrival after empty period a i+1 arrivals in (i+1) th service time... (i+1) th service time t i th departure system is empty (i+1) th departure system may be empty may be not n i = 0 n i+1 = a i+1 Figure 10: Time diagram of the M/G/1 queuing system when n i = 0. In this case we have the following relation between n i and n i+1 : n i+1 = a i+1, n i = 0. (18) Lecture: M/G/1 and M/G/1/K systems: part I 20

21 Summarizing both cases (n i > 0 and n i = 0) we have: n i+1 = n i 1 + a i+1, n i = 1, 2,..., n i+1 = a i+1, n i = 0. (19) Introducing unit step function U(n i ) rewrite: n i+1 = n i U(n i ) + a i+1, n i = 0, 1,..., (20) where unit step function is given by U(n i ) = 1, n i = 1, 2,..., U(n i ) = 0, n i = 0. (21) this expressin gives us the idea how the system evolves between t i and t i+1 ; recall that t i, i = 0, 1,... are imbedded points of the Markov chain. GENERAL RELATION: n i+1 = n i U(n i ) + a i+1, n i = 0, 1,..., (22) Lecture: M/G/1 and M/G/1/K systems: part I 21

22 Consider imbedded Markov chain at steady-state t : let q ij, i, j = 0, 1,..., be transition probabilities of going from i to j; let a k the number of arrivals in the service time; q ik = α k, i = 0, k = 0, 1,..., q ik = α k i+1, i = 1, 2,..., k = i 1, i, i + 1,... (23) Figure 11: Transition probabilities of imbedded Markov chain at steady-state. Lecture: M/G/1 and M/G/1/K systems: part I 22

23 Observe that α k, k = 0, 1,... fully determine the Markov chain. departure k arrivals... service time departure t Figure 12: k arrivals in a service time. Arrivals come from Poisson process α k, k = 0, 1,..., we can get α k : α k = 0 (λx) k e λx b(x)dx, k = 0, 1,.... (24) k! Lecture: M/G/1 and M/G/1/K systems: part I 23

24 4.4. Transform approach vs. direct integration The difference is twofold: in the way we compute α k, k = 0, 1,... α k = 0 (λx) k e λx b(x)dx, k = 0, 1,.... (25) k! first term number of arrivals from Poisson process in time x (Erlang distribution); b(x) service time distribution equals to x; we integrate over all length of t: from 0 to. in the way we solve linear equations of the Markov chain asking Mathlab to do it for us; trying to get closed-form analytical solution. Let us define the following: q i, i = 0, 1,... : number of customers in the system just after departure; Lecture: M/G/1 and M/G/1/K systems: part I 24

25 4.5. Brute-force direct approach What we do: estimate α k directly via integration; use classic way to get steady-state solution for q i, i = 0, 1,... ; works when capacity is limited M/M/1/K. In vector/matrix form: q 00 q 01 q 02 q q 10 q 11 q 12 q qq = q : (q 0, q 1,... ) = (q 0, q 1,... ) 0 q 21 q 22 q q e = 1 : (q 0, q 1,... ) 1 = 1 (26). Lecture: M/G/1 and M/G/1/K systems: part I 25

26 4.6. Other steady-state distributions What we get: PF and moments of the number as seen by departure... Note: we are usually asked to get state distribution at arbitrary time! Let us define the following: p k, k = 0, 1,... : time-averaged (arbitrary time) steady-state distribution of states; a k, k = 0, 1,... : steady-state distribution of states as seen by arrival; d k, k = 0, 1,... : steady-state distribution of states as seen by departure. Question: how to find p k, k = 0, 1,...? Lecture: M/G/1 and M/G/1/K systems: part I 26

27 Two next results gives us a way to use this PF: Kleinrock s result (!!! remember it as well as PASTA): the following holds for those system when state can change at most ±1 in t 0: d k = a k,..., k = 0, 1,.... (27) PASTA property: the following holds when the arrival process is homogenous Poisson: a k = p k,..., k = 0, 1,.... (28) combining both we have for M/G/1 queuing system: a k = p k = d k,..., k = 0, 1,.... (29) What it gives: we obtained steady-state distribution at any instant of time; q k = a k facilitates derivation of waiting time CDF. Lecture: M/G/1 and M/G/1/K systems: part I 27

28 4.7. Performance parameters Mean time spent in the buffer: E[W Q ] Mean number of customers in the buffer: E[N Q ] Mean time spent in the system: E[W ] E[W Q ] = λe[x2 ] 2(1 ρ). (30) E[N Q ] = λe[w Q ] = λ2 E[X 2 ] 2(1 ρ). (31) E[W ] = E[W Q ] + E[X] = λe[x2 ] 2(1 ρ) Mean number of customers in the system: E[N Q ] + E[X]. (32) E[N] = E[W ]λ = λ2 E[X 2 ] 2(1 ρ) + ρ. (33) Lecture: M/G/1 and M/G/1/K systems: part I 28

29 5. Transform approach Consider n i+1 = n i U(n i ) + a i+1 and take means of both sides to get: E[n i+1 ] = E[n i ] E[U(n i )] + E[a i+1 ]. (34) Since we consider the system at steady-state E[n i+1 ] = E[n i ] = E[n]: E[U(n i )] = E[a i+1 ]. (35) Note the following: assuming that the system is empty with probability d 0 we have: E[U(n i )] = 1 d 0. (36) mean number of arrivals in (i + 1) th service time is given by: E[a i+1 ] = 0 λtb(t)dt = λe[x] = ρ. (37) Therefore, in the steady-state we have: d 0 = 1 ρ. (38) Lecture: M/G/1 and M/G/1/K systems: part I 29

30 Introduce PGFs of n i and n i+1 : P i (z) = E[z n i ] = z k P r{n i = k}, k=0 P i+1 (z) = E[z n i+1 ] = z k P r{n i+1 = k}. (39) Considering n i+1 = n i U(n i ) + a i+1 as the sum of RVs n i U(n i ) and a i+1 : To simplify it, consider the following: k=0 P i+1 (z) = E[z n i U(n i ) ]E[z a i+1 ]. (40) since we consider system at steady-state we can drop subscripts i and write: P (z) = E[z n U(n) ]E[z a ]. (41) defining PGF of number of arriving customers in service time by A(z) = E[z a ], then: A(z) = 0 E[z a X = t]b(t)dt = 0 z n (λt)n e λt b(t)dt = n! 0 e λt(1 z) b(t)dt. (42) Lecture: M/G/1 and M/G/1/K systems: part I 30

31 Rewrite the last result as: A(z) = 0 e λt(1 z) b(t)dt = 0 b(t)e (λ λz)t dt. (43) Recall the definition of the Laplace transform: F (s) = 0 f(t)e st dt, (44) Thus, we have for PGF of arriving customers in service time: A(z) = B(λ λz). (45) where B( ) is the LT. Now we can get moments of number of arrivals in service time: A (1) = λb (0) = λe[x] = ρ; A (1) = λ 2 B (0) = λ 2 E[X 2 ]. (46) we used moments generating properties of PGF and Laplace transform. Lecture: M/G/1 and M/G/1/K systems: part I 31

32 Let us get back to P (z) = E[z n U(n) ]E[z a ]: P (z) = A(z)E[z n U(n) ] = A(z) z k U(k) P r{n = k} = k=0 ( ) ( ) = A(z) z 0 d 0 + z k 1 p k = A(z) d z k p k 1 z z d 0 = k=1 k=0 ( 1 = A(z) z P (z) 1 ) z d 0(1 z). (47) Use d 0 = (1 ρ) to get PGF of the number of customers as seen by departure: P (z) = (1 ρ)(1 z)a(z), (48) A(z) z where A(z) = B(λ λz). Substituting A(z) = B(λ λz), we get Pollaczek-Khinchine transform equation: P (z) = (1 ρ)(1 z)b(λ λz). (49) B(λ λz) z Lecture: M/G/1 and M/G/1/K systems: part I 32

33 5.1. Mean number of customers as seen by departure Using using moment generating properties of PGF and LT: we can get moments of the number of customers as seen by departure. To get mean, rewrite (48) as follows: Taking two derivatives we get: P (z)(a(z) z) = (1 ρ)(1 z)a(z). (50) P (z)(a(z) z) + P (z)(a (z) 1) = (1 ρ)(1 z)a (z) (1 ρ)a(z), P (z)(a(z) z) + 2P (z)(a (z) 1) + P (z)a (z) = (1 ρ)(1 z)a (z) (1 ρ)a (z). (51) Let z = 1 and use A(1) = 1, P (1) = 1, A (1) = ρ and A (1) = λ 2 E[X 2 ] to get: 2P (1)(1 ρ) + λ 2 E[X 2 ] = (1 ρ)( 2ρ) (52) Mean number of customers as seen by departing customer: E[N] = λ2 E[X 2 ] 2(1 ρ). (53) Lecture: M/G/1 and M/G/1/K systems: part I 33