S Queueing Theory, II/2007 Exercises. Virtamo / Penttinen. 10th December Discrete and Continuous Distributions 1

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1 S Queueing Theory, II/007 Exercises Virtamo / Penttinen 0th December 007 Contents Discrete and Continuous Distributions Distributions and Markov Chains 5 3 Birth&death- and Poisson-processes, Little s result 8 4 Loss Systems 5 Queuing Systems 6 6 Priority queues, queuing networks 0 ABTEKNILLINEN KORKEAKOULU

2 Helsinki University of Technology Department of Electrical and Communications Engineering Networking Laboratory P.O. Box 3000 FIN-005 HUT

3 Exercise DISCRETE AND CONTINUOUS DISTRIBUTIONS EX : Discrete and Continuous Distributions. There are two coins one of which is normal and the other one has head on both sides. One of the coins is chosen randomly and tossed m times and each time the result is head. What is the probability that the chosen coin is the normal one. Calculate also the numerical value for m,, 3. Denote the possible event with the following symbols: X the normal coin is chosen, P{X} / Y the fake coin is chosen, P{Y } / C m tossing the chosen coin m times gives m heads in a row Then the conditional probabilities for getting m heads in a row are P{C m X} (/) m, P{C m Y }. The asked quantity is P{X C m }, so we can apply the Bayes formula: P{A j B} P{B A j}p{a j } i P{B A i}p{a i }, where i A i S and A i Aj i j. So in this case we get P{X C m } P{C m X}P{X} P{C m X}P{X} +P{C m Y }P{Y } (/) m (/) (/) m (/) + (/) m +. For m,, 3 we get m P{X C m } /3 /5 3 /9. In a Bernoulli trial is obtained with probability p and 0 with probability q p. Thevalue of p is unknown, but it is known that it is drawn from a uniform distribution in (0,), i.e. all values in this interval are aprioriequally likely. The Bernoulli trial is repeated in order to estimate the real value of p. In the trials result 0 occurs n 0 times and result n times. What is the posterior distribution (Bayes) of p according to the trials? Where is the maximum of the distribution? Let n n 0 + n. The probability, that after n trials we have gotten result exactly n times is ( ) n P{n p} p n ( p) n n n Denote with f(p) the conditional pdf of p and with f 0 (p) the aprioripdf of p. THe Bayes formula gives f(p) P{n p}f 0 (p) 0 P{n p}f 0 (p)dp P{n p} 0 P{n p}dp pn ( p)n n 0 pn ( p) n n dp, as f 0 (p) (uniform distribution). Denote C(n,n) 0 pn ( p) n n dp, when f(p) pn ( p) n n. C(n,n)

4 S Queueing Theory, II/007 Virtamo / Penttinen The maximum of distribution is obtained by taking the derivate: f (p) [ n p n ( p) n 0 n 0 p n ( p) n 0 ] C f (p ) 0 n ( p )n 0 p p n. n 0 + n Next we deduce C(n,n): n ( ) n C(n,n)z n n 0 n 0 n 0 0 n ( ) n (zp) n ( p) n n dp n ((zp)+( p)) n dp / ((z )p +)n+ n +z 0 ( z n+ ) n z n, n + z n + 0 ((z )p +) n dp n 0 and comparing the factors of z n one sees that C(n,n) ( n + n ) n!(n n )! n (n +)! [ ( n (n +) n )]. Hence, the conditional distribution of p is ( ) n f(p) (n +) p n ( p) n 0. Furthermore, the expected value of p on condition that n is known, is E[p n ] 3. Apply the conditioning rules 0 n ( ) { }}{ n pf(p) dp (n +) p n+ ( p) n n dp n ) n 0 C(n +,n+) (n +)( n (n +) ( n+ ) (n +)n!(n +)!(n n )! (n +)n n +!(n n )!(n +)! n + n +. E[X] E[E[X Y ]] V[X] E[V[X Y ]] + V [E [X Y ]] to the case X X X N,wheretheX i are independent and identically distributed (i.i.d.) random variables with mean m and variance σ,andn is a positive integer valued random variable with mean n and variance ν. Hint: Condition on the value of N. By applying the given conditioning rules we get E[X] E[E[X N]] E [Nm]nm. V[X] E[V[X N]] + V [E [X N]] E [ σ N ] +V[mN] nσ + m ν.

5 Exercise DISCRETE AND CONTINUOUS DISTRIBUTIONS 4. Five different applications are transmitting packets in a LAN. For each application i, i,...,5, measurements are carried out to determine the mean, m i, and the standard deviation, σ i, of the packet lengths. The measurements provide also a classification of the packets giving the proportion of packets belonging to each application i (denoted by p i ). The results of the measurements are summarized in the following table. Compute the mean and the standard deviation of the lengths of all packets in the network. application p i m i σ i Let the random variable X denote the packet length and Y denote the application in which the packet belongs to. Apply the chain rule of expectation and variance: m i {}}{ E[X] E[ E[X Y ]] p i m i 50.5 i σ i m i {}}{{}}{ V[X] E[ V[X Y ]] + V[ E[X Y ]] i p i σ i + i p i m i ( i p i m i ) Thus the packet length in the network has the mean 50.5 and standard deviation Assume that the traffic process in a LAN is stationary, whence periods of equal length are statistically identical. Denote the amount of traffic (kb) carried in the network in consecutive 0 min periods by X, X and X 3 (random variables). In measurements over a long time one has observed that the variance of the traffic in periods of 0 min, 0 min and 30 min are v, v ja v 3, respectively. In other words, v V[X ]V[X ]V[X 3 ], v V[X + X ] V[X + X 3 ] and v 3 V[X + X + X 3 ]. Determine Cov[X,X ]Cov[X,X 3 ] and Cov[X,X 3 ] in terms of v, v and v 3. Hint: e.g. the previous one can be obtained by expanding v V[X + X ]Cov[X + X,X + X ]. X X X 3 Definition: Cov[A, B] E[(A E[A]) (B E[B])] E [AB] E[A]E[B]. Thus e.g. Cov[A + B,C]E[AC + BC] E[A + B]E[C] Cov[A, C]+Cov[B,C] etc. (i) Stationary Cov[X,X ]Cov[X,X 3 ] v Cov[X + X,X + X ]Cov[X,X ]+Cov[X,X ]+Cov[X,X ] v +Cov[X,X ] Cov[X,X ]v / v. (ii) v 3 Cov[X + X + X 3,X + X + X 3 ] V[X ]+V[X ]+V[X 3 ]+Cov[X,X ]+Cov[X,X 3 ]+Cov[X,X 3 ] 3v +4(v / v )+Cov[X,X 3 ] Cov[X,X 3 ]v / v + v 3 /. 3

6 S Queueing Theory, II/007 Virtamo / Penttinen Hence, measuring variances v, v and v 3 is enough to determine covariances. 6. The length of the packet (in bytes) arriving to a router is assumed to obey geometric distribution. The mean length of packet is 00 bytes. Each packet is first read into an input buffer. How large the input buffer should be in order that an arriving packet fits there with probability of 95 % or higher? Let random variable X denote the length of the packet. The packet length was assumed to obey a geometric distribution with mean 00, and thus P{X n} ( p) n p q n p, where p /00. Let the input buffer length be N. The probability that a random packet overflows is then, N n P{X N} q i p p q i p qn qn q from what N can be obtained: i i0 P{X N} 0.95 q N 0.05 N ln q ln 0.05 (ln q<0) N ln 0.05/ ln q 98., thus when the buffer length is 99 bytes or more, the arriving packet fits in the buffer with probability of 95 % or more. 4

7 Exercise DISTRIBUTIONS AND MARKOV CHAINS EX : Distributions and Markov Chains. Let S X X N,whereX i Exp(), be i.i.d. and N an independent geometrically distributed random variable, P{N k} ( p)p k, k,,... Determine the tail distribution of S, G(x) P{S >x}. First we derive the generating function of geometrical distribution: ( ) p G N (z) z k ( p)p k (pz) k p pz p p pz z p pz. k k Similarly, the generating function of exponential distribution is G X (s) The generating function of random sum is S(s) G N (G X (s)) ( p) 0 e t e st dt +s p +s 0 e (+s)t dt + s. ( p) + s p ( p) ( p) + s. In other words, the sum S obeys the exponential distribution with parameter ( p). Hence, the tail distribution is / P{S >x} ( p)e ( p)t dt e ( p)t e ( p)x. x. Prove (without using the generating function), that the sum of two Poisson random variables, N Poisson(a )andn Poisson(a ), is also Poisson distributed: (N +N ) Poisson(a + a ). Prove the same result with the aid of generating functions. Poisson-distribution: P {N i n} (a i) n n! e a i. x P{N n} P{N + N n} n (a ) j j0 j! n P{N j} P{N n j} j0 e a (a ) n j (n j)! e a e (a+a) n! n j0 n! j!(n j)! (a ) j (a ) n j (a + a ) n e (a +a ), (binomial theorem) n! thus the sum N N + N is Poisson(a + a ). On the other hand, let X Poisson(a). Then the generating function of the random variable X is GX(z) E[z X ] j0 z j aj j! e a e a (az) j e a e az e (z )a. j! Let N(z) denote the generating function of sum N + N, for which we obtain j0 N(z) N (z) N (z) e (z )a e (z )a e (z )(a +a ). Hence, N obeys Poisson distribution with parameter a + a. 5

8 S Queueing Theory, II/007 Virtamo / Penttinen 3. Let X be an exponential random variable. Without any computations, tell which one of the following is correct. Explain your answer. a) E [ X X> ] E [ (X +) ] b) E [ X X> ] E [ X ] + c) E [ X X> ] (+E[X]) Clearly a) is true, as from the lack of memory property of exponential distribution it follows that on condition X > the random variable X is similarly distributed as X The transition probability matrix of a four state Markov chain is P / / B A Draw the state transition diagram of the chain and deduce which states are transient and which are recurrent. Clearly, all the states of the chain are recurrent (process returns to each state with probability of ) and there are no transient states (probability of return is less than ). Formally, let Thus, p n def P{ chain returns to state 3 during 3n steps} p and at the limit n we get, p +( ) + ( ) p n n i ( ) i p ( ) i /. i0 Thus the chain returns to state 3 with probability of and the state is recurrent. Similarly for state Time to absorption. A three state Markov chain (with states i,...,3) has the state transition probability matrix: P 0 /4 / / /4 /4A. 0 0 State 3 is an absorbing state. Let T i denote the average time (number of steps) needed by a system in state i to reach the absorbing state 3 (T 3 0). Write equations for T i, i,, based on the facts that the next transition i j takes one step and due to the Markovian property it takes T j steps on the average to reach the absorbing state from state j. Solve the equations. 6

9 Exercise DISTRIBUTIONS AND MARKOV CHAINS From symmetry it follows that T T, and thus T p ( + T )+p ( + T )+p 3 ( + T 3 ) +p T + p T (symmetry) T T 4. /4 / / /4 /4 6. Define the state of the system at the n th trial of an infinite sequence of Bernoulli(p) trialstobe the number of consecutive succesful trials preceeding and the current trial, i.e. the state is the distance to previous unsuccesful trial. If the n th trial is unsuccesful, then X n 0; if it succeeds but the previous one was unsuccesful, then X n, etc. a) What is the state space of the system? b) Argue that X n forms a Markov chain. c) Write down the transition probability matrix of the Markov chain (give its structure). /4 (a) X n can be any integer from 0 to (b) The experiments are independent: { Xn + with probability of p, X n+ 0 with probability of p, that is the state of the system at time n+depends only on the state at time n. That is the Markov property and hence the system is a Markov chain. (c) p p 0 0 p 0 p 0 P p 0 0 p

10 S Queueing Theory, II/007 Virtamo / Penttinen EX 3: Birth&death- and Poisson-processes, Little s result. Determine the probability distribution in equilibrium for birth-death processes (state space i 0,,,...), which transition intensities are a) λ i λ, i i, b)λ i λ/(i +), i, where λ and are constants. λ i i+ i i+ (i+) Figure : Transition intensities between the states i and i +. λ i+ (a) From fig. it can be seen that (i +)π i+ λπ i π i+ λ i + π i a i + π i. That is Furthermore, i π i,i.e. π aπ 0, π a π a! π 0,. π i ai i! π 0. π i π 0 i So π i ai i! e a. (that is a Poisson-distribution with parameter a λ/) (b) In this case we obtain i a i i! π 0e a π 0 e a. λ i + π i π i+ π i+ a i + π i that is the probability distribution in equilibrium is the same as in (a).. In a game audio signals arrive in the interval (0,T) according to a Poisson process with rate λ, where T > /λ. The player wins only if there will be at least one audio signal in that interval and he pushes a button (only one push allowed) upon the last of the signals. The player uses the following strategy: he pushes the button upon the arrival of the first (if any) signal after a fixed time s T. a) What is the probability that the player wins? b) What value of s maximizes the probability of winning, and what is the probability in this case? 8

11 Exercise 3 BIRTH&DEATH- AND POISSON-PROCESSES, LITTLE S RESULT Player bets on that during the time (s, T ) there is exactly one arrival (what happened during (0,s) has no effect here). Let τ T s, when the number of arrivals obeys Poisson distribution with parameter a λτ. Here, 0 τ T,i.e.0 a λt where λt >. a) p v P{N(T ) N(s) } a! e a ae a, where a λ(t s). b) Maximum can be found by taking the first derivate in relative to a: d da p v(a) e a ae a ( a)e a. The derivate has clearly exactly one root, a, which is also the maximum of the function (first strictly increasing and then stricly decreasing function). a λ(t s) s T /λ, which also lies inside the allowed interval. The maximum probability of winning is hence /e. 3. It has been observed that in the interval (0,t) one arrival has occurred from a Poisson process, i.e. N(0,t). Prove that conditioned on this information, the arrival time τ is uniformly distributed in (0,t). Hint: determine the conditional cumulative distribution function of the arrival time P{τ s one arrival during (0,t)}. P{τ s N(0,t)} P{τ s and N(0,t)} P{N(0,t)} λs (λ(t! e λs s))0 0! e λ(t s) P{N(0,s)and N(s, t) 0} P{N(0,t)} s/t, Thus, τ Uniform(0,t). λt! e λt 4. Customers arrive at the system according to a Poisson process with the intensity λ. Eachcus- tomer brings in a revenue Y (independently of other customers), which is assumed be an integer with the distribution p i P{Y i}, i,,...letx t denote the total revenue gained during the time interval (0,t). a) Derive expressions for E[X t ] and V[X t ]. b) Deduce that X t E +E +3E 3 +,wherethee i are independent random variables with the distributions E i Poisson(p i λt). Denote X t total revenue on the interval (0,t), N t number of customers on the interval (0,t), N t Poisson(λt). (a) Clearly, the system revenue is a random sum X t Y Y Nt, 9

12 S Queueing Theory, II/007 Virtamo / Penttinen where the Y i are i.i.d. random variables. Conditioning on the value of N t we can apply the chain rules for expectation and variance: Variance: E[X t ]E[E[X t N t ]] E [ N t Y ] E[N t ]E[Y ]λt E[Y ]. V[X t ] V[E[X t N t ]] + E [V [X t N t ]] V [ N t Y ] +E[N t V[Y ]] ( ) V[N t ]E[Y ] +E[N t ]V[Y ]λt E[Y ] +V[Y ] λt E [ Y ]. λt t p λt t p λt t Figure : Random splitting of the Poisson process (b) On the interval (0,t) the customers arrive according to the Poisson process N t. Random splitting of this process with probabilities p i leads to new independent Poisson processes E i with the respective parameters p i λt. The customers of "class i" brought in the revenue i. Thus, X t i i E i. 5. Million (0 6 ) data packets per second arrive at a network from different sources. The lengths of routes, defined by the source and destination addresses, vary considerably. The time a packet spends in the network depends on the length of the route, but also on the congestion of the network. The distribution of the time a packet spends in the network is assumed to have the following distribution: ms (90 %), 0 ms (7 %), 00 ms (3 %). How many packets there are in the network on average? etc. transport delay proportion arrival intensity ms 90% /s 0ms 7% /s 00ms 3% /s So arrival intensity and average transport delay is known for for each traffic class. Next Little s result, N λt, can be applied: N ms /s 900, N 0 ms /s 700, N 3 00 ms /s 3000, The process is first split into two processes with the probability p of which the latter is split again with the probability p / X i p i 0

13 Exercise 3 BIRTH&DEATH- AND POISSON-PROCESSES, LITTLE S RESULT 6. The states of some irreducible Markov-process, which steady state probabilities π i are assumed to be known, can be partioned into two disjoint sets, A {,,...,n} and B {n +,n+,...}, so that the transition rates between the sets are non-zero only from state n to state n + and the opposite direction, i.e. q n,n+ λ and q n+,n. Using the Little s result write down an equation for the average time it takes from the transition n n +to the time the system returns back to set A (transition n + n), i.e. for the average time the system spends at set B. Define a new system to which a customer arrives when the original system moves from state n to state n +, i.e. a transition A B is interpreted as an arrival of a customer in a new system. Similarly, the transition B A in the original system is interpreted as a departure of the customer in the new system. Little s formula: N λ T, where λ is the arrival rate of customers. In the new system the average number of customers is clearly, A n λ n+ B N P{B} in+ Similarly, the average sojourn time T in the new system is the time from event A B to event B A, i.e. the quantity asked. The arrival intensity of the customers, λ,is λ λπ n. π i. Thus, T N λ in+ π i λπ n n i π i. λπ n

14 S Queueing Theory, II/007 Virtamo / Penttinen EX 4: Loss Systems. Consider a car-inspection center where cars arrive at a rate of one every 50 seconds and wait for an average of 5 minutes (inclusive of inspection time) to receive their inspections. After an inspection, 0 % of the car owners stay back an average of 0 minutes having their meals at the center s cafeteria. What is the average number of cars within the premises of the inspection center (inclusive of the cafeteria)? Little s theorem: N λw. The average time spent in the system is W 5min min 7min, and thus, N 50s 7min 6 7min min. Consider a communications link with unbounded capacity. At t 0 the link is empty and calls start arriving according to Poisson process with the intensity λ. The call holding times are assumed to be independent and exponentially distributed with the mean /. Consider the number of on-going calls N t at the time instant t 0. Determine the distribution of N t and, in particular, its mean as a function of time t. Hint: The number of on-going calls equals the number of arrivals between (0,t) from a certain inhomogeneous Poisson process, which can be obtained from the original Poisson process with a suitable random selection. p(x) 0 t Figure 3: Random selection from a Poisson process. The original process is a Poisson process with the intensity λ. We make a random selection of the original process by selecting calls with the probability that they are still in the system at the time t. In other words, p(x) P{call arriving at time x is still in the system at time t} e (t x). The resulting inhomogeneous Poisson process has the intensity λ (t) λp(x) λe (t x). According to the lecture notes, the number of arrivals on an interval (0,t) in an inhomogeneous Poisson process obeys the Poisson distribution with the parameter a(t) t 0 / t λe (t x) dx λe t ex λ ( e t e t ) λ ( e t ). 0 The mean of the distribution is the parameter itself, so the mean number of active calls at time t is given by λ ( e t ), which has the limit value λ, as it should. arrival intesity is a deterministic function of time

15 Exercise 4 LOSS SYSTEMS 3. A mail order company has 3 persons serving incoming calls. The calls arrive according to a Poisson process with intensity of /min and the average call duration is min. a) What is the probability that an incoming call is blocked, when blocked calls are lost? b) Is it profitable to hire a 4 th person if the total expenses per person are 00 /h and the average revenue per order is 0? The system is Erlang s loss system with 3 or 4 servers (M/M/K/K-system, where K 3or 4). The offered load a is a λ x /min min. Because we are considering a Poisson process, the call blocking and time blocking are the same. a) The probability of state 3 (K 3) can be obtained by calculating or by looking from a graph approximately: a 3 /3! E(3,a) +a/! + a /! + a 3 /3! b) In this case K 4. The fourth person is useful when there are 4 persons in service. The probability of system being in state 4 is E(4,a) a 4 /4! +a/! + a /! + a 3 /3! + a 4 /4! The blocked traffic lowers, i.e. the served traffic grows, with amount λ (E(3,a) E(4,a)). Thus, the net change is Hence, the 4 th person is worth hiring. λ (E(3,a) E(4,a)) 0 00 /h 38.3 /h. 4. Use recursive Erlang s blocking formula to calculate E(n, 6) for n 0,...,6. Recursive Erlang s blocking formula: E(n, a) Applying the recursive formula gives: a E(n,a), and E(0,a). n + a E(n,a) E(0, 6), E(4, 6) 6 36/ /6 54/5, E(, 6) /7, E(5, 6) 6 54/ /5 34/899, E(, 6) 6 6/7 +6 6/7 8/5, E(6, 6) 6 34/ /899 34/3. E(3, 6) 6 8/ /5 36/6, 5. Consider Erlang s loss system with n servers, where the intensity of the offered traffic is a. a) Show by direct calculations based on the steady state probabilities, that the mean number of customers in the system N is equal to ( E(n, a))a. Explain this by using Little s theorem. b) By using Little s theorem, deduce the mean time the system stays at state N n at a time. (Hint: How often the system moves to state N n? The mean number of customers in a given state is the same as steady state probability of the same state; the system either is in that particular state or not.) Deduce the same result directly from the properties of the exponential distribution. 3

16 S Queueing Theory, II/007 Virtamo / Penttinen a) For the steady state distribution of Erlang s blocking system it holds that π k a k /k! n j0 aj /j! For N one gets, N n k π k k0 n a k /k! n k n j0 aj /j! k0 k ak /k! n j0 aj /j! k0 n k ak /(k )! n j0 aj /j! a a n k0 ak+ /k! n a n k0 ak /k! a n+ /n! j0 aj /j! n j0 aj /j! a n /n! n a( E(n, a)). j0 aj /j! b) Define first a new stochastic process Y t as follows. Let X t denote the state of the original system at time t and let Y t denote { the occupancy of state n:, when Xt n, Y t 0, otherwise. Then, a transition to state n in the original process π X t corresponds to an arrival in Y t, and similarly a n n πn transition away from state n corresponds to a departure in Y t. In the equilibrium the arriving traffic to n n state n is λ Y π n λ. On the hand, the average λ occupancy of process Y t is Figure 4: Transitions to and from state n. N Y 0 ( π n )+ π n π n. For this we can apply Little s result: N λw π n π n λ W Y W Y π n π n λ a n /n! P n j0 aj /j! P a n /n! n j0 aj /j! λ a n /n! a n /n! λ a nλ n. On the other hand, the sojourn time in state n is exponentially distributed with parameter n (no higher states), and thus the mean sojourn time must be equal, W Y n. 6. Consider - and4 -concentrators, where for each input port calls arrive according to independent Poisson-processes with intensities γ. The mean call holding time is denoted by / and the offered load by â γ/ 0.. Compare in these two concentrators the probabilities that a call arriving to a free input port gets blocked because all the output ports are busy. The system corresponds to an Engset s system with n sources and s servers, 0 s n. Letπ j [n], j 0,...,s, denote the steady state probabilities. Similarly, let πj [n], j 0,...,s, denote the state probabilities seen by an arriving customer. 4

17 Exercise 4 LOSS SYSTEMS Time blocking in Engset s system is generally the same as the steady state probability of state s, π s [n] ( n s s k0 ) â s ( n k ) â k ( n s s k0 ) p s ( p) n s ( n k â, where p )p k ( p) n k +â. Similarly, for call blocking it holds that π s [n] π s[n ] ( ) n â s s s (. n )â k k0 k Substituting n and s to above gives (p 0.09) n s time blocking call blocking 6.7% 9.% 4 4.%.3% 5

18 S Queueing Theory, II/007 Virtamo / Penttinen EX 5: Queuing Systems. Assume that there is always one doctor at a clinic (4 hours a day) and that customers arrive according to a Poisson process with intensity λ. The service time of each customer obeys an exponential distribution with mean 30 minutes. Determine the maximum arrival intensity that allows 95% of the customers to be served within 7 hours from (the first) arrival. For the M/M/ queue, we know that (see lectures) which now yields the constraint P{W >t} ρ e ( λ)t, P{W >7} λ/ e ( λ) 7 5/00 λe 7λ e 44 /0. The maximum λ is obtained from the corresponding equality which can be solved numerically (it may be wise to take the logarithm on both sides of the equality first). The solution is λ max Show that in an M/G/-LIFO queue the equilibrium distribution of the queue length, π n ( ρ)ρ n, is insensitive to the holding time distribution. Hint: Reason that a customer who upon arrival finds j customers in the system will be exclusively served when the system is in state N j. Thus, the time the system spends in the state N j is completely composed of the full service times of the customers who arrive to the system in the state N j. Howmany such arrivals occur in a long interval of time T? During a long time interval T there is on average λπ j T arrivals to the state j. Because those customers are only served when the system is in state j, the system remains in that state on average the time λπ j T S, wheres is the average service time of one customer. That must be the same Tπ j,and it follows that Tπ j λπ j T S π j ρπ j, which is the same as with the ordinary M/M/-queue. 3. Consider an M/M//K queue with states 0,,...,K. Find the probability P n that the queue which initially is in the state n becomes empty before flowing over. Hint: Add an imaginary state K +to the system; a transition to the state K +corresponds to the queue flowing over. We have P 0 and P K+ 0. For states n,...,k, write the probability P n in terms of P n ja P n+. Solve the equations. State n is followed by state n with probability /(λ + ) andbystaten +with probability λ/(λ + ). By the Markov property, conditioning on the next state yields P n λ + P n + λ λ + P n+ λ(p n P n+ )(P n P n ). 6

19 Exercise 5 QUEUING SYSTEMS Defining D n P n P n+ allows writing this in the form D n λ D n ρd n D K i ρ i D K. On the other hand, we have K n i0 D K i (P K P K+ )+(P K P K )+...+(P n+ P n+ )+(P n P n+ )P n P K+ K n P n D K which, when written for P 0, allows determining D K : i0 ρ i, D K K i0 ρ i P 0 D K { ρ, ρ K+ ρ K+, ρ. 4. Customers arrive at a two-server system according to a Poisson process having rate λ 5/min. An arrival finding server free will begin service with that server. An arrival finding server busy and server free will enter service with server. An arrival finding both servers busy goes away. Once a customer is served by either server, he departs the system. The service times of the servers are exponential with rates 4/min and /min. a) What is the average time an entering customer spends in the system? b) What proportion of time is server busy? From the Fig. it can obtained the following equations: (π + π 3 )5π (5.) 4(π + π 3 )5(π 0 + π ) (5.) π +4π 5π 0 (5.3) Furthermore, the normalization condition, λ 3 λ λ 0 π 0 + π + π + π 3 (5.4) By combining (5.) and (5.4) one gets 3 4(π + π 3 )5( (π + π 3 )) π + π 3 5/9 π 0 + π 4/9. Similarly from (5.) gives, (π + π 3 )5π (π 0 + π )5π π 0 +7π, 3 Server itself behaves like normal M/M// system. 7

20 S Queueing Theory, II/007 Virtamo / Penttinen and from (5.) and (5.4), From what it follows, π +4π 5π 0 (π 0 + π )+4π 7π 0 8/9+4π 7π π 6π 0 π , π a) The average time spent in system: T 6 + 7π 6 44π π and π b) The server is reserved: π + π Customers arrive at an M/Erlang(k, )/ system according to a Poisson process with rate λ. Find the mean waiting and sojourn times of a customer in the system? Erlang(k, )-distributed random variable is a sum of k Exp()-random variables. Thus, E[X ]k/, V[X ]k/, E [ X ] V[X ]+E[X ] k/ + k / (k + k )/. In this case we have a M/G/-queue so we can apply Pollaczek-Khinchin mean value formulas. The offered load is ρ λx k λ and thus the mean sojourn time is T x + λx ( ρ) k + λ k+k ( kλ) k Similarly, the mean waiting time is ( + ) λ(k +). kλ W λx ( ρ) λ k+k ( kλ) λk(k +) ( kλ). 6. If in a single server system each customer has to pay a fee to the system according to some rule, then the average revenue rate of the system λ (average fee), whereλ is the mean rate of arriving customers. Apply this to the M/G/ system with the following charging rule: each customer in the system pays at rate which is the same as the customer s remaining service time. What is the average fee? Show by equating the above average revenue rate with the average charging rate (time charging) that W λ (X W + X /), where W and X are the waiting and service times. Solve W. What is this result? 8

21 Exercise 5 QUEUING SYSTEMS Let, { W unfinished work in the queue when customer arrives, and U unfinished work in the queue at an arbitary moment. We have a M/G/ queue and the fee rate is equal to the remaining service time. The average fee is X W + X, where X and W are independent random variables (the fee rate during the waiting is equal to the service time, and the fee rate drops linearly to zero when customer is under service) and thus the average revenue rate is λ(x W + X ). On the other hand, this is the same as the virtual waiting time, i.e. the waiting time of the customer if he would arrive to the system at the given moment of time. It follows from the PASTA property that the real waiting times W obey the same distribution, i.e. U W and U W. Thus, from which it can be deduced that W U λ(x W + X ), W ( λx) λx W λx ( λx) λx ( ρ), which is P-K mean value formula. 9

22 S Queueing Theory, II/007 Virtamo / Penttinen EX 6: Priority queues, queuing networks. Carloads of customers arrive at a single-server station in accordance with a Poisson process with rate 4 per hour. The service times are exponentially distributed with mean 3 min. If each carload contains either,, or 3 customers with respective probabilities 4,,, compute the average 4 customer waiting time in the queue. Hint: The waiting time of the first customer of each group can be obtained from an appropriate M/G/ queue. Consider separately the internal waiting time in the group. Consider first the whole batch as one unit and later the waiting time inside batch can be added to it. For the whole batch one can apply M/G/-queue model. It was given that The service time of the batch is, λ 4/h /5min, and / 3min. S X X N, where N is, or 3. For the number of customers, N, in one group it holds that, E[N] (/4) + (/) + 3 (/4) V[N] E [ (N E[N]) ] (/4) + 0 (/) + (/4) /. from which, by using the conditioning rule (or tower property), one gets for the service time S that, E[S]E[E[S N]] E [N 3min] 6min, V[S] E[V[S N]] + V [E [S N]] E [ N 9min ] +V[N 3min] 8min + 9 min 45/min E [ S ] V[S]+E[S] 45/min + 36min 7 min. The average waiting time of batch can be obtained by using the Pollaczek-Khinchin formula, E[W g ] λe [ S ] 7 ( λe[s]) 5min min ( 5min 6min) min 3 min 3min 5s. 4 5 In addition the customer sometimes have to wait within his batch. This is easiest to obtain by determining the total waiting time inside one batch and then dividing that by the average size of the batch: E[W c ](/4 0+/ +/4 3) 3min/ 5/8min min 5.5s. (Note. When batch size is 3, the average waiting time is (0++) 3min 3 3min.) By adding the waiting times together one gets the total average waiting time of customer, which is E[W ]E[W g ]+E[W c ]5min7.5s.. Persons arrive at a copying machine according to a Poisson process with rate /min. The number of copies to be made by each person is uniformly distributed between and 0. Each copy takes 3 s. Find the average waiting time in the queue when a) Each person uses the machine on a first-come first-served basis. b) Persons with no more than copies to make are given non-preemptive priority over other persons. 0

23 Exercise 6 PRIORITY QUEUES, QUEUING NETWORKS a) We will apply the P-K formula. Now S X 3s, where X is the number of copies needed. We have E[S] E[X] 3s 0 + 3s 33 s, E [ S ] E [ X ] 9s 0 0 i i 9s 77 9 s. The load of the system is ρ λ E[S] , and the average waiting time of a customer is W R/( ρ), wherer λe [ S ] / 77 9/( 60)s 3/80s is the average amount of unfinished work (service time) in the server. Thus, W s 3 58 s 3.98s. b) Customers have two priority classes (m 0,n and m,n 8): λ k E[S k ] ρ k 9 class 300 s 3/00 4 class 300 s 3( + 9 ) 39 56/600 3/50 In the prioritized system, the class-specific mean waiting times are W W R ρ.93s, R ( ρ )( ρ ρ ) W 4.04s. ρ ρ (R is the same as above; the order in which customers are served does not affect the average residual work.) The average waiting time is W λ W + λ W λ + λ 3.8s. 3. Consider a priority queue with two classes and preemptive resume priority. Customers arrive according to two independent Poisson processes with intensities λ and λ. Service times in both classes are independent and exponentially distributed with a joint mean /. Determine the mean sojourn times T and T for both classes. For preemptive priority queue it holds that (cf. lecture notes), T ( ρ )S + R ρ, T ( ρ ρ )S + R ( ρ )( ρ ρ ),

24 S Queueing Theory, II/007 Virtamo / Penttinen where R k k i λ i Si. In this case, and thus, from what one obtains, S S /, ρ λ /, S S /, ρ λ /, R λ / λ /, R T ( λ /)(/)+λ / λ / ( λ / + λ / ) (λ + λ )/, λ + λ λ λ, T ( (λ + λ )/)(/)+(λ + λ )/ ( λ /)( (λ + λ )/) ( λ )( λ λ ). λ λ + λ + λ ( λ )( λ λ ) Note that T is equal to the mean sojourn time in M/M/-queue, as it should be. 4. The Pollaczek-Khinchin formula for the Laplace transform of the waiting time W is W (s) s( ρ) s λ + λs (s) where S (s) is the Laplace transform of the service time S and ρ λs. Using this result, rederive the PK mean formula for the waiting time. The mean of random variable W is W Similarly, for the service time S it holds that [ S d ] ds S (s) Taylor serie of S (s) is Thus, S (s) S (0) }{{} W (s) from which it follows that [ d + s0 ] ds S (s) s0 }{{} S [ d ] ds W (s) s0 and S [ ] d ds S (s). s0 s + [ ] d ds S (s) s S s + s0 S s }{{} S s( ρ) s λ + λs (s) s( ρ) s λ + λ( S s + S s ) s( ρ) s( ρ)+ λs s W + λs [ d ] ds W (s) s0 ρ s, λs ρ.

25 Exercise 6 PRIORITY QUEUES, QUEUING NETWORKS 5. Queues and of the open Jackson queueing network depicted in the figure receive Poissonian arrival streams with rates and (customers/s). Service times are exponentially distributed with the given rates (customers/s). Calculate a) customer streams through each of the queues, b) average occupancies of the queues and the average total number of customers in the network, c) mean delays in the network of customers arriving at queues and as well as the delay of an arriving customer chosen at random. queue 3 queue 6 queue 3 0 / / a) Let the traffic stream fed back to system be x customers per second. Then the traffic stream to server 3 is x, from which half goes out. The traffic flow in and out of the system must be equal. Hence, x 3and λ λ 4 λ 3 6 ρ /3 ρ /3 ρ 3 3/5 b) Jacksons theorem states that queues of the network behave like they were independent and the offered traffic were Poissonian, i.e. P{N n} i P{N i n i }, where P{N i n i } ( ρ i )ρ n i i. In this problem, { P{N i} P{N i} 3 ( i 3), P{N 3 i} 5 ( 3 i 5), so the average number of customers in each server are { N N ρ ρ, N 3 ρ N N 3 ρ 3 3/. + N + N 3 /. c) The average sojourn times in each network node is T i N i λ i T T / T 3 /4 A customer arriving to the queue i sojourn time in the network is T i,d, for which it holds that T i,d T i + j q ij T j,d, where q ij is the probability that a customer leaving from queue i enters to queue j. Denote T i,d {a, b, c }. Then the set of linear equations can be written as a +c T,d b + c T,d 3/ c 4 + b T 3,d 3

26 S Queueing Theory, II/007 Virtamo / Penttinen Furthermore, a random customer stays in the network on average (Little s result) T N/λ /6. 6. Consider a cyclic closed network consisting of two queues. The service times in the queues are exponentially distributed with parameters and. There are three customers circulating in the network. a) Draw the state transition diagram of the network (four states). b) Determine the equilibrium probabilities and calculate the mean queue lengths. c) Calculate the customer stream in the network (e.g. the customer stream departing from queue ). d) Rederive the results of b) and c) by means of the mean value analysis (MVA). 3,0,, 0,3 Figure 5: State diagram of the system. a) State diagram of the system is depicted in Fig. 5. b) In steady state it holds that π 0 π π ρπ 0 π π π ρ π 0 π π 3 π 3 ρ 3 π 0 Normalization: The average queue lengths are ( + ρ + ρ + ρ 3 )π 0 π 0 N (3+ρ + ρ )π 0 3+ρ + ρ +ρ + ρ + ρ 3 +ρ + ρ + ρ 3. N (3ρ 3 +ρ + ρ)π 0 3ρ3 +ρ + ρ +ρ + ρ + ρ 3. (N + N 3) c) The traffic flow is λ (π 0 + π + π ) d) Mean value analysis (MVA): Here, N[0] [0, 0] T [] [ / [,ρ] ] N[] +ρ, ρ +ρ +ρ + ρ +ρ + ρ + ρ 3 ρ3 ρ ρ ρ 4 ρ3 ρ 4. T i [k] (+N i [k ]) / N i [k] k P λ it i [k] j λ jt j [k] λ i [k] N i [k]/t i [k] [ T [] / +ρ [ N[] +ρ, +ρ +ρ +ρ+ρ, +ρ ρ ] (+ρ) ρ+ρ +ρ+ρ ] [ ] T [3] / +ρ+ρ ++ρ, +ρ + ρ +ρ+ρ + ρ +ρ +ρ + ρ ρ [ (+ρ+ρ ) 3+ρ + ρ,ρ+ρ +3ρ [ ] 3] 3+ρ+ρ N[3] 3 ρ+ρ 3+3ρ+3ρ +3ρ, +3ρ ρ+3ρ +3ρ 3 [ 3+ρ + ρ,ρ+ρ +3ρ 3] π 0 4 [ +ρ, ρ +ρ ] [ +ρ +ρ+ρ, ] ρ+ρ +ρ+ρ

27 Exercise 6 Similarly, the traffic flow becomes λ λ [3] 3+ρ + ρ ( + ρ + ρ ) +ρ + ρ + ρ 3 3+ρ + ρ ( + ρ + ρ ) +ρ + ρ + ρ 3 ρ3 ρ 4. 5