STOCHASTIC NETWORKS EXAMPLE SHEET 2 SOLUTIONS

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1 STOCHASTIC NETWORKS EXAMPLE SHEET 2 SOLUTIONS ELENA YUDOVINA Exercise. Show that the traffic equatios for a ope migratio process have a uique solutio, ad that this solutio is positive. [Hit: From the irreducibility of the ope migratio process deduce the irreducibility of a certai Markov process o J + states, ad the use the fact that the equilibrium distributio for this process is uique.] Deduce that i a ope migratio process α j λ j is the mea arrival rate at coloy j, coutig arrivals from outside the system ad from other coloies. Proof. Cosider the Markov process o J + states correspodig, loosely, to the case of N = perso i the ope migratio etwork (state J + correspods to rest-of-theuiverse ). The trasitio rates will be q(j, k) = λ jk q(j, J + ) = µ j q(j +, j) = ν j j, k =,..., J j =,..., J j =,..., J I claim that this etwork is irreducible. Ideed, ote that i the origial etwork the states e j = a sigle idividual i coloy j alog with the state 0 = obody i ay coloy all commuicate possibly goig through states with multiple idividuals i the system. I particular, everythig commuicates with state 0. (State e j correspods to j, ad 0 correspods to J +.) Now, to get from e j to 0 we eed the idividual i j to leave somewhere. Let us simply trace his path through the etwork: it must be alog edges with positive rates, ad it must termiate with a trasitio out of the etwork. If we simply look at the correspodig trasitios i the modified Markov process, they will have positive rates too (potetially differet oes if we had a few arrivals ito state j first, the rate of leavig it may have greatly icreased or decreased; however, it could t have become o-zero if the trasitio rate i our process is zero). Thus, we have show that it is possible to get from j to J + for all j =,..., J +. The reverse directio follows by tracig back the path alog which the idividual who is i coloy j whe the system state is e j came ito the etwork. Now, the equilibrium distributio ( α j ) of the modified Markov process satisfies J J α j ( λ jk + µ j ) = α j λ kj + α J+ ν j, j =,..., J α J+ ( J+ j= J ν j ) = α j = J α j µ j

2 2 ELENA YUDOVINA Sice the Markov process is irreducible o a fiite state space, there exists a uique distributio α j with these properties, ad it will the assig positive probabilities to all states (as otherwise the chai would have to be reducible). Now, the solutio to the traffic equatios (α j ) correspods to the first two equatios above, but with a differet ormalisatio, amely, α J+ =. Clearly, we ca go betwee the two: give α j we ca produce α j = α j /( + j α j) (with α J+ = /( + j α j)), ad coversely give α j we ca defie α j = α j /α J+. Thus, we coclude existece, uiqueess, ad positivity of the solutio to the traffic equatios. Fially, the mea arrival rate ito coloy j is π()( q(t jk, )+ν j ) = π()( λ kj φ k ( k +))+ν j = λ kj π()φ k ( k +))+ν j k k k = k α k λ kj + ν j = α j λ j. Exercise 2. Show that the reversed process obtaied from a statioary closed migratio process is also a closed migratio process, ad determie its trasitio rates. Proof. Recall that the form of the equilibrium distributio for a closed migratio process is J α j j π() = B N j r= φ j (r). I that case, the trasitio probabilities for the reversed process are q (, T jk ()) = π(t jk()) π() j= q(t jk (), ) = φ k ( k + )/π j ( j )λ kj φ k ( k + ) = φ j ( j ) α k/α j. λ kj Note that this has the required form, with φ j( j ) = φ j ( j ) ad λ jk = α k α j λ kj. Exercise 3. A restaurat has N tables, with a customer seated at each table. Two waiters are servig them. Oe of the waiters moves from table to table takig orders for food. The time that he speds takig orders at each table is expoetially distributed with parameter µ. He is followed by the wie waiter who speds a expoetially distributed time with parameter µ 2 takig orders at each table. Customers always order food first ad the wie, ad orders caot be take cocurretly by both waiters from the same customer. All times take to order are idepedet of each other. A customer, after havig placed her two orders, completes her meal at rate ν, idepedetly of the other customers. As soo as she fiishes her meal, she departs ad a ew customer takes her place ad waits to order. Model this as a closed migratio process. Show that the statioary probability that both waiters are busy ca be writte i the form G(N 2) ν 2, G(N) µ µ 2 for a fuctio G( ), which may also deped o ν, µ, µ 2, to be determied. I the above model it is assumed that the restaurat is always full. Develop a model i which this assumptio is relaxed: for example, assume that customers eter the restaurat at rate λ while there are tables empty. Agai obtai a expressio for the probability that both waiters are busy.

3 STOCHASTIC NETWORKS EXAMPLE SHEET 2 SOLUTIONS 3 Proof. The closed migratio model will have three coloies: coloy ot-yet-ordered, coloy 2 ordered food but ot wie, ad coloy 3 ordered both. The trasitio rates are λ 2 = µ φ = I[ > 0] λ 23 = µ 2 φ 2 = I[ 2 > 0] λ 3 = ν φ 3 = 3. (You may otice that this looks strikigly familiar from the lectures.) We coclude that α 0 : α : α 2 = µ : µ 2 : ν, ad π(, 2, 3 ) = C(N) >0, 2 >0, =N µ µ 2 2 ν 3 3! for the appropriate ormalisig costat C(N). The probability i which we are iterested is P(both waiters busy) = π(, 2, 3 ) = C(N) µ µ 2 2 ν 3 3! = C(N), 2, =N 2 >0, 2 >0, =N /(µ µ 2 ) µ µ 2 2 ν 3 2 3! = C(N 2). µ µ 2 C(N) Now, this has almost the right shape, but it would be icer to express the aswers as dimesioless for example, if we were to try ad scale the quatities i the problem by a costat factor, it would be ice to be able to easily see that this probability is uchaged. I light of this, we defie ( ) ( ) 2 ν ν G(N) = = ν N C(N), 3! µ µ =N which is obviously dimesioless, ad for which the desired form holds. (Quatities of the form ρ = ν/µ, the load, are very atural i queueig theory.) If we allow customers to leave the system after eatig ad to arrive to empty tables, the we will eed to have a extra coloy, 0 table empty i our system. The trasitio rates become This gives ad therefore λ 0 = λ φ 0 = I[ 0 > 0] λ 2 = µ φ = I[ > 0] λ 23 = µ 2 φ 2 = I[ 2 > 0] λ 30 = ν φ 3 = 3. α 0 : α : α 2 : α 3 = λ : µ : µ 2 : ν π( 0,, 2, 3 ) = C (N) λ 0 µ µ 2 2 ν 3 3!

4 4 ELENA YUDOVINA We are still iterested i summig over the states with > 0 ad 2 > 0, which will still give a expressio of the form C(N 2) C(N) µ µ 2, or the same termial form as before, for G (N) = ν N C ( ν ) ( ) ( ) 2 0 ν ν (N) = 3! λ µ µ =N Note that we ca also defie G (N) = λ N C (N) ad get λ 2 /(µ µ 2 ) as the secod factor. Exercise 4. Show that if the parameters of a statioary ope migratio process are such that there is o path by which a idividual leavig coloy k ca later reach coloy j, the the stream of idividuals movig directly from j to k forms a Poisso process. Proof. Note first of all that this does ot show the system is reducible all states ca still commuicate with the state system empty. Cosider the commuicatig class of coloy j, i.e. the set of coloies j such that a idividual ca get from j to j ad back agai without leavig the system. Observe that if we look oly at the idividuals i these coloies, the we have a Markov process, ad its trasitio rates mea that it is itself a ope migratio process. Cosequetly, the exit stream from coloy j i this process is Poisso. Now, the stream of idividuals movig directly from j to k is a probabilistic thiig of this exit stream, ad therefore also Poisso. Exercise 5. Airlie passegers arrive at a passport cotrol desk i accordace with a Poisso process of rate ν. The desk operates as a sigle-server queue at which service times are idepedet ad expoetially distributed with mea µ (< ν ) ad are idepedet of the arrival process. After leavig the passport cotrol desk a passeger must pass through a security check. This also operates as a sigle-server queue, but oe at which service times are all equal to τ (< ν ). Show that i equilibrium the probability that both queues are empty is ( νµ)( ντ). It if takes a time σ to walk form the first queue to the secod, what is the equilibrium probability that both queues are empty ad there is o passeger walkig betwee them? Proof. To solve the first part of the problem, recall the result that queues i a series of M/M/ queues are idepedet, ad ote that we ca make the last queue i that a /D/ queue without chagig aythig. Thus, the probability that both queues are empty is P(both empty) = P(queue empty)p(queue 2 empty) (Sice the arrivals to both queues are Poisso, the probability that the queue is empty whe a arrival happes is the same as the statioary probability of its beig empty.) For queue we kow this probability exactly to be ( νµ), but i fact it is isesitive to the service time distributio. Ideed, we ca compute the statioary probability that a queue with mea service time τ is empty from Little s law: let the umber i the system be the umber of customers i service (0 or ), i which case the probability that the system is occupied is the expected umber i system L. The sojour time is the service time, so W = τ; the arrival rate is still ν, assumig the queue is stable (the customers must come up for service about as ofte as they come ito the queue). Thus, L = ντ, ad the probability of the server beig idle is ( ντ).

5 STOCHASTIC NETWORKS EXAMPLE SHEET 2 SOLUTIONS 5 Addig a walkig time is like addig a ifiite-server queue with determiistic service time ito the series queue. The evet that the system is empty correspods to the followig: firstly, we eed queues ad 2 to be empty at time 0, ad secodly, we eed o departures from queue betwee σ util 0. Now that the iput stream for queue 2 is simply the output stream from queue shifted by σ; so the state of queue 2 at time 0 is determied by the departures from queue durig the time (, σ]. The umber of people walkig is determied by the departures from queue durig the time ( σ, 0). These are idepedet of each other ad of the state of queue at time 0. Therefore, the desired probability is ( νµ)( ντ)e νσ. Exercise 6. Cars arrive at the begiig of a log road i a Poisso stream of rate ν from time t = 0 owards. A car has a fixed velocity V > 0 which is a radom variable. The velocities of cars are idepedet ad idetically distributed, ad idepedet of the arrival process. Cars ca overtake each other freely. Show that the umber of cars o the first x miles of road at time t has a Poisso distributio with mea [ x ] νe V t. Proof. By the theorem from lecture, we kow that the umber of cars o the first x miles of road at time t has a Poisso distributio with mea M(t, [0, x]) = ν t 0 P (u, [0, x])du, where P (u, [0, x]) is the probability that a car is i the set [0, x] a time u after eterig the system. Now, a car is i the set [0, x] a time u after eterig the system if ad oly if its speed V satisfies V x/u, or equivaletly x/v u. Thus, we have t M(t, [0, x]) = ν P( x [ x ] V u)du = νe V t. 0 (Here, stads for miimum.) To see the equality, recall that for a radom variable X we have E[X] = P(X > x)dx; here we simply use the fact that, for the radom variable 0 X = mi(x/v, t) the probability that X > t will be 0, so we ca trucate the itegral. Exercise 7. Recall the mathematical model for a loss etwork with fixed routig. A etwork cosists of three odes, with each pair of odes coected by a lik. A call i progress betwee two odes may be routed o the direct lik betwee the odes, o o the two lik path through the third ode. A call i progress ca be rerouted if this will allow a additioal arrivig call to be accepted. Describig carefully the modellig assumptios you make, obtai a exact expressio for the probability a arrivig call is lost, ad sketch a etwork with fixed routig which shares the same loss probabilities. Deduce a Erlag fixed poit approximatio for the loss probabilities i the origial etwork, i terms of blockig probabilities across certai cuts i the etwork. Proof. This problem is solved i the lecture otes. We will assume Poisso arrivals of calls (with rates ν ij for calls goig betwee i ad j), expoetial holdig times of rate for each call, ad idepedece of lots of thigs.

6 6 ELENA YUDOVINA Exercise 8. Calls arrive as a Poisso process of rate ν at a lik with capacity C circuits. A call is blocked ad lost if all C circuits are busy; otherwise the call is accepted ad occupies a sigle circuit for a expoetially distributed holdig time with mea oe. Holdig times are idepedet of each other ad of the arrival process. Calculate the mea umber of circuits i use, M(ν, C). Show that, as, M(ν, C) ν C, ad the mea umber of idle circuits satisfies C C M(ν, C) ν C C. Proof. From the Erlag formula, M(ν, C) = C ν k /k! k C k=0 νk /k! The first method of proof is to observe that ad the E(ν, C) = (ν)c /(C)! C k=0 νk /k! = M(ν, C) = ν( E(ν, C)) + C + (C)(C ) C + C = ν (ν) 2 ν ν 2 { 0, C ν C/ν, C < ν (This is mootoe covergece the deomiator coverges to the geometric series termby-term from below.) The M(ν, C) ν C as required. A alterative method of proof is to show that the distributio of the umber of busy circuits is tightly cetered o the smaller of C ad ν. Suppose first that ν > C, ad let C/ν = α <. Cosider k with k < C. The k π(k ) = π(k) ν π(k) C ν = π(k) α π(c) αc k. Summig over all k < ( ɛ)c, we get k<( ɛ)c ( ɛ)c π(k) < π(c) α C k < π(c)α ɛ k=0 k =0 α k = π(c) αɛ α. Note that as this teds to 0, for ay ɛ. Cosequetly, for ay ɛ > 0 ad ay δ > 0, for large eough, the probability that the umber of busy circuits is at least ( ɛ)c will be at least δ. Therefore, M(ν, C) C. Remark. We do ot eed to worry about tightess of measures here we are iterested i the distributio of k, which lives o the compact set [0, C]. We ve show that the probability that it lives outside of [C( ɛ), C] ca be made arbitrarily small, which tells us that its average will coverge to C.

7 STOCHASTIC NETWORKS EXAMPLE SHEET 2 SOLUTIONS 7 O the other had, if ν < C, we cosider the probability that k < ( ɛ)ν or k > ( ɛ) ν. k I the first case, the calculatio will be the same, usig the boud π(k ) = π(k) < ν π(k) ( ɛ) for k ( ɛ)ν. This gives π(k) < π(( ɛ)ν) ( ɛ) ( ɛ)ν k ad the computatio cotiues as above to give a total probability that teds to 0 as ν for ay fixed ɛ. For the upper boud, we ote π(k + ) = π(k) < π(k) ( ɛ) if k+ k > ( ɛ) ν, ad repeat the calculatio. Thus, if ν < C we have show M(ν, C) ν. The case ν = C ca be icorporated ito either of these. For the mea umber of idle circuits, first ote that if ν < C, i.e. ν C = max(ν, C) = C, the M(ν, C)/ ν ad therefore C M(ν, C) = C ν + o(). Thus, we are oly iterested i the case ν C. I this case, as we saw above, the probability that k < ( ɛ)c circuits are occupied ca be made arbitrarily small. Cosequetly, for the system as a whole the arrival rate is ν ad the service rate is betwee ( δ)( ɛ)c ad C. Thus, the free circuits i the system behave (almost) as a M/M/ queue with arrival rate C ad service rate ν. Their mea, therefore, coverges to C ν C = C ν C. (This could be made precise with a couplig argumet: thik of a queue with arrival rate ( δ)( ɛ)c for the lower boud, ad with arrival rate C for the upper boud.) Exercise 9. Show that the solutios of the fixed poit equatio B = E(ν[ + 2B( B)], C) locate statioary poits of the potetial fuctio [ ν e y + e 2y ( + 2 ] 3 e y ) + y 0 U(z, C)dz. Proof. Differetiate the potetial fuctio with respect to y, ad set to 0: Simplifyig: 0 = νe y 2νe 2y ( 2 3 e y ) νe 2y e y + U(y, C) νe y + 2νe 2y 2νe 3y = U(y, C). Lettig B = e y, the left-had side becomes ν ( ( B) + 2( B) 2 2( B) 3) = ν( B)( + 2B( B)), while the right-had side is U( log( B), C). The statemet that precisely expresses the fact that ν( B)( + 2B( B)) = U( log( B), C) B = E(ν[ + 2B( B)], C) : }{{} ν

8 8 ELENA YUDOVINA ideed, U( log( B), C) = U( log( E(ν, C)), C) = ν[ + 2B( B)] ( E(ν )], C) = [+2B( B)]( B). }{{} ν