. Now define y j = log x j, and solve the iteration.

Transcription

1 Problem 1: (Disribued Resource Allocaion (ALOHA!)) (Adaped from M& U, Problem 5.11) In his problem, we sudy a simple disribued proocol for allocaing agens o shared resources, wherein agens conend for resources bu back off in he face of conenion. One example applicaion is in he design of muliple-access proocols in wireless neworks, where his scheme forms he basis of he ALOHA proocol and is varians. Consider he following simplified model of a cell-phone nework: he wireless specrum is divided ino a se of n channels, where every cell-phone can access every channel, bu each channel can suppor a mos one cell-phone conversaion. Now suppose n users are rying o simulaneously download a file using heir cell-phones we wan a way o mach hem o channels in a disribued manner. We sar wih n users and n channels, and assume he sysem evolves over rounds. In every round, users pick from he n channels independenly and uniformly a random. A user who picks a channel no picked by any oher user, is served in ha round (i.e., finishes downloading he file) and removed from consideraion. The remaining users again pick one of he n channels in he nex round. We finish when every user is served. If here are m users a he sar of a round. wha is he expeced number of users a he sar of he nex round? Suppose ha in every round, he number of users removed is exacly he expeced number of users rounded up o he neares ineger. Show ha all he users are served in O(log log n) rounds. Hin: If x j is he expeced number of users lef afer j rounds, show ha x j+1 (x j) 2 n. Now define y j = log x j, and solve he ieraion. Problem 2: (The SIS Epidemic on Random Graphs) Consider an SIS epidemic on an undireced graph G, where neighboring nodes mee a rae λ. In class, we showed he following hresholds for a long infecion lifeime: λ < 1 E[T ex ] = O(1), λ > 1 d max η(k) E[T ex] = Ω (e k) where d max is he maximum degree of nodes in G, and η(k) is he generalized isoperimeric consan, which for any k n is defined as: η(k) = E(S, S) min S V, S k S (where E(S, S) is he number of edges beween S and S = V \ S). 1

2 We now sudy hese hresholds in he case where G is generaed as a G(n, p) random graph in he conneced regime (i.e., wih average degree (n 1)p > ln n) in paricular, we wan o show ha he phase ransiion hreshold is λ = Θ(1/(n 1)p) (w.h.p, as n ). Consider a graph G generaed from he G(n, p) model wih p = 6 ln n/(n 1), and le d = (n 1)p and D max be he average and maximum degrees respecively. Prove ha: P [D max > 2d] 1 n Thus, if we choose λ < 1/2d, hen, wih high probabiliy, an SIS epidemic on G has fas exincion. Hin: For X = X i wih X i [0, 1] i.i.d, recall he Chernoff bounds on he upper and lower ails: P[X > (1 + ɛ)e[x]] < e ɛ2 E[X]/3, P[X < (1 ɛ)e[x]] < e ɛ2 E[X]/2 We now wan o show a similar hreshold for λ o ensure a long-lasing epidemic. As before, consider a graph G generaed from he G(n, p) model, and average degree d = (n 1)p. For any k n, and any ɛ, argue ha: k ( ) n P[η(k) < d/3] = P [δ i < d i/3] i where δ i Binomial(i(n i), p). Hin: For any se S wih S = i, wha is he disribuion of he cu-size E(S, S)? Le k = n/3, and define ɛ i = 1 (n 1) 3(n i) argue ha for any n > 1 and i n/3, we have ha ɛ i (1/2, 1). Now, using he above equaion and a Chernoff bound, show ha: n/3 P[η(n/3) < d/3] ( ) n/3 n e i(n i)pɛ2 i /2 i ( ) n e i(n 1)p/8 i Par (d) (Opional) Now suppose we choose p = 16 ln n/(n 1). Using he inequaliy ( ) n i n i i!, show ha: P[η(n/3) < d/3] n/3 n i i! e 1/n 1 Thus, argue ha choosing λ > 3/d ensures ha as n, wih probabiliy going o 1, he SIS epidemic has E[T ex ] = Ω(e n/3 ). 2

3 Problem 3: (Differenial Equaion Approximaions of CTMCs) In class, for any CTMC X(), we defined he rae marix A (where for i j, A ij = λ ij, he rae of ransiions from sae i o sae j; moreover A ii = j A ij), and ransiion marix P () (where P i,j () = P[X() = j X(0) = i ]). We also discussed ha hese marices saisfy Kolmogorov s dp () forward equaion: = P ()A. We now formally show one example of how we can use his o ge exac differenial equaions for E[X()]. (Warning: This is a somewha roundabou way of deriving he differenial equaion for E[X()], bu i will give you a feel for he more general heory behind CTMC and differenial equaions.) Consider he simple birh process of cellular growh: we sar wih X(0) = 1 cell, and each cell splis ino wo a rae λ (i.e., afer Exponenial(λ) ime). Le X() be he number of cells a ime clearly his is a CTMC, aking values in N = {1, 2,...}. Wrie down he rae marix A and he sae ransiion diagram for X(). Noe ha he sae-ransiion diagram and rae marix are infinie we did no explicily deal wih his in class, bu mos hings we discussed exend o his case. For his quesion, you need o specify he sae-ransiions and rae marix for any single sae i. Le π() be he disribuion of he number of cells X() a ime. Since X(0) = 1, his means π 1 (0) = 1, π k (0) = 0 for k > 0. Moreover, we have π() = π(0)p (). Now argue ha for any k N: [ ] dπ k () dp () = π(0) = kλπ k () + (k 1)λπ k 1 (), k Solve his equaion explicily for π 1 () does he answer make sense? Hin: Noe ha for any n 1 row vecor π and n n marix P, we have [πp ] k = j π jp jk for an infinie vecor {π k } k N and marix {P ij } i,j N, we can define muliplicaion in a similar manner. For any random variable X, recall we defined he generaing funcion φ X (s) = E[s X ]. Le F (s, ) = φ X() (s). Argue ha F (s, ) = k N π k()s k, and also ha E[X()] = (i.e., evaluae he s=1 parial derivaive Par (d) Nex, prove ha: F (s,) a s = 1). F (s,) F (s, ) F (s,) = λ(s 2 F (s, ) s) F (s,) Hin: Wrie ou boh and by differeniaing each erm in he summaion, and compare coefficiens of s k. You do no need o jusify he correcness of operaions on infinie sums. 3

4 Par (e) Finally, use he fac ha 2 F (s,) = 2 F (s,) (i.e., for he funcion F (s, ), you can ake parial derivaives in any order noe ha is no rue in general!), o show ha: de[x()] = λe[x()] Hin: Differeniae he above equaion w.r.. s, and se s = 1. Problem 4: (Designing a Viral Adverising Campaign) The SIS epidemic can be used as a model for brand recall amongs consumers. Suppose we have a populaion of n people, wherein each peron knows exacly d oher people (i.e., he social nework is d-regular), and pairs who know each oher mee a rae λ. A ime 0, n 0 people are aware a cerain brand for example, hey are using a cerain new app. Subsequenly, people who are using he app forge abou i a rae 1, while people who have forgoen abou i are reminded of i when hey mee someone who currenly is using i. Le X() be he number of people using he app a ime in class we discussed ha X() can be sochasically dominaed from above by a birh-deah Markov chain Y () (i.e., Y () X() a all ) wih he following ransiions Y () goes from n n 1 a rae n, and for all oher saes Y () {1,..., n 1}, we have: Y () Y () 1 a rae Y (), Y () Y () + 1 a rae Y () λ d (1) Noe in paricular ha under he above ransiions, Y () = 0 is an absorbing sae. Now we use he differenial equaion mehod o find he phase ransiion for he SIS process. Suppose we ignore he boundary condiion a n, and insead le he ransiions in Equaion 1 hold for all Y () > 0 (noe ha his mainains he sochasic dominance). Argue ha: de[y ()] = (λd 1)E[Y ()], and hence E[Y ()] = Y (0)e (λd 1). From his we see ha if λd < 1, hen all people sop using he app afer consan ime. Noe: You don have o formally prove he las poin however, convince yourself ha his is easy o show using Markov s inequaliy. Suppose now we have some budge o design a viral adverising campaign. To do so, we hire k people (for some fixed k, which is much smaller han n), who each make visis o random people 4

5 a rae 1, and remind hem o use he app. Argue ha X() is now sochasically dominaed from above by a Markov chain Y () wih ransiions Y () Y () 1 a rae Y (), Y () Y () + 1 a rae Y () λ d + k Noe hough a Y () = 0, he above ransiion is no rue, as we wan i o be absorbing. However, we ignore his o ge he following differenial equaion approximaion for he expeced rajecory: de[y ()] Show ha he soluion o his equaion is: E[Y ()] = ( E[Y (0)] + = (λd 1)E[Y ()] + k k ) e (λd 1) λd 1 k λd 1 Using his soluion, observe ha for any fixed k, if λd < 1, hen he epidemic lifeime is sill small. This suggess ha using a fixed number of markeing agens does no help make a brand go viral. Suppose insead we design he following referral scheme each user who is currenly using he app is offered, a rae α, a reward for sending a promoion mail reminding one of heir friends o sar using he app. Moreover, assume ha he social nework graph has η(n/2) = d/2. Argue ha X() is now sochasically dominaed from below by a Markov chain Y () over he saes {0, 1,..., n/2}, wih ransiions: Y () Y () 1 a rae Y (), Y () Y () + 1 a rae Y () (λd/2 + α) This suggess he following differenial equaion for he expeced rajecory (ignoring he upper bound on he sae of Y ()): de[y ()] = (λd/2 + α 1)E[Y ()] Using his as a guide, wha value of α should you choose o ensure he app coninues o be used by people for a long ime? Problem 5: (The Preferenial Aachmen Process) (OPTIONAL) In his quesion, we will use differenial equaion approximaions o sudy he preferenial aachmen neworks. This is a model of nework growh ha is used o explain he occurrence of power-law degree disribuions, which are ofen seen in real neworks. For a nework wih n nodes, he degree disribuion F (k) is he fracion of nodes wih degree k his is said o exhibi power-law-α if: 1 F (k) ck α In oher words, he fracion of nodes wih degree greaer han k is proporional o k α. 5

6 To undersand why power-laws are unexpeced, consider he G(n, p) graph, wih p = λ/(n 1). Argue ha for any node i, is degree D i Binomial(n 1, p) P oisson(λ) for large n. Now since he fracion of nodes wih degree less han k is same as he probabiliy ha a random node has degree less han k, argue ha for large n, 1 F (k) = 1 k i=0 e λ λ i /i!. Noe ha his is exponenial in k (raher han power-law). The preferenial aachmen model is a dynamic model for generaing a graph, which resuls in a power-law degree disribuion. We will sudy a model for forming a direced graph however i can be used for undireced graphs in a similar way. The model proceeds as follows: ime is sloed and nodes arrive one-by-one in he order {1, 2,..., n}. Each arriving node brings a direced edge he node arriving a ime chooses he exising node o link o uniformly a random wih probabiliy 1 p, and wih probabiliy p, i picks an exising node v wih probabiliy proporional o is in-degree D v (). Noe ha a he end of slo, he number of nodes in he sysem is, and so is he number of direced edges (and hence he sum of in-degrees). For node v, argue ha is in-degree D v () has he following dynamics: D v () = 0 if v, and for all > v, D v () ransiions as: D v () = D v ( 1) + 1 wih probabiliy 1 p + pd v(), else D v () = D v ( 1) This suggess ha E[D v ()] has he following differenial equaion approximaion: de[d v ()] 1 p + pe[d v()] Using he boundary condiions E[D v (v)] = 0, solve he above equaion o ge (for > v): (( ) (1 p) p E[D v ()] 1), p v Par (d) An imporan hing o noe from he above characerizaion is ha he (expeced) in-degree of a node depends on when i arrived (in conras o he G(n, p) seing, where all nodes were symmeric). Now afer all n nodes arrive, we can approximae he degree disribuion F (k) as he fracion of nodes v {1, 2,..., n} for which E[D v ()] k. Le γ = 1 F (k) = (γk + 1) 1 p p (1 p) show ha: Thus, he resuling graph has a power-law degree disribuion wih α = 1/p. 6