IEOR 4404: Homework 6

Transcription

1 IEOR 4404: Homework 6. Consider the FIFO GI/GI/ queue as a model for a printer which prints one job at a time according to the distribution G(x) = P (S x) = e x, x 0 ( service time distribution ), and jobs arrive according to a renewal process with iid interarrival times {T n } distributed as A(x) = P (T x) =, x 0. Suppose further that in (+x) 3 addition, there are disasters that occur to the printer according to an indepent Poisson process at rate γ = 0.0 : when a disaster occurs, all jobs in the line (if any) are instantaneously removed (gone), while the job in service (if any) is re-started with a new indepent service time distributed as G. If the system is empty when the disaster occurs, then it has no effect at all on the system. Our objective is to only simulate a copy of L(t ), the total number of jobs in the system at time t > 0. The events are: e (arrival occurs), e 2 (service completion occurs; e.g. departure of a job from service), e 3 (disaster occurs). (a) Confirm that we can generate an interarrival time T via : Generate U, set T = U /3. SOLUTION: Solve y = (+x) 3 to obtain A (y) = ( y) /3 T = ( U) /3. We can replace U by U to get T = U /3.. This yields Confirm that we can generate a service time S via Generate U. Set S = (ln (U)) 2. SOLUTION: Solve y = e x to obtain G (y) = (ln( y)) 2. This yields S = (ln( U)) 2. Similarly we can replace U by U to get S = (ln(u)) 2. We already know that we can generate disaster interarrival times (we will denote by R) since they are exponential at rate γ: Generate U. Set R = γ ln (U). (b) Program up (Pseudo code is provided below) for t = 200, and use your code to generate 0, 000 copies of L(t ) so as to estimate E(l(t )). Redo with γ = PSEUDO CODE INITIALIZE: n denotes L(t), t is time, t A time of next arrival, t D time of next service completion, t R time of next disaster. Set γ = 0.0. Generate T distributed as A and generate R distributed as exponential at rate γ n = 0, t A = T, t D =, t R = R. ARRIVAL IS NEXT (e )(min{t A, t D, t R } = t A and t A < t ): n = n +, t = t A. If n = 0, then place new arrival in service: Generate S distributed as G, set t D = t + S. Generate T distributed as A, set t A = t + T. SERVICE COMPLETION IS NEXT (e 2 )(min{t A, t D, t R } = t D and t D < t ): n = n, t = t D. If n, then place next job (in line) into service: Generate S distributed as G, set t D = t + S. Otherwise, if n = 0, then set t D =.

2 DISASTER IS NEXT (e 3 )(min{t A, t D, t R } = t R and t R < t ): t = t R. If n, then reset n =, and generate a new service time for the job in service: Generate S distributed as G, set t D = t + S. (If n = 0 do nothing.) Schedule next disaster time: generate R exponential at rate γ, set t R = t + R. NEXT EVENT > t (e.g., min{t A, t D, t R } > t ): Stop. Output n. SOLUTION: For γ = 0.0, E(L(t )) ; and for γ = 0.05, E(L(t )) Sample codes in MATLAB as follows: T = 200; n = 0000; % sample size gamma = 0.0; % can change to 0.05 L = zeros(n,); for i = :n % initialization t = 0; n = 0; t_a = (rand)^(-/3)-; t_d = +Inf; t_r = -log(rand)/gamma; % event simulation while(min([t_a t_d t_r]) <= T) if(min([t_a t_d t_r])==t_a) % next event is an arrival n = n + ; t = t_a; if(n-==0) t_d = t + (log(rand))^2; t_a = t + (rand)^(-/3)-; if(min([t_a t_d t_r])==t_d) % next event is a service completion n = n - ; t = t_d; if(n>=) t_d = t + (log(rand))^2; t_d = +Inf; % next event is a disaster t = t_r; if(n>=) n = ; t_d = t + (log(rand))^2; t_r = t - log(rand)/gamma; 2

3 % return sample result L(i) = n; 2. Consider a non-stationary Poisson process {t n } with rate function: λ(t) = 2 + sin(2πt), t 0. Recall our thinning algorithm for simulating such a point process (Lecture Notes on non-stationary Poisson processes). (a) What is the smallest value of λ > 0 that we can use; e.g., that satisfies λ(t) λ, t 0? SOLUTION: The smallest value of λ = 3 since sin(2πt). (b) Program up the algorithm and give 5 outputs (e.g. 5 runs) for (t,..., t 5 ). For each run, also output how many Poisson rate λ arrivals were needed to obtain the (t,..., t 5 ). SOLUTION: Sample outputs are: Sample codes in MATLAB: output num t t t t t t t t t t t t t t t t = 0; lambda_star = 3; num = 0; % number of arrivals for PP(lambda*) n = 0; % number of arrivals for PP(lambda(t)) Arr = []; while(n < 5) t = t - log(rand)/lambda_star; 3

4 num = num + ; if(rand<=(2+sin(2*pi*t))/lambda_star) % accept the arrival n = n + ; Arr = [Arr, t]; (c) Let {N (t) : t 0} denote the counting process of the Poisson process at rate λ. Denote its arrival times by {s n }. Recall (Lecture Notes on the Poisson process) that conditional on the event N (t) = n, the vector of the n arrival times (s,..., s n ) has the same joint distribution as the n order statistics 0 < V () < V (n) < t of n iid unforms over (0, t), V,..., V n. Use this fact together with our thinning algorithm to give an algorithm for simulating our non-stationary Poisson process up to a desired time t. First write it up in Pseudo Code, then run it in MATLAB (or Python) giving output (N(t) and the arrival times (t,..., t N(t) )) for when t = 0. Do 5 runs. SOLUTION: PSEUDO CODE Enter t and λ. i. Generate N = N (t) distributed as Poisson with mean α = λ t. ii. If N = 0, then output There are no arrivals by time t and stop. iii. Otherwise (N (t) > 0): A. Generate N i.i.d. uniforms on (0, t), i.e. V,, V N. B. Sort the V i s in ascing order to get the order statistics V () < < V (N ) and set t i = V (i) for i N. C. Set N = 0 and Arr = []. D. For each i =,, N, generate U Unif(0, ). If U λ(t i )/λ, set N = N + and Arr = [Arr, t i ]. E. If N = 0, then output There are no arrivals by time t and stop. Otherwise (N ) output N and Arr. Sample codes in MATLAB: lambda_star = 3; t = 0; % generate N* ~ Poisson(lambda_star*t) N_star = 0; alpha = lambda_star*t; P = rand; while(p>=exp(-alpha)) N_star = N_star + ; P = P * rand; if(n_star==0) disp( There are no arrivals by time t ); V = sort(rand(n_star,)*t); N = 0; Arr = []; for i = :N_star 4

5 if(rand <= (2+sin(2*pi*V(i)))/lambda_star) N = N + ; Arr = [Arr; V(i)]; if(n==0) disp( There are no arrivals by time t ); disp(n); disp(arr); Sample outputs: Output : N = 8 and the arrival times are (0.3855, ,.72,.3903, , 3.473, , , , 6.263, , 6.833, 7.665, , 7.474, , , 9.440). Output 2: N = 20 and the arrival times are (0.30, , ,.0288,.726, 3.078, , 3.657, 4.579, 4.285, , , , 6.026, , , , 9.96, , 9.832). Output 3: N = 6 and the arrival times are (0.2537, 2.892, , 2.859, , , 3.306, , 5.774, 5.760, , , 8.863, 9.953, , 9.499). Output 4: N = 9 and the arrival times are (2.46, 2.307, , 3.9, 3.688, , , 4.96, , , , 6.279, 6.669, , , , , , 9.823). Output 5: N = 22 and the arrival times are (0.039, , ,.0264,.238, , 3.592, , , 4.096, 4.235, , , , 7.24, 7.235, 7.883, , , , 9.539, 9.949). 3. Electoral College (EC) Vote numbers used in 202, and which will be used in 206 and 2020: In 200, the USA Census was taken and thus the 5 EC numbers slightly changed (but still total to 538). (The Census will be taken again in 2020 at which time new numbers yet again will be determined and used in 2024, and 2028.) Here are the numbers (Alphabetical order by State, including DC= District of Columbia= Washington D.C.): Let n i denote the number for state i, i 52. We know that 5 n i = 538. i= (RECALL LECTURE NOTES 7 FROM OUR COURSE WEBSITE.) By Generating 5 iid Bernoulli (/2) rvs, B,..., B 5, you are to simulate a copy of 5 C = n i I{B i = }. i= C represents the outcome for an election (total EC vote) in which each state is equally likely to go to candidate A versus Candidate B (and indepently among states). If C = 269, then it is a tie. 5

6 Now generate n = million iid copies (sequentially) of C, C,..., C n and use them to estimate the probability of a tie P T : P T ˆP T = n n I{C i = 269}, i= the proportion of runs that led to a tie. Finally, use the above ˆP T to estimate the number of ways there can be a tie, N T, using the fact that P T = N T 2 5, hence N T = 2 5 P T : N T 2 5 ˆPT. SOLUTION: Numerical results ˆP T = and ˆN T = Sample codes in MATLAB as follows: % input EC numbers n = [9,3,,6,55,9,7,3,3,29,6,4,4,20,,6,6,8,8,4,0,,6,... 0,6,0,3,5,6,4,4,5,29,5,3,8,7,7,20,4,9,3,,38,6,3,3,2,5,0,3]; N = 0^6; % sample size C = zeros(n,); for i = :N B = rand(,5)<=/2; C(i) = sum(n.*b); P_T = mean(c==269) N_T = 2^5*P_T 6