h(x) = f(x)/g(x) = (1 + x)e (1/2)x.

Transcription

1 IEOR 4404 Homework 3 1. TEXT, Page 94, Exercise 20. (HINT: Consider g(x) = λe λx, for an appropriate value of λ.) We want to simulate from the density f(x) = 1 2 (1 + x)e x, x > 0. We can use an exponential density, g(x) = λe λx, x > 0 for any 0 < λ < 1; λ = 1/2 for example. For then f(x)/g(x) is bounded. (If λ 1, then f(x)/g(x) ts to as x ). Illustrating with λ = 1/2: h(x) = f(x)/g(x) = (1 + x)e (1/2)x. Solving h (x) = 0 yields x = 1, and thus c = h(1) = 2e (1/2) Program up simulating a Poisson point process {t n : n 1} up to a desired time T > 0 via using the fact that conditional on N(T ) = n, the joint distribution of the n time points (t 1,..., t n ) is the same as the order statistics of n iid uniform numbers on (0, T ). Here is the pseudo code for simulating the Poisson point process up to desired time T : Enter T, enter λ. 1 Generate N = N(T ) distributed as Poisson with mean α = λt. (Use your code from HMWK 2.) 2 If N = 0, then output There are no arrivals by time T and stop. 3 Otherwise (N(T ) > 0): Generate N iid uniforms on (0,T); V 1,..., V N. (For example set V i = T U i, 1 i N.) 4 Sort the V i in ascing order (smallest to largest) to get the order statistics V (1) < V (2) <... < V (N). 5 Set t i = V (i), 1 i N and then stop. 5 Output N and (t 1,..., t N ). Run the code for λ = 3 and T = 5. Do so 10 times so as to see the variation in output. function X = poisson rv(alpha) X = 0; P = rand; while P >= exp( alpha) 1

2 X = X + 1; P = P*rand; function [N, t] = p2(lambda,t) alpha = lambda*t; N = poisson rv(alpha); if (N==0) display('there are no arrivals by time T'); else v = T*rand(1,N); t = sort(v); Python Code def Poisson RV(alpha): X = 0 P = np.random.uniform(0,1) while(p > m.exp( alpha)): X = X + 1 P = P * np.random.uniform(0,1) return X def p2(lambda,t): alpha = T * Lambda N = Poisson RV(alpha) if N == 0: print("there are no arrivals by time T") t = 0 else: v = np.empty(n) for i in range(int(n)): v[i] = T * np.random.uniform(0,1) t = np.sort(v) return(n,t) 3. Recall the FIFO single-server queue recursion for customer delay in queue (line): D n+1 = (D n + S n T n ) +, n 0, where D 0 = 0, T n = t n+1 t n (interarrival times) (with t 0 = 0), and {S n : n 0} are the service times. We will assume here, that the arrival process is a Poisson process at rate λ, e.g., {T n : n 0} is an iid sequence of rvs that have an exponential 2

3 distribution at rate λ, and that indepently the service times {S n } are iid with (general) distribution G(x) = P (S x), x 0. This is denoted as the FIFO M/G/1 queue. In what follows you will be given a specific value for λ and a specific G. Note that, if we use the inverse transform method to generate the T i and the S i, then to obtain 1 copy of D n requires 2n uniforms U. For example D 1 = (S 0 T 0 ) + requires 2, and D 2 = (D 1 + S 1 T 1 ) + requires 2 more for a total of 4. (a) For λ = 3, and G the uniform (0, 0.5) distribution, estimate E(D 15 ) via Monte Carlo simulation: Use the recursion to get a first copy of D 15 (denote by X 1 ), run/repeat indepently over and over to get 1000 iid such copies X 1,..., X 1000 and use the empirical average as the estimate E(D 15 ) X i function ED = p3a(lambda,n,k) % n = 15 for i=1:k ; for j = 1:n T = log(rand)/lambda; S = rand/2; D = max(d + S T, 0); Python code n=1 def p3a(lambda, N, k): Darray = np.empty(k) S = np.random.uniform(0,0.5) T = np.log(np.random.uniform())/lambda D = max(d + S T, 0) D 15 =

4 (b) Once again, use λ = 3, and G as uniform (0, 0.5) distribution, but instead, letting N = N(5) denote the number of arrivals by time 5, you now are to estimate E(D N ). To do so, first use the algorithm in Problem (1) above (with λ = 3 and T = 5) to obtain (t 1,..., t N ). Then define T i = t i+1 t i, 0 i N 1 so as to obtain T 0,..., T N 1. Then use the recursion to obtain D N. That gives you a first copy of D N (denote by X 1 ). Run/repeat indepently over and over to get 1000 iid such copies X 1,..., X 1000 and use the empirical average as the estimate E(D N ) X i NOTE: If N = 0, for a given run X i, then we still count the initial customer who arrived at time t 0 = 0 with D 0 = 0. So the output would be X i = D N = 0 for that run X i. function ED = p3b(lambda,t,k) % T = 5 for i=1:k, ; [N, t] = p2(lambda,t); T arr = t [0 t(1:length(t) 1)]; if (N==0) display('there are no arrivals by time T'); else for j = 1:N S = rand/2; D = max(d + S T arr(j), 0); Python Code n=1 def p3b(lambda, T, k): Darray = np.empty(k) [N,t] = p2(lambda,t) if N == 0: print("there are no arrivals by time T") else: 4

5 Tarray = np.zeros(n) Tarray[0] = t[0] for i in range(1,n): Tarray[i] = t[i] t[i 1] S = np.random.uniform(0,0.5) D = max(d + S Tarray[i], 0) D N = (c) Re-do (a) and (b) in the case when λ = 3 again but G(x) = 1 1 (1 + 4x) 2, x > 0. (Use Inverse Transform to generate S distributed as G.) Note that in both cases (a), (b), E(S) = 1/4, but you should find quite different answers for E(D 15 ) and E(D N ). (What is the variance of S here in (c))? % Redo part (a) function ED = p3c a(lambda,n,k) % n = 15 for i=1:k, ; for j = 1:n T = log(rand)/lambda; S = 1/4 * ( (1/sqrt(rand)) 1 ); D = max(d + S T, 0); % Redo part (b) function ED = p3c b(lambda,t,k) % T = 5 for i=1:k, ; [N, t] = p2(lambda,t); T arr = t [0 t(1:length(t) 1)]; 5

6 if (N==0) display('there are no arrivals by time T'); else for j = 1:N S = 1/4 * ( (1/sqrt(rand)) 1 ); D = max(d + S T arr(j), 0); Python code # Redo Part (a) def p3c a(lambda, N, k): Darray = np.empty(1000) S = ((1/m.sqrt(np.random.uniform(0,1))) 1)/4 T = np.log(np.random.uniform())/lambda D = max(d + S T, 0) # Redo Part (b) def p3c b(lambda, T, k): Darray = np.empty(k) [N,t] = p2(lambda,t) if N == 0: print("there are no arrivals by time T") else: Tarray = np.zeros(n) Tarray[0] = t[0] for i in range(1,n): Tarray[i] = t[i] t[i 1] S = ((1/m.sqrt(np.random.uniform(0,1))) 1)/4 D = max(d + S Tarray[i], 0) D 15 = D N =