ISyE 6644 Fall 2014 Test #2 Solutions (revised 11/7/16)

Transcription

1 1 NAME ISyE 6644 Fall 2014 Test #2 Solutions (revised 11/7/16) This test is 85 minutes. You are allowed two cheat sheets. Good luck! 1. Some short answer questions to get things going. (a) Consider the linear congruential generator X n+1 = (17X n + 7)mod(128). Using X 0 = 5, calculate the first two pseudo-random numbers, U 1 and U 2. Solution: X 1 = (17X 0 + 7)mod(128) = 92, so U 1 = 92/128 = Similarly, U 2 = (b) You may recall the following corollary from the class notes: The LCG X i = (ax i 1 + c)mod(2 n ) (for c and n > 1) has full cycle if c is odd and a = 4k + 1 for some integer k. With this result in mind, again consider the LCG in Question 1a with seed X 0 = 5. Calculate U Solution: For this generator, c = 7 is odd, the modulus is 2 n, where n = 7 > 1, and a = 17 = 4k + 1 for k = 4. Thus, the corollary applies and so the period of this generator is 128. Thus, U 2050 = U 1922 = = U 130 = U 2 = (c) Use the Box-Muller method along with U 1 and U 2 from Question 1a to generate two independent Nor(0, 1) random variables. Solution: We have Z 1 = Z 2 = 2ln(U 1 ) cos(2πu 2 ) = ln(U 1 ) sin(2πu 2 ) = (d) Use your answer in Question 1c to generate two independent Nor(0, 0.25) random variables. Solution: X 1 = 0.25 Z 1 = and X 2 = 0.25 Z 2 =

2 2 (e) Use your answer in Question 1c to generate a χ 2 (2) random variable. Solution: χ 2 (2) = Z Z 2 2 = (f) Consider a Tausworthe generator with r = 2, q = 4, and B 1 = B 2 = B 3 = B 4 = 1. Find B 5 and B 6. Solution: We have B i = (B i r + B i q )mod(2) = B i 2 XOR B i 4. Thus, B 5 = B 3 XOR B 1 = 1 XOR 1 = 0 and B 6 = B 4 XOR B 2 = Consider the following n = 12 pseudo-random numbers We shall assume that the U i s are all Unif(0, 1). But let s conduct a correlation test to test whether or not the U i s are independent. We define the lag-j correlation between the U i s by ρ j Corr(U i, U i+j ). Ideally, this correlation should equal zero for all j. A good estimator for the lag-1 correlation ρ 1 is given by ˆρ 1 [ ] 12 n 2 U 1+k U 2+k 3. n 1 k=0 Under the assumption that the U i s are independent, it turns out that Var(ˆρ 1 ) 13n 19 (n 1) 2. Under the null hypothesis H 0 : ρ 1 = 0, it can be shown that the test statistic C ˆρ 1 Var(ˆρ 1 ) is approximately standard normal. Of course, we reject the null hypothesis if C is too big or small. (a) Calculate ˆρ 1. Solution: ˆρ 1 = 12 (3.1204) 3 =

3 3 (b) Calculate Var(ˆρ 1 ) (using the above approximation). Solution: Var(ˆρ 1 ) (13)(12) = (c) Calculate C. Solution: C = / = (d) Use the above with level α = 0.05 to test whether or not H 0 is true. Solution: Since C < z α/2 = 1.96, we fail to reject H 0. (This is actually a little surprising to me, since ˆρ 1 seems high, but I guess the sample size is too low to merit rejection.) (e) If ρ 1 actually equals 0, does ρ 2 = 0? Solution: Not necessarily. You can easily come up with some weirdo cyclic behavior which gives ρ 2 0. For instance, let U 0 and U 1 be i.i.d. Unif(0,1), and then take U i = U i 2 for i Consider the first-order moving average process [MA(1)], X i+1 = θz i + Z i+1, where the Z i s are i.i.d. Normal(0,1) random variables, and 1 < θ < 1. (a) Calculate the covariance function, R k this process. = Cov(X i, X i+k ), k = 0, 1, 2,..., for Solution: We actually did this in class, but I ll go through the details again.... Since the Z i s are independent, and R 0 = Var(X i ) = Var(θZ i 1 + Z i ) = θ 2 Var(Z i 1 ) + Var(Z i ) = θ 2 + 1, R 1 = Cov(X i, X i+1 ) = Cov(θZ i 1 + Z i, θz i + Z i+1 ) = θvar(z i ) = θ, R 2 = Cov(X i, X i+2 ) = Cov(θZ i 1 + Z i, θz i+1 + Z i+2 ) = 0. Thus, R 0 = 1 + θ 2, R ±1 = θ, and R k = 0 for k 2.

4 4 (b) It turns out that (after some algebra) σ 2 n nvar( X) = (1 + θ) 2 2θ n, where X = n i=1 X i /n is the sample mean. Calculate the limiting quantity σ 2 = lim n σ 2 n. Solution: σ 2 = (1 + θ) 2. (c) For θ > 0, it turns out that the ratio σ 2 n/σ 2 < 1. Suppose that θ = 0.5. Using your answer to part (b), find the smallest value of n such that 0.95 σ 2 n/σ 2. Solution: By (b), σ 2 n/σ 2 which leads to n 8.89, i.e., 9. = (1 + θ)2 2θ n (1 + θ) 2 = /n , 4. Suppose we observe 1000 pseudo-random numbers to obtain the following data. interval [0.0, 0.2) [0.2, 0.4) [0.4, 0.6) [0.6, 0.8) [0.8, 1.0] number observed Conduct a χ 2 goodness-of-fit test to see if these numbers are approximately U(0, 1). Use level of significance α = Here are some table entries that you may need: χ ,3 = 7.81, χ ,4 = 9.49, and χ ,5 = Solution: Let s re-write the table: Then interval [0.0, 0.2) [0.2, 0.4) [0.4, 0.6) [0.6, 0.8) [0.8, 1.0] O i E i χ 2 0 = 5 (O i E i ) 2 = i=1 E i Note that we have k 1 = 4 degrees of freedom. Then since χ 2 0 < χ ,4, we fail to reject uniformity.

5 5 5. Suppose the random variable X has the following p.d.f. 0 if x < 0 or x > 2 f(x) = x if 0 x 1 1/2 if 1 x 2. (a) Give an inverse transform method for generating realizations of X. Solution: After a little algebra, we get the c.d.f., 0 if x < 0 x 2 F (x) = 2 if 0 x 1 x 2 if 1 x 2 1 if x > 2. Case 1: Setting X2 2 = U, we have X = 2U for 0 U 1/2. Case 2: Setting X 2 = U, we have X = 2U for 1/2 U 1. Thus, in order to generate the RV, we use X = { 2U if 0 U 1/2 2U if 1/2 U 1. (b) Use the U(0, 1) random number 0.55 to generate a realization of X. Solution: X = 2(0.55) = Questions on Poisson random variables. (a) Use the acceptance-rejection technique to generate one Poisson(4.5) random variate. Use as many of the following U(0, 1) numbers as is necessary Solution: Define p n n+1 i=1 U i. We ll stop as soon as p n < e 4.5 = Let s make the following convenient table. n U n+1 p n Stop? nope nope nope yup

6 6 So we take N = 3. (b) How many U(0, 1) s would you have expected to use in Question 6a? Solution: From class notes, E[N + 1] = 5.5. (c) What would you have had to do if I had asked you to generate a Poisson(200) random variate? Solution: Use the normal approximation. N = λ + λz 0.5 = Z Joey owns Dogs R Us, a fancy hot dog stand in Smyrna. Customers arrive to buy hot dogs according to a Poisson process at the rate of 10/hour. Two servers work for Joey. Each customer buys X hot dogs, where X Pois(3) + 1. It takes exactly 4 minutes to serve each hot dog. After a customer is served, there is a 20% chance that he will chat for U(4, 8) minutes with his server. The servers take a 30-minute break every 4 hours (although they will finish serving any customer currently in progress). Write a VERY, VERY NEAT Arena program to simulate this system for 40 hours. In particular, (a) Draw an appropriate block diagram. I know that you are not an artist, so label each block indicating what kind of block it is; and indicate any necessary information about the block, such as parameters like arrival rates, or commands such as Seize-Delay-Release. (b) Write out the precise command that you would need to implement the random variable X. Solution: In the Delay information, just enter 4 (POIS(3) + 1). (c) Describe in words how you might schedule the breaks. Solution: Define a resource schedule and fill in the table or chart.

7 7 (d) Describe in words how you would run the simulation for 40 hours. Solution: Go to Run Setup > Replication Parameters and stop the run at 40 hours. 8. More short answer questions. (a) If the number of arrivals to a parking lot follows a nonhomogeneous Poisson process with intensity function λ(t) = t 2, where t [0, 10] denotes hours, find the expected number of customers to show up between t = 4 and 6. Solution: E[N(6) N(4)] = 6 4 λ(t) dt = (b) What is the name of the method we discussed in class to generate NHPP arrivals? Solution: Thinning. (c) Suppose that W Q i+1 is the (i + 1)st customer s waiting time. Write down a simple expression for this time in terms of the previous customer i s waiting time and service time, and customer (i + 1) s interarrival time. Solution: W Q i+1 = max {0, W Q i + S i I i+1 }. (d) Consider the majorizing function t(x) used in an acceptance-rejection algorithm. i. TRUE or FALSE? t(x) f(x) for all x. Solution: TRUE. ii. TRUE or FALSE? For efficiency purposes, t(x) should be as close to the target p.d.f. as possible. Solution: TRUE. iii. TRUE or FALSE? Let c max x t(x). Then t(x)/c is a p.d.f.

8 8 Solution: This problem seems to be a little ambiguous, so let s pretend it never happened! iv. TRUE or FALSE? The p.d.f. corresponding to t(x) should be as easy as possible to generate RV s from. Solution: TRUE. v. TRUE or FALSE? The number of trials before an acceptance takes place is geometric. Solution: TRUE. (e) Suppose that W(t) is a Brownian motion process. Find Cov ( 1 0 W(t) dt, W(1)). Solution: 1 0 Cov(W(t), W(1)) dt = 1 0 t dt = 1/2. (f) What is the process e W(t) called? Solution: Geometric Brownian motion.