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© Manuel D. Rossetti, All Rights Reserved Chapter 2 Solutions for Simulation Modeling and Arena, 2nd Edition, by Manuel D. Rossetti, John-Wiley & Sons. Exercises 2.1 The sequence of random numbers generated from a given seed is called a random number (a) . (a) Stream 2.2 State three major methods of generating random variables from any distribution. (a) __________________ (b) _____________________ (c)_________________ (a) inverse transform, (b) convolution, (c) acceptance/rejection 2.3 Consider the multiplicative congruential generator with (a = 13, m = 64, and seeds X0 = 1,2,3,4) a) Does this generator achieve its maximum period for these parameters? Use Theorem 2.1 to justify your answer. b) Generate a period’s worth of uniform random variables from each of the supplied seeds. a = 13, m = 64, and X0 = 1, 2, 3, 4 a.) The multiplicative linear congruential generator is a special case of the linear congruential generator, therefore the LCG theorem can be applied to check if the generator will achieve its maximum period. The theorem states that an LCG has a full period if and only if the following three conditions hold: 1. The only positive integer that (exactly) divides both m and c is 1 (i.e. c and m have no common factors other than 1). 2. If q is a prime number that divides m then q should divide (a-1) (i.e. (a-1) is a multiple of every prime number that divides m). 3. If 4 divides m, then 4 should divide (a-1) (i.e. (a-1) is a multiple of 4 if m is a multiple of 4). Condition 1 does not hold because c = 0, meaning that m and c have multiple common factors. Condition 2 holds because the prime numbers, q, that divide m = 64 are 1 and 2. (a-1) = 12, and both 1 and 2 divide 12. Condition 3 holds because 4 divides both m = 64 and (a-1) = 12. Also, since m is a power of 2 (m = 64 = 26) and c = 0, the longest possible period is m/4 = 64/4 = 16. 1 © Manuel D. Rossetti, All Rights Reserved b.) Below is a period’s worth of uniform random variables from each of the supplied seeds. For Xo = 1, i Table 4 – A Period’s Worth of Uniform Random Variables Ri Ui 1 13 0.2031 2 41 0.6406 3 21 0.3281 4 17 0.2656 5 29 0.4531 6 57 0.8906 7 37 0.5781 8 33 0.5156 9 45 0.7031 10 9 0.1406 11 53 0.8281 12 49 0.7656 13 61 0.9531 14 25 0.3906 15 5 0.0781 16 1 0.0156 For Xo = 2, Table 5 – A Period’s Worth of Uniform Random Variables i Ri 1 2 3 4 5 6 7 8 Ui 26 18 42 34 58 50 10 2 0.4063 0.2813 0.6563 0.5313 0.9063 0.7813 0.1563 0.0313 For Xo = 3, Table 6 – A Period’s Worth of Uniform Random Variables 2 © Manuel D. Rossetti, All Rights Reserved i Ri 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Ui 39 59 63 51 23 43 47 35 7 27 31 19 55 11 15 3 0.6094 0.9219 0.9844 0.7969 0.3594 0.6719 0.7344 0.5469 0.1094 0.4219 0.4844 0.2969 0.8594 0.1719 0.2344 0.0469 For Xo = 4, i 1 2 3 4 Table 7 – A Period’s Worth of Uniform Random Variables Ri Ui 52 0.8125 36 0.5625 20 0.3125 4 0.0625 2.4 Consider the multiplicative congruential generator with (a = 11, m = 64, and seeds X0 = 1,2,3,4) a) Does this generator achieve its maximum period for these parameters? Use Theorem 2.1 to justify your answer. b) Generate a period’s worth of uniform random variables from each of the supplied seeds. Condition 1 does not hold because c = 0, meaning that m and c have multiple common factors. Thus, it cannot reach its full period. Also, since m is a power of 2 (m = 64 = 26) and c = 0, the longest possible period is m/4 = 64/4 = 16. 3 © Manuel D. Rossetti, All Rights Reserved 4 © Manuel D. Rossetti, All Rights Reserved 2.5 Analyze the following LCG: Xi = (11Xi1 + 5)(mod(16)), X0 = 1 using Theo- rem 2.1. a) What is the maximum possible period length for this generator? Does this generator achieve the maximum possible period length? Justify your answer. b) Generate 2 pseudo-random uniform numbers for this generator. Condition 1 holds because the only positive integer that divides both m = 16 and c = 5 is 1 The prime numbers, q, that divide m = 16 are q = 1 and 2. (a-1) = 11-1=10, and both 1 and 2 divide 10. Thus, condition 2 holds. 5 © Manuel D. Rossetti, All Rights Reserved Condition 3 does not hold because 4 divides m = 16 but not (a-1) = 10. Thus, the LCG does not obtain full period. 6 © Manuel D. Rossetti, All Rights Reserved 2.6 Analyze the following LCG generator: Xi = (13Xi1 + 13)(mod(16)), X0 = 37 using Theorem 2.1. . a) What is the maximum possible period length for this generator? Does this generator achieve the maximum possible period length? Justify your answer. . b) Generate 2 pseudo-random uniform numbers for this generator. a = 13, c = 13, m = 16 Condition 1: 1 is the only common factor of c =13 and m=16 Condition 2: The prime numbers, q, that divide m = 16 are q = 1 and 2. (a-1) = 131=12, and both 1 and 2 divide 12. Thus, condition 2 holds. Condition 3: 4 divides m = 16 and 4 divides (a-1)=12. Thus, condition 3 holds. Thus, LCG obtains the full period of 16 R1 = (13*37+13)mod 16 = 494 mod 16 = 494 – 16 floor(494/16) = 494 – 16(30) = 14 U1 = 14/16 = 0.875 R2 = (13*R1 + 13) mod 16 = (13*14+13)mod 16 = 194 – 192 = 2 U2 = 2/16 = 0.125 2.7 Analyze the following LCG generator: Xi = (4Xi1 + 3)(mod(16)), X0 = 11 using Theorem 2.1. . a) What is the maximum possible period length for this generator? Does this generator achieve the maximum possible period length? Justify your answer. . b) Generate 2 pseudo-random uniform numbers for this generator. a = 4, c = 3, m = 16 Condition 1: 1 is the only common factor of c=3 and m = 16, Condition 1 holds. Condition 2: The prime numbers, q, that divide m = 16 are q = 1 and 2. (a-1) = 4-1=3, since 2 does not divide 3, condition 3 does not hold. No need to check condition 3. 7 © Manuel D. Rossetti, All Rights Reserved R1 = (4*11+3)mod 16 = 47 mod 16 = 47 – 16 floor(47/16) = 15 U1 = 15/16 = 0.9375 R2 = (4*R1 + 3) mod 16 = (4*15+3)mod 16 = 63 – 48 = 15 U2 = 15/16 = 0.9375 This generator is degenerate at 15 2.8 Analyze the following LCG generator: Xi = (8Xi1 + 1)(mod(10)), X0 = 3 using Theorem 2.1. . a) What is the maximum possible period length for this generator? Does this generator achieve the maximum possible period length? Justify your answer. . b) Generate 2 pseudo-random uniform numbers for this generator. a = 8, c = 1, m = 10 Condition 1: 1 is the only common factor of c =1 and m=10 Condition 2: The prime numbers, q, that divide m = 10 are q = 1, 2, 5. (a-1) = 8-1=7, only 1 divides 7, so condition 2 does not hold Condition 3: 4 does not divide 10, thus condition 3 does not hold Thus, LCG does not reach full period R1 = (8*3+1)mod 10 = 25 mod 10 = 25 – 10 floor(25/10) = 25 – 10(2) = 5 U1 = 5/10 = 0.5 R2 = (8*5 + 1) mod 10 = (41)mod 10 = 41 – 10 floor(41/10) = 41 – 40 = 1 U2 = 1/10 = 0.1 8 © Manuel D. Rossetti, All Rights Reserved 2.9 The following results are from a random sample of 100 uniform(0,1) numbers. n 𝑥̅ s minimum 1st Quartile median 3rd Quartile maximum D+ -D 100 0.4615 0.2915 0.0102 0.1841 0.4609 0.7039 0.9687 0.090733 0.080733 . a) Form a 95% confidence interval for the mean. State any assumptions you need in order to make this confidence interval. . b) What sample size would be necessary to estimate the mean to within ±0.01 with 95% confidence? . c) What would you conclude based on the Kolmogorov-Smirnov Test results at the α= 0.01 level? Justify your answer using statistics. a) If n = 100 is large enough to assume that the sample average is normally distributed, we have a 95% confidence interval based on the student-t statistic of: 0.4615 ± 1.66 * 0.2915/10 = 0.4615 ± 0.04839, [0.4131, 0.5099] b) Using the normal approximation, approximately 3265 c) Dn = max(0.0907, 0.0807) = 0.0907, D(0.01) ≈ 1.63/10 = 0.163. Since Dn < D(0.01) do not reject the hypothesis of U(0,1) 2.10 Consider the following sequence of (0,1) random numbers: 0.943 0.498 0.102 0.398 0.528 0.057 0.372 0.272 0.409 0.943 0.899 0.398 0.204 0.294 0.400 0.794 0.156 0.997 Test if the sequence is distributed U (0, 1) using both a K-S test and a Chi-Squared 9 © Manuel D. Rossetti, All Rights Reserved test. The chi-square test will vary based on intervals selected. Five equally spaced intervals: Do not reject Kolmogorov-Smirnov Test Test Statistic = 0.202 Corresponding p-value > 0.15 2.11 Consider the following set of pseudo-random numbers. 0.2379 0.2972 0.9496 0.3045 0.1246 0.3525 0.1489 0.5095 0.5195 0.6536 0.7551 0.8469 0.2268 0.6964 0.842 0.8075 0.5480 0.4047 0.6545 0.3427 0.2989 0.4566 0.8699 0.1709 0.6557 0.9462 0.9537 0.9058 0.1117 0.6653 0.247 0.6146 0.9084 0.3387 0.9672 0.9583 0.9376 0.3795 0.3258 0.7864 0.3237 0.6723 0.5649 0.9804 0.3356 0.3807 0.8364 0.6242 0.8589 0.5824 a) Test the hypothesis that these numbers are drawn from a U (0, 1) at a 95% confidence level using the Chi-squared goodness of fit test using 10 intervals. b) Test the hypothesis that these numbers are drawn from a U (0, 1) at a 95% confidence level using K-S Test. 10 © Manuel D. Rossetti, All Rights Reserved c) Test the hypothesis that these numbers are uniformly distributed within the unit square, {(x, y) : x 2 (0, 1), y 2 (0, 1)} using the 2-D Chi-Squared Test at a 95% confidence level. Use 4 intervals for each of the dimensions. d) Test the hypothesis that these numbers have a lag-1 correlation of zero. Make an autocorrelation plot of the numbers. (a) > myFile = file.choose() #ch2P11data.txt > data = read.table(myFile) > b = seq(0,1, by = 0.1) > h = hist(data$V1, b, right = FALSE) > chisq.test(h$counts) Chi-squared test for given probabilities data: h$counts X-squared = 17.2, df = 9, p-value = 0.04567 Since the p-value = 0.04567 <= 0.5, the hypothesis should be rejected. It is close. b) > ks.test(data, "punif", 0, 1) One-sample Kolmogorov-Smirnov test data: data 11 © Manuel D. Rossetti, All Rights Reserved D = 0.1572, p-value = 0.1516 alternative hypothesis: two-sided Since the p-value = 0.1516 >= 0.05, the hypothesis should not be rejected. c) Run the R code from Listing 2.1 using the supplied data for the problem. > nd = 100 #number of data points > myFile = file.choose() #u01data.txt > data <- read.table(myFile) #read in the data > d = 2 # dimensions to test > n = nd/d # number of vectors > #m = t(matrix(u,nrow=d)) > m = t(matrix(data$V1,nrow=d)) # convert to matrix and transpose > # b = seq(0,1, by = 0.1) > b = seq(0,1, by = 0.25) # setup the cut points > xg = cut(m[,1],b,right=FALSE) # classify the x dimension > yg = cut(m[,2],b,right=FALSE) # classify the y dimension > xy = table(xg,yg) # tabulate the classifications > k = length(b) - 1 # the number of intervals > en = n/(k^d) # the expected number in an interval > vxy = c(xy) # convert matrix to vector for easier summing > vxymen = vxy-en # substract expected number from each element > vxymen2 = vxymen*vxymen # square each element > schi = sum(vxymen2) # compute sum of squares > chi = schi/en # compute the chi-square test statistic > dof = (k^d) - 1 # compute the degrees of freedom > pv = pchisq(chi,dof, lower.tail=FALSE) # compute the p-value > # print out the results > cat("#observations = ", nd,"\n") #observations = 100 > cat("#vectors = ", n, "\n") #vectors = 50 > cat("size of vectors, d = ", d, "\n") size of vectors, d = 2 > cat("#intervals =", k, "\n") #intervals = 4 > cat("cut points = ", b, "\n") cut points = 0 0.25 0.5 0.75 1 > cat("expected # in each interval, n/k^d = ", en, "\n") expected # in each interval, n/k^d = 3.125 > cat("interval tabulation = \n") interval tabulation = 12 © Manuel D. Rossetti, All Rights Reserved > print(xy) yg xg [0,0.25) [0.25,0.5) [0.5,0.75) [0.75,1) [0,0.25) 0 1 1 2 [0.25,0.5) 1 2 2 2 [0.5,0.75) 1 1 2 1 [0.75,1) 1 2 3 3 > cat("\n") > cat("chisq value =", chi,"\n") chisq value = 15.68 > cat("dof =", dof,"\n") dof = 15 > cat("p-value = ",pv,"\n") p-value = 0.4036319 Since the p-value = 0.40 >= 0.5, we do not reject the hypothesis. Caution should be considered here since the number in each interval is less than 5. d) There is nothing unusual looking about the ACF plot. Lag-k estimates are well within testing region rho = acf(data, main="ACF Plot for P2-11", lag.max = 9) > rho Autocorrelations of series ‘data’, by lag 0 1 2 3 4 5 6 7 8 9 1.000 -0.045 -0.138 -0.088 0.078 0.025 0.029 0.016 -0.254 -0.009 13 © Manuel D. Rossetti, All Rights Reserved 2.12 Consider the following discrete distribution of the random variable X whose probability mass function is p(x). . a) Determine the CDF F(x) for the random variable, X. . b) Create a graphical summary of the CDF. See Example 2.9. . c) Create a look-up table that can be used to determine a sample from the discrete distribution, p(x). See Example 2.9. . d) Generate 3 values of X using the sequence of (0,1) random numbers in Exer cise 2.10 (starting with the first row, reading across). U1 = 0.943 U2 = 0.398 U3 = 0.372 X=4 X=1 X=1 14 © Manuel D. Rossetti, All Rights Reserved 2.13 Consider the following uniformly distributed random numbers: . a) Generate an exponentially distributed random number with a mean of 10 using the 1st random number. . b) Generate a random variate from a (12,22) discrete uniform distribution using the 2nd random number. a) X = -10 ln (1 – 0.9396) = 28.0677 b) X = 12 + Floor((22-12+1)*0.1694) = 13 2.14 Consider the following uniformly distributed random numbers: a) Generate a uniformly distributed random number with a minimum of 12 and a maximum of 22 using U8. b) Generate 1 random variate from an Erlang(r = 2, β = 3) distribution using U1 and U2 c) The demand for magazines on a given day follows the following probability distribution Using the supplied random numbers for this problem starting at U1, generate 4 random variates from this distribution. a) X = 22 + 0.3734*(22-12) = 25.734 b) Use convolution to generate 2 exponential random variables X1 = -3ln(1-0.9559) = 9.364 X2 = -3ln(1-0.5814) = 2.612 X = X1 + X2 = 11.976 15 © Manuel D. Rossetti, All Rights Reserved c) Make the table look up as per example 2.9 U1 = 0.9559 X = 80 U2 = 0.5814 X = 50 U3 = 0.6534 X = 60 U4 = 0.5548 X = 50 2.15 Suppose that customers arrive at an ATM via a Poisson process with mean 7 per hour. Determine the arrival time of the first 6 customers using the data given in Exercise 2.10 (starting with the first row). Use the inverse transformation method. Customers arrive at an ATM via a Poisson process with mean 7 per hour (λ = 7). The CDF of the exponential distribution is: F(x) = 0 x<0 -λx 1-e x≥0 The inverse CDF is computed as follows: F(x) = 1-e-λx U = 1-e-λx F-1(U) = -1/λ * ln(1-U) Using the data given in Problem 3-14 and the above inverse CDF, the arrival time for the first six customers can be calculated. Arrival Times of First Six Customers (in hours) Ui Customer 1 2 3 4 5 6 0.943 0.498 0.102 0.398 0.528 0.057 InterArrival Arrival Time time 0.4092 0.4092 0.0985 0.5077 0.0154 0.5231 0.0725 0.5956 0.1073 0.7029 0.0084 0.7113 The demand, D, for parts at a repair bench per day can be described by the following discrete probability mass function: 2.16 16 © Manuel D. Rossetti, All Rights Reserved Generate the demand for the first 4 days using the sequence of (0,1) random numbers in Exercise 2.10 (starting with the first row). To generate the demand for the first four days using the sequence of (0,1) random numbers in Problem 3-14, we first need to find the inverse CDF for the discrete distribution. Table 2 - CDF of the Discrete Distribution xi f(xi) F(xi) 0 0.3 0.3 1 0.2 0.5 2 0.5 1.0 The above CDF can also be written as: 0 if 0 ≤ x ≤ 0.3 F(x) = 1 if 0.3 < x ≤ 0.5 2 if 0.5 < x ≤ 1.0 Using the above function and the random numbers in Problem 3-14, the demand for the first four days is as follows: Table 3 - Demand for the First Four Days Ui Demand Day 1 0.943 2 Day 2 0.498 1 Day 3 0.102 0 Day 4 0.398 1 2.17 The service times for an automated storage and retrieval system has a shifted exponential distribution. It is known that it takes a minimum of 15 seconds for any retrieval. The parameter of the exponential distribution is = 45. Using the sequence of (0,1) random numbers in Exercise 2.10 (starting with the first row) generate 2 service times for this situation. X1 = 15 + -(1/45)ln(1-.943) = 15.064 X2 = 15 + -(1/45)ln(1-0.398) = 15.011 2.18 The time to failure for a computer printer fan has a Weibull distribution with shape parameter α = 2 and scale parameter β= 3. Using the sequence of (0,1) random numbers in Exercise 2.10 (starting with the first row) generate 2 failure times for this situation. U1 = 0.943 X1 = 3[-ln(1-0.943)]^(1/2) = 5.0776 17 © Manuel D. Rossetti, All Rights Reserved U2 = 0.398 X1 = 3[-ln(1-0.398)]^(1/2) = 2.1372 2.19 The time to failure for a computer printer fan has a Weibull distribution with shape parameter α = 2 and scale parameter β= 3. Testing has indicated that the distribution is limited to the range from 1.5 to 4.5. Using the sequence of (0,1) random numbers in Exercise 2.10 (starting with the first row) generate 2 failure times for this this truncated distribution. Notice that the range is truncated. Following example 2.14, we have: F(1.5) = 1-exp(-(1.5/3)^2) = 0.22119 F(4.5) = 1-exp(-(4.5/3)^2) = 0.8946 W = 0.22119 + (0.8946 – 0.22119)*0.943 = 0.8562169 X = 3[-ln(1-0.8562169)]^(1/2) = 4.1779 2.20 The interest rate for a capital project is unknown. An accountant has estimated that the minimum interest rate will between 2% and 5% within the next year. The accountant believes that any interest rate in this range is equally likely. You are tasked with generating interest rates for a cash flow analysis of the project. Using the sequence of (0,1) random numbers in Exercise 2.10 (starting with the first row) generate 2 independent interest rate values for this situation. Equally likely means uniformly distributed: U(a=0.02, b = 0.05), using inverse transform: X = a + (b-a)*U For U1 = 0.943 and U2 = 0.398 X1 = 0.02 + (0.05-0.02)*0.943 = 0.04829 X2 = 0.02 + (0.05-0.02)*0.398 = 0.03194 2.21 Customers arrive at a service location according to a Poisson distribution with mean 10 per hour. The installation has two servers. Experience shows that 60% of the arriving customers prefer the first server. By using the first row of (0,1) random numbers given in Exercise 2.10, determine the arrival times of the first three customers at each server. 18 © Manuel D. Rossetti, All Rights Reserved 2.22 Consider the triangular distribution: a) Derive an inverse transform algorithm for this distribution. b) Using the first row of random numbers from Exercise 2.10 generate 5 random numbers from the triangular distribution with a = 2, c = 5, b = 10. a) The inverse CDF is given on page 687 of the text. b) U1 = 0.943 and U2 = 0.398 (c-a)/(b-a) = 0.375 For U1 = 0.943, since 0.943 > 0.375, we have X = b-SQRT(b-a)(b-c)*(1-U1)) = 8.8304 For U2 = 0.398, since 0.398 > 0.375, we have X = b-SQRT(b-a)(b-c)*(1-U2)) = 6.1989 19 © Manuel D. Rossetti, All Rights Reserved 2.23 Consider the following probability density function: a) Derive an inverse transform algorithm for this distribution. b) Using the first row of random numbers from Exercise 2.10 generate 2 random numbers using your algorithm. a) For 𝑥 < −1, 𝐹(𝑥) = 0 1 For −1 ≤ 𝑥 ≤ 1, 𝐹(𝑥) = 2 (𝑥 3 + 1) For 𝑥 > 1, 𝐹(𝑥) = 1 3 𝐹 −1 (𝑢) = √2𝑢 − 1 b) 𝐹 −1 (0.943) = 0.9604 𝐹 −1 (0.398) = −0.5886765 2.24 Consider the following probability density function: . a) Derive an inverse transform algorithm for this distribution. . b) Using the first row of random numbers from Exercise 2.10 generate 2 random numbers using your algorithm. a) For 𝑥 < 2, 𝐹(𝑥) = 0 For 2 ≤ 𝑥 ≤ 4, 𝐹(𝑥) = 𝑥2 4 −𝑥+1 For 𝑥 > 4, 𝐹(𝑥) = 1 Solve the following equation for x: 𝑥 2 − 4𝑥 + 4(1 − 𝑢) = 0 Using the quadratic equation: 20 © Manuel D. Rossetti, All Rights Reserved −𝑏 ± √𝑏 2 − 4𝑎𝑐 𝑥= 2𝑎 Yields 𝑥= 4 ± √16 − 4 ∗ 4(1 − 𝑢) 2 𝑥= 4 ± 4 √𝑢 = 2 ± 2√𝑢 2 Since the final number must be 2 ≤ 𝑥 ≤ 4, we have 𝑥 = 2 + 2√𝑢 = 2(1 + √𝑢) b) 𝐹 −1 (0.943) = 3.94216 𝐹 −1 (0.398) = 3.2617 2.25 Consider the following probability density function: . a) Derive an inverse transform algorithm for this distribution. . b) Using the first row of random numbers from Exercise 2.10 generate 2 random numbers using your algorithm. a) For 𝑥 < 0, 𝐹(𝑥) = 0 For 0 ≤ 𝑥 ≤ 5, 𝐹(𝑥) = 𝑥2 25 For 𝑥 > 5, 𝐹(𝑥) = 1 2 𝐹 −1 (𝑢) = 5 √𝑢 b) 𝐹 −1 (0.943) = 4.8554 𝐹 −1 (0.398) = 3.1544 21 © Manuel D. Rossetti, All Rights Reserved 2.26 Consider the following probability density function: . a) Derive an inverse transform algorithm for this distribution. . b) Using the first row of random numbers from Exercise 2.10 generate 2 random numbers using your algorithm. a) For 𝑥 ≤ 1, 𝐹(𝑥) = 0 1 For 𝑥 > 1, 𝐹(𝑥) = 1 − 𝑏2 2 1 𝐹 −1 (𝑢) = √ 1−𝑢 b) 𝐹 −1 (0.943) = 4.1885 𝐹 −1 (0.398) = 1.2888 2.27 The times to failure for an automated production process have been found to be randomly distributed according to a Rayleigh distribution: . a) Derive an inverse transform algorithm for generating random variables from this distribution. . b) Using the first row of random numbers from Exercise 2.10 generate 5 random numbers from your algorithm with β = 2. x F ( x) 0 2 2 x 2 xe dx 2 2 u x , du 2 x dx F (u ) e u du e u F ( x) e x2 2 x 0 e x2 2 1 The inverse of the CDF is: 22 © Manuel D. Rossetti, All Rights Reserved F ( x ) e U e x x2 2 1 2 2 ln( 1 U ) 1 x2 2 x 2 2 ln( 1 U ) x 2 ln( 1 U ) Using the inverse CDF from above, with β = 2.0, and the uniform numbers given problem 2-10 yields: u= 𝐹 −1 (𝑢) = 0.943 0.398 0.372 0.943 0.204 0.794 3.385087302 1.424777644 1.36413359 3.385087302 0.955313756 2.513864841 2.28 Using the first two rows of random numbers from Exercise 2.10, generate 5 random numbers from the negative binomial distribution with parameters (r = 4, p = 0.4) using: a) the convolution method b) the number of Bernoulli trials to get 4 successes. a) Convolution method: The negative binomial distribution ( r 4, p 0.4 ) is the sum of 4 geometric random variables with ( p 0.4 ). U GEOM(p=0.4) = 1+floor(ln(1-u)/ln(1-p)) 0.943 6 0.498 2 0.102 1 0.398 1 Answer: 10 trials a) Bernoulli trials: Generate Bernoulli trials until you get 4 successes U Bernoulli trial 1 0.943 0 2 0.498 0 3 0.102 1 4 0.398 1 5 0.528 0 6 0.057 1 7 0.372 1 Answer: 7 trials 23 © Manuel D. Rossetti, All Rights Reserved 2.29 Suppose that the processing time for a job consists of two distributions. There is a 30% chance that the processing time is lognormally distributed with a mean of 20 minutes and a standard deviation of 2 minutes, and a 70% chance that the time is uniformly distributed between 10 and 20 minutes. Using the first row of random numbers from Exercise 2.10 generate two job processing times. Hint: X ~ LN(μ, σ2) if and only if ln(X) ~ N(μ, σ2). Also, note that: This is a mixture distribution. Let 𝐹1 represent the lognormal distribution with 𝜔1 = 0.3. Let 𝐹2 represent the uniform distribution with 𝜔2 = 0.7. Using U1 = 0.943 to pick the distribution implies, X ~ U(10,20) because 0.943 > 0.3 X = a + (b-a)U2 = 10 + 10*0.398 = 13.98 Using U3 = 0.372 to pick the distribution implies, X ~ U(10,20) because 0.372 > 0.3 X = a + (b-a)U4 = 10 + 10*0.943 = 19.43 We “got lucky” and did not have to generate from the lognormal distribution. To generate from the lognormal distribution, we can use the inverse for the normal distribution found in Excel. 𝑚 = 𝐸[𝑋] 𝑣 = 𝑉[𝑋] Then, 𝑚 𝜇 = 𝑙𝑛 ( √1 + 𝜎 2 = 𝑙𝑛 (1 + 𝜇 = 𝑙𝑛 20 4 √ ( 1 + 202 ) 𝜎 2 = 𝑙𝑛 (1 + 𝑣 𝑚2 ) 𝑣 ) 𝑚2 = 2.99076 4 ) = 0.00995 202 24 © Manuel D. Rossetti, All Rights Reserved Generate Y ~ N(𝜇, 𝜎) via NORM.INV(p, mu, sigma), then X = EXP(Y) will be lognormal, where p will be the U(0,1). So to generate a lognormal use the equation: EXP(NORM.INV(RAND(), mu, sigma)). 2.30 Suppose that the service time for a patient consists of two distributions. There is a 25% chance that the service time is uniformly distributed with minimum of 20 minutes and a maximum of 25 minutes, and a 75% chance that the time is distributed according to a Weibull distribution with shape of 2 and a scale of 4.5. Using the first row of random numbers from Exercise 2.10 generate the service time for two patients. This is a mixture distribution. Let 𝐹1 represent the U(20,25) distribution with 𝜔1 = 0.25. Let 𝐹2 represent the Weibull distribution with 𝜔2 = 0.75. Using U1 = 0.943 to pick the distribution implies, X ~ Weibull because 0.943 > 0.25 1 𝑋 = 𝛽[−𝑙𝑛(1 − 𝑢)]𝛼 Using U2 = 0.398 X = 4.5[-ln(1-0.398)]^(1/2) = 3.2057 Using U3 = 0.372 to pick the distribution implies, X ~ Weibull because 0.372 > 0.25 Using U4 = 0.943 X = 4.5[-ln(1-0.943)]^(1/2) = 7.616 Use Z = NORM.S.INV(U) where U is read from the table. Do this for 5 PRN’s and, square and sum the values. Students could also use the z-table in the book. 25 © Manuel D. Rossetti, All Rights Reserved 1 2 3 4 5 1 2 3 4 5 U Z~N(0,1) Z^2 0.943 1.580466818 2.497875364 0.398 -0.258527277 0.066836353 0.372 -0.326560927 0.106642039 0.943 1.580466818 2.497875364 0.204 -0.827418321 0.684621077 sum = 5.853850198 Y1 0.794 0.820379146 0.673021943 0.498 -0.005013278 2.5133E-05 0.528 0.070243314 0.004934123 0.272 -0.606775364 0.368176342 0.899 1.275874179 1.627854921 sum = 2.674012462 Y2 2.32 In the (a) technique for generating random variates, you want the (b) function to be as close as possible to the distribution function that you want to generate from in order to ensure that the (c)_________ is as high as possible, thereby improving the efficiency of the algorithm. (a) acceptance/rejection (b) majorizing c) acceptance probability 2.33 Prove that the acceptance-rejection method for continuous random variables is valid by showing that for any x, Hint: Let E be the event that the acceptance occurs and use conditional probability. Let A be the event that acceptance occurs: f (W ) A U g (W ) f (W ) U g (W ) We want to show that F ( x) PX x x f y dy 26 © Manuel D. Rossetti, All Rights Reserved Now X equals W if and only if the event A occurs. Thus, PX x P{W x | A} P( A {W x}) P( A) x P( A {W x}) P( A {W x} | W w)h(w)dw (Law of total probability by conditioning on h(w) x x P( A {W x}) P( A | W w)h( w)dw P( A | W w) g ( w) dw c f (W ) f (W ) f (W ) because U and W are P( A | W w) PU | W w PU g (W ) g (W ) g (W ) independent and U is uniform(0,1) PX x P{W x | A} P( A {W x}) P( A) x f ( w) g ( w) dw x g ( w) c f ( w)dw 1 c Q.E.D 2.34 Consider the following probability density function: a) Derive an acceptance-rejection algorithm for this distribution. b) Using the first row of random numbers from Exercise 2.10 generate 2 random numbers using your algorithm. Choose g(x) = 3/2. Integrating over [-1, 1] yields c = 3. Thus, w(x) = ½ over [-1,1] Algorithm Repeat Generate W ~ w(x) which is U(-1,1) Generate U ~ U(0,1) Until U*g(W) <= f(W) Return W W = a + (b-a)* U = -1 + (1 - -1)U = 2*U -1 27 © Manuel D. Rossetti, All Rights Reserved U1 = 0.943 W = 2*0.943 -1 = 0.886 U2 = 0.398 Is 0.398*1.5 <= 1.5(0.886)^2? 0.597 < 1.177, therefore accept X = W = 0.886 U1 = 0.372 W = 2*0.372 -1 = -0.256 U2 = 0.943 Is 0.943*1.5 <= 1.5(-0.256)^2? 1.4145 < 0.098304, therefore reject W Continue in this manner until you get the 2nd acceptance. 28 © Manuel D. Rossetti, All Rights Reserved hx ab H x x a 1 a 2 xa b x a bu H (u ) 1 u 1 for x 0 b x for x 0 1 a 2.36 Parts arrive to a machine center with three drill presses according to a Poisson distribution with mean 𝜆. The arriving customers are assigned to one of the three drill presses randomly according to the respective probabilities p1, p2, and p3 where p1 + p2 + p3 = 1 and pi > 0 for i = 1, 2, 3. What is the distribution of the inter-arrival times to each drill press? Specify the parameters of the distribution. Suppose that p1, p2, and p3 equal to 0.25, 0.45, and 0.3 respectively and 𝜆 that is equal to 12 per minute. Generate the first three arrival times using numbers from the first row of random numbers from Exercise 2.10. Because of the splitting rule for Poisson processes, the drill presses each see arrivals according to the following three Poisson processes: 𝜆1 = 𝜆𝑝1 = 12 ∗ 0.25 = 3 𝜆2 = 𝜆𝑝2 = 12 ∗ 0.45 = 5.4 𝜆3 = 𝜆𝑝3 = 12 ∗ 0.3 = 3.6 Since the time between arrivals will be exponential, we have the following first arrival time to each drill press: X1 = -(1/3)ln(1-0.943) = 0.9549 29 © Manuel D. Rossetti, All Rights Reserved X2 = -(1/5.4)ln(1-0.398) = 0.09398 X3 = -(1/3.6)ln(1-0.372) = 0.12923 Alternative solution procedure: Generate inter-arrival times by using 𝜆=12. At each arrival, determine which drill press sees the arrival by using the PMF (0.25, 0.45, 0.3) to pick the drill press. Continue generating until you get the first arrival at each drill press. 30