2905 Queueing Theory and Simulation PART IV: SIMULATION

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1 2905 Queueing Theory and Simulation PART IV: SIMULATION 22 Random Numbers A fundamental step in a simulation study is the generation of random numbers, where a random number represents the value of a random variable uniformly distributed on (0, 1) Pseudo-random Number Generation Subroutines for generating random numbers are called random-number generators and are generally included in scientific packages or simulation programming languages. Most of these random-number generators use algorithms that produce a deterministic string of numbers that have all the appearances of being independent uniform (0, 1) random variables. One of the most common approaches to generating pseudorandom numbers starts with an initial value x 0, called the seed, and then recursively computes successive values x n, n 1, by letting x n = (ax n 1 + c) mod m (eg. m = a = 7 5 ) (1) where a, c and m are given positive integers. Each x n is 0, 1,..., m 1 and the quantity x n m, called a pseudorandom number is taken as an approximation to an uniform (0, 1) random variable. 149

2 This approach specified by equation (1) is called congruential method if c 0. If c = 0, then it is called the multiplicative congruential method. Example: Use the mixed congruential method to generate a sequence of pseudorandom numbers R 1, R 2,... with x 0 = 27, a = 17, c = 43, and m = 100. Solution: x 0 = 27 x 1 = [(17)(27) + 43] mod 100 = 502 mod 100 = 2 R 1 = = 0.02 x 2 = [(17)(2) + 43] mod 100 = 77 mod 100 = 77 R 2 = = We shall not pursue further the question of how to construct good random number generators. Rather, we assume in conducting computer simulation of systems, that we have a black box that gives a random number on request. 150

3 22.2 Using Random Numbers to Evaluate Definite Integrals One of the earliest applications of random numbers was in the computation of integrals. Let f(x) be a function and suppose we wanted to compute the integral I = If U is uniformly distributed over (0, 1) then 1 0 f(x)dx. I = E(f(U)). One of the applications of the random numbers is the computation of definite integrals. Suppose we want to evaluate the integral I = 1 0 f(x)dx. To obtain I we note that in fact I = E(f(U)) where U is the random variable which uniformly distributed over (0, 1). To estimate E(f(U)), suppose U 1, U 2,..., U n are independent random variables uniformly distributed in (0, 1) then f(u 1 ), f(u 2 ),..., f(u n ) 151

4 are independent and identically distributed random variables having mean I. Therefore by Strong Law of Large Numbers, with probability one, n f(u i ) lim n n = I. This means that when n is large I n n i=1 i=1 f(u i ) n will be close to I. Example: Find an approximation of the definite integral I = 1 0 sin(x)dx by using simulation method discussed. The true value of I is The following table shows the results for different number of simulation n. n I n Table 22.1 The Approximation of I for Different n. 152

5 Let us extend the idea of simulation to evaluate I = b a f(x)dx. To apply the result, we perform the following substitution Therefore How about the case when y = dy = dx/(b a) and I = I = 0 (x a) (b a). 1 0 f(a + [b a]y)(b a)dy. f(x)dx? We can apply the substitution Therefore we have and y = (1 + x) 1. dy = dx/(x + 1) 2 = y 2 dx I = 1 0 f( 1 y 1) dy. y 2 153

6 23 Generate Discrete Random Variables (The Inverse Transform Method) Suppose we wanted to generate the value of a discrete random variable X having probability mass function P {X = x j } = p j, j = 0, 1,..., p j = 1. j We generate a random number U and set x 0 if U < p 0 x 1 if p 0 U < p 0 + p 1.. X = j 1 x j if p i U < j i=0 i=0.. since, for we have 0 < a < b < 1, P {a U < b} = b a, P {X = x j } = P and so X has the desired distribution. { j 1 i=1 p i U < p i } j p i = p j i=1 154

7 This method can be written algorithmically as Generate a random number U. If U < p 0 set X = x 0 and stop. If U < p 0 + p 1 set X = x 1 and stop. If U < p 0 + p 1 + p 2 set X = x 2 and stop.. If the x i, i 0 are ordered so that x 0 < x 1 < and if we let F denote the distribution function of x, then j F (x j ) = and so i=0 X = x j if F (x j 1 ) U < F (x j ). p i 155

8 . p 0 + p 1 p 0 x 0 x 1 Figure x 2 x 3 In other words, after generating a random number U we determine the value of X by finding the interval (F (x j 1 ), F (x j )) in which U lies [or, equivalently finding the inverse of F (U)]. Thus this method is called the discrete inverse transform method for generating X. In general, after generating U, one has to go through a table-lookup procedure in order to find X. However, for some cases, one can determine X by an algebraic formula. 156

9 Example: Suppose we want to generate the value of X such that Using the inverse transform method, P {X = j} = 1, j = 1, 2,..., n. n or X = j if j 1 n U < j, j = 1, 2,..., n. n X = nu + 1 where x (called the floor of x) is the greatest integer less than or equal to x. In this example, we have an algebraic formula for determining X in terms of U. 157

10 Example: (Calculating Averages) Suppose we want to approximate n a(i) a = n where n is large and the values a(i), i = 1, 2,..., n are complicated and not easily calculated. Let X be a discrete uniform random variable such that Then E[a(X)] = i=1 P {X = j} = 1, j = 1, 2,..., n. n n a(i)p {X = i} = i=1 n i=1 a(i) n = a. Hence if we generate k random numbers U i and setting X i = nu i + 1, then each of the k random variables a(x i ) will have mean a and so by the strong law of large numbers it follows that when k is large, k a(x i ) a with probability 1. k i=1 158

11 Example: Consider the Geometric random variable X where Since P {X = i} = pq i 1, i 1 and q = 1 p. j 1 i=1 P {X = i} = 1 q j 1, j 1, we can generate the value of X by generating a random number U and setting X equal to that value j for which 1 q j 1 U < 1 q j or, equivalently, for which { X = Min{j : q j < 1 U} = Min{j : j log q < log(1 U)} = Min j : j > log(1 U) }. log q We note that log q and log(1 U) are both negative. Hence log(1 U) X = + 1. (2) log q Since 1 U is also uniformly distributed on (0, 1), Eq. (2) is equivalent to log U X = + 1. log q Thus we again have an algebraic formula for determining X. 159

12 Example: Generating a Poisson random variable. Let X be a Poisson random variable with probability mass function By noting that p i = P {X = i} = e λλi i!, i = 0, 1,.... p i+1 = λ i + 1 p i, i 0, we can write the inverse transform method for generating X as follows: (where F = F (i) = P {X i}) Step 1: Generate a random number U. Step 2: i = 0,p = e λ, F = p. Step 3: If U < F, set X = i and stop. Step 4: p = λp/(i + 1), F = F + p, i = i + 1. Step 5: Go to Step 3. The algorithm successively checks if the Poisson value is 0, then if it is 1, then 2, and so on. Thus the number of comparisons needed will be 1 greater than the generated value of the Poisson. Hence on average, the above algorithm needs to make (1 + λ) searches. 160

13 Example: Generating Binomial random variables. Suppose X is a binomial random variable with probability mass function P {X = i} = n! i!(n i)! pi (1 p) n i, i = 0, 1,..., n. Again we apply the inverse transform method by making use of the recursive relation ( ) ( ) n i p P {X = i + 1} = P {X = i}. i p Step 1: Generate a random number U. Step 2: c = p/(1 p), i = 0, r = (1 p) n, F = r. Step 3: If U < F, set X = i and stop. Step 4: r = [c(n i)/(i + 1)]r, F = F + r,i = i + 1. Step 5: Go to Step 3. In the algorithm with i denoting the value currently under consideration, r = P {X = i} and F = F (i) = P {X i}. The number of searches required is the value of X plus 1. Since E[X] = np, on average, it will take 1 + np searches to generate X. 161

14 24 Generate Discrete Random Variables (The Acceptance-Rejection Technique) Suppose we have an efficient method for simulating a random variable Y having probability mass function {q j, j 0}. We can use this as the basis for simulating another random variable X with probability mass function {p j, j 0}. Let c be a constant such that p j q j c for all j such that p j > 0. Rejection Method (acceptance-rejection method) Step 1: Simulate the value of Y, having probability mass function q j. Step 2: Generate a random number U. Step 3: If U < p Y /cq Y, set X = Y and stop. Otherwise, return to Step 1. Proposition 1 The acceptance-rejection method generates a random variable X such that P {X = j} = p j, j = 0, 1,.... In addition, the number of iterations of the algorithm needed to obtain X is a geometric random variable with mean c. 162

15 Proof: j. Summing over j, Let us first determine the probability that a single iteration produces the accepted value P {Y = j, it is accepted} = P {Y = j}p {accepted Y = j} = q j p j cq j = p j c. P {accepted} = j p j c = 1 c. As each iteration independently results in an accepted value with probability 1 c, we see that the number of iterations needed is geometric with mean c. Also, P {X = j} = n = n P {j accepted on iteration} ( 1 1 ) n 1 p j c c = p j. Remarks: (i) Since j p j c j q j and j p j = j q j = 1, we have c 1. (ii) The number of iterations is smaller if c is smaller. 163

16 Example: Suppose we wanted to generate a random variable X that takes one of the values 1, 2,..., 10 with respective probabilities 0.11, 0.12, 0.09, 0.08, 0.10, 0.08, 0.10, 0.09, 0.11, One way is to use the inverse transform method. However, here let us use the rejection method with Y being the random variable with the uniform discrete density q j = P {Y = j} = 1 10, j = 1, 2,..., 10. Let p j = P {X = j}. Then and the algorithm is given as follows: c = max { pj q j } = 1.2 Step 1: Generate a random number U 1 and set Y = 10U Step 2: Generate a second number U 2. Step 3: If U 2 p Y.12, set X = Y and Stop. Otherwise return to Step

17 24.1 The Composition Approach Suppose we have efficient methods for generating random variables X 1 and X 2 with probability mass functions {p (1) j, j 0} and {p (2) j, j 0} respectively, and we want to generate a random variable X having mass function where 0 < α < 1. The composition algorithm is as follows: P {X = j} = αp (1) j + (1 α)p (2) j, j 0 Step 1: Generate X 1 with mass function {p (1) j, j 0}. Step 2: Generate X 2 with mass function {p (2) j, j 0}. Step 3: Generate a random number U. Step 4: If U < α, set X = X 1. Otherwise, set X = X 2. We can prove that the algorithm gives the correct value by observing P {X = j} = P {U < α and X 1 = j} + P {U α and X 2 = j} = P {U < α}p {X 1 = j} + P {U α}p {X 2 = j} = αp (1) j + (1 α)p (2) j. 165

18 25 Generating Continuous Random Variables (Inverse Transform Algorithm) The inverse transform algorithm for generating continuous random variable is based on the following proposition: Proposition 2 Let U be a uniform (0, 1) random variable. For any continuous distribution function F the random variable X defined by has distribution F. X = F 1 (U) Proof: Let F X denote the distribution function of X = F 1 (U). Then F X (x) = P {X x} = P {F 1 (U) x}. Since F is a monotonic increasing function of x and so F 1 (U) x is equivalent to or F (F 1 (U)) F (x), F X (x) = P {F (F 1 (U)) F (x)} = P {U F (x)} = F (x). 166

19 Example: Let X be an exponential random variable with mean 1 λ and its cummulative probability density function If we let then F (x) = x 0 λe λt dt = 1 e λx. x = F 1 (u), u = F (x) = 1 e λx or x = 1 log(1 u). λ Hence we can generate the exponential random variable by generating a random number U and then setting X = F 1 (U) = 1 log(1 U). λ Since U 1 U (i.e. they have the same distribution), equivalently we can write X = 1 λ log(u). 167

20 Example: (Another algorithm for generating a Poisson random variable) Recall that a Poisson process with rate λ results when the time between successive events are independent exponential random variables with mean 1 λ. For such a process N(1), the number of events by time 1, is Poisson distributed with mean λ. However, if we let X i, i = 1, 2,... denote the successive inter-arrival times, then the nth event will occur at time n X i, and so i=1 { N(1) = Max n : n i=1 } X i 1. Using the result above, { N(1) = Max n : { = Max n : } n 1 λ log U i 1 } n log U i λ i=1 i=1 = Max { n : U 1 U n e λ}. (where U 1, U 2,... are random numbers) 168

21 Based on this result, we have the following algorithm for generating a Poisson random variable N with rate λ: Step 1: N = 0, P = 1. Step 2: Generate random number U. Step 3: Set P = P U. Step 4: If P < e λ, stop. Otherwise, set N = N + 1 and go to Step

22 Example: Suppose we want to generate an n-phase Erlangian random variable X. Since the distribution function is given by F (x) = 1 n 1 i=0 (λx) i e λx. i! It is not possible to give a closed form expression for its inverse. However, X is the sum of n independent exponential random variable, each with mean λ 1. We can use the result of previous example to generate X. Specifically, we generate n random numbers U 1, U 2,..., U n and then set X = 1 λ log U 1 1 λ log U n = 1 λ log(u 1 U n ). 170

23 26 Generating Continuous Random Variables (The Rejection Method) Suppose we have a method for generating a random variable Y having density function g(x). We can use this as basis for generating a random variable X having density function f(x). Let c be a constant such that The Rejection Method Step 1: Generate Y having density g. f(y) g(y) c for all y. Step 2: Generate a random number U. Step 3: If U f(y ) cg(y ), set X = Y. Otherwise, return to Step 1. The method is similar to the rejection method for discrete random variables except that density function replaces mass function. As in the discrete case, we can prove the following result. Proposition 3 The random variable generated by the rejection method has density f. and the number of iterations of the algorithm that are needed is a geometric random variable with mean c. 171

24 Example Let us use the rejection method to generate a random variable having density function Consider g(x) = 1, 0 < x < 1. f(x) = 20x(1 x) 3, 0 < x < 1. To determine c such that f(x)/g(x) c, we use calculus to determine the maximum value of f(x) g(x) d ( f(x) ) dx g(x) = 20x(1 x) 3 = 20[(1 x) 3 3x(1 x) 2 ]. The derivative is zero when x = 1/4, positive when x < 1 4 and negative when x > 1 4. ( ) f(x) Hence max occurs at x = 1/4 and thus 0 x 1 g(x) f(x) g(x) f(1 4 ) ( 1 3 ) 3 g( 4)( 1 4 ) = 20 4 = c. 172

25 Hence f(x) cg(x) = x(1 x)3. Hence the rejection method is as follows: Step 1: Generate random numbers U 1 and U 2 Step 2: If U U 1(1 U 1 ) 3, stop and set X = U 1. otherwise, return to Step 1. The average number of iterations that will be performed is c =

26 Example: Suppose we want to generate a random variable having the density f(x) = Kx 1/2 e x, x > 0 where K = 2/ π. Let us try the rejection method with an exponential random variable with mean λ 1. Hence let g(x) = λe λx, x > 0. Now and We have d ( f(x) ) dx g(x) = K λ d ( f(x) ) dx g(x) or when x = 1 2(1 λ), provided that λ < 1. f(x) g(x) = K λ x1/2 e (1 λ)x, [ 1 2 x 1 2e (1 λ)x (1 λ)x 1 2e (1 λ)x ]. = 0 when 1 2 x 1 2 = (1 λ)x 1 2 Moreover d 2 dx 2(f(x) g(x) ) x= 2(1 λ) 1 = Ke 1/2 < 0. 2λ(2(1 λ)) 1/2 174

27 Hence Algorithm: ( f(x) ) c = Max g(x) = K λ 1 [2(1 λ)] 1/2e 1 2. Step 1: Generate a random number U 1 and set Y = 1 λ log U 1. Step 2: Generate a random number U 2. Step 3: If U 2 < f(y ) cg(y ), set X = Y. Otherwise, return to Step 1. The average number of iterations that will be needed is c. Since c depends on λ, we can try to minimize c by choosing suitable λ. We differentiate c with respect λ, [ c (λ) = Ke λ 2 2 1/2 (1 λ) ( 1 )]. 1/2 λ 2 1/2 (1 λ) 3/2 2 c (λ) = 0 when 1 λ = λ or λ = Hence c = 3K ( 3 1/2e 2 2) 1/2 = 33/2 (2πe) 1/2. Since c is minimized, the number of iterations is minimized also. 175

28 Example: Generating a Normal Random Variable. Let Z be the standard normal random variable, and let X = Z. The density function of X is f(x) = 2 2π e x2 2, 0 < x <. We first generate X by using the rejection method with g being the exponential density function with mean 1, that is, g(x) = e x, 0 < x <. Now f(x) 2 g(x) = π ex x2 2 and so the maximum value of f(x)/g(x) occurs at the value of x that maximizes x x 2 /2. This occurs at x = 1 by simple calculus. So we can take { } f(x) c = max = f(1) x g(x) g(1) = Hence f(x) { cg(x) = exp 2e π. (x } 1)

29 Algorithm for generating the absolute value of a standard normal random variable: Step 1: Generate Y, an exponential random variable with mean 1. Step 2: Generate a random number U. (Y 1)2 Step 3: If U exp{ 2 }, set X = Y. Otherwise, return to Step 1. Once we have simulated the random variable X, we can generate the standard normal random variable Z by letting Z be equally likely to be either X or X. This can be written algorithmically as follows: Step 1: Generate X. Step 2: Generate a random number U. Step 3: If U 1 2, set Z = X. Otherwise set Z = X. 177

30 26.1 Generating Poisson Processes Suppose we wanted to generate the first n event times of a Poisson process with rate λ. Since the times between successive events are independent exponential random variables each with mean λ 1, one way to generate the process is as follows: Step 1: i = 1 and t = 0. Step 2: Generate U and set t = t 1 λ log U. Step 3: Set S(i) = t and i = i + 1. Step 4: If i > n, stop. Otherwise, return to Step 2. The values {S(i)} i=1,2,...,n will be the first n event times. If we wanted to generate the first T time units of the Poisson process, then the algorithm can be written as follows: Step 1: t = 0, I = 0. Step 2: Generate a random number U. Step 3: t = t 1 λ log U. If t > T, stop. Step 4: I = I + 1, S(I) = t. Step 5: Go to Step 2. The final value of I will represent the number of events that occur by time T. 178

31 27 Simulations of Markovian Queues Consider an one-server queuing system of unlimited waiting space, the arrival of customers is a Poisson process with the mean inter-arrival time λ 1 being 10 minutes. The exponential server has a mean service time µ 1 of 6 minutes. We are going to simulate this queuing system. The key step in the simulation is to generate two sequences of exponentially distributed random variable a i (for the inter-arrival process of customers) and s i (service time) having mean 10 and 6 respectively. Recall that 1 log(1 U) λ is an exponentially distributed random variable with parameter λ when U is an uniformly distributed random variable on (0, 1). We determine the length of simulation to be 90 minutes. The following inter-arrival time a i are generated: 8, 5, 8, 12, 7, 8, 9, 8, 9, 5, the last customer arrives at the 79th minute. The service time of each customer s i is generated as follows: 7, 8, 4, 6, 7, 6, 5, 4, 8,

32 One may construct the following table. The arrival time, time into service and departure time are the time when the customer arrives at the system, the time when the customer starts his service and the time when the customer leaves the system respectively. We note that the time into service of a customer will be the maximum of his arrival time and the departure time of the previous customer. The queuing time is the time that the customer spent in queuing. The time of a customer spent in the system is the sum of queuing time and service time. In our the case, the average queuing time of each customer is ( )/10 = 0.7 minute and the average time of each customer spent in the system is ( )/10 = 6.7 minutes. Finally by applying the Little s formula, the expected number of customers in the system is equal to the λw c = (10 1 ) 6.7 = The divisor is 90 because the simulation extends over 87 minutes. The theoretical value of the expected number of customers is given by ρ 1 ρ = = 1.5 where ρ = λ µ = 10 1 = The large deviation can be explained by the random error caused by short simulation time. 180

33 Customer Arrival Time into Service Departure Queuing Time in Number Time Service time Time Time System Table

34 A Summary on Simulation Generation of random numbers/uniformly distributed r.v. by using the congruential method. Approximation of definite integral by random numbers. Inverse transformation method for discrete and continuous r.v.. Acceptance and rejection method for discrete and continuous r.v.. Generation of Poisson and Exponential r.v. 182