Section 7.5: Nonstationary Poisson Processes

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Transcription:

Section 7.5: Nonstationary Poisson Processes Discrete-Event Simulation: A First Course c 2006 Pearson Ed., Inc. 0-13-142917-5 Discrete-Event Simulation: A First Course Section 7.5: Nonstationary Poisson Processes 1/ 20

Section 7.5: Nonstationary Poisson Processes Suppose we want the arrival rate λ to change over time: λ(t) Recall the algorithm to generate a stationary Poisson process: Stationary Poisson Process a 0 = 0.0; n = 0; while (a n < τ) { a n+1 = a n + Exponential(1 / λ); n++; } return a 1, a 2,..., a n 1 ; Above algorithm generates a stationary Poisson process Time interval is 0 t < τ Event times are a 1, a 2, a 3,... Process has constant rate λ Discrete-Event Simulation: A First Course Section 7.5: Nonstationary Poisson Processes 2/ 20

Incorrect Algorithm Change constant λ to function: Incorrect Algorithm a 0 = 0.0; n = 0; while (a n < τ) { a n+1 = a n + Exponential(1 / λ(a n )); n++; } return a 1, a 2,..., a n 1 ; Incorrect: ignores future evolution of λ(t) after t = a n If λ(a n ) < λ(a n+1 ) then a n+1 a n will tend to be too large If λ(a n ) > λ(a n+1 ) then a n+1 a n will tend to be too small Inertia error will be small only if λ(t) varies slowly with t Discrete-Event Simulation: A First Course Section 7.5: Nonstationary Poisson Processes 3/ 20

Example 7.5.1 Piecewise-constant rate function 1 0 t < 15 λ(t) = 2 15 t < 35 1 35 t < 50 Simulated using incorrect algorithm for τ = 50 Process replicated 10000 times Partitioned interval 0 t 50 into 50 bins Counted number of events for each bin, divided by 10000 Result is ˆλ(t), an estimate of λ(t) Discrete-Event Simulation: A First Course Section 7.5: Nonstationary Poisson Processes 4/ 20

Incorrect process generation 2.5 2.0 ˆλ(t) 1.5 λ(t) ˆλ(t) 1.0 0.5 0 15 35 50 t Incorrect algorithm has inertia error: ˆλ(t) under-estimates λ(t) after the rate increase ˆλ(t) over-estimates λ(t) after the rate decrease Use one of 2 correct algorithms instead Discrete-Event Simulation: A First Course Section 7.5: Nonstationary Poisson Processes 5/ 20

Thinning method Due to Lewis and Shedler, 1979 Uses an upper bound λ max λ(t) for 0 t < τ Generates a stationary Poisson process with rate λ max Discards (thins) some events, probabilistically Event at time s is kept with probability λ(s)/λ max λ max λ(s) λ(t). 0 a n s t Efficiency depends on λ max being a tight bound Discrete-Event Simulation: A First Course Section 7.5: Nonstationary Poisson Processes 6/ 20

Algorithm 7.5.1 Algorithm 7.5.1 a 0 = 0.0; n = 0; while (a n < τ) { s = a n ; do { s = s + Exponential(1 / λ max ); u = Uniform(0, λ max ); } while (u > λ(s)); a n+1 = s; n++; } return a 1, a 2,..., a n 1 ; When λ(s) is low, The event at time s is more likely to be discarded The number of loop iterations is more likely to be large Discrete-Event Simulation: A First Course Section 7.5: Nonstationary Poisson Processes 7/ 20

Example 7.5.2 The thinning method was applied to Example 7.5.1, using λ max = 2 Computation time increased by a factor of about 2.2 The algorithm is not synchronized Even if a separate stream is used for Uniform ˆλ(t) 2.5 2.0 1.5 1.0 0.5 λ(t) ˆλ(t) t 0 15 35 50 Discrete-Event Simulation: A First Course Section 7.5: Nonstationary Poisson Processes 8/ 20

Inversion Method Due to Çinlar, 1975 Similar to inversion for random variate generation Requires only one call to Random per event Based upon the cumulative event rate function: Λ(t) = t 0 λ(s) ds 0 t < τ Λ(t) represents the expected number of events in interval [0,t) If λ(t) > 0 then Λ( ) is strictly monotone increasing There exists an inverse Λ 1 ( ) Discrete-Event Simulation: A First Course Section 7.5: Nonstationary Poisson Processes 9/ 20

Algorithm 7.5.2: idea Generates a stationary unit Poisson process u 1,u 2,u 3,... Equivalent to n random points in interval 0 < u i < Λ(τ) Each u i is transformed into a i using Λ 1 ( ). u 4 Λ(t) u 3 u 2 u 1 0 a 1 a 2 a 3 a 4 t Discrete-Event Simulation: A First Course Section 7.5: Nonstationary Poisson Processes 10/ 20

Algorithm 7.5.2: details Algorithm 7.5.2 a 0 = 0.0; u 0 = 0.0; n = 0; while (a n < τ) { u n+1 = u n + Exponential(1.0); a n+1 = Λ 1 (u n+1 ); n++; } return a 1, a 2,..., a n 1 ; The algorithm is synchronized Useful when Λ 1 ( ) can be evaluated efficiently Discrete-Event Simulation: A First Course Section 7.5: Nonstationary Poisson Processes 11/ 20

Example 7.5.3 Use the rate function from Example 7.5.2 1 0 t < 15 λ(t) = 2 15 t < 35 1 35 t < 50 By integration we obtain t 0 t < 15 Λ(t) = 2t 15 15 t < 35 t + 20 35 t < 50 Solving u = Λ(t) for t we obtain u 0 u < 15 Λ 1 (u) = (u + 15)/2 15 u < 35 u 20 35 u < 50 Discrete-Event Simulation: A First Course Section 7.5: Nonstationary Poisson Processes 12/ 20

Example 7.5.3 results 2.5 2.0 ˆλ(t) 1.5 λ(t) ˆλ(t) 1.0 0.5 t 0 15 35 50 Generation time using inversion is similar to the incorrect algorithm For this example, Λ(t) was easily inverted If Λ(t) cannot be inverted in closed form Use thinning if λ max can be found Use numerical methods to invert Λ(t) Use an approximation Discrete-Event Simulation: A First Course Section 7.5: Nonstationary Poisson Processes 13/ 20

Next-Event Simulation Orientation The three algorithms must be adapted for next-event simulation: Given current event time t, generate next event time Incorrect Algorithm arrival = t + Exponential(1 / λ(t)); Thinning Method arrival = t; do { arrival = arrival + Exponential(1 / λ max); u = Uniform(0, λ max); } while (u > λ(arrival)); Inversion Method arrival = Λ 1 (Λ(t) + Exponential(1.0)); Discrete-Event Simulation: A First Course Section 7.5: Nonstationary Poisson Processes 14/ 20

Piecewise-Linear Rate Functions A piecewise-constant rate function is (usually) unrealistic Obtaining an accurate estimate of λ(t) is difficult Requires lots of data see Section 9.3 We will examine piecewise-linear λ(t) functions Can be specified as a sequence of knot pairs (t j, λ j ) λ(t) λ 0 λ 1 λ 4 λ 2 λ 3 λ 5 λ 6. 0 t 0 t 1 t 2 t 3 t 4 t 5 t 6 t Discrete-Event Simulation: A First Course Section 7.5: Nonstationary Poisson Processes 15/ 20

Algorithm 7.5.3 Algorithm 7.5.3 Step 1 Given k + 1 knot pairs (t j,λ j ) with 0 = t 0 < t 1 < < t k = τ λ j 0 Four steps to construct Piecewise-linear λ(t) Piecewise-quadratic Λ(t) Λ 1 (u) Define the slope of each segment s j = λ j+1 λ j t j+1 t j j = 0,1,...,k 1 Discrete-Event Simulation: A First Course Section 7.5: Nonstationary Poisson Processes 16/ 20

Algorithm 7.5.3 step 2/4 Algorithm 7.5.3 Step 2 Define the cumulative rate for each knot point as Λ j = tj 0 λ(t) dt j = 0,1,...,k These can be computed recursively with Λ 0 = 0 Λ j = Λ j 1 + 1 2 (λ j + λ j 1 )(t j t j 1 ) Discrete-Event Simulation: A First Course Section 7.5: Nonstationary Poisson Processes 17/ 20

Algorithm 7.5.3 step 3/4 Algorithm 7.5.3 Step 3 For subinterval t j t < t j+1 λ(t) = λ j + s j (t t j ) Λ(t) = Λ j + λ j (t t j ) + 1 2 s j(t t j ) 2 If s j 0 then λ(t) is linear Λ(t) is quadratic If s j = 0 then λ(t) is constant Λ(t) is linear Discrete-Event Simulation: A First Course Section 7.5: Nonstationary Poisson Processes 18/ 20

Algorithm 7.5.3 step 4/4 Algorithm 7.5.3 Step 4 For subinterval Λ j u < Λ j+1 Λ 1 (u) = t j + λ j + If s j = 0 then the above reduces to 2(u Λ j ) λ 2 j + 2s j (u Λ j ) Λ 1 (u) = t j + (u Λ j) λ j Discrete-Event Simulation: A First Course Section 7.5: Nonstationary Poisson Processes 19/ 20

Algorithm 7.5.4: Inversion with piecewise-linear λ(t) Modified algorithm 7.5.2 Also keeps track of index j, the current segment Algorithm 7.5.4 a 0 = 0.0; u 0 = 0.0; n = 0; j = 0; while (a n < τ) { u n+1 = u n + Exponential(1.0); while ((Λ j+1 < u n+1 ) and (j < k)) j++; a n+1 = Λ 1 (u n+1 ); /* Λ j < u n+1 Λ j+1 */ n++; } return a 1, a 2,..., a n 1 ; Discrete-Event Simulation: A First Course Section 7.5: Nonstationary Poisson Processes 20/ 20