23-Ind-A6 Systems Simulation · December 2017

Question 2 of 8: Non-Stationary Poisson Arrivals: Variate Generation and a Stationarity Test

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — December 2017 — 98-Ind-A6 Systems Simulation. Three-hour, closed-book exam; one of two permitted calculators (Sharp or Casio), two 8.5″×11.0″ aid sheets (both sides). Format: eight questions of equal value (10 marks each); candidates complete SIX of the EIGHT (only the first six as they appear in the answer book are marked). All eight questions are solved below for completeness (the paper's own Q6 is printed with an orphan leading "6." followed by "2. Consider an M/M/1 system…" and is answered as the paper's sixth question). Common Discrete/Continuous Distribution tables, Student-t and Chi-square tables were supplied with the exam; the values below are the same table values obtained by direct computation.

Reference texts: Banks, Carson, Nelson & Nicol, Discrete-Event System Simulation (5th ed., Pearson) — random-number/random-variate generation, input data analysis, output analysis (replications), comparing alternative systems, queueing simulation, and the simulation study life cycle.

Question 2 — Non-Stationary Poisson Arrivals: Variate Generation and a Stationarity Test (10 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Five days of observed counts:

Period	Day 1	Day 2	Day 3	Day 4	Day 5
8–9	3.3	3.5	3.1	3.7	3.5
9–10	3.4	4.1	3.3	3.9	3.4
10–11	3.9	3.8	4.1	3.7	3.8

Find. (a) two Poisson(3) variates via inverse transform; (b) whether the mean arrival rate is unchanged (stationary) from 9–10 vs. 10–11; (c) why the "change $\lambda$ on the fly" heuristic is wrong for a non-homogeneous Poisson process (NHPP).

Approach. (a) inverse-transform the discrete Poisson CDF; (b) since the same 5 days are measured in both hours, use a paired $t$-test on the day-by-day differences; (c) reason from the definition of an NHPP's cumulative-rate function.

(a) Poisson(3) variate generation

Build the Poisson(3) CDF. $$P(X\le0)=e^{-3}=0.0498,\quad P(X\le1)=0.1991,\quad P(X\le2)=0.4232,\ \ldots$$
Invert at $u=0.02$. Since $u=0.02 \le P(X\le0)=0.0498$: $X=\boxed{0}$.
Invert at $u=0.30$. Since $P(X\le1)=0.1991 < 0.30 \le P(X\le2)=0.4232$: $X=\boxed{2}$.

(b) Stationarity test, 9–10 vs. 10–11

Paired differences (same 5 days). $d_i = (\text{9-10})_i-(\text{10-11})_i$: $-0.5,\ 0.3,\ -0.8,\ 0.2,\ -0.4$. $$\bar d = -0.24,\qquad s_d = 0.472.$$
Test statistic. $H_0:\ \mu_{9\text{-}10}=\mu_{10\text{-}11}$ (stationary): $$t = \frac{\bar d}{s_d/\sqrt n} = \frac{-0.24}{0.472/\sqrt5} = \boxed{-1.136}.$$
Critical value and decision. $df=4$, two-sided $\alpha=0.05$: $t_{0.025,4}=2.776$. Since $|t|=1.136 < 2.776$, fail to reject $H_0$ — the 9–11 window IS statistically stationary at the 5% level.

Quantity	Result
(a) Poisson(3) variate at $u=0.02$	$X=0$
(a) Poisson(3) variate at $u=0.30$	$X=2$
(b) paired $t$ (9-10 vs 10-11)	$t=-1.136$, $t_{crit}=2.776$ — stationary (fail to reject)

(c) Why "changing $\lambda$ on the fly" is wrong. The naive recipe generates the next interarrival time as $-\ln(U)/\lambda(\text{now})$, i.e. it treats the process as freshly homogeneous, with the CURRENT instantaneous rate, at every draw. A true NHPP is defined by its cumulative rate function $\Lambda(t)=\int_0^t\lambda(s)\,ds$, not by memoryless behaviour local to the current clock time — an arrival "due" under the real $\lambda(t)$ trajectory can be systematically delayed or advanced whenever $\lambda$ changes mid-interarrival, and the count of arrivals in any interval straddling a rate change is not correctly Poisson-distributed with the right expectation under the naive method. The two textbook-correct techniques are: (1) inverse-transform on $\Lambda(t)$ — generate a unit-rate homogeneous Poisson process and map its arrival times through $\Lambda^{-1}$; or (2) thinning (Lewis–Shedler) — generate candidate arrivals as a homogeneous Poisson process at the maximum rate $\lambda^*=\max_t\lambda(t)$ over the horizon, then accept each candidate at time $t$ with probability $\lambda(t)/\lambda^*$. Both respect the entire $\lambda(t)$ trajectory rather than only its value at the instant of each draw.

Check: part (b) is answered with a paired $t$-test because the same 5 calendar days are sampled in both hourly windows (natural pairing); an unpaired two-sample test would discard that correlation and understate the test's power.

Concept & Theory

Topic. Non-homogeneous Poisson process (NHPP) input modelling: variate generation and testing for stationarity.

Key relations. Poisson CDF inversion $P(X\le k)=\sum_{i=0}^k e^{-\lambda}\lambda^i/i!$; paired $t=\bar d/(s_d/\sqrt n)$; NHPP cumulative rate $\Lambda(t)=\int\lambda(s)\,ds$; thinning acceptance probability $\lambda(t)/\lambda^*$.

Why this works. Arrival processes in real systems (call centres, airports, ambulance dispatch) are rarely stationary across a full day; testing stationarity hour-by-hour before committing to one $\lambda$ per period avoids silently smoothing over a genuine rate change, while the thinning/inverse-transform methods are the only two constructions that provably reproduce a specified $\lambda(t)$.

Common pitfall. Using an unpaired test when the same days/units are measured twice (loses power); or, for NHPP simulation, assuming any exponential-interarrival recipe is safe as long as $\lambda$ is "updated" — only inversion on $\Lambda$ or thinning is unbiased.

Source. Banks, Carson, Nelson & Nicol, Discrete-Event System Simulation, ch. 8 (random variate generation, NHPP), ch. 9 (input data analysis).

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