23-Ind-A6 Systems Simulation · May 2018
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
National Exams — May 2018 — 17-Ind-A6 Systems Simulation. Three-hour, closed-book exam; one of two permitted calculators (Sharp or Casio), one 8.5″×11.0″ aid sheet (both sides). Format: three sections — Section A (four concept questions, candidates choose any two, 10 marks each, 20 marks total), Section B (three methods questions built around one continuing warehouse-simulation case study, candidates choose any two, 15 marks each, 30 marks total), Section C (two applications questions, candidates choose any one, 20 marks each). All nine questions are solved below for completeness. Two source anomalies are flagged where they occur: the front-page summary table states Section A is "Do 2 of 3," while Section A's own instructions and its four printed question sets read "two of the following four" — the printed four-question section is answered in full here; and Part C Question 2's sub-parts (a)–(c) are never printed anywhere in the paper, even though the results text for (d)–(f) explicitly refers back to "the factorial design matrix in (a)." Two-page Normal-distribution 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 modeling and goodness-of-fit testing, output analysis (warm-up, replication length, batch means vs. replication/deletion), and comparing alternative systems; Montgomery, Design and Analysis of Experiments (current ed., Wiley) — single-factor ANOVA, multiple comparisons, and 2k factorial designs with interaction analysis (Part C).
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. Hourly throughput from 3 pilot replications, $t=1,\dots,10$:
| Time $t$ | Rep 1 | Rep 2 | Rep 3 |
|---|---|---|---|
| 1 | 3 | 5 | 4 |
| 2 | 49 | 40 | 37 |
| 3 | 20 | 19 | 19 |
| 4 | 22 | 24 | 24 |
| 5 | 24 | 20 | 20 |
| 6 | 22 | 25 | 25 |
| 7 | 25 | 20 | 22 |
| 8 | 25 | 21 | 24 |
| 9 | 21 | 20 | 24 |
| 10 | 21 | 24 | 25 |
Find. (a) transient vs. steady-state definitions; (b) why a warm-up period matters; (c) the warm-up period from Welch's technique with $w=1$.
Approach. Answer (a)/(b) conceptually; for (c), average across replications at each $t$, apply Welch's shrinking moving-window smoother with $w=1$, and read the warm-up length off where the smoothed series stops trending.
(a) Transient vs. steady-state. In the transient (initialization) phase, the distribution of the model's output at time $t$ still depends on the arbitrary initial conditions used to start the run (here, an empty, idle warehouse) and changes systematically with $t$. In steady-state, that dependence has died out — the output's distribution has settled into one that no longer depends on $t$ or on the starting conditions, so successive observations are statistically representative of the system's long-run behaviour rather than of how it happened to start.
(b) Why the warm-up period matters. Including transient-phase observations in a reported performance estimate biases it — typically low here, since an empty-and-idle warehouse understates the throughput the system achieves once loaded. Deleting a properly chosen warm-up period lets the estimator (mean throughput, utilization, …) reflect steady-state behaviour rather than a blend of ramp-up and steady operation, at the cost of discarding some simulated output and shrinking the effective sample used for the estimate.
(c) Welch's technique, $w=1$.
| $t$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| $\bar Y(t)$ | 4.00 | 42.00 | 19.33 | 23.33 | 21.33 | 24.00 | 22.33 | 23.33 | 21.67 | 23.33 |
| $\bar w(t)$ | 4.00 | 21.78 | 28.22 | 21.33 | 22.89 | 22.56 | 23.22 | 22.44 | 22.78 | 23.33 |
After the sharp transient spike at $t=2$–$3$ (driven almost entirely by Replication 1's own startup surge), the smoothed series settles into a roughly constant $21$–$23$ band from $t=4$ onward and simply oscillates around that level for the rest of the run:
$$\boxed{l\approx4\text{ hours} \;(\text{delete the first 4 observations of every future run})}$$
| Item | Result |
|---|---|
| (a)/(b) | transient = still depends on initial conditions; delete it to avoid biasing the steady-state estimate |
| (c) warm-up period $l$ | $\approx4$ hours |
| steady-state mean (post warm-up, $t=5..10$) | $\approx22.68$ |
