23-Ind-A6 Systems Simulation · December 2013

Question 12 of 14: Question 12 (Part D, Set 2 — Hand Simulation of the Welding Station)

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

Notes on this paper

National Exams — December 2013 — 98-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: 14 sub-questions across four parts — Part A (do 1 of 2, 25 marks), Part B (do 3 of 5, 15 marks), Part C (do 1 of 3, 10 marks), Part D (do 2 of 4, 20 marks); 7 questions, 70 marks constitute a complete paper. All fourteen sub-questions are solved below for completeness (the source restarts its own numbering at 1 within each Part). Statistical tables (Normal, t, chi-square, F) 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) — simulation study design, random-number generation, input/output data analysis, variance reduction, verification & validation, queueing simulation; Montgomery, Design and Analysis of Experiments (9th ed., Wiley) — factorial designs and ANOVA.

Question 12 (Part D, Set 2 — Hand Simulation of the Welding Station) (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. Arrival rate $\lambda=6$/hr (Poisson) ⇒ mean interarrival time $=10$ min; two identical servers, each exponential service, mean $=15$ min; the 15 supplied random numbers (start of list, working down): 0.27, 0.71, 0.36, 0.13, 0.21, 0.17, 0.30, 0.65, 0.20, 0.63, 0.70, 0.49, 0.61, 0.86, 0.51.

Find. Average queue size $L_q$, average wait $W_q$, average time in system $W$, and combined-server utilization $\rho$, from a 5-customer by-hand simulation.

Approach. Inverse-transform each exponential draw as $t=-\beta\ln(U)$; use the random numbers in order, alternating interarrival draw then service-time draw per customer; assign each arrival to whichever of the two servers frees up earliest (FIFO single queue feeding both servers); tabulate arrival/start/end times for 5 customers, then compute the four required statistics from the table.

Check: the exam gives one pooled list of 15 random numbers with no explicit rule for which draws are interarrival times vs. service times; this solution adopts the natural convention — alternate (interarrival, service) per customer, using numbers 1–10 of the 15 for the first 5 customers — and states it explicitly since a different pairing convention would shift the individual times (though not the qualitative pattern of the answer).

Interarrival and service times (inverse transform, $t=-\beta\ln U$). Interarrival draws (numbers 1,3,5,7,9): 0.27, 0.36, 0.21, 0.30, 0.20 $\Rightarrow$ 13.09, 10.22, 15.61, 12.04, 16.09 min. Service draws (numbers 2,4,6,8,10): 0.71, 0.13, 0.17, 0.65, 0.63 $\Rightarrow$ 5.14, 30.60, 26.58, 6.46, 6.93 min.
Cumulative arrival times. $13.09,\ 23.31,\ 38.92,\ 50.96,\ 67.05$ min.
Simulation table. Assign each arrival to the earlier-freeing of the two servers:

Cust. Arrival Server Start End Wait $(W_q)$ System $(W)$

1 13.09 1 13.09 18.23 0.00 5.14

2 23.31 2 23.31 53.91 0.00 30.60

3 38.92 1 38.92 65.50 0.00 26.58

4 50.96 2 53.91 60.37 2.96 9.42

5 67.05 2 67.05 73.98 0.00 6.93

Only Customer 4 arrives while both servers are still busy (Server 1 free at 65.50, Server 2 free at 53.91) and must wait $53.91-50.96=2.96$ min.
Average wait and average time in system. $$W_q = \frac{0+0+0+2.96+0}{5} = \boxed{0.591\ \text{min}}, \qquad W = \frac{5.14+30.60+26.58+9.42+6.93}{5} = \boxed{15.73\ \text{min}}.$$
Average queue size and system size (Little's Law). Effective arrival rate over the simulated horizon (last departure at 73.98 min, 5 customers): $\lambda_{eff}=5/73.98=0.0676$/min: $$L_q = \lambda_{eff}W_q = 0.0676\times0.591 = \boxed{0.040\ \text{customers}}, \qquad L = \lambda_{eff}W = 0.0676\times15.73 = \boxed{1.063\ \text{customers}}.$$
Combined server utilization. Total server-busy time is the sum of the 5 service times, spread over 2 servers for the 73.98-minute simulated horizon: $$\rho = \frac{\sum s_i}{2\times T} = \frac{5.14+30.60+26.58+6.46+6.93}{2\times73.98} = \boxed{51.2\%}.$$

Cust.	Arrival	Server	Start	End	Wait $(W_q)$	System $(W)$
1	13.09	1	13.09	18.23	0.00	5.14
2	23.31	2	23.31	53.91	0.00	30.60
3	38.92	1	38.92	65.50	0.00	26.58
4	50.96	2	53.91	60.37	2.96	9.42
5	67.05	2	67.05	73.98	0.00	6.93

Quantity	Result
Average time in queue $W_q$	0.591 min
Average time in system $W$	15.73 min
Average queue size $L_q$	0.040
Average number in system $L$	1.063
Combined server utilization $\rho$	51.2%

Concept & Theory

Topic. Hand simulation of a multi-server (M/M/2-type) queue via the next-event method.

Key relations. Inverse-transform exponential generation $t=-\beta\ln U$; Little's Law $L=\lambda_{eff}W$, $L_q=\lambda_{eff}W_q$; utilization = busy server-time / (servers × horizon).

Why this works. A hand simulation of only 5 customers is a genuine (if noisy) sample path of the system, not a steady-state estimate — here only one of five customers happens to wait at all, which is a plausible small-sample outcome for a system whose theoretical utilization $\rho=\lambda/(2\mu)=6/8=0.75$ would predict more queueing over a longer run; the point of the exercise is the METHOD (event sequencing, server assignment, Little's Law bookkeeping), not convergence to the theoretical M/M/2 formulas at $n=5$.

Common pitfall. Computing utilization per-server instead of for "the pair as a group" as explicitly asked; and computing $L_q,L$ as simple time-averages of the WAIT/SYSTEM columns rather than via Little's Law using the actual simulated arrival rate over the observed horizon.

Source. Banks, Carson, Nelson & Nicol, Discrete-Event System Simulation, ch. 2–3 (hand simulation, next-event mechanics).

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