Question 10 of 10: Monte Carlo Simulation — Machine Breakdown Repair Work
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2014 — 98-Ind-A1 Operations Research. Three-hour, open-book exam (any non-communicating calculator permitted); the paper totals 150 marks across 10 questions and only 100 marks are required, so a candidate would normally answer a subset — all ten are solved below for completeness.
Reference texts: Hillier & Lieberman, Introduction to Operations Research (11th ed., McGraw-Hill) — linear/integer programming, network optimization (CPM), dynamic programming, decision analysis, Markov chains and queueing theory; Nahmias, Production and Operations Analysis (7th ed.) — EOQ and inventory-control models.
Question 10: Monte Carlo Simulation — Machine Breakdown Repair Work (15 marks)
Given data — repair-time and breakdown-count distributions, with assigned 2-digit random-number ranges
Repair time (hr)
Probability
RN range
1
0.30
00–29
2
0.50
30–79
3
0.20
80–99
Given data — breakdowns per day
Breakdowns/day
Probability
RN range
0
0.30
00–29
1
0.30
30–59
2
0.40
60–99
Find. (a) A flowchart for the daily-work Monte Carlo procedure; (b) the simulated average daily repair work over 2 days, compared with the theoretical average.
Approach. Build cumulative-probability random-number ranges for each distribution (table above), then for each simulated day draw one RN for the breakdown count and one further RN per breakdown for its repair time, summing to that day's total work.
Flowchart for the daily-repair-work Monte Carlo simulation (N = number of days simulated).
Theoretical averages. Mean repair time/breakdown $=1(0.30)+2(0.50)+3(0.20)=0.3+1.0+0.6=1.9$ hr. Mean breakdowns/day $=0(0.30)+1(0.30)+2(0.40)=0+0.3+0.8=1.1$/day. So theoretical mean daily repair work is
$$\boxed{1.9\times1.1=2.09\text{ hr/day}.}$$
Simulate Day 1. Chunk the given digit string into consecutive 2-digit numbers: 13, 51, 60, 48, 66, 29, .... Day 1 draws RN$=13$ for breakdown count: $13\in[00,29]\Rightarrow0$ breakdowns, so no repair-time draw is needed and $W_1=0$ hr.
Simulate Day 2. Next RN$=51$ for breakdown count: $51\in[30,59]\Rightarrow1$ breakdown. Draw the next RN$=60$ for that breakdown's repair time: $60\in[30,79]\Rightarrow2$ hr. So $W_2=2$ hr.
Simulated average and comparison. Over the 2 simulated days, total work $=0+2=2$ hr, so
$$\boxed{\text{Simulated average}=2/2=1.0\text{ hr/day},}$$
noticeably below the theoretical $2.09$ hr/day. A 2-day run is a very small sample (only one of the two days even had a breakdown), so this gap is expected sampling variation, not an error — by the law of large numbers the simulated average would converge toward 2.09 hr/day as more days are simulated.
