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23-Ind-A1 Operations Research · Undated paper

Question 7 of 10: Decision Analysis — Oil Pipeline Weld Inspection

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

Notes on this paper

National Exams — May 2019 — 17-Ind-A1 Operations Research. Three-hour, open-book exam (any non-communicating calculator permitted); the paper totals 175 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 programming & the simplex method (ch. 3–4), duality & sensitivity analysis (ch. 6), dynamic programming (ch. 11), network optimization & CPM/PERT project crashing (ch. 9–10), queueing theory incl. finite-source (machine-repair) models (ch. 17), decision analysis & the value of information (ch. 15–16), Markov chains (ch. 16), Monte Carlo simulation (ch. 20). Nahmias, Production and Operations Analysis — deterministic EOQ inventory models with and without planned shortages.

Question 7: Decision Analysis — Oil Pipeline Weld Inspection (20 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. 1000 seams; rework cost $1,200/defective seam; defect-rate prior: $P(5\%)=0.30$, $P(10\%)=0.50$, $P(20\%)=0.20$; cleanup-team flat cost $130,000 (covers full inspection + repair of all 1000 seams); x-ray inspection of 1 randomly selected completed weld, cost $2,000.

Find. (a) The expected-value-optimal decision (team vs. no team). (b) Whether the value of the extra information from inspecting one weld exceeds its $2,000 cost.

Approach. (a) Compare the expected cost of reworking seams as they occur (no team) against the team's flat fee, using the prior over defect rates. (b) This is a pre-posterior (Bayesian) decision-analysis question: find how a single weld's inspection outcome (defective / good) would update the defect-rate belief, re-decide under each posterior, and compare the resulting expected cost (with the option to act on the sample) against the $2,000 sampling cost — this is the Expected Value of Sample Information (EVSI).

  1. (a) Expected defect rate and expected rework cost with no team: $$E[\text{rate}]=0.05(0.30)+0.10(0.50)+0.20(0.20)=0.015+0.050+0.040=0.105\ (10.5\%)$$ $$E[\text{cost, no team}]=1000\times 0.105\times\$1{,}200=\boxed{\$126{,}000}$$ Compared with the team's flat $130,000: $$\boxed{\$126{,}000\ <\ \$130{,}000\ \Rightarrow\ \text{do NOT hire the cleanup team}}$$ (reworking as-you-go is expected to be $4,000 cheaper).
  2. (b) Marginal probability the sampled weld is defective (law of total probability, using the same prior — the sampled weld's defect chance IS the unknown overall rate): $$P(D)=E[\text{rate}]=0.105,\qquad P(G)=1-0.105=0.895$$
  3. Posterior defect-rate distribution given the sampled weld is Defective (Bayes' rule, $P(\text{rate}\mid D)\propto P(\text{rate})\times\text{rate}$):
    Posterior after observing the sampled weld is defective
    Rate5%10%20%
    Posterior prob.0.14290.47620.3810
    $$E[\text{rate}\mid D]=0.05(0.1429)+0.10(0.4762)+0.20(0.3810)=0.1310\ (13.10\%)$$ $$E[\text{cost, no team}\mid D]=1000(0.1310)(1200)=\$157{,}142.86\ >\ \$130{,}000$$ — if the sample comes back defective, the optimal follow-up action flips to hire the team, at $130,000.
  4. Posterior given the sampled weld is Good ($P(\text{rate}\mid G)\propto P(\text{rate})\times(1-\text{rate})$):
    Posterior after observing the sampled weld is good
    Rate5%10%20%
    Posterior prob.0.31840.50280.1788
    $$E[\text{rate}\mid G]=0.05(0.3184)+0.10(0.5028)+0.20(0.1788)=0.1020\ (10.20\%)$$ $$E[\text{cost, no team}\mid G]=1000(0.1020)(1200)=\$122{,}346.37\ <\ \$130{,}000$$ — if the sample comes back good, the optimal action stays no team, at $122,346.37.
  5. Expected cost WITH the option to sample (act optimally after seeing the result): $$E[\text{cost}\mid\text{sample}]=P(D)\min(157{,}142.86,\,130{,}000)+P(G)\min(122{,}346.37,\,130{,}000)$$ $$=0.105(130{,}000)+0.895(122{,}346.37)=13{,}650.00+109{,}500.00=\$123{,}150.00$$
  6. Value of sample information vs. its cost: $$EVSI=\underbrace{\$126{,}000}_{\text{best without sampling}}-\underbrace{\$123{,}150.00}_{\text{expected cost with sampling}}=\$2{,}850.00$$ $$\boxed{EVSI=\$2{,}850.00\ >\ \$2{,}000\text{ inspection cost}\ \Rightarrow\ \text{YES, worth inspecting (net gain \$850)}}$$
Final results — Question 7
ItemValue
Prior expected defect rate10.5%
(a) Expected cost, no team$126,000
(a) DecisionDo not hire cleanup team (saves $4,000)
Posterior rate | defective sample13.10% → hire team ($130,000)
Posterior rate | good sample10.20% → no team ($122,346.37)
Expected cost with sampling$123,150.00
EVSI$2,850.00
(b) DecisionWorthwhile (EVSI $2,850 > $2,000 cost)