City Guilds Model Questions Vol1 2020 Engineering Question 32

Practice objective / multiple choice question 32 from the 2020 City Guilds Model Questions Vol1 Engineering examination.

City Guilds Model Questions Vol1 2020 Engineering Objective / Multiple Choice Hard Difficulty
Question 32 CITY_GUILDS_MODEL_QUESTIONS_VOL1 • 2020 • ENGINEERING • objective

A machine component fails according to an exponential distribution with constant failure rate \(\lambda = 2\times10^{-4}\ \text{h}^{-1}\). Corrective maintenance after a failure costs \(C_c = 8{,}000\) N and incurs a downtime of 5 h. Preventive maintenance performed at a fixed interval \(T\) costs \(C_p = 1{,}500\) N and requires 2 h downtime. Assuming the age‑replacement policy, the expected cost per operating hour for a given \(T\) is\n\[\nE(T)=\frac{C_p + C_c\bigl(1-e^{-\lambda T}\bigr)}{T + 2 + 5\bigl(1-e^{-\lambda T}\bigr)}\n\]\nCalculate \(E(T)\) for \(T = 2{,}000\) h and \(T = 3{,}500\) h, and state which interval gives the lower expected cost per hour.

Answer Options

A. E(2 000 h) ≈ 2.50 N/h; E(3 500 h) ≈ 1.58 N/h. The 3 500 h preventive interval yields the lower expected cost per hour.
B. E(2 000 h) ≈ 2.06 N/h; E(3 500 h) ≈ 2.06 N/h. Both intervals give the same expected cost per hour.
C. E(2 000 h) ≈ 1.90 N/h; E(3 500 h) ≈ 2.10 N/h. The 2 000 h preventive interval yields the lower expected cost per hour.
Correct Answer D. E(2 000 h) ≈ 2.06 N/h; E(3 500 h) ≈ 1.58 N/h. The 3 500 h preventive interval yields the lower expected cost per hour.
Correct Answer
D
Correct Option:
E(2 000 h) ≈ 2.06 N/h; E(3 500 h) ≈ 1.58 N/h. The 3 500 h preventive interval yields the lower expected cost per hour.

About This Question

This is City Guilds Model Questions Vol1 2020 Engineering Question 32. It is one of the objective questions from the 2020 City Guilds Model Questions Vol1 Engineering examination.

Difficulty level: Hard .

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