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Question 1Engineering Economics

Opportunity cost is best defined as:

  • A.The cost of lost production due to machine breakdown
  • B.The foregone return from the next best alternative use of resources
  • C.The cost of raw materials not purchased
  • D.The difference between fixed and variable costs
Show solution

Answer · B

Opportunity cost is the value of the best foregone alternative. If capital is invested in Project A, the opportunity cost is the return that could have been earned in Project B (next best option). This concept underlies MARR selection.

Question 2Human Factors, Ergonomics, and Safety

OSHA permissible noise exposure at 90 dBA is 8 hours. At 95 dBA, the permissible duration is:

  • A.6 hours
  • B.4 hours
  • C.2 hours
  • D.1 hour
Show solution

Answer · B

OSHA 29 CFR 1910.95 (5 dB exchange rate): 90 dBA = 8 hr, 95 dBA = 4 hr, 100 dBA = 2 hr, 105 dBA = 1 hr. Each 5 dB increase halves the permissible duration. Noise dose D = Σ(C_i/T_i) ≤ 1.0. OSHA 5 dB exchange rate (permissible durations): 90 dBA=8 hr, 95 dBA=4 hr, 100 dBA=2 hr, 105 dBA=1 hr, 110 dBA=0.5 hr. NIOSH uses a stricter 3 dB exchange rate with an 85 dBA criterion: 85 dBA=8 hr, 88 dBA=4 hr. Noise dose D = Σ(Cᵢ/Tᵢ) ≤ 1.0; when D>1.0, engineering or administrative controls are required. Engineering controls (barriers, enclosures) are preferred over hearing protection.

Question 3Manufacturing, Service, and Other Production Systems

Safety stock is maintained to protect against:

  • A.Predictable demand during lead time
  • B.Variability in demand and/or lead time
  • C.Excess inventory costs
  • D.Order processing delays that are constant
Show solution

Answer · B

Safety stock = Z × σ_d × √(LT) for variable demand, constant lead time. Z = service level factor (Z=1.65 for 95% service level). Higher service level → more safety stock → more holding cost. Reorder Point = demand during LT + safety stock. Safety stock formula for variable demand: SS = Z × σ_d × √LT. For 95% service level: Z=1.65; 99%: Z=2.33. Reorder Point (ROP) = average demand × LT + SS. Higher service level dramatically increases safety stock (and holding cost) — there's a trade-off between service level and inventory investment.

Question 4Mathematics

Evaluate ∫₀² (3x² + 2x) dx.

  • A.10
  • B.12
  • C.16
  • D.8
Show solution

Answer · B

∫₀² (3x² + 2x) dx = [x³ + x²]₀² = (8+4) - (0+0) = 12. Power rule: ∫xⁿ dx = x^(n+1)/(n+1) + C. Definite integral: evaluate antiderivative at upper minus lower limit. Verify by differentiation: d/dx(x³ + x²) = 3x² + 2x ✓. Definite integration computes net area under the curve—negative area occurs where f(x) < 0, and positive where f(x) > 0.

Question 5Modeling and Quantitative Analysis

The knapsack problem is a classic:

  • A.Linear programming problem with continuous variables and resource capacity constraints
  • B.Integer programming problem — maximize value within capacity constraint
  • C.Queuing theory problem involving customer service times and wait line analysis
  • D.Demand forecasting problem involving prediction of future market consumption patterns
Show solution

Answer · B

0-1 Knapsack: maximize Σv_i × x_i subject to Σw_i × x_i ≤ W, x_i ∈ {0,1}. v_i = item values, w_i = item weights, W = capacity. NP-hard in general. Dynamic programming gives pseudo-polynomial solution. Applications: resource allocation, capital budgeting, cargo loading. Knapsack is NP-hard in the general case, but a pseudo-polynomial DP solution runs in O(nW) time. Greedy (highest value/weight ratio first) gives optimal solution for the fractional knapsack, but not 0-1 knapsack. Applications beyond packing: portfolio selection (capital budgeting), resource allocation with budget constraints.

Question 6Probability and Statistics

A 2³ full factorial experiment involves:

  • A.2 factors each at 3 levels
  • B.3 factors each at 2 levels (8 runs)
  • C.3 replications of a 2-factor experiment
  • D.8 factors at 2 levels
Show solution

Answer · B

2³ factorial: 3 factors (A, B, C), each at 2 levels (+/-). Total runs = 2³ = 8. Allows estimation of 3 main effects, 3 two-way interactions (AB, AC, BC), and 1 three-way interaction (ABC). Full factorial estimates all effects without confounding. 2ᵏ factorial: k factors at 2 levels (+/-). 2³ = 8 runs. Effects estimated: 3 main effects (A, B, C), 3 two-factor interactions (AB, AC, BC), 1 three-factor interaction (ABC) = 7 effects + intercept. For k=4: 2⁴=16 runs; k=5: 32 runs. Fractional factorial 2^(k-p): 2^(3-1)=4 runs (half-fraction) — used when full factorial is too expensive.

Question 7Modeling and Quantitative Analysis

Little's Law states that L = λW where:

  • A.L=service rate, λ=queue length, W=waiting time in queue
  • B.L=average number in system, λ=throughput rate, W=average time in system
  • C.L=lot size, λ=lead time per unit, W=work in process inventory
  • D.L=labor cost per unit, λ=equipment utilization, W=wage rate
Show solution

Answer · B

Little's Law: L = λW relates average inventory in system to throughput rate and average time in system. Holds for any stable system regardless of arrival/service distribution. In manufacturing: WIP = throughput × cycle time. In service: average customers in system = arrival rate × average service time. Universal applicability is its power.

Question 8Quality

The DMAIC methodology in Six Sigma stands for:

  • A.Define, Measure, Analyze, Improve, Control
  • B.Design, Manufacture, Assemble, Inspect, Complete
  • C.Define, Monitor, Assess, Implement, Check
  • D.Detect, Measure, Assign, Improve, Close
Show solution

Answer · A

DMAIC: Define (problem, scope, goals), Measure (baseline performance, data collection), Analyze (root causes), Improve (solutions, pilot), Control (sustain gains, monitoring plan). Used for improving existing processes. Each phase has distinct tools: Define uses project charter and SIPOC; Measure uses capability studies and MSA; Analyze uses fishbone, regression, and hypothesis tests; Improve uses DOE and piloting; Control uses control charts and OCAP. Contrast with DMADV (Define-Measure-Analyze-Design-Verify), used for designing new processes.

Question 9Engineering Sciences

Newton's Second Law applied to a particle states:

  • A.For every action there is an equal and opposite reaction
  • B.ΣF = ma (net force equals mass times acceleration)
  • C.A body at rest stays at rest unless acted upon
  • D.Work equals force times displacement
Show solution

Answer · B

Newton's 2nd Law: ΣF = ma (vector equation). In x, y, z: ΣFx=max, ΣFy=may, ΣFz=maz. This is the fundamental equation of dynamics. For statics: a=0 → ΣF=0. 1st Law: F=0 → a=0. 3rd Law: F_AB = -F_BA. Newton's Second Law is a vector equation—x, y, z components must each balance independently: ΣFx = max, ΣFy = may. For constant acceleration, use kinematic equations: v = v₀ + at and x = v₀t + ½at². Newton's Third Law action-reaction pairs act on different bodies and never cancel each other in the same free-body diagram.

Question 10Engineering Sciences

A 2 m steel rod stretches 1.5 mm under load. The axial strain is:

  • A.0.00075
  • B.0.0015
  • C.0.75
  • D.3.0 × 10⁻⁴
Show solution

Answer · A

Axial strain ε = δ/L = 1.5 mm / 2000 mm = 0.00075 (dimensionless). Percent elongation = 0.075%. Strain is unitless (mm/mm or in/in). Strain is dimensionless (mm/mm or in/in) and is often expressed as a percentage or in microstrain (με = 10⁻⁶ strain). True (logarithmic) strain ε = ln(L/L₀) is used for large deformations; engineering strain δ/L₀ is valid in the linear elastic range. Elastic strain is fully recoverable; plastic strain beyond the yield point is permanent.

Question 11Systems Engineering, Analysis, and Design

The primary goal of systems engineering is to:

  • A.Minimize system cost regardless of performance requirements per cost management
  • B.Ensure a system meets stakeholder needs through entire lifecycle
  • C.Design and optimize only hardware subsystems per mechanical engineering discipline
  • D.Maximize system feature complexity and technical sophistication across all domains
Show solution

Answer · B

Systems engineering: interdisciplinary approach to enable the realization of successful systems through: requirements analysis, functional analysis, design synthesis, verification/validation. Key processes: INCOSE defines in SE Handbook. Considers technical and non-technical aspects across the full lifecycle.

Question 12Manufacturing, Service, and Other Production Systems

Value Engineering (VE) focuses primarily on:

  • A.Reducing product weight only
  • B.Improving function at minimum cost
  • C.Increasing product features
  • D.Maximizing manufacturing speed
Show solution

Answer · B

Value Engineering analyzes functions to achieve required performance at the lowest overall cost. Value = Function/Cost. VE asks 'what does it do?' and 'what does it cost?' to find alternatives that deliver the same function more economically.

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