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NCEES Engineering Licensure

FE Industrial and Systems Practice Exam

FE Industrial and Systems exam practice — the NCEES discipline exam, your first step toward the PE.

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Built for FE Industrial and Systems candidates

  • NCEES FE Industrial and Systems topic mix: engineering economics, probability & statistics, operations research, manufacturing & production, quality, and ergonomics
  • Answerable from the NCEES FE Reference Handbook + first principles
  • Mix of calculation and conceptual items, in SI and US customary units
  • Step-by-step solutions on every question (first 2 free; all with Premium)
  • Follow-up study chat to dig into any concept that trips you up

The FE Industrial and Systems exam is one of the seven NCEES Fundamentals of Engineering discipline exams — a 110-question, computer-based test, taken near graduation, that is the first step toward becoming a licensed Professional Engineer (PE). Every question is meant to be answered from the FE Reference Handbook and engineering fundamentals, so practice that drills the right Industrial and Systems topic mix, in the right units, with worked solutions is exactly how you build exam-day speed.

Crameleon builds your practice exam on demand — pick the pathway and module you're studying, and we generate a fresh paper with step-by-step solutions (the first 2 free; all of them with Premium). Every question comes with a follow-up study chat, so you can dig into any concept that trips you up without leaving the page. The generation takes under 2 minutes and your first exam is free.

Sample FE Industrial and Systems exam questions

Three original questions on FE Industrial and Systems exam topics, each with a worked solution.

Question 1Multiple choiceQuality

A machining process produces shafts whose diameters are normally distributed with a mean of 50.20 mm and a standard deviation of 0.40 mm. The specification limits for the diameter are 49.00 mm to 51.00 mm, and the process is in statistical control. What is the process capability index CpkC_{pk}?

  1. A.

    0.33

  2. B.

    0.67

  3. C.

    0.83

  4. D.

    1.00

Worked solution
Answer:B

CpkC_{pk} measures how much room the process has between its mean and the nearer specification limit, in units of three standard deviations. It is the smaller of the two one-sided indices, because the side closer to the mean is the one that produces nonconforming parts:

Cpk=min⁡[USL−μ3σ, μ−LSL3σ]C_{pk} = \min\left[\frac{\text{USL} - \mu}{3\sigma},\ \frac{\mu - \text{LSL}}{3\sigma}\right]

Upper side:

51.00−50.203(0.40)=0.801.20=0.667\frac{51.00 - 50.20}{3(0.40)} = \frac{0.80}{1.20} = 0.667

Lower side:

50.20−49.003(0.40)=1.201.20=1.00\frac{50.20 - 49.00}{3(0.40)} = \frac{1.20}{1.20} = 1.00

The upper side governs because the mean sits above the centre of the tolerance:

Cpk=min⁡(0.667, 1.00)=0.667C_{pk} = \min(0.667,\ 1.00) = 0.667

A value below 1 means the process is not capable: the upper limit is only 0.80/0.40=2.00.80/0.40 = 2.0 standard deviations above the mean, so about 2.3% of shafts exceed 51.00 mm.

CpkC_{pk} ≈ 0.67 (option B)

The most tempting wrong answer is C (0.83), which is Cp=(USL−LSL)/(6σ)=2.00/2.40C_p = (\text{USL} - \text{LSL})/(6\sigma) = 2.00/2.40. CpC_p measures the spread of the process relative to the tolerance width only; it ignores where the mean is and would equal CpkC_{pk} only if the process were centred at 50.00 mm. Option D (1.00) takes the lower-side index, but CpkC_{pk} is governed by the nearer limit. Option A (0.33) divides the one-sided distance by 6σ6\sigma instead of 3σ3\sigma.

Question 2Short answerEngineering Economics

A company wants to deposit a single sum today in an account earning 6% per year, compounded annually, so that it can withdraw $2,500 at the end of each year for the next 8 years, leaving the account empty after the last withdrawal. How much must be deposited today? Use (P/A,6%,8)=6.2098(P/A, 6\%, 8) = 6.2098 or the factor formula.

Worked solution
Answer:

P ≈ $15,525

The withdrawals form a uniform series of A=2,500A = 2{,}500 at the end of each of n=8n = 8 years, and the question asks for the single present amount that is equivalent to that series at i=6%i = 6\%. That is exactly what the uniform-series present-worth factor converts:

P=A (P/A,i,n)P = A\,(P/A, i, n)

The factor can be read from the 6% interest table or computed from its formula, which is the sum of the eight individual present-worth factors (1+i)−t(1+i)^{-t} in closed form:

(P/A,6%,8)=(1+i)n−1i(1+i)n=(1.06)8−10.06(1.06)8=1.5938−10.06(1.5938)=0.59380.09563=6.2098(P/A, 6\%, 8) = \frac{(1+i)^n - 1}{i(1+i)^n} = \frac{(1.06)^8 - 1}{0.06(1.06)^8} = \frac{1.5938 - 1}{0.06(1.5938)} = \frac{0.5938}{0.09563} = 6.2098

Multiply by the annual withdrawal:

P=2,500(6.2098)=15,524.5P = 2{,}500(6.2098) = 15{,}524.5

The eight withdrawals total $20,000, so the deposit is about $4,475 less than the money paid out; the difference is the interest the balance earns while it is being drawn down. A deposit of $20,000 would be the answer only at 0% interest.

P ≈ $15,525

Question 3Worked problemModeling and Quantitative Analysis

Customers arrive at a single-server help desk according to a Poisson process at an average rate of 8 per hour. Service times are exponentially distributed with a mean of 6 minutes. There is no limit on queue length, customers are served first-come first-served, and the system is in steady state (an M/M/1 queue).

(a) Determine the server utilization.

(b) Determine the average number of customers in the system, LL, and the average number waiting in the queue, LqL_q.

(c) Determine the average time a customer spends in the system, WW, and the average time waiting in the queue, WqW_q, in minutes.

(d) Determine the probability that an arriving customer finds at least 3 customers already in the system.

Worked solution
Answer:

(a) ρ = 0.80; (b) L = 4.0, LqL_q = 3.2; (c) W = 30 min, WqW_q = 24 min; (d) P(n ≥ 3) = 0.512

(a) Utilization. Convert the service time to a rate so that both rates are in the same units: μ=60/6=10\mu = 60/6 = 10 customers per hour. The utilization is the arrival rate divided by the service rate:

ρ=λμ=810=0.80\rho = \frac{\lambda}{\mu} = \frac{8}{10} = 0.80

Since ρ<1\rho < 1 the queue is stable and the steady-state formulas apply.

(b) Numbers in system and in queue. For an M/M/1 queue,

L=ρ1−ρ=0.800.20=4.0,Lq=ρ21−ρ=0.640.20=3.2L = \frac{\rho}{1 - \rho} = \frac{0.80}{0.20} = 4.0, \qquad L_q = \frac{\rho^2}{1 - \rho} = \frac{0.64}{0.20} = 3.2

The difference L−Lq=0.80=ρL - L_q = 0.80 = \rho is the average number in service, as it should be for a single server.

(c) Times in system and in queue. Little's law, L=λWL = \lambda W, converts the counts to times, or equivalently

W=1μ−λ=110−8=0.50 h=30 min,Wq=λμ(μ−λ)=810(2)=0.40 h=24 minW = \frac{1}{\mu - \lambda} = \frac{1}{10 - 8} = 0.50\,\text{h} = 30\,\text{min}, \qquad W_q = \frac{\lambda}{\mu(\mu - \lambda)} = \frac{8}{10(2)} = 0.40\,\text{h} = 24\,\text{min}

The two differ by the mean service time, 30−24=630-24 = 6 min, as expected. A customer spends four times longer waiting than being served, a typical consequence of running a single server at 80% utilization.

(d) Probability of at least 3 in the system. The steady-state probability of nn customers is Pn=(1−ρ)ρnP_n = (1 - \rho)\rho^n. Summing the geometric series from n=3n = 3 upward:

P(n≥3)=∑n=3∞(1−ρ)ρn=ρ3=0.803=0.512P(n \ge 3) = \sum_{n=3}^{\infty} (1-\rho)\rho^n = \rho^3 = 0.80^3 = 0.512

By the PASTA property (Poisson arrivals see time averages), this is also the probability an arriving customer finds at least 3 others present.

ρ = 0.80; L = 4.0, LqL_q = 3.2; W = 30 min, WqW_q = 24 min; P(n ≥ 3) = 0.512

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