Training Systems Engineering Placement Test Practice — Systems Engineering
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Placement Test Practice — Systems Engineering

25 min Systems Engineering

Placement Test Practice — Systems Engineering

These problems cover reliability, feedback systems, decision analysis, and Markov chains.

Practice Test — 25 Questions

1. Four components in series, each $R = 0.95$. Find system reliability.
2. Two components in parallel, $R_1 = 0.9$, $R_2 = 0.8$. Find system reliability.
3. $\lambda = 0.002$/hr. Find MTBF and $R(200)$.
4. Availability: MTBF = 1{,}000 hrs, MTTR = 10 hrs. Find availability.
5. $G(s) = 15/(s+3)$, unity feedback. Find $T(s)$ and the closed-loop pole.
6. Steady-state error for step input with $G(s) = 20/(s+4)$, unity feedback.
7. Two blocks in series: $G_1 = 4$, $G_2 = 1/(s+2)$. Find the combined transfer function.
8. Decision matrix: Options A(7,8,6) and B(9,5,8), weights (0.3, 0.4, 0.3). Which is better?
9. EMV: 40% chance of \$200K, 60% chance of -\$50K. Find EMV.
10. A Markov chain: $P(A→A) = 0.7$, $P(A→B) = 0.3$, $P(B→A) = 0.5$, $P(B→B) = 0.5$. Steady state?
11. For the Markov chain in #10, starting in A, what is $P(B \text{ after 1 step})$?
12. A 2-out-of-3 system with $R = 0.9$ each component. Find system reliability.
13. If MTBF of a component is 10{,}000 hrs, find $\lambda$.
14. A standby system: primary $R=0.95$, backup $R=0.95$. Perfect switching. Find $R_{\text{sys}}$.
15. $G(s) = K/(s+a)$, unity feedback, DC gain $= 0.8$. If $a=5$, find $K$.
16. Block diagram: $G$ in forward path, $H=0.5$ feedback. $G=10$. Find closed-loop gain.
17. Normalize weights: importance ratings 8, 6, 4, 2. Find weights summing to 1.
18. Risk priority number: probability × severity × detection. P=3, S=8, D=5. Find RPN.
19. A system has 50 identical components in series, each $R=0.999$. Find system $R$.
20. Mean time to failure for exponential model with $R(100) = 0.9$. Find $\lambda$.
21. Cost-benefit: project A costs \$300K with \$900K benefit. Project B costs \$150K with \$400K benefit. Which has better BCR?
22. A simple Markov chain: component fails with P=0.05 each day, repairs with P=0.3 each day. Long-run availability?
23. Sensitivity: in the decision matrix of #8, how much must weight on criterion 1 increase for B to win?
24. Little's Law: $L = 5$ customers, $\lambda = 10$/hr. Find average time in system.
25. A system engineer must choose between 3 architectures. The Pugh matrix gives net scores +3, -1, +1. Which is recommended?
Show Answer Key

1. $R = 0.95^4 = 0.8145$

2. $R = 1 - (0.1)(0.2) = 0.98$

3. MTBF $= 500$ hrs; $R(200) = e^{-0.4} = 0.670$

4. $A = 1000/1010 = 0.9901 = 99.01\%$

5. $T(s) = 15/(s+18)$; pole at $s = -18$

6. $e_{ss} = 1/(1+20/4) = 1/6 = 0.167$

7. $G = 4/(s+2)$

8. $S_A = 2.1+3.2+1.8 = 7.1$; $S_B = 2.7+2.0+2.4 = 7.1$. Tied!

9. $EMV = 0.4(200K) + 0.6(-50K) = 80K - 30K = \$50K$

10. $0.3\pi_A = 0.5\pi_B$; $\pi_A = 5/8 = 0.625$, $\pi_B = 3/8 = 0.375$

11. $P(B|\text{start A}) = 0.3$

12. $R = 3(0.9)^2(0.1) + (0.9)^3 = 0.243 + 0.729 = 0.972$

13. $\lambda = 10^{-4}$/hr

14. $R = 0.95 + (0.05)(0.95) = 0.9975$

15. DC gain $= K/(K+a) = 0.8$; $K = 0.8K + 4$; $0.2K = 4$; $K = 20$

16. $T = 10/(1+10 \times 0.5) = 10/6 = 1.667$

17. Sum $=20$; weights: $0.4, 0.3, 0.2, 0.1$

18. $RPN = 3 \times 8 \times 5 = 120$

19. $R = 0.999^{50} = e^{-0.05} = 0.951$

20. $e^{-100\lambda} = 0.9$; $\lambda = -\ln(0.9)/100 = 0.00105$/hr

21. $BCR_A = 900/300 = 3.0$; $BCR_B = 400/150 = 2.67$. Project A is better.

22. $0.05\pi_W = 0.3\pi_F$; $\pi_W = 6\pi_F$; $7\pi_F = 1$; $\pi_W = 6/7 = 85.7\%$

23. Currently tied. If $w_1$ increases slightly, B (score 9 on criterion 1) wins.

24. $W = L/\lambda = 5/10 = 0.5$ hrs $= 30$ min

25. Architecture 1 with net score +3 is the strongest recommendation.

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