Training Industrial Engineering Placement Test Practice — Industrial Engineering
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Placement Test Practice — Industrial Engineering

25 min Industrial Engineering

Placement Test Practice — Industrial Engineering

These problems cover linear programming, quality control, inventory, and queuing theory.

Practice Test — 25 Questions

1. Max $Z = 4x + 3y$ s.t. $x + y \le 10$, $2x + y \le 16$, $x,y \ge 0$. Find the optimal solution.
2. How many corner points does the feasible region in #1 have?
3. EOQ: $D = 8{,}000$, $S = \$60$, $H = \$3$. Find $Q^*$.
4. For #3, find the total annual cost (ordering + holding).
5. Number of orders per year for #3.
6. ROP with $d = 30$/day, $L = 6$ days, SS $= 40$ units.
7. $\bar{x}$-chart: $\mu = 75$, $\sigma = 3$, $n = 9$. Find UCL and LCL.
8. $C_p$: USL $= 82$, LSL $= 68$, $\sigma = 2$. Find $C_p$.
9. $C_{pk}$ if $\bar{x} = 77$ in #8.
10. M/M/1: $\lambda = 12$/hr, $\mu = 16$/hr. Find $\rho$.
11. Average number in system for #10.
12. Average wait time in system for #10.
13. Verify Little's Law for #10.
14. 80% learning curve: $T_1 = 60$ hrs. Find $T_4$.
15. Find cumulative time for units 1–4 with the 80% learning curve (approximate sum).
16. Normal time: observed = 8 min at 95% rating. Find NT.
17. Standard time: NT $= 7.6$ min, allowance $= 12\%$. Find ST.
18. Line balancing: tasks are 3, 5, 2, 4, 6 min. Cycle time = 6 min. How many stations needed minimum?
19. Efficiency of the line in #18 with 4 stations.
20. Min $C = 5x + 4y$ s.t. $x + y \ge 6$, $x \ge 2$, $y \ge 1$. Find optimal.
21. If ordering cost $S$ triples, how does EOQ change?
22. Probability of 0 customers in M/M/1: $P_0 = ?$ for $\lambda = 12$, $\mu = 16$.
23. A process has 3% defect rate. In a sample of 200, expected defects?
24. $\bar{R}$-chart: $\bar{R} = 6$, $D_4 = 2.004$ ($n=5$). Find UCL for range.
25. Which is more important for $C_{pk}$: reducing variability or centering the process?
Show Answer Key

1. Corners: $(0,0),(8,0),(6,4),(0,10)$. $Z(6,4)=36$, $Z(8,0)=32$, $Z(0,10)=30$. Max $Z=36$ at $(6,4)$.

2. 4 corner points

3. $Q^* = \sqrt{2(8000)(60)/3} = \sqrt{320{,}000} = 566$ units

4. $TC = (8000/566)(60) + (566/2)(3) = 848 + 849 = \$1{,}697$

5. $N = 8000/566 = 14.1$ orders/year

6. $ROP = 30(6) + 40 = 220$ units

7. $UCL = 75 + 3(3/3) = 78$; $LCL = 72$

8. $C_p = 14/12 = 1.17$

9. $C_{pu} = (82-77)/6 = 0.833$; $C_{pl} = (77-68)/6 = 1.50$; $C_{pk} = 0.833$

10. $\rho = 12/16 = 0.75$

11. $L = 0.75/0.25 = 3$

12. $W = 1/(16-12) = 0.25$ hrs $= 15$ min

13. $L = 12 \times 0.25 = 3$ ✓

14. $T_4 = 60 \times 0.8^2 = 38.4$ hrs (doubling from 1→2→4)

15. $T_1=60, T_2=48, T_3\approx43.5, T_4=38.4$; sum $\approx 189.9$ hrs

16. $NT = 8 \times 0.95 = 7.6$ min

17. $ST = 7.6 \times 1.12 = 8.51$ min

18. $\sum t = 20$; min stations $= \lceil 20/6 \rceil = 4$

19. $\text{Eff} = 20/(4 \times 6) = 83.3\%$

20. Corners: $(2,4),(5,1),(2,1)$ — check feasibility: $(2,4)$: $C=26$; $(5,1)$: $C=29$; $(2,4)$ is optimal, $C=26$.

21. $Q^* \propto \sqrt{S}$; tripling $S$ → EOQ $\times \sqrt{3} \approx 1.73$

22. $P_0 = 1 - 0.75 = 0.25 = 25\%$

23. $200 \times 0.03 = 6$ defects

24. $UCL = 2.004 \times 6 = 12.02$

25. Both matter, but centering ($C_{pk}$) addresses the more immediate form of waste. A well-centered process with moderate variation can still be capable.

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