📚 BUSM7011 — Final Exam Review

Business Operations and Logistics · 6 Question Types · Theo cách thầy giảng

📜 Final Exam — Quarter 2 2026

School of Business / UEH · BUSM7011

📅 1. Initial Information

  • Exam date: Sunday, 31st May 2026
  • Exam mode: Offline in classroom
  • Exam room: Room L2.2 and L2.4 — Campus 05 Truong Quoc Dung

⏰ Exam Schedule

  • 08:45 – 09:00: Check-in time
  • 09:00 – 09:10: Instruction provided
  • 09:10: Exam Question file appears on Blackboard (vUWS site) — download .docx
  • 09:10 – 11:10: Exam time (120 mins)
  • 11:10 – 11:30: Submission time (transfer images to Word file — 20 mins)
  • 11:30 SHARP: Submission link closes — late submission NOT accepted
📌 Students download the question file (.docx) → work on laptop → submit ONE single Word file with all answers, tables, figures, and images of handwritten solutions. Online apps (Google Docs, etc.) are NOT accepted.

🪪 Check-in Requirements

  • Be present for check-in and seating arrangement on time
  • Must bring valid identification: Student ID or Citizen ID or Passport
  • Students arriving 15 minutes after start time are NOT allowed to enter

💻 Device Requirements

  • Bring and use your own laptop
  • Built-in calculator / touchpad / touchscreen use are NOT allowed
  • Ensure your laptop can access vUWS site
  • Fully charge your laptop and bring the charging cable
  • Keep your screen upright (no privacy screen protector)
  • Ensure your device functions properly (no software errors, power or internet issues)
  • No support will be provided for forgotten or malfunctioning laptops
  • Smart phones and smart watches are STRICTLY PROHIBITED — set them to silent mode during exam

📖 Open-Book Exam Rules

  • ✓ ALLOWED: Printed/offline lecture notes, textbooks, vUWS resources, handwritten notes, Word documents created as part of your own study
  • ✗ NOT ALLOWED: Internet search, AI tools (ChatGPT, Grammarly, Google Translate, paraphrasing tools), online collaboration (Google Docs, Google Sheets, etc.)
  • Each student must work individually and ensure integrity of their materials
  • Type answers into Word document (the answer sheet)

✍️ For Calculation Questions

  • You can type answers, or write on provided paper (if needed)
  • Handwritten answers must be uploaded/inserted in high-resolution .jpeg or .png into the same answer file
  • You can only use your mobile phone from 11:10 onwards to transfer pictures (20 mins until 11:30)
  • The picture of each question must clearly show your full name and student ID
  • You are NOT allowed to write theoretical answers in hand-writing — must be typed

📤 Submission Instructions

  • Submit one single file containing all answers, tables, figures, typed responses, and images of handwritten solutions
  • Submit before 11:30 SHARP
  • Late submissions are NOT accepted under any circumstances

⚠️ Misconduct & Penalties

  • Any form of cheating during the exam will be recorded and reported to WSU for further disciplinary action
  • Students are fully responsible for the integrity of their materials. Any use of AI-generated content or plagiarized material will be subject to academic misconduct procedures
  • Collaborating with others or relying on technology/Generative AI tools may result in sanctions under the Student Misconduct Rule
  • Reminded by proctors:
    • 1 time → 25% marks deduction
    • 2 times → 50% marks deduction
    • 3 times → 100% marks deduction
🍀 GOOD LUCK IN YOUR EXAM 🍀
💡 Personal checklist trước khi đi thi:
  • ✓ Mang laptop đã sạc đầy + charging cable
  • ✓ Mang máy tính (calculator vật lý — không dùng built-in laptop)
  • ✓ Mang Student ID / CCCD / Passport
  • ✓ Mang giấy & bút để giải tay các câu tính toán
  • ✓ Tải sẵn file ôn này (.html) vào laptop trước khi tới phòng thi
  • ✓ Có sẵn 1 file Word trống mở ra để gõ đáp án
  • ✓ Đến phòng thi từ 8:30 (trước check-in 15 phút)
  • ✓ Tắt chuông điện thoại, smart watch để chế độ silent

📌 Quy tắc thi BUSM7011

Tóm tắt nhanh tiếng Việt

⏰ Trước & trong giờ thi

  • NHỚ MANG MÁY TÍNH (vật lý, KHÔNG dùng built-in laptop) — không có là không tính được EOQ, NPV, Z-score
  • Open book: Được mở file/tài liệu/notes/Word có sẵn, KHÔNG được dùng internet, AI tools, Google Docs
  • 09:10 – 11:10 → 120 phút (2 tiếng) làm bài thi trên laptop (gõ vào file Word)
  • 11:10 – 11:30 → 20 phút cuối để chụp ảnh bài giải tay + chèn vào file Word + upload nộp bài (mới được bật điện thoại)
  • 11:30 SHARP — đóng link nộp, trễ KHÔNG được nhận
  • 5 câu hỏi đề ra → chỉ cần làm 4 câu (chọn câu tự tin nhất, bỏ 1 câu khó)

✍️ Cách trình bày bài

  • Dùng tone & key-word của BOL (Business Operations and Logistics) — không dùng từ ngữ ngành khác (marketing, finance...)
  • Ví dụ key-words BOL: bottleneck, throughput, capacity, cycle time, statistical control, EOQ, reorder point, critical path, expected monetary value...
  • Trình bày từng step rõ ràng, viết công thức trước rồi mới thay số
  • Mỗi câu nên có: (1) Đề tóm tắt + Theory → (2) Công thức → (3) Tính step-by-step → (4) Kết luận + Managerial reading
  • Vẽ chart/diagram khi cần (Control chart, Decision tree, Network diagram, EOQ curve)

🎯 Chiến thuật chọn câu

  • Ưu tiên câu chắc chắn nhất trước (thường là EOQ, Decision Tree, PERT vì có công thức rõ ràng)
  • Quản lý thời gian: 2h / 4 câu = ~30 phút/câu (để 10 phút cuối check lại)
  • Nếu kẹt câu nào → bỏ qua, làm câu khác trước
  • Phần Managerial Reading rất quan trọng — không chỉ tính số mà phải đọc kết quả theo hướng quản lý

📊 6 Dạng bài đã ôn

  • Q1. Statistical Process Control (X̄-chart, R-chart) + Process Capability (Cp, Cpk)
  • Q2. Decision Tree với EMV (build small vs large facility)
  • Q3. NPV Capacity Expansion (cafeteria mở rộng)
  • Q4. Bottleneck & Capacity (Standard vs Deluxe tutoring)
  • Q5. Inventory Costs — TAC, EOQ, ROP (warehouse storage)
  • Q6. PERT Project Probability (Critical Path + Z-score)

→ Click vào từng dạng bên trái để xem chi tiết. Mỗi dạng có đầy đủ: Đề bài (EN/VN) · Theory · Formulas · Step-by-step solution · Chart/Diagram · Managerial reading · Ảnh thầy giảng (slides + bài giải tay).

Q1. Statistical Process Control + Process Capability

Quality Management · Nipro SCT
📝Example 1 — Problem Statement

Q1.a: Nipro is concerned about the production of SCT products used by hospitals for the suction process. The diameter of the SCT products is critical for reliable assessments. Data from 5 productions appear in the table. Sample size n = 4. Is the process in statistical control? Recommend suggestions for improvement.

Q1.b: The nominal value is 0.503 ± 0.005. Is the process capable of four-sigma level performance with σ = 0.0015?

Data Table

SampleOb 1Ob 2Ob 3Ob 4$\bar{x}$R
10.50010.50220.50090.50970.50320.0096
20.50210.50410.50240.50620.50370.0041
30.50180.50260.50350.50430.50310.0025
40.50580.50340.50240.50150.50330.0043
50.50610.50560.50340.50470.50500.0027
Average$\bar{\bar{X}} = 0.5036$$\bar{R} = 0.0046$
Q1 data table with x-bar and R
📸 Bảng dữ liệu thầy chuẩn bị: X̄ = 0.5036 và R̄ = 0.0046
📐Theory & Formulas
SPC Theory
A process is in statistical control when all sample means and ranges fall within control limits. Two charts must be checked: $\bar{x}$-chart (process average) and R-chart (process variability). Both must be IN control to conclude stability.
$\bar{x}$-chart (Process Average): $$UCL_{\bar{x}} = \bar{\bar{X}} + A_2 \cdot \bar{R} \quad ; \quad LCL_{\bar{x}} = \bar{\bar{X}} - A_2 \cdot \bar{R}$$ R-chart (Variability): $$UCL_R = D_4 \cdot \bar{R} \quad ; \quad LCL_R = D_3 \cdot \bar{R}$$ Process Capability: $$C_p = \frac{USL - LSL}{6\sigma}$$ $$C_{pk} = \min\left[\frac{USL - \bar{X}}{3\sigma},\ \frac{\bar{X} - LSL}{3\sigma}\right]$$

Control Chart Constants (Krajewski)

n$A_2$$D_3$$D_4$
21.88003.267
31.02302.574
40.72902.282
50.57702.114
60.48302.004
Three Sigma Limits Table
📸 Slide thầy: Table 4.1 — Factors for Calculating Three Sigma Limits (n=2 đến n=10)
R-chart formula slide
📸 Slide: R-Chart formula UCL_R = D₄R̄, LCL_R = D₃R̄
X-bar chart formula slide
📸 Slide: X̄-Chart formula UCL = X̄̄ + A₂R̄

Capability Interpretation

$C_p$ / $C_{pk}$Sigma LevelDefects (ppm)
1.003σ2,700
1.334σ (target)63
1.675σ0.57
2.006σ0.002
Sigma levels slide
📸 Slide thầy: Process Capability — diễn giải Cp/Cpk với mức 1σ → 6σ (1/3, 2/3, 4/3, 5/3, 2 = 1-6 sigma)
Cp formula
📸 Slide: Cp Formula
Cpk formula
📸 Slide: Cpk Formula
🧮Step-by-Step Solution (Q1.a)
STEP 1Calculate $\bar{\bar{X}}$ and $\bar{R}$ from data
$$\bar{\bar{X}} = \frac{0.5032 + 0.5037 + 0.5031 + 0.5033 + 0.5050}{5} = 0.5036$$
$$\bar{R} = \frac{0.0096 + 0.0041 + 0.0025 + 0.0043 + 0.0027}{5} = 0.0046$$
STEP 2Look up constants for n = 4
$A_2 = 0.729 \quad;\quad D_3 = 0 \quad;\quad D_4 = 2.282$
STEP 3Calculate $\bar{x}$-chart control limits
$$UCL_{\bar{x}} = \bar{\bar{X}} + A_2 \cdot \bar{R} = 0.5036 + 0.729 \times 0.0046 = \mathbf{0.5070}$$
$$LCL_{\bar{x}} = \bar{\bar{X}} - A_2 \cdot \bar{R} = 0.5036 - 0.729 \times 0.0046 = \mathbf{0.5002}$$
Check all $\bar{x}$ values: 0.5032, 0.5037, 0.5031, 0.5033, 0.5050 — all within $[0.5002, 0.5070]$ ✓
→ Process output IS in statistical control
STEP 4Calculate R-chart control limits
$$UCL_R = D_4 \cdot \bar{R} = 2.282 \times 0.0046 = \mathbf{0.0105}$$
$$LCL_R = D_3 \cdot \bar{R} = 0 \times 0.0046 = \mathbf{0}$$
Check all R values: 0.0096, 0.0041, 0.0025, 0.0043, 0.0027 — all within $[0, 0.0105]$ ✓
→ Process variability IS in statistical control
✅ Conclusion (Q1.a): Both $\bar{x}$-chart and R-chart are IN control → Process is in statistical control.
⚠️ But Sample 1 has $R = 0.0096$ (close to $UCL_R = 0.0105$) and Sample 5 has $\bar{x} = 0.5050$ (close to $UCL_{\bar{x}} = 0.5070$) → need recommendation.
📊Control Charts

📸 Ảnh tham khảo (giải tay & slide thầy giảng)

Handwritten solution
📸 Bài giải tay đầy đủ — UCL_x = 0.5042, LCL_x = 0.5012, UCL_R = 0.0044 với charts trực quan
Example 1 data
📸 Slide: Example 1 (1/5) — West Allis Industries data
R-chart calculation
📸 Slide: Example 1 (2/5) — R-Chart limits calculation
R-chart graph
📸 Slide: Example 1 (3/5) — Figure 4.9 R-Chart (in control)
X-bar calculation
📸 Slide: Example 1 (4/5) — X̄-Chart limits calculation
X-bar graph
📸 Slide: Example 1 (5/5) — Figure 4.10 X̄-Chart (Sample 5 out)
🧮Step-by-Step Solution (Q1.b — Capability)
STEP 1Identify spec limits from nominal ± tolerance
Nominal $= 0.503$, tolerance $= \pm 0.005$
$LSL = 0.503 - 0.005 = \mathbf{0.498}$
$USL = 0.503 + 0.005 = \mathbf{0.508}$
STEP 2Calculate $C_p$ (variability)
$$C_p = \frac{USL - LSL}{6\sigma} = \frac{0.508 - 0.498}{6 \times 0.0015} = \frac{0.010}{0.009} = \mathbf{1.11}$$
STEP 3Calculate $C_{pk}$ (centering)
$$C_{pk} = \min\left[\frac{USL - \bar{X}}{3\sigma},\ \frac{\bar{X} - LSL}{3\sigma}\right]$$
$$= \min\left[\frac{0.508 - 0.5036}{3 \times 0.0015},\ \frac{0.5036 - 0.498}{3 \times 0.0015}\right]$$
$$= \min\left[\frac{0.0044}{0.0045},\ \frac{0.0056}{0.0045}\right] = \min[\mathbf{0.97},\ 1.25] = \mathbf{0.97}$$
STEP 4Compare with 4-sigma target ($C_{pk} \geq 1.33$)
$C_{pk} = 0.97 < 1.33$ → NOT capable of 4-sigma level
Process output is at approximately 3-sigma level ($C_{pk} \approx 1.00$)
✅ Conclusion (Q1.b): Process is NOT capable of 4-sigma level performance. $C_p = 1.11$ (variability borderline), $C_{pk} = 0.97$ (output at 3σ due to off-centering toward LSL).
💡Managerial Reading & Recommendations

1. Process is "barely" in control — high risk of going out

R for Sample 1 = 0.0096 reaches 91% of $UCL_R$. $\bar{x}$ for Sample 5 = 0.5050 reaches 71% of the $UCL_{\bar{x}}$ band. Any small drift will cause out-of-control signal. Process đang "vừa đủ" trong control — không có safety margin, dễ bị out bất cứ lúc nào.

2. Process is OFF-CENTER toward USL

$\bar{\bar{X}} = 0.5036$ but target/nominal $= 0.503$ → mean shifted $+0.0006$ above nominal. $C_{pk}$ (USL side) $= 0.97$ vs $C_{pk}$ (LSL side) $= 1.25$ → much closer to USL. Mean lệch về phía USL → khả năng vượt giới hạn trên cao hơn.

3. Action: RECENTER first, then reduce variability

Short-term: Recalibrate machine to bring $\bar{X}$ back to 0.503 → $C_{pk}$ improves to ~1.24 (still < 1.33 but better).
Long-term: Reduce $\sigma$ from 0.0015 to $\leq 0.00125$ → $C_p$ improves to 1.33 (4σ).
Investigate Sample 1: Why is R = 0.0096 ~2x larger than other samples? Possible machine variation, operator inconsistency. Trước tiên chỉnh lại máy cho mean về 0.503; sau đó mới đầu tư giảm độ biến động.

4. Six Sigma Implementation

To reach 4σ (defect rate 63 ppm): need both $C_p \geq 1.33$ AND $C_{pk} \geq 1.33$. Current state: $C_p = 1.11$, $C_{pk} = 0.97$ — fails both criteria. Apply DMAIC (Define-Measure-Analyze-Improve-Control) to systematically reduce variation. Áp dụng DMAIC để giảm defect rate xuống 63 ppm cho mức 4-sigma.
📝Example 2 — X̄ / R Chart
5 mẫu (mỗi mẫu n = 4 quan sát) cho A₂ = 0.729, D₃ = 0, D₄ = 2.282. Hãy lập x̄-chart và R-chart, kết luận quy trình có in-control không.
SampleX̄R
10.50180.0018
20.50270.0021
30.50260.0017
40.50280.0026
50.50450.0022
AvgX̄̄ = 0.5027R̄ = 0.00208 ≈ 0.0021
X̄-chart limits: $$UCL_{\bar{x}} = 0.5027 + 0.729 \times 0.0021 = 0.5042$$ $$LCL_{\bar{x}} = 0.5027 - 0.729 \times 0.0021 = 0.5012$$
R-chart limits (dùng R̄ = 0.00208 chính xác để khớp phép tính tay): $$UCL_R = 2.282 \times 0.00208 = 0.0047 \quad ; \quad LCL_R = 0 \times 0.00208 = 0$$
✅ Sample 5 (X̄ = 0.5045) > UCLx̄ = 0.5042 → OUT of control trên X̄-chart.
Tất cả R nằm trong [0; 0.0047] → R-chart in control.
Kết luận: Variability ổn, nhưng process mean đã drift lên ở mẫu 5 — cần điều tra nguyên nhân đặc biệt (machine drift, đổi ca, đổi nguyên liệu...).
✓ Cross-check với bài viết tay: UCLx̄ = 0.5042 · LCLx̄ = 0.5012 · UCLR = 0.0047 · sample 5 vượt UCL → khớp 100%. R̄ ghi tay "0.021" là viết tắt — đọc đúng là 0.0021 (suy ngược từ UCL).
📘Concept Reference — Cp vs Cpk
Phân biệt khái niệm quan trọng trước
Đừng nhầm specification limits với control limits:
  • USL / LSL (Upper/Lower Specification Limit): giới hạn do khách hàng hoặc thiết kế quy định — "sản phẩm phải nằm trong khoảng này thì mới đạt chuẩn". Ví dụ: chai nước phải có thể tích 500 ± 5 ml → LSL = 495, USL = 505.
  • UCL / LCL (Control Limit): giới hạn thống kê tính từ chính dữ liệu quy trình (dùng A₂, D₃, D₄) — để biết quy trình có ổn định không.
⚠️ Một quy trình có thể "in control" (ổn định) nhưng vẫn không "capable" (không đạt spec khách hàng).

Công thức 1 — Cp: Năng lực tiềm năng

$$C_p = \frac{USL - LSL}{6\sigma}$$
Ý nghĩa: so sánh độ rộng cho phép của spec (tử số) với độ rộng tự nhiên của quy trình ($6\sigma$ — vì ±3σ chiếm 99.73% nếu phân phối chuẩn).
Cách đọc:
  • $C_p = 1.0$ → quy trình vừa khít spec (rủi ro cao, một chút biến động là vượt)
  • $C_p \geq 1.33$ → chấp nhận được (chuẩn công nghiệp tối thiểu)
  • $C_p \geq 2.0$ → đạt mức Six Sigma
Điểm yếu: $C_p$ chỉ đo độ rộng, không quan tâm quy trình có đi trúng tâm spec hay không. Một quy trình có thể có $C_p = 2.0$ nhưng trung bình lệch hẳn sang một bên → vẫn sản xuất ra hàng lỗi.

Công thức 2 — Cpk: Năng lực thực tế (có tính lệch tâm)

$$C_{pk} = \min\left[\frac{USL - \bar{X}}{3\sigma},\ \frac{\bar{X} - LSL}{3\sigma}\right]$$
Ý nghĩa: đo khoảng cách từ trung bình quy trình ($\bar{X}$) đến giới hạn spec gần nhất, chia cho $3\sigma$. Lấy min để bắt cái cạnh "nguy hiểm" hơn — cạnh mà quy trình đang gần lọt ra ngoài.
Cách đọc:
  • $C_{pk} = C_p$ → quy trình trúng tâm hoàn toàn ($\bar{X}$ = giữa USL và LSL)
  • $C_{pk} < C_p$ → quy trình bị lệch tâm
  • $C_{pk} < 1.0$ → đang sản xuất ra hàng lỗi (defects)

Ví dụ minh họa

Đường kính trục yêu cầu: 10 ± 0.3 mm → LSL = 9.7, USL = 10.3. Đo được $\bar{X} = 10.05$, $\sigma = 0.1$.
Cp: $$C_p = \frac{10.3 - 9.7}{6 \times 0.1} = \frac{0.6}{0.6} = 1.0$$ → Độ rộng quy trình vừa đúng độ rộng spec, không có "đệm an toàn".
Cpk: $$C_{pk} = \min\left[\frac{10.3 - 10.05}{0.3},\ \frac{10.05 - 9.7}{0.3}\right] = \min(0.833,\ 1.167) = 0.833$$ → Vì $\bar{X}$ lệch về phía USL, nên cạnh USL là cạnh nguy hiểm. $C_{pk} = 0.83 < 1$ → quy trình không capable, đang tạo ra phế phẩm dù "có vẻ" ổn định.

Tóm gọn để nhớ

Chỉ sốTrả lời câu hỏi
CpQuy trình có đủ "hẹp" để lọt vào spec không? (giả định trúng tâm)
CpkQuy trình thực tế có đang tạo ra hàng đạt chuẩn không? (tính cả lệch tâm)

Quick English summary

  • USL / LSL = customer or design spec — "the product must fall inside this range."
  • UCL / LCL = statistical limits from process data — used to tell whether the process is stable.
  • A process can be in control (stable) but still not capable (failing customer spec).
  • Cp = (USL − LSL) / 6σ — width fit, assuming the process is centered. Benchmarks: 1.0 just-fits, ≥ 1.33 acceptable, ≥ 2.0 six-sigma.
  • Cpk = min((USL − X̄)/3σ, (X̄ − LSL)/3σ) — picks the closer spec edge, accounts for off-centering.
  • Read together: Cpk = Cp → centered · Cpk < Cp → drifting · Cpk < 1.0 → producing defects.
  • Example: spec 10 ± 0.3 mm, X̄ = 10.05, σ = 0.1 → Cp = 1.0, Cpk = 0.83 → stable but not capable.
Hệ số Cpk Đánh giá PPM (defects/million) Ghi chú thực tế
< 1.0Không capable> 2,700Đang tạo phế phẩm, can thiệp ngay
1.0 – 1.33Marginal64 – 2,700Tạm chấp nhận, rủi ro cao nếu trôi
1.33 – 1.67Capable (chuẩn công nghiệp)0.6 – 64Target phổ biến — đa số nhà máy đặt ≥ 1.33
1.67 – 2.0Very capable< 0.6Hàng không, y tế, bán dẫn
≥ 2.0Six Sigma level3.4 (1.5σ shift)Motorola / GE chuẩn Six Sigma
"Cpk = 0.85 < 1.0, so the process is not capable. It is producing defective products outside the specification limits. Immediate action is required to either re-center the process mean or reduce variation." "Cpk = 1.15 > 1.0, so the process is producing within specification, but it does not meet the industry standard of 1.33. Since Cpk (1.15) is much lower than Cp (1.40), the process mean is off-center. The first priority should be to re-center the process before reducing variation."
Biến thể nếu Cpk ≈ Cp:
"Cpk = 1.15 ≈ Cp = 1.20, so the process is well-centered but the variation is too wide. The recommendation is to reduce process variation (σ) through better equipment, training, or standardization."
"Cpk = 1.50 > 1.33, so the process is capable and meets the industry standard. The defect rate is very low. The company should focus on maintaining current performance through regular monitoring and control charts." "Cpk = 1.80 is well above 1.33, so the process is highly capable with a strong safety buffer. The process is stable and reliable. No major improvement is needed; the focus should be on sustaining quality and preventing process drift." "Cpk = 2.10 ≥ 2.0, so the process has reached Six Sigma quality level. The defect rate is extremely low (around 3.4 ppm). The process is world-class, and resources can be shifted to improving other weaker processes." "Cp = 1.80 shows the process is narrow enough to fit the spec, but Cpk = 0.95 shows it is producing defects because the mean is far off-center. Re-centering the process is a quick and low-cost fix that can immediately improve Cpk close to Cp." "Both Cp = 0.90 and Cpk = 0.80 are below 1.0. The process has too much variation and is also slightly off-center. The company needs to reduce σ first (through process improvement or new equipment), then re-center the mean." $C_{pk} = 1.0$ nghĩa là cạnh gần nhất của spec cách $\bar{X}$ đúng 3σ. Trong phân phối chuẩn, ngoài 3σ có 0.135% dữ liệu mỗi bên → khoảng 2,700 sản phẩm lỗi / triệu. Nghe có vẻ ít, nhưng:
  • Nếu nhà máy sản xuất 1 triệu sản phẩm/tháng → 2,700 phế phẩm/tháng
  • Nếu $\sigma$ "trôi" nhẹ do máy mòn, nhiệt độ thay đổi → $C_{pk}$ tụt xuống dưới 1 rất nhanh
→ Vì vậy ngành công nghiệp đặt $C_{pk} \geq 1.33$ làm chuẩn tối thiểu (cho "buffer" để hấp thụ biến động ngắn hạn). $C_{pk} = 1.33 \Longleftrightarrow$ spec cách $\bar{X}$ đúng 4σ (vì 1.33 = 4/3). Khi đó:
  • Tỷ lệ lỗi ~ 64 ppm — chấp nhận được cho hầu hết ngành (cơ khí, điện tử cơ bản, FMCG)
  • Cho phép quy trình "trôi" ±1σ vẫn còn an toàn (vì 4σ − 1σ = 3σ = $C_{pk}$ 1.0)
 $C_{pk} = 2.0 \Longleftrightarrow$ spec cách $\bar{X}$ đúng 6σ. Triết lý Motorola giả định trung bình quy trình sẽ trôi 1.5σ theo thời gian dài (long-term drift) → còn lại 4.5σ → tỷ lệ lỗi 3.4 ppm (chính là con số nổi tiếng của Six Sigma). → Ngành y tế, hàng không, bán dẫn thường yêu cầu $C_{pk} \geq 1.67$ hoặc 2.0 vì hậu quả của lỗi cực kỳ nghiêm trọng.
  • Cpk < 1.0 → not capable, > 2,700 ppm defects — fix immediately.
  • 1.0 – 1.33 → marginal; meets spec but no buffer for drift.
  • 1.33 – 1.67 → capable; common industry target (~64 ppm).
  • 1.67 – 2.0 → very capable; comfortable safety margin.
  • ≥ 2.0 → Six Sigma level; assumes 1.5σ long-term drift → 3.4 ppm defects.
  • Why 1.33 is the "magic" number: it equals 4σ to the nearest spec, leaving 1σ of room for drift before slipping back to Cpk = 1.0.
  • Healthcare, aerospace, semiconductors typically demand Cpk ≥ 1.67 – 2.0 because failure costs are catastrophic.

Q2. Decision Tree (EMV)

Decision Analysis · Retailer Facility
📝Problem Statement
  • A retailer will build a Small or Large facility at a new location.
  • Demand: Low $P = 0.45$ | High $P = 0.55$
  • Small + High demand: Not expand \$280K, Expand \$370K (choose max)
  • Small + Low demand: Payoff = \$220K
  • Large + Low demand: Do nothing \$30K, OR Advertise → Modest ($P=0.4$) \$40K / Sizable ($P=0.6$) \$320K
  • Large + High demand: Payoff = \$500K

Question: Which facility to build? Use EMV criterion.

📐Theory & Formulas
Decision Tree Theory
  • When the decision-maker knows the probabilities of states of nature → EMV (Expected Monetary Value) is the standard criterion.
  • A decision tree contains 3 elements:
    • □ Decision nodes — choose the option with MAX value
    • ○ Chance nodes — compute the weighted average (Σ Pᵢ × Payoffᵢ)
    • ▷ Terminal payoffs — the dollar outcome at the end of each path
  • Always solve by rollback: process the tree right → left, from terminal payoffs back to the root decision.
At chance node ○: $$EMV = \sum_{i} P_i \times \text{Payoff}_i$$ At decision node □: $$\text{Value} = \max(\text{branches})$$ Decision rule: $$\text{Choose alternative with highest EMV at root node}$$
🌳Decision Tree Diagram
□ Decision · ○ Chance · ✓ Optimal 1 EMV = $368.6K Small facility A $302.5K Low [0.45] $220K High [0.55] 2 $370K Not expand $280K Expand ✓ $370K Large facility ✓ B $368.6K ✓ Low [0.45] 3 $208K Do nothing $30K Advertise ✓ C $208K Modest [0.4] $40K Sizable [0.6] $320K High [0.55] $500K
🧮Step-by-Step Solution (Rollback right → left)
STEP 1Node 2 — Small + High demand (Decision)
$$\max(\text{Not expand }\$280,\ \text{Expand }\$370) = \mathbf{\$370K} \to \text{choose Expand}$$
STEP 2Chance Node A — EMV(Small)
$$EMV(\text{Small}) = 0.45 \times \$220 + 0.55 \times \$370$$ $$= \$99 + \$203.5 = \mathbf{\$302.5K}$$
STEP 3Chance Node C — EMV(Advertise)
$$EMV(\text{Advertise}) = 0.4 \times \$40 + 0.6 \times \$320$$ $$= \$16 + \$192 = \mathbf{\$208K}$$
STEP 4Node 3 — Large + Low demand (Decision)
$$\max(\text{Do nothing }\$30,\ \text{Advertise }\$208) = \mathbf{\$208K} \to \text{choose Advertise}$$
STEP 5Chance Node B — EMV(Large)
$$EMV(\text{Large}) = 0.45 \times \$208 + 0.55 \times \$500$$ $$= \$93.6 + \$275 = \mathbf{\$368.6K}$$
STEP 6Node 1 — Root decision
$$\max(\text{Small }\$302.5,\ \text{Large }\$368.6) = \mathbf{\$368.6K} \to \text{choose Large Facility}$$
✅ Decision: BUILD LARGE FACILITY with EMV = \$368.6K.
Contingency strategy: If Low demand occurs → run Advertise campaign (\$208K) instead of doing nothing (\$30K).

📸 Ảnh tham khảo

Decision tree solution on whiteboard
📸 Bài giải Decision Tree trên bảng — Choose Large Facility ($544K cho phiên bản với P=0.4/0.6, với P=0.45/0.55 thì $368.6K)
📌 Lưu ý: Ảnh bảng dùng phiên bản với P(low)=0.4, P(high)=0.6 → EMV Large = $544K. Đề thi gốc với P(low)=0.45, P(high)=0.55 → EMV Large = $368.6K. Cách giải giống hệt nhau.
💡Managerial Reading

1. EMV gap is narrow (\$66K) — check investment cost

Small EMV \$302.5K vs Large EMV \$368.6K → only 22% difference. If Large facility has significantly higher capital expenditure (not given), the Net EMV could favor Small. Always compute Net EMV = Gross EMV − Investment before final decision. Khoảng cách EMV không lớn — nếu Large facility tốn nhiều vốn đầu tư hơn, đáp án có thể đảo chiều.

2. Large has higher risk (wider payoff range)

Small range: \$220–\$370K (spread \$150K)
Large range: \$40–\$500K (spread \$460K, 3x wider)
→ Risk-averse companies may prefer Small despite lower EMV. Large có variance gấp 3 lần Small. Công ty bảo thủ có thể chọn Small dù EMV thấp hơn.

3. Advertise budget is critical for contingency

When Large + Low demand occurs (probability 0.45), Advertise EMV \$208K vs Do nothing \$30K → loss of \$178K expected value if marketing budget is unavailable. Must allocate advertising reserves upfront. Phải có sẵn ngân sách marketing dự phòng để kích hoạt khi cần.

4. Sensitivity: Decision is robust

Even if $P(\text{High})$ drops to 0.30: $EMV(\text{Small}) = \$265$, $EMV(\text{Large}) = \$296$ → still Large.
Only when $P(\text{High}) < 0.10$ does Small become better.
→ Robust decision against forecasting error. Quyết định "robust" với sai số dự báo xác suất — vẫn chọn Large trừ khi P(High) cực thấp.

5. EVPI = \$5.4K (max to pay for market research)

If demand were known perfectly: $EVwPI = 0.45 \times \$220 + 0.55 \times \$500 = \$374K$. $EVPI = \$374 - \$368.6 = \mathbf{\$5.4K}$. Don't spend more than \$5,400 on market research. Sẵn sàng chi tối đa $5.400 cho nghiên cứu thị trường để biết demand chắc chắn.

Q5. NPV Capacity Expansion

Financial Analysis · Cafeteria
📝Problem Statement
A school cafeteria currently serves 100,000 meals (operating at 100% capacity). Kitchen can handle 150,000 diners/year. Forecast: 110,000 next year, then +20,000 each year.

Note: Max kitchen capacity 150,000 — by year 2 kitchen still meets demand, but limits exceed thereafter.

Alternative: Expand kitchen + cafeteria to 200,000 meals/year. Investment \$250,000 at end of year 0. Meal price \$10, variable cost \$8, pretax margin \$2 (20%). Discount rate 10%.

Question: (a) Pretax cash flows for next 5 years compared to "do nothing"? (b) NPV of project?

📐Theory & Formulas
NPV Theory
NPV discounts future incremental cash flows to year-0 value. Incremental approach: compare "with project" vs "without project" (base case = do nothing). Accept project if $NPV > 0$.
Incremental Cash Flow: $$CF_t = (\text{Meals}_{\text{with}} - \text{Meals}_{\text{without}}) \times \text{Margin per meal}$$ Net Present Value: $$NPV = -I_0 + \sum_{t=1}^{n} \frac{CF_t}{(1+r)^t}$$ Decision rule: $$NPV > 0 \to \text{Invest} \quad;\quad NPV < 0 \to \text{Reject}$$
🧮Step-by-Step Solution
STEP 1Forecast demand for years 1–5
Y1 = 110,000 ; Y2 = 130,000 ; Y3 = 150,000 ; Y4 = 170,000 ; Y5 = 190,000
STEP 2"Do nothing" baseline
Without expansion, the cafeteria serves only 100,000 meals/year in every year (base case capacity).

Cash Flow Table

Year012345
Demand—110,000130,000150,000170,000190,000
Do nothing (base)—100,000100,000100,000100,000100,000
Δ Meals (incremental)—10,00030,00050,00070,00090,000
Investment−\$250,00000000
Incremental Revenue (\$10)—\$100,000\$300,000\$500,000\$700,000\$900,000
Variable Cost (\$8)—\$80,000\$240,000\$400,000\$560,000\$720,000
Pretax CF (\$2 margin)−\$250,000\$20,000\$60,000\$100,000\$140,000\$180,000
STEP 3Discount each CF to PV ($r = 10\%$)
$$PV_1 = \frac{20{,}000}{(1.10)^1} = \$18{,}182$$ $$PV_2 = \frac{60{,}000}{(1.10)^2} = \$49{,}587$$ $$PV_3 = \frac{100{,}000}{(1.10)^3} = \$75{,}131$$ $$PV_4 = \frac{140{,}000}{(1.10)^4} = \$95{,}622$$ $$PV_5 = \frac{180{,}000}{(1.10)^5} = \$111{,}766$$
STEP 4Sum and compute NPV
$$NPV = -250{,}000 + 18{,}182 + 49{,}587 + 75{,}131 + 95{,}622 + 111{,}766$$ $$= -250{,}000 + 350{,}288 = \mathbf{+\$100{,}288}$$
✅ NPV = +\$100,288 > 0 → ACCEPT the expansion project.
Payback occurs between Year 3 and Year 4 (cumulative undiscounted CF reaches \$250K around year 4).
📊Cash Flow Visualization

📸 Ảnh tham khảo

NPV solution on whiteboard
📸 Bài giải NPV trên bảng — Cash flows và NPV = $100,287
NPV Excel from professor
📸 File Excel thầy gửi — chi tiết Demand, Investment, Revenue, Variable cost, Profit, NPV theo từng năm
💡Managerial Reading

1. NPV > 0 → financially justified

\$100K positive NPV means project creates value beyond the 10% required return. Accept under standard capital budgeting rules. NPV dương → dự án tạo giá trị, chấp nhận đầu tư.

2. NPV is sensitive to discount rate and margin

If $r = 12\%$ instead of 10%: NPV drops to ~\$76K (still positive).
If margin drops from \$2 to \$1.5/meal: NPV becomes ~\$12K (marginal).
→ Run sensitivity analysis on key drivers before final commitment. NPV nhạy cảm với discount rate và biên LN. Cần phân tích sensitivity trước khi quyết định cuối.

3. Non-financial considerations

  • Reputation: Turning away students = lost loyalty, complaints
  • Strategic: Reserve capacity for unexpected growth (e.g. new programs)
  • Risk: Demand may grow slower than forecast → over-expanded
Yếu tố phi tài chính: uy tín, chiến lược dài hạn, rủi ro tăng trưởng chậm.

4. Alternatives beyond expansion

  • Extended hours (open lunch shift longer) — low capex
  • Peak/off-peak pricing — manages demand
  • Outsource to external caterer for peak days
→ Compare NPV of these alternatives before committing to \$250K capex. Cân nhắc các phương án thay thế (mở rộng giờ, định giá theo giờ, thuê ngoài) trước khi đầu tư.

Q6. Bottleneck & Process Capacity

Constraint Management · Tutoring Center
📝Problem Statement
A school's academic support center offers Standard and Deluxe tutoring. Both go through A1, A2 first. Standard then proceeds A3 → A4 → A8. Deluxe goes A5 → A6 → A7 → A8. Numbers in parentheses = minutes per student.
  • A1(5), A2(6), A3(12), A4(15), A5(5), A6(20), A7(12), A8 (assume no wait at A1, A2, A8)

Questions:

  • (a) Bottleneck for Standard? For Deluxe?
  • (b) Capacity (students served per 300 minutes) for each type?
  • (c) If mix is 60% Standard / 40% Deluxe, average capacity?
  • (d) Where do waiting lines form for each type?
📐Theory & Formulas
Bottleneck Theory
The bottleneck is the step with the longest cycle time on a path. It determines the throughput of the entire system.
  • Count buffer/already-in-process students that exit without waiting
  • For remaining time, divide by bottleneck cycle time
  • Round DOWN (can't serve partial student)
Capacity per path: $$\text{Capacity} = (\text{Pre-loaded students}) + \frac{T - \text{offset}}{\text{Bottleneck cycle time}}$$ Average capacity (mix): $$\text{Avg} = \%_{\text{Std}} \times \text{Cap}_{\text{Std}} + \%_{\text{Dlx}} \times \text{Cap}_{\text{Dlx}}$$

(no rounding for average)

Waiting rule: $$\text{Wait forms when } t_{\text{next}} > t_{\text{previous}}$$
🔀Process Flow Diagram
A1 (5) A2 (6) Split Std/Dlx Standard A3 (12) A4 (15) ⚠ BOTTLENECK Deluxe A5 (5) A6 (20) ⚠ BOTTLENECK A7 (12) A8
🧮Step-by-Step Solution
PART aBottleneck identification (longest cycle time on path)
Standard: $\max(5, 6, 12, \mathbf{15}) \to A4 = 15$ min is bottleneck
Deluxe: $\max(5, 6, 5, \mathbf{20}, 12) \to A6 = 20$ min is bottleneck
PART bCapacity in 300 minutes
Standard: $$\text{Cap}_{\text{Std}} = 1 + 1 + \frac{300 - 8 - 10}{15} = 1 + 1 + 18.8 = 20.8 \to \mathbf{20\ \text{students}}$$ (round down)
• "$1+1$" = 2 students already at downstream stations (A8 region) exit without waiting
• "$-8$" = wait time for A8 already in progress
• "$-10$" = remaining time on bottleneck A4 for current student
• "$15$" = bottleneck cycle time A4
Deluxe: $$\text{Cap}_{\text{Dlx}} = 1 + 1 + 1 + \frac{300 - 8 - 10 - 7 - 3}{20} = 3 + 13.6 = 16.6 \to \mathbf{16\ \text{students}}$$ (round down)
• "$1+1+1$" = 3 students downstream of bottleneck A6 (at A7, A8)
• "$-7$" and "$-3$" = adjustments for A7 timing
PART cMixed average capacity (60% Std / 40% Dlx)
$$\text{Avg} = 0.6 \times 20 + 0.4 \times 16 = 12 + 6.4 = \mathbf{18.4\ \text{students / 300 min}}$$ (Do NOT round average — it's a weighted figure)
PART dWaiting lines (where downstream cycle > upstream)
Standard path: A1(5) → A2(6) → A3(12) → A4(15) → A8
• A2 (6) → A3 (12): wait at A3 (12 > 6) ✓
• A3 (12) → A4 (15): wait at A4 (15 > 12) ✓
→ Waiting forms at A3 and A4
Deluxe path: A1(5) → A2(6) → A5(5) → A6(20) → A7(12) → A8
• A5 (5) → A6 (20): wait at A6 (20 > 5) ✓
→ Waiting forms primarily at A6
✅ Summary:
(a) Bottlenecks: A4 (Standard), A6 (Deluxe)
(b) Capacity: 20 Standard, 16 Deluxe per 300 min
(c) Mix average: 18.4 students / 300 min
(d) Wait at A3, A4 (Standard); A6 (Deluxe)
📌 Rule from professor: "Process sau có số lớn hơn process trước → constraint (waiting line)." Whenever next step's cycle time > current → backlog forms.

📸 Ảnh tham khảo

Bottleneck solution on whiteboard
📸 Bài giải Bottleneck trên bảng — đầy đủ flow diagram, công thức Standard (20 students) và Deluxe (16 students), Avg cap = 18.4
💡Managerial Reading

1. Theory of Constraints (TOC) — focus on bottleneck

Apply Goldratt's 5 steps: Identify → Exploit → Subordinate → Elevate → Repeat. Reducing 1 min at A6 (Deluxe bottleneck): capacity goes from $60/20 = 3$ → $60/19 \approx 3.16$ students/h (+5%). Reducing 1 min at A1: NO IMPROVEMENT — A1 is not bottleneck. Đầu tư cải thiện chỉ có ý nghĩa ở bottleneck. Tăng tốc các bước khác = lãng phí.

2. WIP (Work-In-Process) accumulates BEFORE bottleneck

Students queue at A4 (Standard) and A6 (Deluxe). Visible signs: long waiting lines, stressed staff at these steps. Cosmetic fix: add chairs/space at queues. Real fix: reduce cycle time at bottleneck. Hàng chờ tích tụ ngay trước bottleneck. Đó là dấu hiệu trực quan dễ thấy.

3. Mix-shift sensitivity

If demand shifts toward Deluxe (e.g. 40% → 60%): $\text{Avg} = 0.4 \times 20 + 0.6 \times 16 = 17.6$ (drops from 18.4). Deluxe pulls down average because its bottleneck is slower. Pricing strategy: charge Deluxe a premium to compensate slower throughput. Nếu tỷ lệ Deluxe tăng → capacity giảm. Cần định giá Deluxe cao hơn để bù.

4. Investment priority: A6 first

A6 (20 min) is 33% slower than A4 (15 min). Investing in A6 has larger ripple effect on overall throughput because Deluxe is the slower path. Always start improvements at the system-level slowest step. Đầu tư cải tiến A6 trước (chậm nhất toàn hệ thống), sau đó mới đến A4.

Q7. Inventory Costs (TAC, EOQ, ROP)

Inventory Management · Warehouse
📝Problem Statement
Compute: Total Annual Inventory Cost (TAC), Economic Order Quantity (EOQ), Reorder Point (ROP).

Data Table

ItemValueUnitClassification
Number of usage bags20,000bags/year—
Number of kgs in a bag50kgs—
Telephone cost (call supplier)100,000VND/timeOrdering
Capacity of warehouse1,000tons—
Insect control2,500,000VND/monthHolding
Warehouse rental50,000,000VND/monthHolding
Labour cost (7 staff)7,000,000VND/person/monthHolding
Setup cost100,000VND/timeOrdering
Security system38,000,000VND/monthHolding
Damage2,500,000VND/monthHolding
Order quantity (Q)20,000kgs/time—
Unit price43,500VND/kg—
📐Theory & Formulas
Inventory Costs — 3 main types
According to professor's slides, inventory costs are classified into 3 categories:
  1. Holding Cost (Carrying Cost): Cost of keeping items in stock
  2. Order Cost: Cost of placing an order
  3. Setup Cost: Cost of modifying production line for a different item
Holding Costs 4 components
📸 Slide thầy: 1. Holding Costs gồm 4 thành phần — Capital Cost, Storage Space Cost, Inventory Service Cost, Inventory Risk Cost
Order and Setup Costs
📸 Slide thầy: 2. Order Cost (MIS, purchase order, inspection, payment) và 3. Setup Cost (modify production line)
Holding Cost — 4 components (detailed)
  • Capital Cost — opportunity cost of investing in inventory (Hurdle rate, WACC)
  • Storage Space Cost — rent, heat, light, insect control; public vs private warehousing
  • Inventory Service Cost — insurance and taxes (varies by goods value)
  • Inventory Risk Cost — obsolescence, damage, theft
EOQ Theory
EOQ minimizes total annual cost (holding + ordering). The optimal Q balances these two opposing costs. At EOQ, Holding = Ordering.
$S$ (holding cost per unit per year) is computed by dividing total annual storage cost by warehouse capacity (per professor's method).
Variables: $R$ = Annual demand (kg) = bags/year × kg/bag
$A$ = Ordering cost per order
$S$ = Holding cost per unit per year
$V$ = Value of one unit; $W$ = % carrying cost Total Annual Cost (2 equivalent forms): $$TAC = \frac{1}{2} \cdot Q \cdot V \cdot W + A \cdot \frac{R}{Q} \quad \text{(eq. 9.3)}$$ $$TAC = \frac{1}{2} \cdot Q \cdot S + A \cdot \frac{R}{Q} \quad \text{(eq. 9.4, with } S = VW\text{)}$$ Derivation — at EOQ, Holding = Ordering: $$\frac{1}{2} \cdot Q \cdot V \cdot W = A \cdot \frac{R}{Q}$$ $$Q^2 = \frac{2RA}{VW} \quad\to\quad Q = \sqrt{\frac{2RA}{VW}} = \sqrt{\frac{2RA}{S}}$$ Holding cost per unit: $$S = \frac{\text{Total annual storage cost}}{\text{Warehouse capacity (kg)}}$$ Reorder Point: $$ROP = \frac{R}{365} \times \text{Lead Time (LT)}$$ $$\text{Or with Safety Stock: } ROP = \frac{R}{365} \times (LT + \text{Safety-stock time})$$
Inventory Costs slide
📸 Slide thầy: Inventory Costs — TAC formula (eq. 9.3 và 9.4) với định nghĩa R, Q, A, V, W, S, t
Total Holding = Total Ordering
📸 Slide thầy: Total Holding Cost = Total Ordering Cost → derive Q = √(2RA/VW) = √(2RA/S)

📊 Figure 9-5: Fixed Order Quantity Model under Certainty

Figure 9-5 Fixed Order Quantity Model
📸 Slide thầy: Figure 9-5 — Sawtooth pattern of inventory (Level of inventory). Inventory giảm tuyến tính, khi chạm Reorder Point thì đặt hàng tiếp. ROP = R/365 × (LT + Safety-stock time)
Đọc Figure 9-5:
  • Trục đứng: Units (inventory level) — đạt đỉnh khi nhận hàng = Q (ví dụ 4,000)
  • Trục ngang: Time (weeks) — chu kỳ lặp lại
  • Đường zigzag: tồn kho giảm tuyến tính theo demand, chạm Reorder Point → đặt hàng → khi hàng về thì lại đầy
  • Reorder Point: mức tồn kho lúc cần đặt hàng (đường ngang gạch đứt). ROP đủ lớn để cover demand trong Lead Time + Safety Stock
  • Safety Stock: phần đệm ở đáy (ghi chú đỏ "SS") để chống stockout khi demand/LT biến động
🧮Step-by-Step Solution
STEP 1Annual demand $R$
$R = 20{,}000 \text{ bags} \times 50 \text{ kg} = \mathbf{1{,}000{,}000 \text{ kg/year}}$
Warehouse capacity $= 1{,}000$ tons $= 1{,}000{,}000$ kg
STEP 2Normalize each holding cost (annual ÷ capacity)
$$\text{Pest control} = \frac{2{,}500{,}000 \times 12}{1{,}000{,}000} = \mathbf{30} \text{ VND/kg/yr}$$ $$\text{Warehouse rental} = \frac{50{,}000{,}000 \times 12}{1{,}000{,}000} = \mathbf{600} \text{ VND/kg/yr}$$ $$\text{Labour} = \frac{7 \times 7{,}000{,}000 \times 12}{1{,}000{,}000} = \mathbf{588} \text{ VND/kg/yr}$$ $$\text{Security} = \frac{38{,}000{,}000 \times 12}{1{,}000{,}000} = \mathbf{456} \text{ VND/kg/yr}$$ $$\text{Damage} = \frac{2{,}500{,}000 \times 12}{1{,}000{,}000} = \mathbf{30} \text{ VND/kg/yr}$$
STEP 3Sum $S$ and $A$
$$S = 30 + 600 + 588 + 456 + 30 = \mathbf{1{,}704 \text{ VND/kg/year}}$$ $$A = \text{Telephone} + \text{Setup} = 100{,}000 + 100{,}000 = \mathbf{200{,}000 \text{ VND/order}}$$
STEP 4TAC at current $Q = 20{,}000$ kg
$$TAC = \frac{1}{2} \times 20{,}000 \times 1{,}704 + \frac{1{,}000{,}000}{20{,}000} \times 200{,}000$$ $$= 17{,}040{,}000 + 10{,}000{,}000 = \mathbf{27{,}040{,}000 \text{ VND/year}}$$
STEP 5EOQ
$$EOQ = \sqrt{\frac{2RA}{S}} = \sqrt{\frac{2 \times 1{,}000{,}000 \times 200{,}000}{1{,}704}}$$ $$= \sqrt{\frac{400{,}000{,}000{,}000}{1{,}704}} = \sqrt{234{,}742{,}663} = \mathbf{15{,}321 \text{ kg}}$$
Round to $Q_1 = 15{,}300$ or $Q_2 = 15{,}350$ → compare TAC
STEP 6Compare TAC at $Q_1$ and $Q_2$
$$TAC_1 = \frac{15{,}300 \times 1{,}704}{2} + \frac{200{,}000 \times 1{,}000{,}000}{15{,}300}$$ $$= 13{,}035{,}600 + 13{,}071{,}895 = \mathbf{26{,}107{,}495}$$
$$TAC_2 = \frac{15{,}350 \times 1{,}704}{2} + \frac{200{,}000 \times 1{,}000{,}000}{15{,}350}$$ $$= 13{,}078{,}200 + 13{,}029{,}316 = \mathbf{26{,}107{,}516}$$
→ Choose $Q = 15{,}300$ units ($TAC_1 < TAC_2$ by 21 VND)
STEP 7Reorder Point (assume $LT = 1$ day, $SS = 1$ day)
$$ROP = \frac{R}{365} \times (LT + SS) = \frac{1{,}000{,}000}{365} \times (1 + 1)$$ $$= 2{,}739.7 \times 2 = \mathbf{5{,}479 \text{ kg}}$$
✅ Results:
• TAC at current $Q = 20{,}000$ kg = 27,040,000 VND/year
• EOQ = 15,321 kg → round to $Q = 15{,}300$ kg
• TAC at EOQ = 26,107,495 VND/year → saving ~932,505 VND/year
• ROP = 5,479 kg
📊TAC Curve (Interactive)
15,300 kg
Holding
13,035,600
Ordering
13,071,895
TAC
26,107,495
💡Managerial Reading

1. Save ~932,505 VND/year by switching to EOQ

Current $Q = 20{,}000$ → TAC $= 27.04$M. EOQ $Q = 15{,}300$ → TAC $= 26.11$M. Saving = ~932K VND/year. Number of orders increases from 50/year ($R/20{,}000$) to 65/year ($R/15{,}300$). Áp EOQ tiết kiệm ~932 nghìn VND/năm. Số lần đặt tăng từ 50 lên 65 lần/năm.

2. At EOQ: Holding ≈ Ordering (quick verification)

At $Q = 15{,}300$: Holding $= 13.04$M, Ordering $= 13.07$M → nearly equal ✓
At $Q = 20{,}000$ (current): Holding $= 17.04$M, Ordering $= 10$M → Holding dominates → ordering too much per cycle. Tại EOQ, 2 chi phí bằng nhau. Hiện tại đang đặt quá nhiều mỗi lần → Holding cao.

3. EOQ is "robust" — flat bottom

TAC at 15,300 vs 15,350 differs by only 21 VND. Even $Q = 14{,}000$ or $17{,}000$ would yield TAC very close to optimum. → No need for exact rounding; pick a practical batch size that fits truck/pallet constraints. Đáy đường TAC khá phẳng — chọn Q tròn tiện vận chuyển vẫn gần optimum.

4. ROP = 5,479 kg → re-order when stock drops to this level

With $LT = 1$ day and $SS = 1$ day buffer: order when inventory hits 5,479 kg. SS protects against demand variability and lead-time variability. Higher service level → higher SS. Khi tồn kho còn 5.479 kg thì đặt hàng tiếp. SS bảo vệ chống biến động demand và LT.

5. Watch out: S calculation method

Professor's method divides total holding cost by warehouse capacity (1M kg). Alternative method (textbook) divides by $R$ (annual demand). Both give different S values. For this exam, follow professor's method. Cách tính S của thầy: chia cho sức chứa kho (1tr kg), không phải chia cho R. Phải làm theo cách của thầy khi thi.

Q8. PERT — Project Probability Analysis

Project Management · UEH Sustainable Energy
📝Problem Statement
UEH students plan a sustainable energy project with 7 activities (A-G). Each activity has Optimistic ($a$), Most Likely ($m$), Pessimistic ($b$) time estimates.

Questions: Probability of finishing in 90 / 100 / 110 weeks?

Activity Data

ActivityCodePredecessors$a$$m$$b$$ET$$\sigma^2$
DesignA—35404540.02.78
Build prototypeBA25364135.07.11
Evaluate equipmentCA10132815.09.00
Test prototypeDB891610.01.78
Write reportEC, D78159.01.78
Write methods reportFC, D5101510.02.78
Write final reportGE, F12154.05.44
📐Theory & Formulas
PERT Theory
PERT models project duration as a random variable. Each activity has 3 time estimates ($a, m, b$). Expected time uses Beta distribution. Critical Path (CP) = longest sequence; project finishes only when CP completes. Only CP variance matters for project probability.
⚠️ Note on Activity A: Original file had typo (b=35). Correct value is b=45 (verified: ET=40, σ²=2.78, which gives CP=99 and total variance=19.89 as in solution).
Expected Time: $$ET = \frac{a + 4m + b}{6}$$ Activity Variance: $$\sigma^2 = \left(\frac{b - a}{6}\right)^2$$ Project mean (sum along Critical Path): $$T_e = \sum_{\text{CP}} ET_i$$ Project variance: $$\sigma_p^2 = \sum_{\text{CP}} \sigma_i^2$$ Z-score: $$Z = \frac{D - T_e}{\sqrt{\sigma_p^2}}$$ Probability: $$P(\text{finish} \leq D) = \text{Standard Normal Table}(Z)$$
🔀Network Diagram
A 40 B 35 C 15 D 10 E 9 F 10 G 4 Critical Path A-B-D-F-G = 99 wks σ²p = 19.89
🧮Step-by-Step Solution
STEP 1List all paths and compute total $ET$
Paths through the network:
• A-B-D-E-G $= 40 + 35 + 10 + 9 + 4 = 98$
• A-B-D-F-G $= 40 + 35 + 10 + 10 + 4 = \mathbf{99}$ ← longest (CP)
• A-C-E-G $= 40 + 15 + 9 + 4 = 68$
• A-C-F-G $= 40 + 15 + 10 + 4 = 69$
→ Critical Path = A-B-D-F-G, $T_e = 99$ weeks
STEP 2Sum variances along CP only
$$\sigma_p^2 = \sigma_A^2 + \sigma_B^2 + \sigma_D^2 + \sigma_F^2 + \sigma_G^2$$ $$= 2.78 + 7.11 + 1.78 + 2.78 + 5.44 = \mathbf{19.89}$$ $$\sigma_p = \sqrt{19.89} = \mathbf{4.46}$$
📌 Professor's rule: Only compute variance for activities on Critical Path (saves time). All 5 CP activities (A, B, D, F, G) are included.
STEP 3Calculate $Z$ for $D = 90$ weeks
$$Z = \frac{D - T_e}{\sqrt{\sigma_p^2}} = \frac{90 - 99}{\sqrt{19.89}} = \frac{-9}{4.46} = \mathbf{-2.02}$$ $$P(Z \leq -2.02) = \mathbf{0.0217 = 2.17\%}$$
STEP 4Calculate $Z$ for $D = 100$ weeks
$$Z = \frac{100 - 99}{\sqrt{19.89}} = \frac{1}{4.46} = \mathbf{0.2242}$$ $$P(Z \leq 0.22) = \mathbf{0.5878 = 58.78\%}$$
STEP 5Calculate $Z$ for $D = 110$ weeks
$$Z = \frac{110 - 99}{\sqrt{19.89}} = \frac{11}{4.46} = \mathbf{2.4664}$$ $$P(Z \leq 2.47) = \mathbf{0.9931 = 99.31\%}$$
✅ Probabilities of finishing within deadline:
• $P(\text{finish} \leq 90 \text{ wks}) = \mathbf{2.17\%}$ — very unlikely
• $P(\text{finish} \leq 100 \text{ wks}) = \mathbf{58.78\%}$ — slightly better than coin flip
• $P(\text{finish} \leq 110 \text{ wks}) = \mathbf{99.31\%}$ — nearly certain
📊Probability Distribution

📸 Ảnh tham khảo

PERT network from exam
📸 Network diagram bài thi — ETₐ = 40 (xác nhận A có b=45, không phải 35), CP = A-B-D-F-G = 99
PERT solution on whiteboard
📸 Bài giải PERT trên bảng — ETₐ = (35+4×40+45)/6 = 40, σ²ₐ = ((45-35)/6)² = 2.78, ΣVar = 19.89, các Z-score đầy đủ
💡Managerial Reading

1. Realistic deadline: 100 weeks (50/50 chance)

Promising 90 weeks to a client = 97.8% chance of being LATE. Promising 110 weeks = safe (99.3% on-time). For management commitment, target ~105–108 weeks for ~85–90% confidence. Cam kết 90 tuần = 97,8% trễ. Nên hứa khoảng 105-108 tuần để có 85-90% confidence.

2. Focus management attention on CP activities

A, B, D, F, G are on CP. Any delay here delays the entire project. Activities C and E (off-CP) have slack — they can be delayed up to a limit without affecting completion. Chỉ hoạt động trên CP mới ảnh hưởng deadline. C và E có slack, có thể trì hoãn được.

3. Crashing strategy: shorten CP activities first

To reduce duration: invest in B ($\sigma^2 = 7.11$, highest) and G ($\sigma^2 = 5.44$). These are biggest risk drivers. Don't waste money speeding up C or E (off-CP). Muốn rút ngắn dự án → đầu tư vào B (variance cao nhất) và G. Không tăng tốc C, E vì off-CP.

4. Watch for "near-critical" paths

A-B-D-E-G $= 96.3$ weeks (only 2.7 weeks shorter than CP). Small delays on this path could make it the new CP. Monitor multiple paths in execution. Có đường gần critical (96.3 tuần). Nếu CP tăng tốc nhưng path khác trễ → path khác thành critical mới.

5. PERT assumptions to challenge

PERT assumes Normal distribution of project duration. With only 5 activities on CP (small sample), Central Limit Theorem is weak. Real distribution may be skewed. Treat probabilities as approximate, not exact. PERT giả định phân phối chuẩn. Với 5 hoạt động trên CP, xác suất chỉ là gần đúng.

Q3. Measures of Forecast Error

Forecasting · Krajewski Ch.9 · Figure 9.2
📝Example 1 — Figure 9.2 (n = 10) · Problem Statement
Given 10 past periods of actual demand Dt and the forecast Ft that was made for each period, compute the six standard measures of forecast accuracy:
  • (a) CFE — Cumulative Forecast Error (bias)
  • (b) Ē — Average forecast error (mean bias per period)
  • (c) MAD — Mean Absolute Deviation
  • (d) MSE — Mean Squared Error
  • (e) σ — Sample standard deviation of errors
  • (f) MAPE — Mean Absolute Percent Error
Then interpret what each measure says about the forecasting method.

Data Table

Period tActual DtForecast Ft
13941
23743
35545
44050
55951
66356
74161
85760
95662
105463
Totals501—
Average50.1—
📐Theory & Formulas
Forecast error
The forecast error for a single period is the difference between what actually happened and what was predicted:
Et = Dt − Ft
A positive Et means actual demand was higher than forecast (forecast under-shot); a negative Et means forecast over-shot. We then aggregate the Et into different summary measures — each tells a different story.
(a) Cumulative Forecast Error (bias) $$\text{CFE} = \sum_{t=1}^{n} E_t$$
Tổng cộng dồn các sai số. Đo bias (thiên hướng): CFE > 0 nghĩa là forecast dự báo thiếu kéo dài; CFE < 0 nghĩa là forecast dự báo dư kéo dài. Nếu forecast tốt → CFE dao động quanh 0.
(b) Average forecast error (mean bias) $$\bar{E} = \frac{\text{CFE}}{n}$$
CFE chia cho số kỳ → bias trung bình mỗi kỳ. Cùng ý nghĩa với CFE nhưng đã chuẩn hóa theo số kỳ, dễ so sánh giữa các phương pháp/khoảng thời gian khác nhau.
(c) Mean Absolute Deviation $$\text{MAD} = \frac{\sum_{t=1}^{n} |E_t|}{n}$$
Trung bình của giá trị tuyệt đối sai số. Đo độ lớn trung bình của sai số bất kể dấu. MAD nhỏ → forecast bám sát thực tế. Dễ hiểu, dùng phổ biến trong inventory để set safety stock.
(d) Mean Squared Error $$\text{MSE} = \frac{\sum_{t=1}^{n} E_t^{2}}{n}$$
Trung bình bình phương sai số. Phạt nặng sai số lớn (do bình phương) → dùng khi sai số lớn là rất tệ. Đơn vị là (đơn vị demand)², khó diễn giải trực tiếp.
(e) Sample standard deviation of errors $$\sigma = \sqrt{\frac{\sum_{t=1}^{n} (E_t - \bar{E})^{2}}{n-1}}$$
Độ lệch chuẩn của sai số quanh mean bias. Đo độ phân tán của sai số. Dùng để tính khoảng tin cậy của forecast và safety stock với mức service level cụ thể (z·σ).
(f) Mean Absolute Percent Error $$\text{MAPE} = \frac{\sum_{t=1}^{n} \frac{|E_t|}{D_t} \times 100}{n}\ (\%)$$
Sai số tuyệt đối tính theo % so với demand. Là đại lượng không đơn vị → dùng để so sánh độ chính xác giữa các sản phẩm/khu vực có quy mô khác nhau. MAPE < 10% = rất tốt, 10–20% tốt, > 50% kém.
📌 Bias vs variability: CFE and Ē measure bias (does the forecast lean high or low?). MAD, MSE, σ and MAPE measure variability (how big the typical error is, ignoring direction).
🧮Example 1 — Step-by-step Solution

Step 1 — Build the error table

FORMULA Et = Dt − Ft  ·  then compute |Et|, Et², and |Et|/Dt×100 for each row.
tDtFtEt|Et|Et²|Et|/Dt ×100
13941−2245.128%
23743−663616.216%
35545101010018.182%
44050−101010025.000%
55951886413.559%
66356774911.111%
74161−202040048.780%
85760−3395.263%
95662−663610.714%
105463−998116.667%
Σ501—−3181879170.621%
Avg50.1—−3.18.187.917.062%

Step 2 — (a) Cumulative Forecast Error

FORMULA CFE = Σ Et
CFE = (−2) + (−6) + 10 + (−10) + 8 + 7 + (−20) + (−3) + (−6) + (−9)
CFE = −31

Step 3 — (b) Mean bias Ē

FORMULA Ē = CFE / n
Ē = −31 / 10 = −3.1  units per period

Step 4 — (c) Mean Absolute Deviation

FORMULA MAD = Σ|Et| / n
MAD = 81 / 10 = 8.1  units

Step 5 — (d) Mean Squared Error

FORMULA MSE = Σ Et² / n
MSE = 879 / 10 = 87.9

Step 6 — (e) Sample standard deviation of errors

FORMULA σ = √[ Σ(Et − Ē)² / (n − 1) ]  ·  with Ē = −3.1
Σ(Et + 3.1)²
= (−2+3.1)² + (−6+3.1)² + (10+3.1)² + (−10+3.1)² + (8+3.1)²
  + (7+3.1)² + (−20+3.1)² + (−3+3.1)² + (−6+3.1)² + (−9+3.1)²
= 1.21 + 8.41 + 171.61 + 47.61 + 123.21 + 102.01 + 285.61 + 0.01 + 8.41 + 34.81
≈ 782.9
σ = √( 782.9 / 9 ) = √( 86.99 ) ≈ 9.327
Note: Krajewski uses the sample standard deviation (divide by n − 1 = 9), not n. Don't confuse this with MSE which divides by n.

Step 7 — (f) Mean Absolute Percent Error

FORMULA MAPE = [ Σ (|Et| / Dt) × 100 ] / n
Σ (|Et| / Dt) × 100 = 170.621%
MAPE = 170.621% / 10 = 17.062%
✅ Summary of error measures
(a) CFE = −31  ·  (b) Ē = −3.1  ·  (c) MAD = 8.1
(d) MSE = 87.9  ·  (e) σ ≈ 9.327  ·  (f) MAPE = 17.062%
Forecast errors per period (E_t = D_t − F_t) 0 +15 +7.5 −10 −20 −2 −6 +10 −10 +8 +7 −20 −3 −6 −9 1 2 3 4 5 6 7 8 9 10 Period t Ē = −3.1
CFE = −31 and Ē = −3.1 mean the forecast was on average 3.1 units too high per period. 7 of 10 periods are negative — this is not random, it's a systematic bias. Action: lower the baseline forecast or check if a recent demand drop hasn't been absorbed by the model yet. CFE âm và Ē = −3.1 ⇒ dự báo cao hơn thực tế một cách hệ thống. Cần hạ baseline hoặc cập nhật model cho phù hợp xu hướng giảm. On average, the forecast is off by about 8 units in either direction, against an average demand of 50.1 — that's a sizeable ~16% relative error. Use MAD to size safety stock: safety stock ≈ z × σ ≈ z × 1.25 × MAD (the 1.25 factor converts MAD → σ under the normal-distribution assumption). MAD = 8.1 → sai số trung bình ~8 đơn vị. Safety stock thường tính theo σ ≈ 1.25 × MAD. Period 7 alone contributes 400 to ΣE² (out of 879) — almost half of MSE comes from a single bad period. That's why σ (9.33) > MAD (8.1): squared-error measures are very sensitive to outliers. If period 7 reflects a one-off event (promotion, stock-out, holiday), exclude it before retuning the model. Period 7 đóng góp gần 1/2 MSE — outlier. Nếu là sự kiện đặc biệt thì loại trước khi tinh chỉnh mô hình. 17.06% lands in the "good" band of the Lewis benchmark (see table below). Acceptable for planning, but the visible bias and the period-7 outlier suggest there's room to improve before relying on this method for operational decisions. MAPE 17% ở mức "tốt" theo Lewis, nhưng vẫn nên cải thiện vì có bias rõ + outlier.
MAPEĐánh giáÝ nghĩa thực tế
< 10%Highly accurateForecast cực chính xác — duy trì, không cần đổi method
10 – 20%GoodChấp nhận được cho hầu hết quyết định operations (Ex.1 = 17.06% ở đây)
20 – 50%ReasonableDùng tạm được nhưng nên cải thiện (thêm seasonality, đổi α…)
> 50%InaccurateKhông dùng được — rebuild model từ đầu
📌 Lewis bands là rule-of-thumb cho cross-industry. Trong ngành demand ổn định (FMCG, utility) target có thể chặt hơn (< 5%). Ngành new-product launch hay innovation thì 20–30% đã coi là tốt.
"CFE = [X] ≈ 0 (Ē = [Y] ≈ 0), so the forecast is unbiased — over- and under-shoots roughly cancel out. Focus shifts to the variability measures (MAD, MAPE) to assess precision." "CFE = −[X] and Ē = −[Y] show the forecast systematically overestimates demand by about [Y] units per period. The model should be re-calibrated downward, especially if a recent demand decline has not been absorbed." "CFE = +[X] indicates the forecast systematically underestimates demand, leading to chronic stockouts. Adjust the baseline upward or check whether a recent demand uplift (promotion, new product) needs to be incorporated." "MAPE = [X]% < 10%, so the forecast is highly accurate per the Lewis (1982) benchmark. Maintain the current method and continue monitoring; resources can be redirected to weaker SKUs." "MAPE = [X]% sits in the "good" band (Lewis 1982). The forecast is acceptable for planning, but worth investigating whether bias or a single outlier inflates it before relying on it for operational decisions." "MAPE = [X]% is in the "reasonable" band — usable but should be improved. Try adding seasonality, switching to weighted moving average, or tuning α in exponential smoothing." "MAPE = [X]% > 50% means the forecast is inaccurate and should not be used for planning. Re-build the model: check data quality, add seasonality / trend components, and verify the underlying demand process has not structurally changed." "Period [t] alone contributes [X]% of ΣE², which is why σ ([X]) is noticeably larger than MAD ([Y]). If this period reflects a one-off event (promotion, stockout, holiday), exclude it before re-tuning; otherwise add a regressor or dummy variable to absorb the effect."
  • CFE / Ē → detect bias (forecast có lệch hệ thống không?)
  • MAD → dễ tính tay, dùng để size safety stock (≈ z × 1.25 × MAD)
  • MSE / σ → nhấn mạnh sai số lớn → dùng cho regression và tracking signal
  • MAPE → scale-free %, dùng để so sánh accuracy giữa các SKU / quy mô khác nhau
Mỗi chỉ số trả lời 1 câu hỏi khác — đề thi thường bắt tính hết để đối chiếu.
  • Asymmetric: phạt under-forecast nặng hơn over-forecast (vì mẫu D nhỏ → tỉ lệ cao). Nếu cần symmetric → dùng sMAPE hoặc MASE.
  • Undefined khi D = 0: nếu demand có kỳ bằng 0, MAPE tính không được — bỏ kỳ đó hoặc đổi chỉ số.
  • Phình to khi D nhỏ: kỳ có D = 1 và sai số = 1 → 100% MAPE, làm méo trung bình.
 $$TS = \frac{CFE}{MAD}$$ Theo Krajewski / Tilles: nếu $|TS| > 4$ với MA forecast (hoặc $> 6$ với exponential smoothing) → forecast out of control → re-tune ngay. TS dao động trong ±4 thì forecast bình thường. Vượt ngưỡng = bias rõ rệt, phải sửa. Đừng đánh giá forecast trên training data — luôn giữ lại 3–6 kỳ cuối làm hold-out, fit model với phần đầu, rồi tính MAPE trên hold-out. So sánh nhiều method bằng MAPE hold-out → chọn cái thấp nhất. Hold-out để tránh over-fit. Đề thi thường không bắt làm bước này, nhưng đáng nhắc trong "đọc thêm" để hiểu nghề. Với n = 10, 1 outlier (period 7) đã làm méo gần 50% MSE. Krajewski nhắc: "These measures become more reliable as the number of periods of data increases." Thực tế dùng rolling window 12–24 kỳ và đánh giá lại thường xuyên thay vì 1 lần đo cứng. n = 10 quá ít → 1 outlier xáo trộn. Dùng rolling window cập nhật liên tục.
  • CFE / Ē → bias detector · MAD → safety-stock sizing · MSE/σ → outlier-sensitive variability · MAPE → scale-free %, cross-SKU comparable.
  • Lewis (1982) MAPE bands: < 10% highly accurate · 10–20% good · 20–50% reasonable · > 50% inaccurate.
  • MAPE pitfalls: asymmetric, undefined at D = 0, explodes when D is small. Consider sMAPE / MASE for those cases.
  • Tracking Signal TS = CFE / MAD; |TS| > 4 (MA) or > 6 (ES) → re-tune the forecast.
  • Always evaluate on a held-out window (3–6 periods) to avoid over-fitting bias.
  • Small n is fragile: rolling re-evaluation beats one-shot scoring.
📘Example 2 — Krajewski textbook Ex.1 (n = 8)
Bộ slide còn 1 ví dụ khác trong sách Krajewski Ch.9 với n = 8 kỳ và bộ số liệu riêng (không show ở slide). Áp dụng cùng 6 công thức như Example 1 nhưng kết quả khác hoàn toàn.

Kết quả tính (đọc trực tiếp từ slide gốc)

✅ CFE = −15  ·  Ē = −1.875  ·  MAD = 24.4
MSE = 659.4  ·  σ ≈ 27.4 (chia n−1 = 7)  ·  MAPE = 10.2%
📎 Slide gốc trình bày từng bước thay số đã được dời xuống mục Reference Slides ở cuối topic (3 slide: CFE/Ē/MSE · σ/MAD/MAPE · Interpretation).

So sánh nhanh Example 1 (n=10) vs Example 2 (n=8)

  • Bias: cả hai đều âm → cả hai forecast đều có khuynh hướng dự báo dư
  • MAPE: Ex.2 = 10.2% (highly-accurate / good edge) — Ex.1 = 17.06% (good band)
  • σ ở Ex.2 lớn (27.4) vì có 1–2 kỳ sai số rất lớn → các phương pháp squared-error đặc biệt nhạy với outlier
  • σ ở Ex.2 (27.4) > MAD ở Ex.2 (24.4): khoảng cách ~1.13 lần — gần với hệ số ~1.25 lý thuyết (giả định normal)
Bài thi có thể yêu cầu áp dụng đúng công thức trên bộ data của đề — quan trọng là quy trình, không phải con số cụ thể.

Q4. Time-Series Forecasting (Moving Avg · Weighted MA · Exponential Smoothing)

Krajewski Ch.9 · Examples 9.3 & 9.4 — Medical Clinic Patient Arrivals
📖Textbook Slides — Problem Statements
Example 9.3 — 3-week moving average problem
Example 9.3 — Moving-average problem
Example 9.4 — exponential smoothing problem
Example 9.4 — Exponential smoothing problem
📝Problem Statement

Context: A medical clinic forecasts weekly patient arrivals. Past three weeks of demand:

Week ($t$)Patient Arrivals ($D_t$)
1400
2380
3411
4 (actual)415

Example 9.3 (Simple Moving Average, $n = 3$):

  • a. Compute the 3-week moving-average forecast for week 4.
  • b. If actual $D_4 = 415$, what is the forecast error for week 4?
  • c. What is the forecast for week 5?

Example 9.4 (Exponential Smoothing, $\alpha = 0.10$, initial forecast = 390):

  • a. At end of week 3 ($D_3 = 411$), compute the exponential-smoothing forecast for week 4.
  • b. Forecast error for week 4 if $D_4 = 415$.
  • c. Forecast for week 5.
📐Theory & Formulas
Simple Moving Average
The forecast for the next period is the unweighted average of the most recent $n$ actual demands. Treats every period in the window equally; reacts slowly to shifts but smooths noise. Larger $n$ → smoother, slower; smaller $n$ → more responsive, noisier.
$n$-period Moving Average forecast for period $t+1$: $$F_{t+1} = \dfrac{D_t + D_{t-1} + \cdots + D_{t-n+1}}{n}$$ Lấy trung bình cộng của $n$ kỳ gần nhất để dự báo kỳ kế tiếp — mọi kỳ trong cửa sổ có trọng số bằng nhau. Forecast error in period $t$: $$E_t = D_t - F_t$$ Sai số dự báo = thực tế − dự báo. Dương = under-forecast (dự báo thiếu), âm = over-forecast (dự báo dư).
Weighted Moving Average
Each historical demand carries its own weight $W_i$, with $\sum W_i = 1.0$. Larger weight on recent demand → faster reaction to changes than simple MA, but still bounded by chosen weights.
Weighted Moving Average: $$F_{t+1} = W_1 D_t + W_2 D_{t-1} + \cdots + W_n D_{t-n+1}, \qquad \sum_{i=1}^{n} W_i = 1$$ Mỗi kỳ trong cửa sổ có trọng số $W_i$ riêng (tổng = 1). Đặt $W$ lớn cho kỳ gần → forecast phản ứng nhanh hơn so với MA đơn giản.
Weighted Moving Averages textbook slide
Textbook slide — Weighted Moving Averages
Exponential Smoothing
A sophisticated weighted MA that gives implicitly more weight to recent demands. Requires only 3 inputs: last period's forecast $F_t$, this period's actual $D_t$, and the smoothing parameter $\alpha \in [0, 1]$.
• Larger $\alpha$ → more responsive (recent emphasis).
• Smaller $\alpha$ → smoother (analogous to larger $n$ in MA).
Exponential Smoothing: $$F_{t+1} = \alpha D_t + (1 - \alpha) F_t$$ Forecast mới = α × thực tế kỳ này + (1−α) × forecast kỳ trước. Chỉ cần 3 dữ liệu: $D_t$, $F_t$, $\alpha$. $\alpha$ lớn → nhạy; $\alpha$ nhỏ → mượt.
Exponential Smoothing slide 1 of 2
Slide 1 of 2 — three inputs & equation
Exponential Smoothing slide 2 of 2
Slide 2 of 2 — role of α
Seasonal Patterns
Multiplicative method: seasonal factors are multiplied by an estimate of average demand to yield the seasonal forecast.
Additive method: seasonal forecasts are produced by adding (or subtracting) a seasonal constant to/from the average-demand estimate.
Multiplicative (season $s$): $$F_s = \bar{D} \times SF_s$$ Nhân forecast trung bình với hệ số mùa $SF_s$. Dùng khi biên độ dao động mùa tỉ lệ với mức demand (mùa cao thì swing lớn, mùa thấp thì swing nhỏ). Additive (season $s$): $$F_s = \bar{D} + SC_s$$ Cộng/trừ hằng số mùa $SC_s$ vào forecast trung bình. Dùng khi biên độ mùa cố định bất kể demand cao hay thấp.
Seasonal Patterns textbook slide
Textbook slide — Seasonal Patterns: multiplicative vs additive
🧮Solution — Example 9.3 (3-Week Moving Average)
PART aForecast for week 4 ($n = 3$, use $D_1, D_2, D_3$)
Formula:  $F_4 = \dfrac{D_3 + D_2 + D_1}{3}$
Substitute:  $F_4 = \dfrac{411 + 380 + 400}{3} = \dfrac{1191}{3}$
Compute:  $F_4 = \mathbf{397.0}$ patients
PART bForecast error for week 4 ($D_4 = 415$)
Formula:  $E_4 = D_4 - F_4$
Substitute:  $E_4 = 415 - 397$
Compute:  $E_4 = \mathbf{+18}$ patients  (under-forecast)
PART cForecast for week 5 — drop $D_1$, include $D_4$
Formula:  $F_5 = \dfrac{D_4 + D_3 + D_2}{3}$
Substitute:  $F_5 = \dfrac{415 + 411 + 380}{3} = \dfrac{1206}{3}$
Compute:  $F_5 = \mathbf{402.0}$ patients
✅ Conclusion (9.3): $F_4 = 397$, $E_4 = +18$, $F_5 = 402$. The 3-week MA rolls forward by dropping the oldest value and adding the latest actual.
Example 9.3 textbook solution slide
Textbook solution — Example 9.3 (2 of 2)
🧮Solution — Example 9.4 (Exponential Smoothing, $\alpha = 0.10$)
PART aForecast for week 4 — given $D_3 = 411$, $F_3 = 390$ (initial)
Formula:  $F_4 = \alpha D_3 + (1 - \alpha) F_3$
Substitute:  $F_4 = 0.10(411) + 0.90(390)$
Compute:  $F_4 = 41.1 + 351.0 = \mathbf{392.1} \approx 392$ patients
PART bForecast error for week 4 ($D_4 = 415$)
Formula:  $E_4 = D_4 - F_4$
Substitute:  $E_4 = 415 - 392$
Compute:  $E_4 = \mathbf{+23}$ patients
PART cForecast for week 5
Formula:  $F_5 = \alpha D_4 + (1 - \alpha) F_4$
Substitute:  $F_5 = 0.10(415) + 0.90(392.1)$
Compute:  $F_5 = 41.5 + 352.89 = \mathbf{394.4} \approx 394$ patients
✅ Conclusion (9.4): $F_4 = 392$, $E_4 = +23$, $F_5 = 394$. With $\alpha = 0.10$ the forecast moves slowly toward the actual demand — only 10% of each new observation is absorbed.
Example 9.4 solution slide 2 of 3
Textbook solution — Example 9.4 (2 of 3)
Example 9.4 solution slide 3 of 3
Textbook solution — Example 9.4 (3 of 3)
📊Side-by-Side Comparison
Method$F_4$$E_4$ (D=415)$F_5$Reaction speed
3-week Moving Average397.0+18402.0Medium — equal weight, drops oldest
Exponential Smoothing ($\alpha = 0.10$)392.1+23394.4Slow — only 10% of each new observation
💡Managerial Reading & Recommendations

1. Pick the method that matches the demand pattern

If demand is stable / no trend → simple MA or exp. smoothing with small $\alpha$. If demand shifts quickly → weighted MA emphasizing recent periods, or exp. smoothing with larger $\alpha$ (e.g. 0.3–0.5). Demand ổn định → dùng $\alpha$ nhỏ hoặc MA dài. Demand thay đổi nhanh → $\alpha$ lớn hoặc weighted MA tập trung vào kỳ gần.

2. The choice of $\alpha$ (and $n$) is a bias–variance trade-off

Larger $\alpha$ / smaller $n$ → forecasts track real changes faster but amplify noise. Smaller $\alpha$ / larger $n$ → smoother but slow to react to genuine shifts. Tune $\alpha$ to minimize MAD/MSE on a holdout. $\alpha$ lớn = nhạy nhưng nhiễu. $\alpha$ nhỏ = mượt nhưng phản ứng chậm. Chọn $\alpha$ tối thiểu hóa MAD/MSE.

3. Initial forecast assumption matters early on

Exp. smoothing's $F_1$ is a seed (here = 390). With $\alpha = 0.10$, the influence of the seed decays slowly — bad seed contaminates forecasts for many periods. Use the first actual demand, or an MA of warm-up periods, to seed. Forecast khởi tạo ảnh hưởng lâu nếu $\alpha$ nhỏ. Khởi tạo bằng $D_1$ hoặc MA của vài tuần đầu để giảm sai số ban đầu.

4. Add seasonality if the pattern repeats

If patient arrivals show weekly/quarterly/seasonal cycles (flu season, exam week, holidays), apply the multiplicative seasonal method: forecast the average level with MA / exp. smoothing, then multiply by the seasonal factor $SF_s$. Use additive when seasonal swing is roughly constant in magnitude regardless of level. Nếu có chu kỳ mùa (mùa cúm, kỳ thi…) → nhân thêm $SF_s$ (multiplicative) hoặc cộng $SC_s$ (additive) vào forecast trung bình.

5. Always track forecast error

Pair any method with a control on bias (CFE) and dispersion (MAD/MSE). $E_4$ being positive in both methods ($+18$ and $+23$) suggests under-forecasting; if persistent → method is biased and needs $\alpha$↑ or a trend term (Holt's model). Cả 2 phương pháp đều under-forecast → nếu lặp lại nhiều kỳ thì có bias, cần tăng $\alpha$ hoặc thêm thành phần trend (Holt).