Decision Modeling & Optimization
Data, Models & Decisions (Bertsimas & Freund) Ch 5, 7, 8, 9 + Deciding How to Decide + DICE. This is the fastest objective to learn, but it's worth about 10–20 points, and almost every sitting has one genuinely hard formulate-and-solve question. The skill that gets tested: read a messy business story → recognize the model type → write decision variables, objective and constraints correctly → interpret the output in business terms.
DMD overview: recognize the problem typeData, Models & Decisions — all chapters (XP key points)
Decision guide: which model is this?
Linear (Ch 7)
- Production mix, blending, sourcing, staffing, transport
- Objective & constraints = Σ (constant × variable)
- Optimum at a corner of the polygon feasible region
- Shadow prices valid over an allowable range
Nonlinear (Ch 8)
- Portfolio: min variance (wiwjσij) for a target return
- Pricing/production: (P − c)·D(P) with demand depending on price
- Location: minimize weighted √ distances
- Optimum can be interior; there may be local optima
Discrete (Ch 9)
- Binary = on/off: select projects, open sites
- Integer: whole units (planes, trucks, staff)
- Mixed: binary + integer + continuous (fixed-charge)
- Solved by branch-and-bound; much harder than LP
Simulation (Ch 5)
- Several random variables combined (price × demand × yield)
- Output = distribution, P(event), mean, sd
- Run once per candidate strategy, then compare
- No optimizing ability
- Formulate from a long description (spot that a binary is needed; spot a sunk cost that must be left out)
- Solve small problems by hand or graphically (no Excel Solver in the exam)
- Interpret solver output and shadow prices → make a recommendation
- Critique a colleague's model (missing constraint, wrong type, sunk cost included, linear when it should be nonlinear)
- The DMD case modules are on the syllabus too
Self-test: classify these 5 situations
- An airline chooses fares for two cabins, and demand in each cabin falls linearly with its fare. → Nonlinear (revenue = P × (a − bP) is quadratic).
- An insurer picks which 3 of 8 IT projects to fund within a $5M budget. → Binary.
- A coffee chain decides pounds of beans to buy from 4 suppliers to meet a blend-quality average at min cost. → Linear (quality % constraint can be linearized).
- A tire maker wants the distribution of next year's profit given random rubber price, demand and FX. → Simulation.
- Decide whether to open each of 3 warehouses (fixed cost if open) and how many units to ship from each. → Mixed-integer (binary open/close + continuous shipments, x ≤ M·y).
Ch 5 — Simulation Modeling: Concepts & PracticeBertsimas & Freund, Ch 5
Why simulate?
- Forecast the consequences of management decisions before making them
- Combining random variables (sum, product…) rarely gives a convenient distribution. The exception is a sum of normals, which is still normal
- Uses: manufacturing operations, service queues (banking, air travel, food service), pricing complex derivatives
- Textbook example: random fish price × demand at the port
What it gives you (and doesn't)
- ✔ Estimated shape of the distribution of quantities of interest
- ✔ Estimated probabilities of events (e.g. P(loss))
- ✔ Means and standard deviations
- ✘ No optimal-decision capability, unlike a decision tree. With several strategies, run a simulation for each and compare
Generating sample data
Discrete distribution (3 steps)
- Divide [0,1] into non-overlapping intervals, one per possible value, width = probability
- Generate uniform U(0,1) random numbers
- Assign each U the value whose interval contains it
Continuous distribution (inverse CDF)
- Generate U ~ Uniform(0,1)
- Solve F(y) = P(Y ≤ y) = U for y, and use y as the draw
e.g. U = 0.8413 → Φ−1 = 1 → y = μ + σ. (Exam on paper: use the normal table. XP flags the p.210 normal example as exam-like.)
Mini demo: map random numbers to a discrete distribution
Normal inverse-CDF draw
Interpreting simulation output (Example 5.8 style)
| Asked for | How to get it from n simulated trials |
|---|---|
| Expected value | sample mean x̄ |
| Variability / risk | sample std dev s, percentiles, histogram shape (skew, fat tails) |
| P(event), e.g. P(profit < 0) | (# trials where event occurred) / n |
| Precision of the mean estimate | 95% CI ≈ x̄ ± 1.96 · s / √n (more trials → narrower) |
| Which strategy is better? | Compare each strategy's mean and risk (sd, P(loss), worst percentile) against management's objective and risk appetite |
Gaining managerial confidence in a simulation model depends on 3 things
- A good understanding of the underlying management problem
- Using probability and statistics correctly
- Communicating these concepts effectively
- Build and quickly calculate a simple simulation by hand from given random numbers (past exams have done this)
- 2-point "interpret this output" question: P(event), mean, recommend a strategy
- Outline how you would set up a simulation: which variables are random, what distributions, what output you'd track, how many trials
- Applied to case-study style businesses (queues, operations, pricing)
- Claiming the simulation "found the optimal strategy." It didn't. You compared the strategies you ran
- Using the average of each input and plugging it in once. The average of a function ≠ function of averages (this is exactly why you simulate)
- Ignoring correlation between inputs (e.g. price and demand)
- Reporting only the mean when the question is about risk
Self-test: Demand is 10 (p=.3), 20 (p=.5), 30 (p=.2). Price per unit is N(5, 1). Udemand = 0.25, 0.62, 0.91. Uprice = 0.5, 0.8413, 0.1587. Simulate revenue.
Intervals: 10 → [0, .3), 20 → [.3, .8), 30 → [.8, 1).
| Trial | Ud → Demand | Up → Φ−1 → Price | Revenue |
|---|---|---|---|
| 1 | .25 → 10 | .5 → 0 → 5 | 50 |
| 2 | .62 → 20 | .8413 → 1 → 6 | 120 |
| 3 | .91 → 30 | .1587 → −1 → 4 | 120 |
Sample mean revenue = 96.7. With 3 trials this is far too few to rely on, so say so and recommend more trials (and a CI).
Self-test: 1,000 trials of strategy A give mean profit 2.1M, s = 1.5M, and 180 trials with a loss. Strategy B gives mean 1.8M, s = 0.6M, 40 losses. Recommend.
P(loss): A = 18%, B = 4%. 95% CI for A's mean: 2.1 ± 1.96·1.5/√1000 = 2.1 ± 0.09. A has a higher expected profit (~+0.3M, statistically meaningful) but much higher risk. Recommendation depends on risk appetite: a capital-constrained or risk-averse firm (e.g. one near a solvency threshold) should choose B. A well-capitalized firm maximizing expected value might choose A. State the trade-off explicitly and tie it to the company in the question.
Ch 7 — Linear OptimizationBertsimas & Freund, Ch 7 (incl. sensitivity & LP under uncertainty)
Vocabulary
Objective
What to optimize (min cost, max profit)Decision variables
The unknowns you control (tons of coal from each mine). State units.Objective function
Objective written in terms of decision variablesConstraints
Limits on the variables (≤, ≥, =). Put the variables on the LHSNon-negativity
x ≥ 0. Easy to forget!Feasible / optimal solution
Feasible = satisfies all constraints. Optimal = feasible with the best objective valueBinding vs non-binding
Binding = holds with equality at the optimum (line passes through it). Non-binding = has slackSpecial outcomes
Multiple optima (isoquant parallel to a binding edge), unbounded (objective → ∞), infeasible (no point satisfies everything → relax constraints)Formulation rules of thumb
Variables & objective
- "What decisions must I make?" → one variable per decision
- Typical objective: max revenue − cost, or min cost − revenue
- Count every cost and revenue exactly once
- Leave out sunk and fixed costs: they don't change with the decision (unless a binary decides whether the cost is incurred)
- There is often no unique formulation. Equality vs inequality takes judgment
4 constraint types
- Resource: can't use more steel or labor than is available
- Capacity: plant can make at most N units
- Quality: average content must meet a standard (linearize it! see below)
- Policy: a management rule, e.g. ≥ 50% of coal from union mines
Multiply through by the (positive) denominator so the constraint stays linear. Policy constraints like "at least 50% from union mines" work the same way: xU ≥ 0.5(xU + xN).
Graphical method: 5 steps (2 variables only)
Worked example
Max profit 30x + 40y (x = chairs, y = tables). Labor 2x + 4y ≤ 40 hrs · Machine 3x + 2y ≤ 36 hrs · Demand x ≤ 10 · x, y ≥ 0
| Corner | How found | Profit |
|---|---|---|
| (0, 0) | origin | 0 |
| (10, 0) | demand ∩ y=0 | 300 |
| (10, 3) | demand ∩ machine | 420 |
| (8, 6) | labor ∩ machine | 480 ★ |
| (0, 10) | labor ∩ x=0 | 400 |
Binding: labor (2·8 + 4·6 = 40 ✓) and machine (3·8 + 2·6 = 36 ✓). Non-binding: demand (x = 8 < 10, slack 2), so its shadow price = 0.
Fast exam trick: just evaluate the objective at every corner of the feasible polygon. The LP optimum is always at one (simplex walks corner to corner, "pivoting" to a better one until no better neighbor exists).
Sensitivity analysis & shadow prices
Range information
Solver output gives an allowable increase / decrease for each RHS. The shadow price holds only within that range. Beyond it, the binding set changes and you must re-solve.Non-binding ⇒ shadow price = 0
There's already slack, so more of that resource adds nothing. The reverse needs care: a binding constraint usually has a non-zero shadow price.| Sensitivity report for the worked example | |||||
|---|---|---|---|---|---|
| Constraint | Final LHS | RHS | Shadow price | Allow. increase | Allow. decrease |
| Labor (hrs) | 40 | 40 | $7.50 / hr | 32 | 8 |
| Machine (hrs) | 36 | 36 | $5.00 / hr | 4 | 16 |
| Demand (chairs) | 8 | 10 | $0 | ∞ | 2 |
Check: 7.5×40 + 5×36 + 0×10 = 480 = optimal profit. (Duals solve 2a+3b = 30, 4a+2b = 40.) Labor stays valid while the binding intersection stays feasible: 32 ≤ L ≤ 72. Machine: 20 ≤ M ≤ 40.
Self-test: interpret the sensitivity report (3 management proposals)
- Hire 10 temp labor hours at $6/hr? 10 ≤ allowable increase 32, so the shadow price holds: gain 10 × 7.50 = $75 vs cost $60 → +$15, do it. Breakeven wage = $7.50/hr.
- Rent 8 extra machine hours at $7/hr? Shadow price $5 < $7 → each hour loses money, so no. Also, only the first 4 hours are covered by the $5 shadow price. Beyond 4, re-solve (the marginal value drops, here to $0 once demand x≤10 binds).
- Marketing campaign to raise chair demand to 14? Demand is non-binding (shadow price 0) → no profit gain. Capacity, not demand, is the bottleneck. Spend on labor instead.
Linear optimization under uncertainty (two-stage)
Think of it as a decision tree inside an LP: today's decisions, then one set of recourse variables per future state, and the objective is the expected value across states.
Self-test: formulate. Buy capacity K now at $4/unit. Next year demand is 100 (p=0.6) or 60 (p=0.4). Each unit sold earns $10 margin; you can't sell more than K or than demand.
Variables: K = capacity bought now (stage 1). sH, sL = units sold in high/low demand state (stage 2).
Max −4K + 0.6(10 sH) + 0.4(10 sL)
s.t. sH ≤ K, sH ≤ 100, sL ≤ K, sL ≤ 60, K, sH, sL ≥ 0.
Solution logic: capacity from 0–60 is used in both states (marginal value 10 − 4 = +6). Capacity from 60–100 is used only in the high state (0.6×10 − 4 = +2 > 0). → K = 100, expected profit = −400 + 600 + 240 = 440. If the cost were $7, the 60–100 tranche would be worth 6 − 7 < 0 → K = 60.
- Turn a complex case paragraph into an LP (often with a policy or quality % constraint)
- Graph and solve a 2-variable LP quickly
- High-cognitive: apply shadow-price data to a specific proposal (buy more resource? at what price? how much?) — XP: "a classic high-cognitive level question"
- Practice: Example 7.2, Section 7.5.4 / Table 7.8 (explain every shadow price), Q&A p.364, quality constraint p.366
- Forgetting non-negativity
- Including sunk costs (already-paid leases, past R&D) in the objective
- Using a shadow price outside its allowable range
- Leaving a ratio constraint non-linear instead of multiplying through
- Double-counting costs (e.g. per-unit cost and also total cost)
- Saying a non-binding resource is "valuable." Its shadow price is 0
Ch 8 — Nonlinear OptimizationBertsimas & Freund, Ch 8
① Portfolio (risk vs return)
Min Var = ΣiΣj wiwjσij = Σ wi²σi² + 2Σi<j wiwjσij
s.t. Σ wi = 1 · Σ wiμi ≥ target (e.g. 11%) · wi ≥ 0 (+ extra policy limits, e.g. ≥ x% tech)
Re-solve for many targets → the efficient frontier (lowest risk for each return).
② Production & pricing
Max (PW − cW)·W + (PP − cP)·P
where demand depends on price, e.g. W ≤ 565 − 0.5·PW and P ≤ 325 − 0.25·PP, plus capacity constraints.
Price × quantity gives a quadratic objective. Optimum is often interior.
③ Facility location
Min Σi 2 · ni · √((x − ai)² + (y − bi)²)
ni = daily trips to site i, 2 = there and back, (ai, bi) = site coordinates, s.t. allowed-zone constraints on (x, y).
Graphical insights
- The optimum need not be at a corner, nor even on the boundary
- If it's on the boundary, it's where the feasible region just touches the best isoquant (tangency)
- Isoquants can be circles (distance), ellipses (variance)…
Local vs global
- Local optimum: best among nearby feasible points, but maybe not the global best
- Calculus-based solvers usually find only a local optimum
- General NLPs can be very hard even with a computer. Try several starting points
- Book picture: "kidney" (XP: "baby seal") shaped region, local min A, global min B
When does local = global? (the easy classes)
| Convex minimization | Concave maximization | |
|---|---|---|
| Objective | Convex (∪ bowl) to be minimized | Concave (∩ hill) to be maximized |
| Equality constraints | must be linear | |
| "≤" constraints | must be convex functions | |
| "≥" constraints | must be concave functions | |
| Result | every local optimum is a global optimum. Linear constraints always qualify (linear is both convex and concave) | |
Portfolio variance (min) is convex, so portfolio problems are easy. Revenue P·(a − bP) is concave, so the max-pricing problem is easy.
- Recognize that a case situation is nonlinear and formulate it (use the 3 patterns as templates)
- Explain why the solver's answer might not be the true optimum, and what you'd do
- Interpret NLP shadow prices with caution
- Practice: Examples 8.1–8.3, Exercise 8.6 (XP video)
- Assuming the NLP optimum is at a corner (that's LP only)
- Using NLP shadow prices for large changes. They're valid only for very small Δ
- Forgetting Σw = 1 or w ≥ 0 (no shorting unless allowed) in portfolio problems
- Writing a variance without the covariance (cross) terms
Self-test: single product, demand D = 200 − 2P, unit cost 20. Optimal price?
Profit = (P − 20)(200 − 2P) = −2P² + 240P − 4000, which is concave, so the local max is the global max. dπ/dP = −4P + 240 = 0 → P = 60, D = 80, profit = 40 × 80 = 3,200. If capacity were 60 units, the constraint binds: D = 60 → P = 70, profit = 3,000 (on the boundary, not at the unconstrained optimum).
Ch 9 — Discrete OptimizationBertsimas & Freund, Ch 9 (special constraints pp.455–456)
Integer model
All variables integer (linear objective & constraints). Small production runs where you can't make 2.4 planes.Binary model
All variables 0 or 1: on/off switches. Project selection, site opening.Mixed-integer
Binary + integer + continuous. Fixed-charge / setup-cost problems.Graphical view
- Feasible region = a set of discrete points, not a polygon
- No "corner points". Border points needn't be where two constraints meet
- Rounding the LP answer can be infeasible or sub-optimal
Solvability
- Much harder than LP
- No system of linear equations whose solution is the optimum
- Solved by branch-and-bound
Binary modelling cheat table
| Condition (in words) | Constraint (xi ∈ {0,1}) |
|---|---|
| If project A is chosen then B must be chosen | xB ≥ xA (i.e. xB − xA ≥ 0). Book: "if 4 then 3" → x3 ≥ x4 |
| A only if both B and C | xA ≤ xB and xA ≤ xC |
| A and B are mutually exclusive (at most one) | xA + xB ≤ 1 |
| Exactly one of A, B, C | xA + xB + xC = 1 |
| At most k of n / at least k of n | Σ xi ≤ k / Σ xi ≥ k |
| A and B together or not at all | xA = xB |
| Budget | Σ costi xi ≤ Budget |
| Fixed charge / setup cost: pay F only if you produce any x | Objective: … − F·y − c·x Constraint: x ≤ M·y, y ∈ {0,1} (M = max possible x, e.g. capacity) |
| If produced, at least m units (minimum batch) | m·y ≤ x ≤ M·y |
| Either constraint 1 or constraint 2 must hold | g1(x) ≤ b1 + M·z and g2(x) ≤ b2 + M(1 − z) |
| Open warehouse j before shipping from it | Σi shipij ≤ Capj·yj |
Branch-and-bound (binary model)
- Formulate a project-selection / site-selection problem with logical conditions and a budget
- Recognize a fixed cost needs a binary linked to a continuous variable (big-M)
- Explain branch-and-bound at a high level (XP isn't sure B&B calculations will be asked)
- Practice: Examples 9.1–9.3 (9.3 uses several variable types and is a great final review), Exercise 9.4 (XP video), pp.455–456
- Writing "if A then B" backwards (xA ≥ xB)
- Rounding the LP relaxation and calling it optimal
- Forgetting to link a fixed cost to the quantity (x ≤ M·y), which lets the model produce without paying setup
- Including a sunk cost of a project already started
Self-test: formulate. 5 projects with NPV vi and cost ci, budget $10M. At most 2 of projects 1–3. Project 5 requires project 2. Projects 3 and 4 can't both be done.
xi = 1 if project i is undertaken, 0 otherwise.
Max Σ vixi s.t. Σ cixi ≤ 10 · x1+x2+x3 ≤ 2 · x2 ≥ x5 · x3 + x4 ≤ 1 · xi ∈ {0,1}.
(If NPV already nets out money spent to date on any project, don't also subtract it. It's sunk.)
Self-test: product needs a $5,000 setup if made at all, $12 margin/unit, capacity 800, and you must make ≥ 100 if you make any.
x = units (continuous/integer), y = 1 if produced. Max 12x − 5000y s.t. 100y ≤ x ≤ 800y, y ∈ {0,1}, x ≥ 0. (Produce only if 12x > 5000, i.e. x > 417.)
The Hard Side of Change Management (DICE), quick reminderCP311-101-25, listed under Objectives 1 & 3. Full treatment: Objective 1 → DICE
Deciding How to DecideCP311-111-26 (Courtney, Lovallo & Clarke, HBR), also in Objective 4
Q1 · Do I know what it takes to succeed?
Do you have a causal model, i.e. which critical success factors and economic conditions, in what combination, lead to success? Test: can you confidently write "if-then" statements ("If the new process cuts cost X% and we win Y% share, then invest")? Do you know the metrics that must be met, almost a recipe?Q2 · Can I predict the range of outcomes?
Single outcome with reasonable certainty? A range of outcomes with probabilities? Or can't specify the range or its probabilities with any precision?Q3 (aggregation): Is the needed information centralized or dispersed across experts/the crowd?
The 5 decision-support tool families
| Tool | What it is | Needs |
|---|---|---|
| Conventional capital budgeting | DCF, NPV, expected rate of return on estimated incremental cash flows | Causal model + predictable outcome; stable, well-understood setting |
| Quantitative multiple-scenario | Monte Carlo (repeated random sampling), decision analysis (scenarios + probabilities → best choice for an objective), real options (value of flexibility, e.g. a pilot) | Causal model + range of outcomes with probabilities |
| Qualitative scenario analysis | A set of representative (not exhaustive) narratives of how the present may evolve, and the decision's consequences under each | Causal model, but outcomes can't be fully specified |
| Case-based decision analysis ★ | Collect analogous past cases, record their results, rate their similarity → choose the option maximizing the similarity-weighted average result. The authors' most-advocated, under-used tool | Works even without a causal model |
| Information aggregation | Gather dispersed knowledge (4 tools below) | Info is decentralized. Used largely independent of Q1/Q2 |
The 5 McDonald's situations: diagnostic matrix
Overlay information aggregation on any cell where the knowledge is dispersed. As uncertainty rises (moving right/down), shift from quantitative tools to analogies and, ultimately, iterative experimentation.
5 complicating factors
1. Executives don't know what they don't know
The diagnosis depends on judging the uncertainty correctly, and most underestimate it2. Cognitive bias creeps in
Biased assessment of uncertainty points to the wrong tool3. Organizational processes get in the way
Need general decision protocols, because politics and behavioral pitfalls are rife when money or power is at stake4. Reliance on a single tool
Usually conventional capital budgeting. Big decisions often need a combination5. Not considering the option to delay
"Deciding when to decide is as important as deciding how." In very uncertain settings, use learning-based, iterative experimentation- Given a case-company decision (e.g. a new country, a new product, a disruptive regulation): diagnose causal model / outcome predictability, then recommend and justify tool(s)
- Critique management for using only DCF in a highly uncertain context
- Recommend an aggregation tool and note its limits (prediction markets need a specifiable outcome set; confidentiality)
- Recommending NPV alone for a new business model
- Forgetting that a pilot = real option (situation 2)
- Saying "prediction market" when the outcomes can't even be enumerated
- Not justifying with the two diagnostic questions applied to the facts
Self-test: a life insurer (think case-study Darwin/Snappy) is considering launching a fully digital, AI-underwritten product in a new country. Which tools?
Causal model: partly. Insurance economics are understood (mortality, expenses, lapse), but digital distribution + new-country customer behavior + regulation are not. Outcomes: can't credibly put probabilities on take-up or regulatory response. → Closest to situation 3 (or 4 for the new digital model): qualitative scenario analysis (take-up, competitor, regulator scenarios) + case-based analysis of analogous digital launches. Use information aggregation (Delphi with local distribution and regulatory experts) because the knowledge is dispersed. Consider a pilot (real option) and the option to delay/stage the investment. DCF alone would understate uncertainty.
Exam playbook for DMD questionsXP answering-skills guidance + standard formulation discipline
Time management
- XP: students struggle to finish DMD solving in time. Budget time in proportion to the points (exam minutes ÷ total points) and stick to it
- Short on time? Explain HOW you would solve (e.g. "evaluate the objective at each corner of the feasible region; the max is optimal") for big part marks
- Formulation marks usually come before solution marks, so write them first
Interpretation = marks
- Translate numbers into the case's language ("an extra pilot hour is worth $X up to Y hours, so hire contractors below $X")
- Name binding constraints (the bottlenecks) and slack resources
- Flag model limitations (data quality, linearity, uncertainty → suggest sensitivity, simulation or scenarios)
- Book guidance: think through the issues, good data is essential, study the data, make many "what-if" runs
- No non-negativity · sunk cost included · undefined variables/units · a shadow price applied beyond its range · "optimal" claimed from a simulation · answering a different question than asked (e.g. formulating when asked to critique)