Logistics 2 — Complete Exam Summary TI

Oct 2, 2026 · @Willem

At a glance

All eight decks are covered below: every concept, formula, step-by-step method and worked example, with the slide answers checked. Numbers marked "(computed here)" are not on the slides but follow directly from slide data — useful for practising.

L1.2 — Forecasting (Reid & Sanders ch. 8)

Forecasting predicts future values (demand, capacity, transport times, external factors like population or income) from past data; in logistics it sizes production/transport capacity, inventory, maintenance and the offer.

Principles

• Forecasts are rarely perfect (stochasticity).
• Forecasts rely on data.
• More accurate for grouped data than for individual items.
• More accurate for shorter than longer horizons (temperature plot: next-day forecast tracks actuals best, 10-day worst).

Five steps: decide what to forecast → evaluate appropriate data → select and test model → generate forecast → monitor accuracy and change model if needed.

Types of methods

• Qualitative (subjective): executive opinion, market research, Delphi method (panel of experts).
• Quantitative (formulas): time series (future follows past patterns; all info is in the history) and causal models (cause-and-effect with leading indicators, e.g. regression).

Time series = pattern + random variation. Patterns: level (horizontal around mean), trend (increasing/decreasing), seasonality (regular repeat within a year), cycle (like seasonal but longer horizon, start/end hard to predict). Random variation cannot be forecast.

Level models

\text{Naive: } F_{t+1} = A_t

\text{Simple mean: } F_{t+1} = \frac{\sum_t A_t}{n}

\text{Moving average: } F_{t+1} = \frac{1}{k}\sum_{i=t-k+1}^{t} A_i

\text{Weighted MA: } F_{t+1} = \sum C_t A_t, \quad \sum C_t = 1

\text{Exponential smoothing: } F_{t+1} = \alpha A_t + (1-\alpha) F_t, \quad 0 \le \alpha \le 1

• Naive and simple mean suit level patterns.
• MA drops the oldest and adds the newest point; more responsive to trends than the mean. WMA puts more weight on recent data, so it reacts faster than MA.
• ES needs only last forecast, last actual and α — very popular. No previous forecast? Use the naive one. Higher α = more weight on the last period (and on its random noise); lower α = smoother.

Worked problem (actuals 300, 315, 290, 345, 320, 360, 375 for periods 1–7):

(Table computed here.) The slide's ES comparison starts from F2 = 300 (naive): α = 0.2 gives 303, 300, 309, 311, 321, 332 for periods 3–8; α = 0.9 gives 314, 292, 340, 322, 356, 373 — high α follows the actuals closely, low α lags.

Trend-adjusted exponential smoothing

Corrects the lag a level model shows on a trend. Three steps:

S_t = \alpha A_t + (1-\alpha)(S_{t-1} + T_{t-1})

T_t = \beta (S_t - S_{t-1}) + (1-\beta) T_{t-1}

FIT_{t+1} = S_t + T_t

Example (lawn care): A_July = 62, previous level S = 57, trend T = 15, α = 0.2, β = 0.1. S_July = 0.2·62 + 0.8·(57+15) = 70; T_July = 0.1·(70−57) + 0.9·15 = 14.8; FIT_August = 70 + 14.8 = 84.8 gallons.

Seasonality (seasonal indices)

1. Average demand per season for each year = yearly total / number of seasons.
2. Seasonal index per year per season = season demand / that year's average.
3. Average the indices per season over the years.
4. Forecast next year's total (any method) and divide by number of seasons.
5. Season forecast = new average × average index.

TU Delft enrollment example (forecast total 90,000 → average 22,500):

(Spring and Summer computed here; the four forecasts sum to 90,000.)

Causal model: linear regression

y = a + bx with x independent, y dependent.

b = \frac{\sum XY - n\bar{X}\bar{Y}}{\sum X^2 - n\bar{X}^2}, \qquad a = \bar{Y} - b\bar{X}

Goodness of fit

r = \frac{n\sum XY - \sum X \sum Y}{\sqrt{n\sum X^2 - (\sum X)^2}\,\sqrt{n\sum Y^2 - (\sum Y)^2}}

• r = correlation coefficient: direction and strength of the linear relation (closer to ±1 = better fit).
• r² = coefficient of determination: share of variation in y around its mean explained by the line (close to 1 desired).
• Slide example: n = 4, ΣX = 189, ΣY = 589, ΣXY = 28,202, ΣX² = 9,253, ΣY² = 87,165 → r = 0.982, r² = 0.964. (Computed here from the same data: b ≈ 1.152, a ≈ 92.8.)

Forecast error and accuracy

E_t = A_t − F_t. Over-forecast → negative error; under-forecast → positive error.

CFE = \sum_t E_t \qquad MAD = \frac{\sum_t |E_t|}{n} \qquad MSE = \frac{\sum_t E_t^2}{n} \qquad TS = \frac{CFE}{MAD}

• CFE measures bias; MAD total error regardless of sign; MSE penalizes large errors; TS controls bias (large |TS| = biased forecast).
• Example (Jan–May): Method A errors 2, 1, 0, −1, 1 → CFE 3, MAD 1, MSE 1.4. Method B errors 3, 1, 3, 2, 2 → CFE 11, MAD 2.2 (slide rounds to 2), MSE 5.4. Method A is better on every measure.

Selecting a model

Consider: amount and type of data available; degree of accuracy required (more accuracy = more data); length of horizon (3 months vs 10 years need different models); data patterns present (a level model on a trend lags). Other methods: machine learning, combining forecasts (often more accurate), CPFR (Collaborative Planning, Forecasting and Replenishment between buyers and sellers).

L2 — Clustering (Zaki & Meira ch. 13–14)

Clustering partitions n items into K groups that are as homogeneous as possible, i.e. members are close in distance. Key choices: the number of clusters and which features to use. In logistics it serves organisation and consolidation (containers, warehouse slotting, store layout, networks).

Notation and distances

• Data matrix D is n × d: each row x_i is an element, each column X_j an attribute. A vector is written vertically; the row form is its transpose (T).
• Dot product aᵀb = Σ a_i b_i (vectors of the same size).

\|a\| = \sqrt{a^T a} = \sqrt{\sum_i a_i^2} \qquad \delta(a,b) = \|a-b\| = \sqrt{\sum_i (a_i-b_i)^2}

Centroid of cluster C_i (its representative point):

\mu_i = \frac{1}{n_i}\sum_{x_j \in C_i} x_j

Objective — sum of squared errors (to minimize):

SSE(\mathcal{C}) = \sum_{i=1}^{k}\sum_{x_j \in C_i} \|x_j - \mu_i\|^2

Why algorithms? Enumerating all K-partitions is impossible; the count is the Stirling number of the second kind, order k^n / k!:

S(n,k) = \frac{1}{k!}\sum_{t=0}^{k}(-1)^t \binom{k}{t}(k-t)^n

For k = 3 the slide table grows from about 9,800 (n = 10) to about 3·10^14 (n = 32). Both algorithms below are greedy (myopic: take the best nearby option), so they do not guarantee the optimum. An algorithm = steps that build a solution and stop after a fixed number of steps or when a condition holds.

K-means

1. Step 0: generate k initial centroids at random (uniform).
2. Step 1 (assignment): put each point in the cluster of the nearest centroid: j* = argmin_i ‖x_j − μ_i‖².
3. Step 2 (update): recompute each centroid as the mean of its points.
4. Step 3: stop if the stopping criterion holds, else go to Step 1. Criteria: assignments no longer change, or Σ‖μ_i^t − μ_i^{t−1}‖² ≤ ε.

1-D slide example, data {2, 3, 4, 10, 11, 12, 20, 25, 30}, k = 2, start μ1 = 2, μ2 = 4:

Tip: in 1-D, a point goes to C1 if it lies below the midpoint (μ1+μ2)/2.

Hierarchical (agglomerative) clustering

Starts with n singleton clusters and repeatedly merges the closest pair until one cluster remains (or k clusters). The result is a rooted binary tree, the dendrogram (leaves = items, root = everything). Consecutive clusterings are nested, C_{t−1} ⊂ C_t: A = {A1..Ar} is nested in B = {B1..Bs} if r > s and every A_i lies inside some B_j.

1. Step 0: n clusters, compute the distance matrix.
2. Step 1: find the closest pair C_i, C_j and merge: C_ij = C_i ∪ C_j.
3. Step 2: update C = (C \ {C_i, C_j}) ∪ C_ij.
4. Step 3: stop when k clusters (or one) remain; else update the distance matrix and repeat.

Cluster distances (linkage)

Lance–Williams update for the distance of merged C_ij to another cluster C_r:

\delta(C_{ij},C_r) = \alpha_i\,\delta(C_i,C_r) + \alpha_j\,\delta(C_j,C_r) + \beta\,\delta(C_i,C_j) + \gamma\,|\delta(C_i,C_r) - \delta(C_j,C_r)|

Check: single link gives ½a + ½b − ½|a−b| = min(a,b); complete link gives max(a,b).

Single-link example (A–B 1, A–C 3, A–D 2, A–E 4, B–C 3, B–D 2, B–E 3, C–D 1, C–E 3, D–E 5): merge AB at 1 → AB–C 3, AB–D 2, AB–E 3; merge CD at 1 → AB–CD 2, CD–E 3; merge ABCD at 2; merge with E at 3 → ABCDE.

Group-average Lance–Williams example: merge [x1,x2,x3] (n=3) and [x5,x7] (n=2) at distance 1; distance to [x4,x6] = (3/5)·3 + (2/5)·2 = 2.6.

Evaluating clusters with labelled data (supervised)

With known classes (e.g. Iris: setosa, versicolor, virginica; 150 flowers, 4 attributes) you know k and can compare clusters to the true classes. Each cluster is matched to the class it holds most of.

• Positive = the reference class; negative = all others.
• TP = item of the class put in its cluster; TN = other-class item correctly outside; FP = other-class item in the cluster; FN = class item outside the cluster.

\text{Accuracy} = \frac{TP+TN}{P+N} \qquad \text{Precision} = \frac{TP}{TP+FP} \qquad \text{Recall} = \frac{TP}{P}

• Slide example (6 points, C1 = 2 setosa + 1 versicolor, C2 = 1 setosa + 2 versicolor, positive = setosa): TP 2, TN 2, FP 1, FN 1 → accuracy 4/6 = 0.67, precision 2/3, recall 2/3 (computed here).
• Iris contingency table (computed here): C1 = 50 setosa → precision 1, recall 1; C2 = 36 versicolor + 1 virginica → versicolor precision 36/37 = 0.97, recall 36/50 = 0.72; C3 = 49 virginica + 14 versicolor → virginica precision 49/63 = 0.78, recall 49/50 = 0.98.

Large data sets are clustered in software, e.g. KNIME Analytics Platform.

L3 & L4 — Classification (Zaki & Meira ch. 2, 3, 18)

Classification predicts the class of a new point, given a data set whose points already have classes. Logistics uses: warehouse/inventory (affinity with already clustered products), bundling freight and allocating new cargo to bundles, trash management, classifying a new product to minimize setup times. Other uses: medicine (prognosis), biology.

Probabilistic approach (4 steps): 1) select attributes from a data set with defined classes; 2) fit a probability distribution per class; 3) compute the probability that the item's attributes are produced by each distribution; 4) assign the item to the class with the highest probability.

Bivariate statistics

• Centered data matrix Z = D − 1·μᵀ (subtract each column's mean). Slide example: D = [2 3; 5 4; 2 5], μ = [3 4] → Z = [−1 −1; 2 0; −1 1].
• Attributes are assumed i.i.d.; bivariate analysis finds a joint distribution for both attributes (ideally both normal → bell shape).
• Mean vector μ = (μ1, μ2)ᵀ. Variance σ_j² = (1/n)Σ(x_ij − μ_j)²; total variance = σ1² + σ2².
• Covariance σ12 = (1/n)Σ(x_i1 − μ1)(x_i2 − μ2): how the two variables vary together.
• Correlation ρ12 = σ12 / (σ1 σ2).

\Sigma = E[(X-\mu)(X-\mu)^T] = \frac{1}{n} Z^T Z = \begin{pmatrix} \sigma_1^2 & \sigma_{12} \\ \sigma_{21} & \sigma_2^2 \end{pmatrix}, \quad \sigma_{12} = \sigma_{21}

(For the slide's D, computed here: Σ = (1/3)[6 1; 1 2] = [2 0.333; 0.333 0.667].) The slides use 1/n, not 1/(n−1).

Multivariate (bivariate, d = 2) normal density:

f(x \mid \mu, \Sigma) = \frac{1}{(\sqrt{2\pi})^d \sqrt{|\Sigma|}} \exp\left\{ -\frac{(x-\mu)^T \Sigma^{-1} (x-\mu)}{2} \right\}

\det\begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad - bc, \qquad \begin{pmatrix} a & b \\ c & d \end{pmatrix}^{-1} = \frac{1}{ad-bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}

One-dimensional version: f(x) = (1/√(2πσ²)) · exp(−(x−μ)² / (2σ²)).

Probability theory

• Axioms: P(A) ≥ 0; P(S) = 1; P(A ∪ B) = P(A) + P(B) for disjoint A, B.
• Conditional probability: P(A|B) = P(AB) / P(B) (A restricted to the new space B). Disjoint → P(A|B) = 0; A ⊂ B → P(A|B) = P(A)/P(B) ≥ P(A); B ⊂ A (AB = B) → P(A|B) = 1; independent → P(A|B) = P(A).
• Dice example: A = sum > 8 (P = 10/36), B = first die is 6 (P = 1/6), P(AB) = 4/36 → P(A|B) = (4/36)/(1/6) = 0.67.
• Law of total probability (A1…Am partition S): P(B) = Σ P(A_i) P(B|A_i).
• Bayes' theorem (P(A_i) = prior, P(B|A_i) = likelihood, P(A_i|B) = posterior):

P(A_i \mid B) = \frac{P(B \mid A_i)\,P(A_i)}{\sum_j P(B \mid A_j)\,P(A_j)}

Doctor example: P(E|H1) = 0.9, P(E|H2) = 0.1, P(E|H3) = 0.3; priors 0.03, 0.7, 0.27. Products 0.027, 0.07, 0.081, sum 0.178 → P(H1|E) = 0.15, P(H2|E) = 0.39, P(H3|E) = 0.45 (most likely).

Full Bayes classifier

\hat{y} = \arg\max_{c_i} \frac{P(x \mid c_i) P(c_i)}{P(x)} = \arg\max_{c_i} f_i(x)\,P(c_i)

P(x) is the same for every class, so skip it. Prior P(c_i) = n_i / n.

Algorithm (numeric attributes): for each class i: 1) filter D for class c_i; 2) prior = n_i/n; 3) likelihood = bivariate normal density f_i(x) with that class's μ_i and Σ_i; 4) store likelihood × prior; 5) after the last class, choose the largest.

Numeric example — classify x = (6.75, 4.25); C1 = (5,9), (6,10), (5,6), (8,25); C2 = (8,6), (10,20), (9,10), (7,4). Slide results are in a Brightspace Excel file; values below computed here with the slides' 1/n covariance:

Categorical attributes

Same formula and prior; the likelihood = (number of class-c_i items with exactly x's combination) / n_ci.

Pseudo-count when the combination never occurs in c_i (m_j = number of possible values of attribute j):

\text{Full Bayes: } P(x \mid c_i) = \frac{1}{n_{c_i} + \prod_{j=1}^{d} m_j} \qquad \text{Naive Bayes: } P(x_j \mid c_i) = \frac{1}{n_{c_i} + m_j}

Data set (C1: long-heavy, short-light, short-light, long-heavy, long-light; C2: short-light, short-heavy, long-heavy):

Naive Bayes

Assumes all attributes are independent → covariances = 0, Σ is diagonal. Likelihood = product of one-dimensional likelihoods per attribute (1-D normal for numbers, per-attribute frequency for categories):

P(x \mid c_i) = \prod_{j=1}^{d} P(x_j \mid c_i)

(Final scores computed here; the slides give the factors.)

Exam-style categorical example (last slide)

Class c1 has 50 items, c2 has 100 → priors 1/3 and 2/3. X1 ∈ {very short, short, long, very long} (m1 = 4), X2 ∈ {short, medium, long} (m2 = 3).

Answers (computed here):

L5 — Advanced inventory management (Reid & Sanders ch. 12; Nahmias & Olsen ch. 4)

Inventory balances service level (right product, quantity, time) against cost (ordering + holding, while avoiding stockouts). Learning objectives: explain continuous and periodic review policies; apply optimization; illustrate the effect of uncertainty.

Why hold inventory: meet anticipated demand, smooth production, decouple operations, protect against stockouts, economies of scale, hedge against price increases, quantity discounts. Poor management → stockouts (lost sales) or overstock (excess cost); it affects operations, marketing and finance.

Continuous vs periodic review

EOQ (Economic Order Quantity)

Assumptions: constant known demand (no safety stock), constant known lead time, no quantity discounts, fixed order/setup cost, all demand met, whole order arrives at once, perfect quality.

TC(Q) = \frac{D}{Q} S + \frac{Q}{2} H \qquad Q^* = \sqrt{\frac{2DS}{H}} \qquad TC(Q^*) = \sqrt{2DSH}

• D = demand (per period), S = ordering/setup cost per order, H = holding cost per unit per period. All costs in the same time unit.
• Small Q: many orders, low holding; large Q: few orders, high holding. At Q*, ordering cost = holding cost.
• Reorder point R = d · L (d = demand rate per time unit, L = lead time in the same unit).

Example: D = 10,000/yr, price €30, H = 20% · 30 = €6, S = €75, L = 5 days, 250 working days. Q* = √(2·10,000·75/6) = 500 units; TC = (10,000/500)·75 + (500/2)·6 = €3,000; d = 10,000/250 = 40/day → R = 200 units.

Resource-constrained multi-item EOQ

\min TC = \sum_{i=1}^{n} \left( \frac{D_i S_i}{Q_i} + \frac{Q_i H_i}{2} \right) \quad \text{s.t.} \quad \sum_{i=1}^{n} c_i Q_i \le C

c_i = cost or space per unit of item i; C = capacity (budget/space).

1. Compute each independent EOQ_i = √(2D_iS_i/H_i).
2. If Σ c_i EOQ_i ≤ C, the constraint is not binding → Q_i* = EOQ_i.
3. Otherwise it binds. Special case c_1/H_1 = c_2/H_2 = … (resource use proportional to holding cost): Q_i* = m · EOQ_i with m = C / Σ c_i EOQ_i.

Why the special case is reasonable: with a budget constraint c_i is the unit cost and H_i = I_i c_i, so c_i/H_i = 1/I_i; equal ratios = same interest rate for all items.

Example (budget €30,000, I = 25%):

Σ c_i EOQ_i = 50·172 + 350·63 + 85·61 = 35,835 > 30,000 → m = 30,000/35,835 = 0.8372.

EPQ (Economic Production Quantity)

A production run instead of an order; stock builds up gradually at rate p − d during uptime T1 (assume p > d); downtime T2; cycle T = T1 + T2.

T_1 = \frac{Q}{p} \qquad I_{max} = (p-d)T_1 = Q\left(1 - \frac{d}{p}\right) \qquad TC = \frac{D}{Q}S + \frac{I_{max}}{2}H

Q^* = EPQ = \sqrt{\frac{2DS}{H(1 - d/p)}}

EOQ is the special case p → ∞.

Example (chairs): D = 18,000/yr, d = 1,500/month, p = 2,500/month, S = €800, H = €18/yr, 20 days/month, L = 5 days. Q* = √(2·18,000·800 / (18·0.4)) = 2,000; I_max = 2,000·(1 − 1500/2500) = 800; TC = (18,000/2,000)·800 + (800/2)·18 = €14,400; daily demand = 1500/20 = 75 → R = 375.

Safety stock (demand uncertainty)

R = dL + SS \qquad SS = z\,\sigma_{dL}

σ_dL = standard deviation of demand during lead time; z from the normal table for the service level (95% service = 5% risk → z = 1.645; general table values: 90% → 1.28, 97.5% → 1.96, 99% → 2.33).

Periodic review system

Order every RP (review period, can be derived from EOQ, e.g. EOQ/d) up to target inventory TI; order Q = TI − OH (on-hand). Must cover demand over RP + L.

TI = d(RP + L) + SS \qquad SS = z\,\sigma_t \sqrt{RP + L} \qquad Q = TI - OH

σ_t = standard deviation of demand per time interval (σ_{RP+L} = σ_t√(RP+L)).

Example (pharmacy): d = 6,000/week, σ_t = 500, RP = 1 week, L = 3 weeks, 95% → z = 1.645. SS = 1.645·500·√4 = 1,645; TI = 6,000·4 + 1,645 = 25,645; with OH = 19,000 → Q = 6,645 tablets.

Single-period model (newsvendor)

For perishable/seasonal goods: sold at full price in one period only, highly variable demand with a known distribution, salvage value < cost. Balances cost of excess vs cost of shortage. With discrete demand use an expected value (payoff) matrix:

• Order = Demand: payoff = D · (price − cost).
• Order > Demand: payoff = D · (price − cost) − (Order − D) · (cost − salvage).
• Order < Demand: payoff = Order · (price − cost).

T-shirt example: price €20, cost €8, salvage €2; P(demand 80, 90, 100, 110, 120) = 0.2, 0.25, 0.3, 0.15, 0.1.

Order 100: 0.2·840 + 0.25·1,020 + 0.3·1,200 + 0.15·1,200 + 0.1·1,200 = €1,083.

L7 — Mathematical modelling (Bombelli et al., From theORy to application)

A mathematical (linear) programming model translates a problem — a situation with constraints whose improvement is measurable — into variables, parameters, constraints and an objective. Slide analogy: passing the course = maximize grade with minimum effort, subject to limited time, other courses, scheduling, grade ≥ 6.

Ingredients

• Variables: unknown values to find (e.g. quantity to buy).
• Parameters: known upfront (unit cost, starting inventory).
• Constraints: equalities/inequalities limiting the variables (money, time).
• Objective function: minimize or maximize.
• Linear: only terms αx — no variable × variable, no logs, no square roots.

Standard format: objective function → constraints → domain of the variables (+ list of parameters and variables). Compact form:

\min(\max) \sum_{j=1}^{n} c_j x_j \quad \text{s.t.} \quad \sum_{j=1}^{n} a_{ij} x_j \ (\ge \text{ or } \le) \ b_i \ \ \forall i = 1..m, \qquad x_j \in \ldots \ \forall j

Classification by domain: x ∈ ℝ → linear programming (LP, relatively easy to solve); x ∈ ℕ or ℤ → integer linear programming (ILP, hard); x ∈ {0,1} → binary/integer linear programming. A solution (a vector x) is feasible if it meets all constraints and optimal if it is feasible and the best of all. Solved with algorithms, trial-and-error, or solvers (CPLEX, GUROBI).

LP example — fertilizer

L-fert per ton: 2 A, 1 B, 1 C, revenue €50. H-fert per ton: 1 A, 1 B, 0 C, revenue €15. Stock: 1500 A, 1200 B, 500 C.

\max 50x_1 + 15x_2 \quad \text{s.t.} \quad 2x_1 + x_2 \le 1500, \quad x_1 + x_2 \le 1200, \quad x_1 \le 500, \quad x_i \ge 0

(Optimum computed here: x1 = 500, x2 = 500, revenue €32,500 — chemical A and C are binding, B has 200 spare.)

Knapsack (ILP)

Items i with weight w_i and value v_i, one knapsack of capacity C:

\max \sum_{i \in I} v_i x_i \quad \text{s.t.} \quad \sum_{i \in I} w_i x_i \le C, \quad x_i \in \{0,1\}

Slide illustration: w = (10, 2, 1, 3, 0.25, 1), v = (5, 2, 1, 1, 4, 0), example solution x = (1, 0, 1, 0, 1, 0) → weight 11.25, value 10.

Multi-knapsack

Two or more bins j with capacity C; each item to at most one bin:

\max \sum_{i \in N}\sum_{j} v_i x_{ij} \quad \text{s.t.} \quad \sum_{j} x_{ij} \le 1 \ \forall i, \qquad \sum_{i \in N} w_i x_{ij} \le C \ \forall j, \qquad x_{ij} \in \{0,1\}

Trap: the first slide version omits Σ_j x_ij ≤ 1 ("model is missing something") — then one item can be packed in both bins and its value counted twice.

Feasibility and logical constraints

• x1 + (1−x2) + (1−x3) + x4 = 4 with binary x → only x = (1, 0, 0, 1).
• … ≤ 3 → forbids exactly that one combination.
• x1 + x3 = 1 → exactly one of the two is in the solution (one in, the other out).

Either-or regions F: 2 ≤ x1 ≤ 4, 4 ≤ x2 ≤ 6 and G: 5 ≤ x1 ≤ 8, 2 ≤ x2 ≤ 3. Listing all 8 bounds is wrong — F and G don't overlap, so it is infeasible. Correct: one binary y and big-M:

\begin{aligned}
&x_1 \ge 2 - My, \ x_1 \le 4 + My, \ x_2 \ge 4 - My, \ x_2 \le 6 + My && (F, \text{ active if } y=0)\\
&x_1 \ge 5 - M(1-y), \ x_1 \le 8 + M(1-y), \ x_2 \ge 2 - M(1-y), \ x_2 \le 3 + M(1-y) && (G, \text{ active if } y=1)
\end{aligned}

Check point (3, 6) with M = 1000: all F constraints hold with y = 0; G's hold because they're relaxed by 1000 → y = 0 (F). For multiple regions use one binary per region, y_F + y_G (+ …) = 1, and relax region k by M·(1 − y_k) (the slide writes this with M·y_F etc.).

The big-M construct

Either f(x) ≥ k or f(x) ≤ γ:

f(x_1..x_n) \ge k - M(1 - y_j) \qquad f(x_1..x_n) \le \gamma + M y_j

• M is a large number that disables a constraint; it always comes with a binary variable.
• A ≤ constraint is relaxed toward +M; a ≥ constraint toward −M.
• y = 0 → the second holds and the first is disabled (≥ something far below 0).

Fixed-charge cost (objective ↔ constraints)

Producing x1 costs fixed f plus p per unit: h(x1) = 0 if x1 = 0; f + p·x1 if x1 > 0.

Attempt 3 check: (x1, y) = (0,0) feasible + correct; (>0, 0) infeasible; (0, 1) feasible for the constraints but not for the problem (never optimal because it costs f for nothing); (>0, 1) correct.

Linking variables and parameters

Binary X_ij = 1 if container i goes on train j (or item i in knapsack j).

• Train departs after container is available, only if on board: t_j ≥ a_i − (1 − X_ij)M, or equivalently t_j ≥ a_i X_ij (both work).
• Train departs before due time, only if on board: t_j ≤ d_i + (1 − X_ij)M.
• If deadlines may be exceeded, add slack (delay) variable d̄_i: t_j ≤ d_i + d̄_i + (1 − X_ij)M (and typically penalize d̄_i in the objective).

Bin packing (and with time)

Like multi-knapsack, but all items must be allocated and each used bin has a cost c_j (y_j = 1 if bin j used):

\min \sum_{j \in B} c_j y_j \quad \text{s.t.} \quad \sum_{j \in B} x_{ij} = 1 \ \forall i \in I, \qquad \sum_{i \in I} w_i x_{ij} \le CAP_j\, y_j \ \forall j \in B

The y_j on the right-hand side both caps capacity and forces y_j = 1 when a bin is used (combining multi-knapsack with the fixed-charge idea).

With time (item availability A_i, deadline D_i, bin departure time t_j, travel time T_j):

t_j \ge A_i x_{ij} \ \ \forall i, j \qquad t_j + T_j \le D_i + (1 - x_{ij})M \ \ \forall i, j

(The first version on the slides uses t_j ≤ D_i + (1 − x_ij)M, i.e. departure before deadline; the "visual" version adds travel time so arrival t_j + T_j meets the deadline.)

L8a — Classical mathematical models (Fazi notes ch. 3)

Learning objectives: formulate classical transport/logistics problems as (MI)(I)LPs; formulate an (MI)LP for an unseen problem; apply Dijkstra. Approach for any formulation exercise: write indices/sets → parameters → decision variables → objective → constraints → variable domains.

Coverage (set covering) problem

Demand locations i ∈ N, candidate hospital sites j ∈ S, opening cost c_j, distance d_ij, coverage radius K. Open sites so every location is within K of at least one open site, at minimum cost. x_j = 1 if site j is opened.

Wrong constraint: x_j d_ij ≤ K ∀i, j — if x_j = 1 it forces every location to be within K of that site. Fix with an incidence parameter a_ij = 1 if d_ij ≤ K, else 0:

\min \sum_{j \in S} c_j x_j \quad \text{s.t.} \quad \sum_{j \in S} a_{ij} x_j \ge 1 \ \ \forall i \in N, \qquad x_j \in \{0,1\}

Set partitioning (exact cover): each location covered exactly once → Σ_j a_ij x_j = 1. Every partitioning solution is feasible for covering, not vice versa (smaller feasible space), so for minimization:

OBJ_{covering} \le OBJ_{partitioning}

Assignment problem

Drivers i ∈ N, passengers j ∈ M, cost c_ij; x_ij = 1 if driver i is assigned to passenger j.

\min \sum_{i \in N}\sum_{j \in M} c_{ij} x_{ij} \quad \text{s.t.} \quad \sum_{i \in N} x_{ij} = 1 \ \forall j \in M, \qquad \sum_{j \in M} x_{ij} = 1 \ \forall i \in N, \qquad x_{ij} \in \{0,1\}

A solution is a permutation matrix: exactly one 1 per row (each driver one passenger) and per column (each passenger one driver).

Shortest path problem

Graph G = (V, E), non-negative weights w_ij, source s, target t; x_ij = 1 if edge (i,j) is on the path. Flow-balance formulation:

\min \sum_{(i,j) \in E} w_{ij} x_{ij} \quad \text{s.t.} \quad \sum_{j:(i,j) \in E} x_{ij} - \sum_{j:(j,i) \in E} x_{ji} = \begin{cases} 1 & i = s \\ -1 & i = t \\ 0 & \text{otherwise} \end{cases} \quad \forall i \in V

Out-flow minus in-flow: the source sends one unit, the target absorbs it, every other node passes it on. Applications: navigation (road/rail/air), telecom networks, delivery services. Single-source version: distance from s to every node plus a predecessor tree.

Dijkstra's algorithm (non-negative weights only)

1. Mark all nodes unvisited.
2. Tentative distance 0 for the source, ∞ for all others.
3. Current node = unvisited node with the smallest tentative distance (always the source first).
4. For each unvisited neighbour: distance via current node = current distance + edge weight; if smaller, update it and set its predecessor.
5. Mark the current node visited (its distance is now final).
6. Repeat 3–5 until all nodes are visited.
7. Output: shortest distance to every node; read a path backwards via the predecessors.

Slide example — edges A–B 2, A–D 8, B–D 5, B–E 6, D–E 3, D–F 2, E–F 1, E–C 9, F–C 3; source A:

Final: A 0, B 2 (A), D 7 (B), E 8 (B), F 9 (D), C 12 (F). Backtrack C ← F ← D ← B ← A: A → B → D → F → C, weight 12.

Exercise 1 — single-machine scheduling

Tasks j ∈ J with processing time p_j and cost w_j per unit of completion time; one task at a time. Variables: start time t_j ≥ 0; y_ij = 1 if task j is before i.

(a) Minimize weighted completion times without overlap. Writing both t_i + p_i ≤ t_j and t_j + p_j ≤ t_i is wrong (both can't hold → infeasible). Use big-M either-or:

\min \sum_{j \in J} w_j (t_j + p_j) \quad \text{s.t.} \quad t_i + p_i \le t_j + M y_{ij}, \quad t_j + p_j \le t_i + M(1 - y_{ij}) \quad \forall i < j

(b) Due dates d_j and a fixed penalty z_j if late (regardless of delay); x_j = 1 if task j is late:

\min \sum_{j} w_j (t_j + p_j) + \sum_{j} z_j x_j \quad \text{add:} \quad t_j + p_j \le d_j + M x_j \ \ \forall j, \quad x_j \in \{0,1\}

L9 — Succinct models and logistics costs (Daganzo ch. 1–2)

Learning objectives: understand the difference between analytical and mathematical models; analyze logistics system costs with closed formulas.

Mathematical vs analytical models

Analytical models write each cost component as a formula and find the best value via derivatives.

Logistics cost categories

• Motion costs (movement): transportation, handling.
• Holding costs (time): rent (storage space), waiting (delay of items, e.g. capital interest).

First model and cumulative diagrams

One production site → transport → one destination; production rate D' = consumption rate; the company decides how much to accumulate before shipping. Four cumulative curves over time: (i) production (slope D'), (ii) shipments (steps), (iii) arrivals (steps, shifted by travel time t_m), (iv) consumption. Headways H_i = times between shipments.

In-class exercise — at time t1 read the diagram as:

• Vertical gap production − shipments = items waiting for transport.
• Vertical gap shipments − arrivals = items in transit.
• Vertical gap arrivals − consumption = items waiting for consumption.
• Horizontal gap production → consumption = waiting time of an item.

Holding costs

\overline{Wait} = H_1 + t_m, \quad H_1 = \max_i H_i \qquad \text{Max accumulation} = D' H_1

\text{rent/year} = c_r \cdot D' H_1 \qquad \text{rent/item} = \frac{c_r \cdot D' H_1}{D'} = c_r H_1

Rent is proportional to the maximum number of items stored (= space needed), so it grows with the maximum headway.

Example BiB: D' = 400/day, c_r = €0.1, H = {10, 12, 8, 15, 14, 12, 16, 10} → H1 = 16; max accumulation = 6,400; rent = 0.1 · 6,400 = €640/year; per item 640/400 = €1.6.

Waiting cost (c_i = cost per item per unit time of delay):

\text{wait/year} = c_i D'(H_1 + t_m) = c_i D' H_1 + c_i D' t_m \qquad \text{wait/item} = c_i H_1 + c_i t_m

• c_i H1 = stationary inventory cost; c_i t_m = pipeline (in-transit) inventory cost.
• If inventory at the destination can be ignored (passengers), the slide's average wait is t_m + ½(H̄·0.5 + H1·0.5).

Transportation costs

\text{shipment cost} \approx c_f + c_v v \qquad \text{cost for } n \text{ shipments} = \sum_{i=1}^{n}(c_f + c_v v_i) = c_f n + c_v V, \quad V = \sum_i v_i

v = shipment size, c_f = fixed cost per shipment, c_v = rate per item. For fixed V, total cost depends only on the number of shipments n.

\text{transport/item} = c_f \frac{n}{V} + c_v = \frac{c_f}{\bar{v}} + c_v = \frac{c_f}{D' \bar{H}} + c_v \quad (\bar{v} = D'\bar{H})

Key contrast: transport cost per item falls with the average headway; holding cost rises with the maximum headway — so regular (equal) headways are best.

Distance-dependent version: c_f = c_s + c_d d and c_v = c'_s + c'_d d:

\text{cost for } n \text{ shipments} = c_s n + c_d n d + c'_s V + c'_d V d

c_s = cost per trip/stop, c_d = cost per vehicle-km, c'_s = extra cost per item carried (loading/unloading), c'_d = cost per item-km. With n_s delivery stops per trip: ≈ c_s(1 + n_s)n + c_d n d + c'_s V.

Example TraLog (Delft → Amsterdam): d = 70 km, capacity 200, V = 600/day → n = 3; c_s = 75, c_d = 1, c'_s = 0.2, c'_d = 0.001. Cost = 75·3 + 1·3·70 + 0.2·600 + 0.001·600·70 = 225 + 210 + 120 + 42 = €597; per item €0.995.

Size quiz: the per-shipment cost f_t(v) satisfies (a) f_t(x1 + x2) ≤ f_t(x1) + f_t(x2) (subadditive: one combined shipment pays the fixed cost once).

Optimal shipment size (EOQ with capacity)

Cost per item = holding (∝ v, since H = v/D') + transport c_f/v (+ c_v constant):

\min_v \ A v + \frac{B}{v} \ \ \text{s.t. } v \le v_{max}, \qquad A = \frac{c_h}{D'}, \ B = c_f, \ c_h = c_r + c_i \qquad v^* = \min\left\{\sqrt{B/A}, \ v_{max}\right\}

Example Star (Amsterdam → Bremen, 355 km, 4h20 = 260 min = 0.542 working day of 8 h): price €25, D' = 2,000/day, v_max = 250, c_f = €75, c_v = €2, c_r = €0.5/week = €26/year, c_i = 2% · 25 = €0.5/unit/year → c_h = €26.5. Slide: A = 26.5/2000 = 0.01325, v* = min{√(75/0.01325), 250} = min{75.2, 250} = 75.2 (graph: motion cost per item falls, holding rises; optimum at 75.2).

Units trap (checked here): the slide divides a per-year holding cost by a per-day demand. With consistent units (D' = 730,000/year) √(B/A) ≈ 1,437 > 250, so v* = v_max = 250. Know the slide's method; always state your units.

Motion costs (transport + handling)

Handling adds a fixed part c'_f per handling unit (capacity v'_max) and c'_v per item; the step function is approximated by the line PQ:

f_m(v) \approx c_f + \left(c_v + c'_v + \frac{c'f}{v'{max}}\right) v \qquad c''_f = c_f, \quad c''_v = c_v + c'_v + c'f / v'{max}

So handling is folded into the transport function. Motion cost per item f_m(v)/v falls with v (sawtooth, lower bound from Eq. 2.7); holding cost per item rises with v.

Stochastic effects

• So far travel time and production/consumption rates were constant. If stochastic: thanks to EOQ robustness, lot size and dispatch frequency should not change.
• Order strategy: when on-hand + backordered items = v0 (reorder point), request a shipment of size v.
• Demand: rate D', standard deviation γ; lead time T_ℓ with mean t_ℓ, s.d. σ_ℓ. Items at destination when the shipment arrives = v0 + v − PQ (PQ = demand during lead time):

E(PQ) = D' t_\ell \qquad var(PQ) = D'^2 \sigma_\ell^2 + D' \gamma t_\ell

• Optimal v: balance motion and inventory cost. Long run: motion cost is the same with or without randomness (same number of shipments); holding costs are larger with stochastic effects.
• Safety stock: R = dL + SS, SS = z σ_dL. Order-cycle service level = probability that demand during lead time does not exceed on-hand inventory; 95% (5% stockout risk) → z = 1.645. Slide graph: Q = 400, SS = 50, R = 250.

L10 — One-to-one distribution (Daganzo ch. 3)

Learning objectives: explain EOQ robustness; analyze lot sizing with variable demand (two special cases); explain continuous approximation (CA); apply CA to lot sizing and freight terminal location.

Lot sizing with constant demand (recap)

D' = demand rate, c_f = setup/transport cost, c_h = holding cost per item, v = shipment size.

z = \min_v \left\{ A v + \frac{B}{v} \right\} \text{ s.t. } v \le v_{max}, \qquad A = \frac{c_h}{D'}, \ B = c_f

v^* = \sqrt{B/A}, \quad z^* = 2\sqrt{AB} \ \ (\text{if } v_{max} = \infty) \qquad v^* = \min\left\{\sqrt{B/A}, v_{max}\right\}

At the unconstrained optimum, holding cost = shipping cost. If √(B/A) > v_max it is infeasible → set v* = v_max.

Robustness of EOQ

(a) Decision off by factor γ (v0 = γ v*):

\frac{z_0}{z^*} = \frac{1}{2}\left(\gamma + \frac{1}{\gamma}\right)

γ = 0.8 (20% error in v) → ½(0.8 + 1.25) = 1.025: only 2.5% cost penalty.

(b) Input off by factor δ:

A' = A\delta \Rightarrow v^{*\prime} = v^* \delta^{-1/2} \qquad B' = B\delta \Rightarrow v^{*\prime} = v^* \delta^{1/2}, \quad z^{*\prime} = z^* \delta^{1/2}

You can rough in costs and stay near-optimal.

Example PedLog: B = €100, A = €1.15, transport cost +10% (δ = 1.1). v* = √(100/1.15) = 9.33 → 9.33·√1.1 = 9.78; z* = 2√(100·1.15) = 21.45 → 21.45·√1.1 = 22.48. Both change by ≈ 5% for a 10% input change.

(c) Model error: if the true holding cost z_h(v) stays within a vertical band of width Δ around Av (Av − Δ/2 ≤ z_h(v) ≤ Av + Δ/2), the extra cost cannot exceed Δ/2. The EOQ form is forgiving.

Lot sizing with variable demand

D(t) = cumulative demand from 0 to t, D'(t) = variable demand rate, c_f per vehicle, c_h per item, v_max, horizon t_max. Decide shipment sizes v0 … v_{n−1} and receipt times t0 = 0, t1 … t_{n−1} to minimize shipping + holding cost. Split c_h = c_r (rent) + c_i (waiting/interest).

Special case 1 — rent dominates (c_i ≪ c_r, c_h ≈ c_r). Rent depends on peak inventory, not on when it is held; shipping cost is n·c_f regardless of sizes. So:

1. Minimize the largest shipment → make all shipments equal: v0 = … = v_{n−1} = D(t_max)/n.
2. Times: choose t_i with D(t_i) = (i/n) D(t_max), i = 0 … n−1 — dispatch barely enough to cover demand until the next shipment.
3. Choose n by minimizing cost; it reduces to EOQ with v = D(t_max)/n and average rate D' = D(t_max)/t_max:

\frac{cost}{time} = c_r \frac{D(t_{max})}{n} + c_f \frac{n}{t_{max}} \qquad \frac{cost}{item} = \frac{c_r}{D'} \cdot \frac{D(t_{max})}{n} + c_f \frac{n}{D(t_{max})}

(So v* = √(c_f D'/c_r) and n* ≈ D(t_max)/v*, rounded to an integer — summary computed here.)

Special case 2 — rent negligible (c_i ≫ c_r, c_h ≈ c_i; small but expensive items). Holding cost ∝ item-hours in inventory = shaded area between the arrivals step curve and D(t); no longer depends on n alone.

• Optimality condition: each line P_iP_{i+1} (top of a step to the next) must be parallel to the tangent of D(t) at the receiving time.
• Graphical procedure: 1) choose P1; 2) move across horizontally to T1 on D(t); 3) from P1 draw a line parallel to the tangent to D(t) at T1; 4) draw a vertical line from T1; 5) the intersection is P2; repeat until past t_max. If you don't end exactly at (t_max, D(t_max)), perturb P1 and redo. Insightful but tedious by hand (Exercise P3 asks you to improve an initial feasible solution this way).

Continuous approximation (CA)

CA replaces the search for discrete t_i with a search for a continuous function (e.g. headway H(t)) that then yields the t_i. Benefits: continuous functions are easier to optimize; works for large systems. Requirement: discrete elements vary slowly, D'(t_i) ≈ D'(t_{i+1}).

CA for lot sizing (case 2). Intervals I_i = [t_{i−1}, t_i], headway step function H_s(t) = t_i − t_{i−1} on I_i. Cost per interval = c_f + c_i·area_i with area_i = ½(t_i − t_{i−1})² D'(t'_i). Rewriting as an integral and replacing H_s(t) by a smooth H(t) and D'(t'_i) by D'(t):

cost \approx \int_0^{t_{max}} \left( \frac{c_f}{H(t)} + \frac{c_i H(t)}{2} D'(t) \right) dt

The integrand is an EOQ at every t, so:

H^*(t) = \sqrt{\frac{2 c_f}{c_i D'(t)}} \qquad cost^* \approx \int_0^{t_{max}} \sqrt{2 c_i c_f D'(t)}\, dt \qquad \frac{cost^*}{item} \approx \frac{\int_0^{t_{max}} \sqrt{\frac{2 c_i c_f}{D'(t)}}\, D'(t)\, dt}{\int_0^{t_{max}} D'(t)\, dt}

From H*(t) to real times: plot H*(t); draw a 45° line from t0; draw a horizontal line from a point P1 so that the area under the continuous H*(t) equals the area under the step; t1 is where the horizontal meets the 45° line; repeat from t1 for t2, etc.

CA for freight terminal location (space instead of time)

A line from 0 to d_max (corridor/country); terminals at d1, d2, …; a truck collects at the terminals and hauls to a depot at d̃ ≥ d_max. D(d) = cumulative flow up to d, D'(d) = demand density (items/km), c_s = fixed cost per terminal stop, c'_d = access cost per item-km.

• Line-haul cost = c_s·n_s; access cost = c'_d × item-km travelled to the nearest terminal.
• Same trade-off as EOQ: many stops → low access, high line-haul; few stops → the reverse.
• Terminal i serves the segment between midpoints m_{i−1} and m_i (m_i = (d_i + d_{i+1})/2); spacing s(d_i) = m_i − m_{i−1}. Access item-km in segment i ≈ ¼ s² D'(d_i) (average access distance is s/4).

\text{Total cost} \approx \int_0^{d_{max}} \left( \frac{c_s}{s(d)} + \frac{c'_d\, s(d)}{4} D'(d) \right) dd \qquad s^*(d) = 2\sqrt{\frac{c_s}{c'd D'(d)}} \qquad cost^* \approx \int_0^{d{max}} \sqrt{c_s c'_d D'(d)}\, dd

Denser zones (large D') → terminals closer together. Cost per item = cost* / D(d_max).

Example (FR–BE–NL): c_s = €850, c'_d = €2; cumulative demand 5,000 at 200 km, 15,000 at 400 km, 40,000 at 500 km → D' = 25 (FR), 50 (BE), 250 (NL) items/km.

cost^* \approx \sqrt{850 \cdot 2}\left(200\sqrt{25} + 200\sqrt{50} + 100\sqrt{250}\right) = 41.23 \times 3{,}995.3 \approx 164{,}732.6

(Computed here: optimal spacing s* = 8.2 km in FR, 5.8 km in BE, 2.6 km in NL; cost per item ≈ 164,733/40,000 ≈ €4.12.)

Accuracy of CA

Best when D'(t) or D'(d) varies slowly over one headway/spacing; still good with moderate variation (objective is convex in the local decision); can over- or under-predict in unfavourable cases.

Exam traps and checklist

The errors below are the ones the slides themselves warn about, or that their worked examples make easy.

• Error sign: E = A − F; over-forecasting gives a negative error. CFE shows bias; MAD and MSE do not.
• Exponential smoothing start: with no previous forecast, use the naive one (F2 = A1).
• Trend-adjusted ES: update S first, then T using the new S, then FIT = S + T.
• Seasonal index: divide by that year's average, then average the indices over the years.
• Covariance: the slides use 1/n. The determinant must be computed per class; the C2 likelihood dominates when x lies near C2's points.
• Pseudo-count: full Bayes uses 1/(n_ci + Π m_j); Naive Bayes uses 1/(n_ci + m_j) per attribute. Only apply it when the count is 0.
• Precision vs recall: precision = TP/(TP + FP), the purity of the cluster; recall = TP/P, how much of the class was captured.
• Lance–Williams: check that single link gives the min and complete link gives the max.
• EOQ units: D, H and S must all be on the same time basis. Reorder point R = d · L uses the daily rate if L is in days. In the Star example on the slides, c_h is per year but D' is per day.
• Constrained multi-item EOQ: first check whether Σ c_i EOQ_i ≤ C. Scale by m only if the constraint binds.
• EPQ: the holding term uses I_max = Q(1 − d/p), not Q.
• Periodic review: safety stock covers RP + L (√(RP+L) · σ_t), not only L.
• Newsvendor: build the full payoff matrix and weight each row by the demand probabilities.
• Modelling: list sets, parameters, variables, objective, constraints and domains. Either-or conditions need big-M with a binary variable. Never list two conflicting constraint sets together, as in the F/G regions or the no-overlap scheduling constraints.
• Fixed charge: min fy + p x combined with x ≤ My. Leaving out the linking constraint lets y = 0.
• Multi-knapsack: don't forget Σ_j x_ij ≤ 1. Bin packing uses = 1 together with CAP_j y_j.
• Coverage: a parameter a_ij multiplies x_j, never d_ij · x_j ≤ K. Covering gives a lower-or-equal objective compared with partitioning.
• Dijkstra: take the smallest unvisited tentative distance, update only unvisited neighbours, and record predecessors so you can backtrack the path.
• Cumulative diagrams: vertical distance = number of items; horizontal distance = time.
• Daganzo costs: rent scales with the maximum headway; transport cost per item falls with the average headway. Equal headways are best.
• v = min{√(B/A), v_max}:* always check capacity.
• EOQ robustness: z0/z* = ½(γ + 1/γ). An input changed by a factor δ changes v* and z* by √δ.
• Variable demand: when rent dominates, use equal shipment sizes timed at D(t_i) = (i/n)D(t_max). When rent is negligible, segment P_iP_{i+1} is parallel to the tangent at the receipt time.
• CA formulas: H*(t) = √(2c_f / (c_i D'(t))) for lot sizing; s*(d) = 2√(c_s / (c'_d D'(d))) for terminal location. Note the factor 2 and the ¼ from the average access distance of s/4.