Quick Revision

Compressed to what you must be able to reproduce from memory. Read top to bottom in about fifteen minutes.

Unit 1 — Linear Programming

Core idea

Allocate limited resources among competing activities to optimise one linear objective. Four blocks, every time: decision variables → objective function → constraints → non-negativity.

Must reproduce
Optimise Z = c₁X₁ + … + cₙXₙ s.t. Σ aᵢⱼXⱼ (≤,=,≥) bᵢ , Xⱼ ≥ 0
Four assumptions

Proportionality · Additivity · Divisibility (fails ⟹ ILP) · Certainty (tested by sensitivity analysis).

Keywords

feasible regionbasic feasible solutioncorner pointslacksurplusartificial variable

Graphical method — 6 steps

Equations → two points per line → shade feasible region → find corner points → evaluate Z at each in a table → pick best. Solve intersections simultaneously, never read them off the graph.

Simplex — the loop

Entering = most positive Cⱼ − Zⱼ. Leaving = smallest non-negative replacement ratio. New pivot row = old ÷ pivot element; every other row = old − (its pivot-column entry × new pivot row). Max stops at all Cⱼ − Zⱼ ≤ 0.

Five special cases — the identification signal
CaseSignal
Multiple optimaNon-basic variable with Cⱼ − Zⱼ = 0
UnboundedNo positive replacement ratio
InfeasibleArtificial variable stays basic at non-zero value
DegenerateBasic variable with Quantity = 0 (tie in ratios)
Duality

Max ↔ Min. m constraints → m dual variables; n variables → n dual constraints. RHS ↔ objective coefficients. Transpose the matrix. Max primal ⟹ dual constraints ≥.

Equality splits into two dual variables (+Y₃ − Y₄); a wrong-direction inequality is multiplied by −1, flipping every sign including the RHS.

Sensitivity

ΔZ = shadow price × ΔRHS, valid inside the allowable range. Binding = slack 0 and shadow price ≠ 0. Reduced cost 0 ⟹ variable is basic. A coefficient inside its range keeps the quantities but changes Z.

shadow pricereduced costallowable rangebinding constraintpost-optimality

Unit 2 — Integer Linear Programming

Core idea

LP plus an integrality restriction — because you cannot make 2.5 cars or select 0.4 of a project. It is the divisibility assumption of Unit 1 failing.

Three types

Pure (all integer) · Mixed (some integer) · Binary (0–1 decisions).

Logical constraints
Not both i and j Xᵢ + Xⱼ ≤ 1 Exactly one Xᵢ + Xⱼ = 1 j requires i Xⱼ ≤ Xᵢ At least k ΣXᵢ ≥ k Open-to-ship link Σⱼ Xᵢⱼ − C·Yᵢ = 0

Never omit the closing lineXᵢ are binary (0,1) or … and are integers. It is a full mark.

pure ILPmixed ILPbinaryfixed chargecutting stocklinking constraint

Unit 3 — Transportation & Assignment

Telling them apart

Supply and demand given ⟹ transportation. One-to-one matching of rows to columns ⟹ assignment.

The counting block
Variables = m × n Constraints = m + n (excluding NNC) Balanced? Assignment: m = n · Transportation: Σ supply = Σ demand To balance: add a DUMMY row/column, all costs 0, quantity = the shortfall
Constraint directions

Transportation — supply , demand =. Assignment — rows and columns both = 1.

Profit variant

Profit = Selling price(j) − Plant cost(i) − Transport cost(i,j), objective becomes Max.

balanceddummy rowcost matrixidle time matrixm + nm × n

⚠ Not in your notes

Unit 3's solution algorithms — North-West Corner Rule, Least Cost Method, VAM, MODI, and the Hungarian Method — are named in the session plan but are not covered anywhere in your notes, and neither paper asked for them. See the Unit 3 scope note.

The last hour before the exam

  1. Rehearse the duality conversion once with an equality constraint. It is Q1A in every paper.
  2. Say the counting block out loud: variables multiply, constraints add.
  3. Re-read the special-case identification table — especially degenerate vs multiple optima.
  4. Re-read the sensitivity four-step template: name the rule, quote the number, do the arithmetic, state the meaning.
  5. Remind yourself: "Formulate" means stop at the model.

Then go in. Units 1–3 alone can build a complete 50-mark paper.