STPM Management Mathematics 950 cheat sheet

Standard form: maximise z = cᵀx subject to Ax ≤ b, x ≥ 0; convert ≥ constraints by multiplying by −1, and free variables by x = x⁺ − x⁻ with x⁺, x⁻ ≥ 0.

A basic feasible solution sets n − m variables (non-basic) to zero and solves the m remaining equations; vertices of the feasible region correspond to basic feasible solutions.

Simplex optimality test: a maximisation is optimal when every reduced cost c̄ⱼ = cⱼ − c_BᵀB⁻¹Aⱼ ≤ 0; otherwise enter the variable with the most positive c̄ⱼ (Dantzig rule).

Ratio test for the leaving variable: divide each positive entry of the entering column into the current RHS and choose the smallest ratio; a column with no positive entry means an unbounded objective.

Duality: primal max cᵀx, Ax ≤ b, x ≥ 0 has dual min bᵀy, Aᵀy ≥ c, y ≥ 0; the optimal values are equal (strong duality).

Complementary slackness: xⱼ > 0 implies the j-th dual constraint is tight, and yᵢ > 0 implies the i-th primal constraint is tight.

Shadow price of resource i equals the optimal dual variable yᵢ - the rate of change of z per unit increase in bᵢ, valid until the basis changes (range of feasibility).

Transportation problem: a basic solution has m + n − 1 occupied cells; use the MODI method (uᵢ + vⱼ = cᵢⱼ on occupied cells) and improve if any cᵢⱼ − uᵢ − vⱼ < 0.

Assignment problem is a special transportation problem with unit supplies; the Hungarian method subtracts row minima then column minima and covers zeros with the minimum number of lines.

Forward pass: earliest start ES = max(EF of predecessors), earliest finish EF = ES + duration; project duration equals the maximum EF over terminal activities.

Backward pass: latest finish LF = min(LS of successors), latest start LS = LF − duration; total float = LS − ES = LF − EF.

Critical path is the longest path through the network; every activity on it has zero total float, and any delay there delays the project.

Free float = ES(successor) − EF(activity); it is the delay possible without affecting any successor's earliest start.

Expected activity time in PERT: tₑ = (a + 4m + b)/6 with variance σ² = ((b − a)/6)², where a, m, b are optimistic, most likely, pessimistic.

PERT project variance is the sum of variances of critical activities only; P(T ≤ X) = Φ((X − Σtₑ)/√(Σσ²)).

Crashing: cost slope = (crash cost − normal cost)/(normal time − crash time); crash the cheapest-slope critical activity first, but re-identify the critical path after each reduction.

A dummy activity (zero duration, dashed arrow) enforces logical precedence without consuming time or resources.

Resource levelling shifts non-critical activities within their float to flatten peak resource demand; it does not change project duration.

Expected monetary value: EMV(alternative) = Σ (payoff × probability); choose the alternative with the highest EMV when probabilities are known.

Maximin (pessimist), maximax (optimist) and minimax regret criteria apply under uncertainty; minimax regret uses an opportunity-loss table where regret = column best − payoff.

Expected value of perfect information: EVPI = EV with perfect information − maximum EMV; it is the upper bound on what is worth paying for information.

Decision tree rollback: evaluate terminal payoffs, fold back at chance nodes using expected value, and choose the maximum at decision nodes.

Expected value of sample information = EV with sample information − EV without it; it never exceeds EVPI.

A risk-averse decision maker has a concave utility function for money; risk-neutral is linear and risk-seeking is convex.

Sensitivity analysis on a payoff p: EMV(A) = EMV(B) at the indifference probability, found by setting p·payoff₁ + (1−p)·payoff₂ equal for both alternatives.

Bayes' theorem revises prior probabilities after a survey: P(θ|S) = P(S|θ)P(θ) / Σ P(S|θᵢ)P(θᵢ).

EOQ = √(2DS/H), where D is annual demand, S is ordering cost per order, and H is holding cost per unit per year.

Number of orders per year = D/Q; total annual cost TC = (D/Q)S + (Q/2)H, and at the EOQ the ordering cost equals the holding cost.

Reorder level = demand per unit time × lead time (plus safety stock if demand or lead time is uncertain).

If H is given as a fraction i of unit cost C, then H = iC and EOQ = √(2DS/(iC)).

With planned shortages, EOQ = √(2DS/H · (H + B)/B) where B is the shortage cost per unit per year.

Quantity discounts: compare total cost including purchase cost DC at each break quantity with the EOQ-based cost; a discount is worthwhile only if the saving exceeds the extra holding cost.

Production run model (finite replenishment): EPQ = √(2DS/(H(1 − d/p))), where d is demand rate and p is production rate.

ABC classification: a small proportion of items (A) accounts for most of inventory value and warrants tight control; C items get loose control.

Kendall notation A/B/c/K/N/D describes arrival process, service process, number of servers, system capacity, population and discipline.

For M/M/1 with arrival rate λ and service rate μ: utilisation ρ = λ/μ < 1, L = ρ/(1 − ρ), Lq = ρ²/(1 − ρ).

Little's law: L = λW and Lq = λWq; also W = Wq + 1/μ, so Lq = L − ρ.

M/M/1 probabilities: P₀ = 1 − ρ and Pₙ = (1 − ρ)ρⁿ; probability the server is busy is ρ.

M/M/1 waiting time: Wq = ρ/(μ − λ) and W = 1/(μ − λ).

M/M/c: use ρ = λ/(cμ); P₀ = [Σ_{n=0}^{c−1} (λ/μ)ⁿ/n! + (λ/μ)^c/(c!(1 − ρ))]⁻¹ and Lq = P₀(λ/μ)^c ρ / (c!(1 − ρ)²).

M/M/c requires λ < cμ; as ρ approaches 1, Lq and Wq grow without bound, so queue length is highly sensitive near saturation.

M/G/1 (Pollaczek–Khinchine): Lq = (λ²σ² + ρ²)/(2(1 − ρ)), which reduces to ρ²/(1 − ρ) when service is exponential.

A saddle point exists if max over rows of row minima equals min over columns of column maxima; the value of the game is that common number.

A pure strategy saddle point is optimal for both players; if no saddle point exists, use mixed strategies.

Dominance: delete a row that is dominated by another row (all entries ≤, at least one <, for the maximising player) and a column dominated for the minimising player.

2×2 game with no saddle point: value v = (ad − bc)/(a + d − b − c) for payoff matrix [[a, b], [c, d]], with row player's probabilities p₁ = (d − c)/(a + d − b − c).

A zero-sum game has payoffs summing to zero; value v > 0 favours the row player, v < 0 favours the column player.

Mixed strategy probabilities are found by equalising the opponent's expected payoffs across the strategies they use.

A game is fair when its value is zero; a strictly determined game has a saddle point in pure strategies.

Graphical method solves 2×n or m×2 games by plotting expected payoffs against one player's probability and taking the lower (or upper) envelope.

Additive decomposition: Y = T + S + C + I; multiplicative: Y = T × S × C × I, with seasonal index typically averaging 100 (multiplicative) or 0 (additive).

Simple moving average of order k: forecast = mean of the last k observations; larger k gives smoother but slower-reacting forecasts.

Exponential smoothing: F_{t+1} = αYₜ + (1 − α)Fₜ, with 0 < α < 1; a higher α weights recent data more heavily.

Holt's linear trend: level Lₜ = αYₜ + (1 − α)(L_{t−1} + T_{t−1}); trend Tₜ = β(Lₜ − L_{t−1}) + (1 − β)T_{t−1}; forecast = Lₜ + hTₜ.

Mean absolute deviation MAD = Σ|eₜ|/n; mean squared error MSE = Σeₜ²/n; MAD relates to standard deviation by σ ≈ 1.25 MAD.

Tracking signal = running sum of forecast errors / MAD; values outside roughly ±4 signal a biased or out-of-control forecast.

Seasonal index (ratio-to-moving-average): index = (actual / centred moving average) × 100, then average by season and adjust so indices sum to 400 for quarterly data.

Naive forecast: F_{t+1} = Yₜ; it is the benchmark any forecasting model must beat.

Laspeyres price index = (Σpₙq₀ / Σp₀q₀) × 100, using base-year quantities; it tends to overstate price rises.

Paasche price index = (Σpₙqₙ / Σp₀qₙ) × 100, using current-year quantities; it tends to understate price rises.

Fisher's ideal index = √(Laspeyres × Paasche); it satisfies both the time-reversal and factor-reversal tests.

Marshall–Edgeworth index uses average quantities (q₀ + qₙ)/2 in both numerator and denominator.

Chain index: link relatives multiplied together, or index = (current link relative × previous chain index)/100.

Deflating: real value = (nominal value / price index) × 100, converting money values to base-year terms.

Weighted aggregate quantity index: Σp₀qₙ / Σp₀q₀ × 100 (base-year prices as weights).

Base shifting: new index = (old index / old index of new base year) × 100; splicing joins two series by rescaling the newer one to the old base.

Simple average of price relatives = (Σ(pₙ/p₀) × 100)/n, which ignores quantity weights and is therefore less representative.

Compound amount: A = P(1 + i)ⁿ; present value P = A(1 + i)⁻ⁿ, where i is the rate per period and n the number of periods.

Effective annual rate from nominal rate r compounded m times per year: EAR = (1 + r/m)^m − 1; continuous compounding gives eʳ − 1.

Annuity present value: PV = R[1 − (1 + i)⁻ⁿ]/i; future value: FV = R[(1 + i)ⁿ − 1]/i, for payments R at each period end.

Annuity due multiplies the ordinary annuity value by (1 + i); a perpetuity has PV = R/i.

Sinking fund payment: R = FV · i/[(1 + i)ⁿ − 1], the periodic deposit needed to accumulate a target sum.

Loan amortisation: each payment splits into interest (i × opening balance) and principal repayment; the outstanding balance equals the PV of remaining payments.

Net present value: NPV = Σ Cₜ(1 + i)⁻ᵗ − C₀; accept a project when NPV > 0.

Internal rate of return is the i making NPV = 0; a project is acceptable when IRR exceeds the cost of capital.

Depreciation: straight line gives annual charge (C − S)/n; reducing balance gives book value C(1 − r)ⁿ with rate r = 1 − (S/C)^(1/n).

Break-even: contribution per unit = price − variable cost; break-even quantity = fixed costs / contribution per unit.

Markov chain steady state: solve π = πP with Σπᵢ = 1; the long-run share of state j is πⱼ.

Input–output (Leontief) model: gross output x satisfies x = Ax + d, so x = (I − A)⁻¹d, where A is the technical coefficient matrix.

Learning curve: time for the nth unit = T₁n^b with b = log(learning rate)/log 2; an 80% curve gives b ≈ −0.322.

Simple and weighted aggregate indices, moving averages and ratios are the standard tools for building management models from raw operating data.

Elasticity of demand: E = (dQ/dP)(P/Q); revenue is maximised where |E| = 1.

Cost–volume–profit: profit = (price − variable cost) × quantity − fixed costs; margin of safety = (actual sales − break-even sales)/actual sales.

Simulation modelling: generate random variates from a cumulative distribution, then average outcomes over many trials to estimate expected performance.