Unit 3: Probability distributions, LP and game theory
Quantitative Techniques notes · PTU syllabus (MCOP103-18)
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Unit summary
Probability distributions model uncertainty; linear programming and game theory support optimal decisions. This unit covers the Binomial, Poisson and Normal distributions, formulation of LP problems, graphical and simplex (Big-M) solutions, degeneracy, duality and post-optimality analysis, and two-person zero-sum games with pure and mixed strategies, dominance and graphical solutions.
After this unit you can
- Apply Binomial, Poisson and Normal distributions
- Formulate and solve LPPs by graphical and simplex methods
- Explain degeneracy, duality and post-optimality (sensitivity) analysis
- Solve two-person zero-sum games including graphical solutions
PTU syllabus topics
- Binomial
- Poisson and Normal distributions with properties and applications
- formulation of linear programming problems
- graphic and Simplex (including Big-M) solution methods
- degeneracy
- duality
- post-optimality analysis
- two-person zero-sum games
- pure and mixed strategies
- dominance rule
- graphic solution to games
- 1Standard form
Add slack, surplus, artificial variables
- 2Initial table
Basic feasible solution
- 3Key column
Most negative Cj − Zj (max problem)
- 4Key row
Minimum ratio test
- 5Pivot and iterate
Until no improvement is possible
Topic 1
Binomial, Poisson and normal distributions
Use
Fixed n trials, two outcomes, constant p
Rare events in an interval
Formula
P(r) = nCr pʳ qⁿ⁻ʳ
P(r) = e⁻ᵐ mʳ / r!
Mean
np
m
Variance
npq
m
Normal distribution: a continuous, bell-shaped, symmetric distribution with mean = median = mode. Probabilities are found with Z = (X − μ)/σ and the normal table. About 68.27% of values lie within μ ± 1σ, 95.45% within μ ± 2σ and 99.73% within μ ± 3σ.
Example
Marks are normal with μ = 60 and σ = 10. For X = 75, Z = 1.5; the area from 0 to 1.5 is 0.4332, so P(X > 75) = 0.5 − 0.4332 = 0.0668 (about 6.7%).
Example
Binomial: 10% of items are defective; in a sample of 5, P(exactly one defective) = 5C1 (0.1)(0.9)⁴ = 5 × 0.1 × 0.6561 = 0.328. Poisson: on average 2 calls a minute; P(no call) = e⁻² = 0.135.
Topic 2
Linear programming: formulation and graphical method
Linear programming (LP) optimises a linear objective function subject to linear constraints and non-negativity.
- Components: decision variables, objective function, constraints, non-negativity.
- Assumptions: linearity (proportionality, additivity), divisibility, certainty, finiteness.
Example
A firm makes chairs (x) and tables (y). Profit ₹40 and ₹60. Carpentry: 2x + 3y ≤ 120 hours; painting: 2x + y ≤ 80 hours. Maximise Z = 40x + 60y.
- 1Plot each constraint as a line
- 2Identify the feasible region
- 3Find corner points
- 4Evaluate Z at each corner
- 5Choose the best value
- Corner points: (0, 0) Z = 0; (40, 0) Z = 1,600; (0, 40) Z = 2,400; intersection of 2x + 3y = 120 and 2x + y = 80 → y = 20, x = 30 → Z = 1,200 + 1,200 = 2,400. Two corners give the same Z — multiple optimal solutions (objective is parallel to the carpentry constraint).
- Special cases: multiple optima, unbounded solution, infeasible problem, redundant constraint.
Topic 3
Simplex method, artificial variables and Big-M
- Convert inequalities to equations: add slack (≤), subtract surplus and add artificial variables (≥ or =).
- 1
Standard form with slack/surplus/artificial variables
- 2
Initial basic feasible solution table
- 3
Compute Cj − Zj (net evaluation)
- 4
Entering variable
Most positive Cj − Zj (maximisation)
- 5
Leaving variable
Minimum ratio (b ÷ key column, positive only)
- 6
Pivot and update the table
- 7
Repeat until all Cj − Zj ≤ 0 (maximisation)
- Big-M method: artificial variables are given a very large penalty −M in a maximisation objective (+M for minimisation) so they leave the basis; if an artificial variable remains positive in the final table, the problem is infeasible.
- Two-phase method: alternative to Big-M — first minimise the sum of artificial variables.
- Duality: every LP (primal) has a dual; optimal values are equal; dual values give shadow prices of resources.
Topic 4
Degeneracy, duality and post-optimality analysis
- Degeneracy: a basic variable takes the value zero — ties in the minimum ratio; may cause cycling (resolved by perturbation or Bland's rule).
- Duality: every primal LP has a dual; if the primal maximises with ≤ constraints, the dual minimises with ≥ constraints; the objective values are equal at the optimum; dual variables = shadow prices (value of one more unit of a resource).
Variables
n decision variables
m dual variables (one per constraint)
Constraints
m constraints (≤)
n constraints (≥)
Objective coefficients
Become right-hand sides of the dual
Right-hand sides of the primal
Optimum
Max Z
Min W = Max Z
- Post-optimality (sensitivity) analysis: how the optimal solution changes with changes in objective coefficients, right-hand-side values (resources) or constraint coefficients — ranges of optimality and feasibility.
Example
If the shadow price of machine hours is ₹30, buying an extra hour for less than ₹30 increases profit — useful for capacity decisions.
Topic 5
Game theory: two-person zero-sum games
A game is a competitive situation where the gain of one player is the loss of the other (zero-sum).
- Terms: players, strategies (pure or mixed), payoff matrix (from the row player's view), value of the game, saddle point.
- Saddle point: an element that is the minimum of its row and the maximum of its column — maximin = minimax; both players use pure strategies.
- 1Check for a saddle point
Pure strategies if maximin = minimax
- 2Apply dominance
Remove dominated rows/columns
- 32 × 2 without saddle point
Odds/algebraic method for mixed strategies
- 42 × n or m × 2
Graphical method or sub-games
- 5Larger games
Linear programming
Probability A plays row 1
p = (d − c) ÷ [(a + d) − (b + c)]
Probability B plays column 1
q = (d − b) ÷ [(a + d) − (b + c)]
Value of the game
V = (ad − bc) ÷ [(a + d) − (b + c)]
Example
Matrix [3 −1; −2 4]: no saddle point. Denominator = (3 + 4) − (−1 − 2) = 10. p = (4 + 2) ÷ 10 = 0.6; q = (4 + 1) ÷ 10 = 0.5; V = (12 − 2) ÷ 10 = 1.
- Dominance rule: a row that is ≤ another row in every column (for the maximising player) can be deleted; a column ≥ another column (for the minimising player) can be deleted.
Graphical solution of 2 × n and m × 2 games
- For a 2 × n game, plot each column's expected payoff as a line against the row player's probability p (0 to 1); the highest point of the lower envelope (maximin) gives the optimal p and value; the two lines meeting there form a 2 × 2 sub-game solved by the formula.
- For an m × 2 game, use the lowest point of the upper envelope (minimax).
Key terms
- Poisson distribution
- Distribution for the number of rare events in an interval
- Shadow price
- Increase in the objective value from one extra unit of a resource
- Degeneracy
- A basic variable at zero value in an LP solution
- Dual
- The associated LP problem derived from the primal
- Dominance
- Removing inferior strategies to simplify a game
Quick revision
- Binomial np, npq; Poisson m, m; normal Z = (X − μ)/σ.
- LP: formulation, graphical corner points, simplex and Big-M.
- Duality: equal optimum values; shadow prices.
- Sensitivity analysis: ranges of optimality and feasibility.
- Games: saddle point, dominance, 2 × 2 formula, graphical method.
Important exam questions
Practice questions written to the PTU exam pattern for this unit's syllabus: short answers (Section A style) and long answers (Sections B and C style).
Short-answer questions
- Q1.State the properties of the normal distribution.
- Q2.When is the Poisson distribution used?
- Q3.What is a shadow price?
- Q4.What is degeneracy in LP?
- Q5.What is post-optimality analysis?
- Q6.What is a saddle point?
Long-answer questions
- Q1.Explain Binomial, Poisson and Normal distributions with examples.
- Q2.Formulate and solve an LPP by the simplex method.
- Q3.Explain duality and sensitivity analysis in LP.
- Q4.Explain the solution of two-person zero-sum games including the graphical method.
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