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Decision Theory and Network Analysis

Operations Research

Decision Theory and Network Analysis

Syllabus tag: KASNEB CPA | Foundation Level | CA15 Quantitative Analysis | Topic 10 Decision Theory and Network Analysis

Lesson objectives

By the end of this topic, you will be able to:

  • Construct a payoff table and apply the expected value criterion
  • Apply the maximax, maximin and minimax regret rules
  • Compute the value of perfect information
  • Construct a network and identify the critical path
  • Compute float and interpret it

Why this matters

Decision theory supplies a defensible method where outcomes are uncertain, and network analysis supplies one where activities compete for time. Both convert an argument into a calculation.

One payoff table throughout

Three options, three states of demand, payoffs in KES thousands:

Low (0.30)Medium (0.45)High (0.25)
A407090
B2085120
C556065

Expected value

EV = Σ (probability × payoff)

OptionWorkingEV
A(0.3 × 40) + (0.45 × 70) + (0.25 × 90)66.00
B(0.3 × 20) + (0.45 × 85) + (0.25 × 120)74.25
C(0.3 × 55) + (0.45 × 60) + (0.25 × 65)59.75

On expected value, choose B.

This criterion assumes the decision-maker is risk-neutral and that the decision is repeatable. For a one-off decision that could threaten the business, B's low payoff of 20 may matter more than its higher average.

Decision rules without probabilities

Where probabilities cannot be assigned, three rules apply — and they disagree, which is the point of learning all three.

Maximax — the optimist. Take the best of the best payoffs. Best of A = 90, B = 120, C = 65. Choose B.

Maximin — the pessimist. Take the best of the worst payoffs. Worst of A = 40, B = 20, C = 55. Choose C.

Minimax regret — the regret-averse. Build a regret table showing what is lost by not having chosen the best option for each state.

Best payoff in each state: Low 55, Medium 85, High 120.

RegretLowMediumHighMaximum
A15153030
B350035
C0255555

Choose the option with the smallest maximum regret: A, with 30.

Three rules, three different answers — B, C and A. None is more correct than the others; each expresses a different attitude to risk. An examiner expects the candidate to compute all three and say what each reveals about the decision-maker.

:::checkpoint A board applies maximin and chooses C. A shareholder objects that C has the lowest expected value. Explain how both can be right and what the board would need to justify its choice. :::

The value of perfect information

If the state of demand were known in advance, the best option for each state could be chosen every time:

EV with perfect information = (0.3 × 55) + (0.45 × 85) + (0.25 × 120) = 16.5 + 38.25 + 30 = 84.75

EVPI = 84.75 − 74.25 = KES 10,500

That is the maximum worth paying for a perfect forecast. Since no forecast is perfect, real research is worth less — which is the practical use of the figure. A market study quoted at 15,000 can be declined without further analysis.

Network analysis

A network shows activities, their durations and their dependencies.

Activities: A (2 weeks), then B (4) and C (3) in parallel; B is followed by E (2) and C by D (5).

PathDuration
A → B → E8 weeks
A → C → D10 weeks

The critical path is A → C → D, at 10 weeks. It is the longest path, because every activity on it must finish before the project can. The shortest possible project duration is the longest path — which sounds contradictory until you see why.

Float is the time an activity can be delayed without delaying the project:

Float on path A → B → E = 10 − 8 = 2 weeks

Activities on the critical path have zero float. A one-week delay to C delays the whole project by a week; a one-week delay to B delays nothing.

Crashing means shortening the project by adding resources to critical activities. Two rules follow: spending on a non-critical activity is wasted, and once a critical activity is shortened enough, a different path becomes critical and further spending there achieves nothing.

:::checkpoint A manager proposes to speed up activity B by hiring extra staff, at a cost of KES 80,000. Using the network above, explain what this would achieve. :::