Unit 07 β Decision Theory & AHP
Choosing among actions when the future is uncertain: the five uncertainty criteria, EMV and the value of perfect information, decision trees with and without survey information, and AHP priority weights β 15 of the 50 marks in each 2025-26 paper.
This is the highest-scoring unit of the Term End portion. In each 2025-26 paper it supplied Q1C (AHP, 2.5), Q1D (uncertainty criteria, 2.5), Q6A (decision tree, 5) and Q6B (Hurwicz, 5) β 15 of the 50 marks Final QP pp. 1β5 Re-Exam QP pp. 1β5.
- Q1C β AHP priorities from a pairwise matrix (concept 7). Both solved β
- Q1D β build the payoff matrix, apply Maximin/Maximax or Laplace (concepts 1β2). Solved β
- Q6A β draw a decision tree, compute EMVs; the Re-Exam version adds survey information (concepts 5β6). Solved β
- Q6B β Hurwicz with a given Ξ± (concept 3). Solved β
1. Decision Environments and the Payoff Table Core syllabus concept
1Understand the Concept
Every decision problem has three ingredients: the courses of action (acts, strategies β what you control), the states of nature (events β what you do not control: demand, the economy, a strike), and the payoff of each actβevent pair Lecture p. 4. Arrange them as a table with acts as rows and states as columns.
2Three environments β which tells you which tool
| Environment | What is known about the states | Tools |
|---|---|---|
| Certainty | Which state will occur | Pick the act with the best payoff in that state (LP is decision-making under certainty). |
| Uncertainty | The possible states, but no probabilities | Laplace Β· Maximax/Minimin Β· Maximin/Minimax Β· Savage (regret) Β· Hurwicz β concept 2 |
| Risk | The states and their probabilities | EMV Β· EOL Β· EPPI Β· EVPI β concept 4; decision trees β concept 5 |
No probabilities given βΉ uncertainty criteria. Probabilities given βΉ EMV. That single check tells you which half of this unit the question is from.
3Building the payoff table from a paragraph β Final Q1D part 1
Three strategies β mass production, custom manufacturing, limited edition β against high, moderate and low demand. High: 120, 100, 80. Moderate: 70, 85, 60. Low: β40 (a loss), 20, 40 (βΉ lakh). Construct the payoff matrix.
- Rows = acts, columns = states
Course of action High (Sβ) Moderate (Sβ) Low (Sβ) Mass production (Aβ) 120 70 β40 Custom manufacturing (Aβ) 100 85 20 Limited edition (Aβ) 80 60 40 The paragraph is written by state ("if demand is highβ¦"), but the table is laid out by act β so each sentence fills a column. A loss goes in as a negative number.
2. Decision under Uncertainty β the Five Criteria Asked in the 2025-26 papers
1The rules β profit table and cost table side by side
| Criterion | Attitude | Profit (payoff) table | Cost table |
|---|---|---|---|
| Laplace (equal probability) | Neutral β every state equally likely | Average each row; choose the highest | Average each row; choose the lowest |
| Maximax / Minimin | Optimist | Best (max) of each row; choose the max of the maxima | Best (min) of each row; choose the min of the minima |
| Maximin / Minimax | Pessimist | Worst (min) of each row; choose the max of the minima | Worst (max) of each row; choose the min of the maxima |
| Savage β minimax regret | Avoid regret | Regret = column max β payoff; choose the act with the smallest maximum regret | Regret = cost β column min; same rule |
| Hurwicz (Ξ± = optimism) | In between | Ξ± Γ row max + (1 β Ξ±) Γ row min; choose the highest | Ξ± Γ row min + (1 β Ξ±) Γ row max; choose the lowest |
2The lecture example, worked through every criterion
| N1 | N2 | N3 | N4 | |
|---|---|---|---|---|
| S1 | 1000 | 1500 | 750 | 0 |
| S2 | 250 | 2000 | 3750 | 3000 |
| S3 | β500 | 1250 | 3000 | 4750 |
| S4 | β1250 | 500 | 2250 | 4000 |
Treat it as a profit table and apply all five criteria (Ξ± = 0.6).
- Laplace β average of each row S1 = (1000 + 1500 + 750 + 0)/4 = 812.5 S2 = 2250 S3 = 2125 S4 = 1375 Choose S2 (2250)
- Maximax β the optimist Row maxima: 1500, 3750, 4750, 4000 Choose S3 (4750)
- Maximin β the pessimist Row minima: 0, 250, β500, β1250 Choose S2 (250)
- Savage β build the regret table (column max β entry)
Choose S3 β smallest maximum regret (1500)Regret N1 N2 N3 N4 Max regret S1 0 500 3000 4750 4750 S2 750 0 0 1750 1750 S3 1500 750 750 0 1500 S4 2250 1500 1500 750 2250 - Hurwicz, Ξ± = 0.6 S1 = 0.6(1500) + 0.4(0) = 900 S2 = 0.6(3750) + 0.4(250) = 2350 S3 = 0.6(4750) + 0.4(β500) = 2650 S4 = 0.6(4000) + 0.4(β1250) = 1900 Choose S3 (2650)
- Read the result as a manager
Different attitudes pick different acts: the pessimist and the Laplace decision-maker choose S2, the optimist, the regret-minimiser and Hurwicz choose S3. The examiner wants the criterion named with each answer β "by the Maximin criterion, choose S2".
- Computing regret row-wise. It is column-wise: subtract from the best value in that state.
- Applying a profit rule to a cost table. For costs the optimist picks the minimum of the minima. Decide which kind of table it is before you start β the steel-strike problem in Assignment 8 is a cost table.
- Forgetting to state ties. In Re-Exam Q1D two strategies share the highest Laplace average (60) and the key accepts both β say so Key p. 2.
- Reporting the number without the act. "Maximax = 120" earns half; "choose mass production (maximax payoff βΉ120 lakh)" earns it all.
3. The Hurwicz Criterion in Detail Asked in the 2025-26 papers
1Understand the Concept
Hurwicz lets the decision-maker say how optimistic they are, with a coefficient Ξ± between 0 and 1. Ξ± close to 1 is optimistic, close to 0 pessimistic. Your lecture says to take Ξ± = 0.6 if none is given Lecture p. 13.
2Worked Example β Final Exam Q6B
| Strategy | High | Moderate | Low |
|---|---|---|---|
| Social media (M1) | 120 | 70 | 20 |
| Email (M2) | 100 | 80 | 30 |
| Influencer (M3) | 150 | 60 | 10 |
Ξ± = 0.6. Which strategy does the Hurwicz criterion choose?
- Step 1 β Row max and min, then WO
Strategy Max Min WO = 0.6Β·Max + 0.4Β·Min M1 120 20 72 + 8 = 80 M2 100 30 60 + 12 = 72 M3 150 10 90 + 4 = 94 - Step 2 β Decide
Choose M3, Influencer Partnerships (WO = 94)
β Matches the key Key p. 7.
Re-Exam Q6B gives options U1 (180, 90, 40), U2 (140, 100, 60), U3 (160, 80, 50) with Ξ± = 0.7 QP p. 5. The key's Max/Min columns are (120, 20), (100, 30), (150, 10) β those are the Final paper's numbers β and so it picks U3 with 108 Key p. 6. With the Re-Exam's own data: U1 = 0.7(180) + 0.3(40) = 138, U2 = 0.7(140) + 0.3(60) = 116, U3 = 0.7(160) + 0.3(50) = 127. The correct choice is U1, upgrading machinery.
For a cost table the optimistic weight belongs on the lowest cost: WO = Ξ± Γ (row min) + (1 β Ξ±) Γ (row max), choose the smallest. Your lecture slide keeps Ξ± on the row maximum even for costs Lecture p. 14 β on every problem in your assignments both versions pick the same act, but write the formula you use so the examiner can follow it.
4. Decision under Risk β EMV, EOL, EPPI, EVPI Core syllabus concept
1The four quantities
2Worked Example β the lecture's risk table
| N1 | N2 | N3 | |
|---|---|---|---|
| S1 | 65 | 60 | 50 |
| S2 | 60 | 70 | 80 |
| S3 | 75 | 55 | 60 |
| Probability | 0.2 | 0.3 | 0.5 |
Find EMV, EOL, EPPI and EVPI.
- Step 1 β EMV of each act S1 = 65(0.2) + 60(0.3) + 50(0.5) = 13 + 18 + 25 = 56 S2 = 12 + 21 + 40 = 73 S3 = 15 + 16.5 + 30 = 61.5 Best EMV = 73 βΉ choose S2
- Step 2 β EPPI and EVPI Best in each state: N1 β 75, N2 β 70, N3 β 80 EPPI = 75(0.2) + 70(0.3) + 80(0.5) = 76 EVPI = 76 β 73 = 3
- Step 3 β EOL from the regret table
Least EOL = 3 (S2) β the same act as the best EMV, and equal to the EVPI βRegret N1 N2 N3 EOL S1 10 10 30 2 + 3 + 15 = 20 S2 15 0 0 3 + 0 + 0 = 3 S3 0 15 20 0 + 4.5 + 10 = 14.5
The slide gives EOL(S2) = 7.5 Lecture p. 17. S2's regrets are 15, 0, 0, so EOL = 15 Γ 0.2 = 3 (7.5 would need the 0.5 probability against the 15). The quick check catches it: the least EOL must equal the EVPI, which is 3.
5. Decision Trees Asked in the 2025-26 papers
1Understand the Concept
A decision tree draws the same problem as a picture, left to right in time order. It is the natural tool once decisions come in stages ("survey first, then decide"), but examiners also ask for it on a one-stage problem just to see that you can draw it.
| Symbol | Meaning | Value written on it |
|---|---|---|
| β Decision node | You choose one branch | The best branch's value; cut the others (β₯) |
| β Chance node | Nature chooses a branch with given probabilities | Its EMV = Ξ£ payoff Γ probability |
| End of branch | Final outcome | The payoff |
2The method β draw forward, solve backward
- Draw the tree left to right: decision node, one branch per act, a chance node on each, one branch per state with its probability, the payoff at each tip.
- Roll back from right to left: at every chance node write its EMV; at every decision node keep the best branch and mark the others as cut.
- The value at the first decision node is the answer; the uncut path is the decision.
3Worked Example β Final Exam Q6A
A glass factory with a backlog can subcontract (S1), run overtime (S2) or build new facilities (S3). Demand low/medium/high has probabilities 0.1, 0.5, 0.4. Profits: S1 = 20, 60, 60; S2 = β10, 70, 100; S3 = β140, 30, 210. Draw the decision tree, find each EMV and recommend.
- Step 1 β EMV at each chance node (roll back) S1: 20(0.1) + 60(0.5) + 60(0.4) = 2 + 30 + 24 = 56 S2: β10(0.1) + 70(0.5) + 100(0.4) = β1 + 35 + 40 = 74 S3: β140(0.1) + 30(0.5) + 210(0.4) = β14 + 15 + 84 = 85
- Step 2 β Decision node
Choose S3, construct new facilities β EMV 85
Add one line on risk for the evaluation mark: S3 has the highest EMV but also the only large loss (β140 if demand is low); with only a 10% chance of that, the EMV rule still favours it. β Matches the key Key p. 6.
6. Decision Trees with Sample Information (Bayesian Analysis) Core syllabus concept
1Understand the Concept
A market survey costs money and gives an imperfect signal β favourable or unfavourable. If it comes back favourable, demand is more likely to be high, so the probabilities you use on that branch are the conditional (posterior) ones, P(High | F) and so on. The question is whether the survey's information is worth its fee.
2Worked Example β the lecture's make-or-buy problem
Manufacture or purchase a component. Payoffs (βΉ'000): manufacture 200 / 60 / β30, purchase 140 / 80 / 20 for high / medium / low demand, prior probabilities 0.3, 0.4, 0.3. Market research costs βΉ10,000 and is favourable with P(F) = 0.4. P(H|F) = 0.45, P(M|F) = 0.40, P(L|F) = 0.15; P(H|U) = 0.20, P(M|U) = 0.40, P(L|U) = 0.40. Should the company do the research?
- Step 1 β Without research (prior probabilities) Manufacture = 200(0.3) + 60(0.4) β 30(0.3) = 60 + 24 β 9 = 75 Purchase = 140(0.3) + 80(0.4) + 20(0.3) = 42 + 32 + 6 = 80 Best without research: purchase, EMV 80
- Step 2 β If research is favourable (use P(Β·|F)) Manufacture = 200(0.45) + 60(0.40) β 30(0.15) = 90 + 24 β 4.5 = 109.5 Purchase = 140(0.45) + 80(0.40) + 20(0.15) = 63 + 32 + 3 = 98 βΉ manufacture, 109.5
- Step 3 β If research is unfavourable (use P(Β·|U)) Manufacture = 200(0.2) + 60(0.4) β 30(0.4) = 40 + 24 β 12 = 52 Purchase = 140(0.2) + 80(0.4) + 20(0.4) = 28 + 32 + 8 = 68 βΉ purchase, 68
- Step 4 β Value of the research EV with research = 0.4(109.5) + 0.6(68) = 43.8 + 40.8 = 84.6 EVSI = 84.6 β 80 = 4.6 (βΉ4,600) Net of its cost: 84.6 β 10 = 74.6 < 80
- Step 5 β Decision
The research is worth only βΉ4,600 but costs βΉ10,000 β do not commission it; purchase the component (EMV βΉ80,000). Consistency check: 0.4(0.45) + 0.6(0.2) = 0.30 = P(High) β, so the conditional probabilities agree with the prior.
- Using the prior probabilities on the survey branches. After a favourable report, use P(H|F), not P(H).
- Forgetting to subtract the survey cost β or subtracting it twice. Subtract it once, at the very first decision.
- Answering only "survey: yes/no". The full decision is a rule: "no survey; purchase" β or "survey; manufacture if favourable, purchase if not".
7. Analytic Hierarchy Process (AHP) Asked in the 2025-26 papers
1Understand the Concept
AHP (Saaty) turns judgements of the form "A is moderately more important than B" into numerical priority weights. Instead of scoring every option on an absolute scale, the decision-maker compares them two at a time β which people find much easier Lecture p. 19.
| Intensity | Definition |
|---|---|
| 1 | Equal importance |
| 3 | Moderate importance |
| 5 | Strong importance |
| 7 | Very strong importance |
| 9 | Extreme importance |
| 2, 4, 6, 8 | Intermediate values ("equally to moderately" = 2) |
| 1/3, 1/5, β¦ | The inverse comparison |
2The method your keys use
- Build the pairwise matrix. Diagonal = 1. If row A vs column B = k, then row B vs column A = 1/k (the reciprocal rule) β this is how you "complete the matrix".
- Add each column.
- Normalise: divide every entry by its column total (each column now sums to 1).
- Average each row of the normalised matrix β these are the priority weights.
- Rank the options by weight and interpret: highest = most preferred.
The weights must add to 1 (allow Β±0.001 for rounding), and the ranking should agree with the raw matrix β the row with the biggest numbers comes first.
3Worked Example β Final Exam Q1C
| Ahmedabad | Hyderabad | Delhi | |
|---|---|---|---|
| Ahmedabad | 1 | 5 | 7 |
| Hyderabad | 1 | 3 | |
| Delhi | 1 |
Complete the matrix, find the priority weights, and identify the most and least preferred city.
- Step 1 β Fill the lower half with reciprocals, then add the columns
Ahmedabad Hyderabad Delhi Ahmedabad 1 5 7 Hyderabad 1/5 = 0.200 1 3 Delhi 1/7 = 0.143 1/3 = 0.333 1 Column total 1.343 6.333 11 - Step 2 β Normalise and average each row
Check: 0.724 + 0.193 + 0.083 = 1.000 βAhmedabad Hyderabad Delhi Priority Ahmedabad 0.745 0.789 0.636 0.724 Hyderabad 0.149 0.158 0.273 0.193 Delhi 0.106 0.053 0.091 0.083 - Step 3 β Interpret
Most preferred: Ahmedabad (0.724). Least preferred: Delhi (0.083)
Hyderabad's row reads 1/5 against Ahmedabad and 3 against Delhi: it is strongly less desirable than Ahmedabad but moderately more desirable than Delhi β exactly what part 1 asks you to state. β Matches the key Key pp. 1β2.
4Consistency ratio β beyond the lecture, but worth one line
AHP can also check whether the judgements contradict each other. Multiply the original matrix by the weight vector, divide each result by its weight and average to get Ξ»max; then CI = (Ξ»max β n)/(n β 1) and CR = CI / RI, with RI = 0.58 for n = 3 and 0.90 for n = 4. CR below 0.10 is acceptable. For Final Q1C: Ξ»max = 3.066, CI = 0.033, CR = 0.057 β consistent.
The Re-Exam's flavour panel is a useful contrast: FizzUp is judged 3Γ CoolMist and CoolMist 5Γ Zestora, which implies FizzUp β 15Γ Zestora β yet the panel rated FizzUp only 2Γ Zestora. Its CR is 0.41, so those judgements should be revisited. The key does not ask for this, but one sentence on it shows real understanding Re-Exam QP pp. 1β2.
5Overall ranking with several criteria
When there are several criteria, repeat the method twice: once to weight the criteria (the lecture's car example gives Price 0.525, Comfort 0.27, Fuel efficiency 0.173, Colour 0.04 Lecture p. 25), and once per criterion to score the options. Each option's overall score is then Ξ£ (criterion weight Γ option's score on that criterion) β highest wins. Illustration: with quality weighted 0.7 and cost 0.3, supplier scores of (0.274, 0.639, 0.087) on quality and (0.5, 0.2, 0.3) on cost give overall 0.342, 0.507, 0.151 β supplier 2 wins.
- Putting k instead of 1/k in the mirror cell. If A vs B = 5, then B vs A = 1/5 = 0.2, never 5.
- Averaging the original rows instead of the normalised ones.
- Normalising by row totals. Divide by the column total.
- Reading the verbal scale wrongly β "equally to moderately" is 2, "moderately to strongly" is 4 (Re-Exam Q1C prints its own 1β7 table; use the one given in the question).