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QUANTITATIVE TECHNIQUESβ€ΊASSIGNMENTSβ€ΊASSIGNMENT 8

Assignment 8 β€” Decision Theory

All five uncertainty criteria on profit and cost tables, EMV/EPPI/EVPI on a stocking problem, a three-option decision tree with a risk analysis, and two survey (Bayesian) decision trees β€” every tree drawn and rolled back.

PROBLEMS: 7 UNIT: UNIT 07 β€” DECISION THEORY Original: Assignment 8
β—† The method for every problem on this page
  1. No probabilities ⟹ uncertainty criteria. Decide first whether the table is profit or cost.
  2. Profit: Maximax (optimistic) = max of row maxima Β· Maximin (pessimistic) = max of row minima Β· Savage = build regret column-wise (column max βˆ’ entry), choose the smallest maximum regret Β· Laplace = highest row average Β· Hurwicz = Ξ±Β·max + (1 βˆ’ Ξ±)Β·min, highest.
  3. Cost: minimin, minimax, regret = entry βˆ’ column min, lowest average, and Hurwicz with Ξ± on the lowest cost β€” always choose the lowest.
  4. Probabilities given ⟹ EMV = Ξ£ payoff Γ— probability (choose highest); EPPI = Ξ£ best-in-column Γ— probability; EVPI = EPPI βˆ’ best EMV; EOL from the regret table (least EOL = EVPI).
  5. Trees: draw left to right, roll back right to left β€” EMV at every chance node, best branch at every decision node; with a survey, use the conditional probabilities on each survey branch and compare EV with the survey (minus its cost) against EV without.

Theory and lecture examples: Unit 7 Β· Decision Theory.

β˜… Tricks that save time in the exam
  • Name the criterion with every answer β€” "by the Savage criterion choose C" β€” the criterion name carries the mark.
  • Regret tables always have a 0 in every column (the best act for that state). No zero in a column ⟹ you subtracted from the wrong value.
  • EVSI below the survey cost ⟹ skip the survey. EVSI = EV with survey (before its cost) βˆ’ EMV without it.
  • Ξ± not given? Your lecture's default is 0.6.

Q1 β€” Mr Senthil's β‚Ή10,000 β€” five criteria

Question
StrategyInflationRecessionNo change
A200012001500
B30008001000
C250010001800

Find the optimistic, pessimistic, regret, Laplace and Hurwicz (Ξ± = 0.7) decisions.

Assignment 8 p. 1

Solution
  1. Step 1 β€” Row max, row min, averages, Hurwicz
    StrategyInflationRecessionNo changeRow maxRow minAverage (Laplace)Hurwicz (Ξ± = 0.7)
    A200012001500200012001566.671760
    B30008001000300080016002340
    C250010001800250010001766.672050

    Hurwicz: A = 0.7(2000) + 0.3(1200) = 1760; B = 0.7(3000) + 0.3(800) = 2340; C = 0.7(2500) + 0.3(1000) = 2050.

  2. Step 2 β€” Regret table (column max βˆ’ entry: 3000, 1200, 1800)
    RegretInflationRecessionNo changeMax regret
    A100003001000
    B0400800800
    C5002000500
  3. Step 3 β€” Decisions
    CriterionRuleChoice
    Optimistic (maximax)max of row maximaB (3000)
    Pessimistic (maximin)max of row minimaA (1200)
    Regret (Savage)min of max regretsC (500)
    Laplacehighest averageC (1766.67)
    Hurwicz Ξ± = 0.7highest weighted valueB (2340)

    An optimist (or Hurwicz with Ξ± = 0.7) backs B for its big inflation return; a cautious investor takes A; C is the balanced choice that minimises regret and has the best average.

Q2 β€” Steel company and a possible strike β€” a COST table

Question
StrategyStrikeNo strike
A: stockpile20,00020,000
B: do not stockpile60,0000

The figures are costs. Find the optimistic, pessimistic, regret, Laplace and Hurwicz (Ξ± = 0.4) decisions.

Assignment 8 p. 1

Solution
  1. Step 1 β€” It is a cost table, so every criterion picks the LOWEST
    StrategyStrikeNo strikeRow maxRow minAverage (Laplace)Hurwicz (Ξ± = 0.4)
    A: stockpile200002000020000200002000020000
    B: do not stockpile6000006000003000036000

    Hurwicz for costs puts Ξ± on the best (lowest) cost: A = 0.4(20,000) + 0.6(20,000) = 20,000; B = 0.4(0) + 0.6(60,000) = 36,000.

  2. Step 2 β€” Regret = cost βˆ’ column minimum (20,000 and 0)
    RegretStrikeNo strikeMax regret
    A: stockpile02000020000
    B: do not stockpile40000040000
  3. Step 3 β€” Decisions
    CriterionRule (cost)Choice
    Optimistic (minimin)min of row minimaB (0)
    Pessimistic (minimax)min of row maximaA (20,000)
    Regret (Savage)min of max regretsA (20,000)
    Laplacelowest averageA (20,000 vs 30,000)
    Hurwicz Ξ± = 0.4lowest weighted costA (20,000 vs 36,000)
    Stockpile (A) under four of the five criteria

    Only the pure optimist gambles on no strike. Your lecture slide applies Ξ± to the row maximum even for costs (B = 0.4 Γ— 60,000 = 24,000) β€” that version also picks A, so the answer does not change.

Q3 β€” Furniture production scale

Question
StrategyHigh demandModerateLow
Large-scale12060βˆ’20
Moderate-scale905510
Small-scale604025

Best strategy by the optimistic, pessimistic, regret, Laplace and Hurwicz (Ξ± = 0.55) criteria.

Assignment 8 p. 1

Solution
  1. Step 1 β€” Row values
    StrategyHighModerateLowRow maxRow minAverage (Laplace)Hurwicz (Ξ± = 0.55)
    Large12060βˆ’20120βˆ’2053.3357
    Moderate905510901051.6754
    Small604025602541.6744.25

    Hurwicz: Large = 0.55(120) + 0.45(βˆ’20) = 66 βˆ’ 9 = 57; Moderate = 49.5 + 4.5 = 54; Small = 33 + 11.25 = 44.25.

  2. Step 2 β€” Regret (column max 120, 60, 25)
    RegretHighModerateLowMax regret
    Large004545
    Moderate3051530
    Small6020060
  3. Step 3 β€” Decisions
    CriterionChoice
    OptimisticLarge (120)
    PessimisticSmall (25)
    RegretModerate (max regret 30)
    LaplaceLarge (53.33)
    Hurwicz Ξ± = 0.55Large (57)

    Large-scale wins on three criteria; the cautious manager picks small-scale; moderate-scale is the regret-minimising compromise.

Q4 β€” Produce or license a new drug β€” with a forecast study

Question
AlternativeHighMediumLow
Produce260100βˆ’50
License18012030
Probability0.350.350.30

A clinical forecast study costing β‚Ή20,000 predicts favourable (F) or unfavourable (U): P(F) = 0.4, P(U) = 0.6; P(H|F) = 0.5, P(M|F) = 0.35, P(L|F) = 0.15; P(H|U) = 0.2, P(M|U) = 0.3, P(L|U) = 0.5. Decide whether to do the study, and what to do. (Payoffs taken in β‚Ή'000, as in your lecture's version of this problem, so the study costs 20.)

Assignment 8 p. 1

Solution
  1. Step 1 β€” Without the study Produce = 260(0.35) + 100(0.35) βˆ’ 50(0.30) = 91 + 35 βˆ’ 15 = 111 License = 180(0.35) + 120(0.35) + 30(0.30) = 63 + 42 + 9 = 114 ⟹ License, EMV 114
  2. Step 2 β€” If the study is favourable Produce = 260(0.5) + 100(0.35) βˆ’ 50(0.15) = 130 + 35 βˆ’ 7.5 = 157.5 License = 90 + 42 + 4.5 = 136.5 ⟹ Produce, 157.5
  3. Step 3 β€” If unfavourable Produce = 260(0.2) + 100(0.3) βˆ’ 50(0.5) = 52 + 30 βˆ’ 25 = 57 License = 36 + 36 + 15 = 87 ⟹ License, 87
  4. Step 4 β€” Is the study worth it? EV with study = 0.4(157.5) + 0.6(87) = 63 + 52.2 = 115.2 EVSI = 115.2 βˆ’ 114 = 1.2 (β‚Ή1,200) Net of cost: 115.2 βˆ’ 20 = 95.2 < 114
  5. Step 5 β€” Decision tree and answer
    No studyProduceH .35260M .35100L .30βˆ’50111LicenseH .35180M .35120L .3030114114Study (βˆ’20)F 0.4ProduceH .50260M .35100L .15βˆ’50157.5LicenseH .50180M .35120L .1530136.5157.5U 0.6ProduceH .20260M .30100L .50βˆ’5057LicenseH .20180M .30120L .50308787115.2114
    Rolled back: the study branch is worth 115.2 βˆ’ 20 = 95.2, so it is cut; license without the study
    Do not commission the study; license the drug β€” EMV β‚Ή1,14,000

    Note: these conditional probabilities imply P(High) = 0.4(0.5) + 0.6(0.2) = 0.32, not the 0.35 printed for the no-study branch β€” the question's numbers are not fully consistent. Use them as given, as above.

Q5 β€” How many cups to stock β€” EMV, EPPI, EVPI

Question

Cost β‚Ή5, price β‚Ή8, salvage β‚Ή2 per unsold cup; demand 15, 16, 17, 18 with probabilities 0.1, 0.2, 0.4, 0.3 (profit matrix given). Find EMV, EVPI and EPPI.

Assignment 8 p. 2

Solution
  1. Step 1 β€” Check one cell of the profit matrix Stock 18, demand 15: 15 Γ— 8 + 3 Γ— 2 βˆ’ 18 Γ— 5 = 120 + 6 βˆ’ 90 = 36 βœ“
  2. Step 2 β€” EMV of each stocking level
    StockD = 15161718EMV
    15 cups4545454545
    16 cups4248484847.4
    17 cups3945515148.6
    18 cups3642485447.4
    Probability0.10.20.40.3
    EMV(17) = 39(0.1) + 45(0.2) + 51(0.4) + 51(0.3) = 3.9 + 9 + 20.4 + 15.3 = 48.6 ⟹ Stock 17 cups, EMV β‚Ή48.60
  3. Step 3 β€” EPPI and EVPI Best profit in each demand column: 45, 48, 51, 54 (stock = demand) EPPI = 45(0.1) + 48(0.2) + 51(0.4) + 54(0.3) = 4.5 + 9.6 + 20.4 + 16.2 = β‚Ή50.70 EVPI = EPPI βˆ’ best EMV = 50.70 βˆ’ 48.60 = β‚Ή2.10
  4. Check with EOL
    StockRegret 15161718EOL
    1503695.7
    1630363.3
    1763032.1
    1896303.3

    Least EOL = 2.1, at 17 cups β€” the same decision, and equal to the EVPI βœ“. Perfect demand information would be worth at most β‚Ή2.10 a day.

Q6 β€” EV manufacturer β€” decision tree and the risk trade-off

Question
OptionHigh (0.40)Moderate (0.35)Low (0.25)
S1 New plant18080βˆ’70
S2 Expand13065βˆ’20
S3 Outsource504025

(1) Draw the complete decision tree. (2) EMV of each option by the backward pass. (3) Analyse the trade-off between expected profit and risk, and recommend.

Assignment 8 p. 2

Solution
  1. (1) Decision tree
    S1 New plantH .40180M .3580L .25βˆ’7082.5S2 ExpandH .40130M .3565L .25βˆ’2069.75S3 OutsourceH .4050M .3540L .252540.2582.5
    One decision node (β– ), three chance nodes (●) with their probabilities, payoffs at the tips (β‚Ή crore)
  2. (2) Backward pass S1 = 180(0.40) + 80(0.35) βˆ’ 70(0.25) = 72 + 28 βˆ’ 17.5 = 82.5 S2 = 52 + 22.75 βˆ’ 5 = 69.75 S3 = 20 + 14 + 6.25 = 40.25 Highest EMV: S1, β‚Ή82.5 crore
  3. (3) Profit against risk
    OptionEMVRange of outcomesStd. deviationRisk per β‚Ή of EMV (CV)
    S182.5βˆ’70 to 18098.081.19
    S269.75βˆ’20 to 13058.940.84
    S340.2525 to 509.810.24

    S1 has the highest expected profit but also the widest spread and the only large loss (βˆ’70 if demand is low). S3 never loses money but earns half as much. S2 sits in between: 85% of S1's EMV with a much smaller worst case. Recommendation: S1 for a firm that decides on EMV (it can absorb a β‚Ή70 crore loss); a risk-averse firm would choose S2, which gives up 12.75 of EMV to cut the worst case from βˆ’70 to βˆ’20.

Q7 β€” Develop software in-house or outsource β€” with a market survey

Question
AlternativeHighMediumLow
In-house300120βˆ’40
Outsource20015060
Probability0.400.250.35

A survey costing β‚Ή15,000 predicts positive (P) or negative (N): P(P) = P(N) = 0.5; P(H|P) = 0.6, P(M|P) = 0.25, P(L|P) = 0.15; P(H|N) = 0.1, P(M|N) = 0.3, P(L|N) = 0.6. (Payoffs in β‚Ή'000; survey = 15.)

Assignment 8 p. 3

Solution
  1. Step 1 β€” Without the survey In-house = 300(0.4) + 120(0.25) βˆ’ 40(0.35) = 120 + 30 βˆ’ 14 = 136 Outsource = 80 + 37.5 + 21 = 138.5 ⟹ Outsource, 138.5
  2. Step 2 β€” Positive survey In-house = 180 + 30 βˆ’ 6 = 204 Outsource = 120 + 37.5 + 9 = 166.5 ⟹ In-house, 204
  3. Step 3 β€” Negative survey In-house = 30 + 36 βˆ’ 24 = 42 Outsource = 20 + 45 + 36 = 101 ⟹ Outsource, 101
  4. Step 4 β€” Value of the survey EV with survey = 0.5(204) + 0.5(101) = 152.5 EVSI = 152.5 βˆ’ 138.5 = 14 (β‚Ή14,000) Net: 152.5 βˆ’ 15 = 137.5 < 138.5
  5. Step 5 β€” Tree and answer
    No surveyIn-houseH .40300M .25120L .35βˆ’40136OutsourceH .40200M .25150L .3560138.5138.5Survey (βˆ’15)P 0.5In-houseH .60300M .25120L .15βˆ’40204OutsourceH .60200M .25150L .1560166.5204N 0.5In-houseH .10300M .30120L .60βˆ’4042OutsourceH .10200M .30150L .6060101101152.5138.5
    The survey is worth β‚Ή14,000 but costs β‚Ή15,000 β€” a close call that the tree settles
    Do not survey; outsource the software β€” EMV β‚Ή1,38,500

    If the survey cost less than β‚Ή14,000 the answer would flip to: survey, develop in-house if positive, outsource if negative.