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Introductory Problem
DECISION THEORY
DECISION MAKING UNDER RISK
Suppose an electrical goods merchant buys, for resale purposes in a
market, electric irons in the range of 0 to 4. His resources permit him
to buy nothing or 1 or 2 or 3 or 4 units. These are his alternative
courses of action or strategies. The demand for electric irons on any
day is something beyond his control and hence is a state of nature.
Let us presume that the dealer does not know how many units will
be bought from him by the customers. The demand could be
anything from 0 to 4. The dealer can buy each unit of electric iron @
Rs.40 and sell it at Rs.45 each, his margin being Rs.5 per unit.
Assume the stock on hand is valueless. Portray in a payoff table and
opportunity loss table the quantum of total margin (loss), that he
gets in relation to various alternative strategies and states of nature.
Payoff Matrix
Courses of Action
0 1 2 3 4
States
of
Nature
0 0–0=0
1 0–0=0
2 0–0=0
3 0–0=0
4 0–0=0
Payoff Matrix
Courses of Action
0 1 2 3 4
States
of
Nature
0 0–0=0 0–40=–40
1 0–0=0 45–40=5
2 0–0=0 45–40=5
3 0–0=0 45–40=5
4 0–0=0 45–40=5
Payoff Matrix
Courses of Action
0 1 2 3 4
States
of
Nature
0 0–0=0 0–40=–40 0–2×40=–80 0–3×40=–120 0–4×40=–160
1 0–0=0 45–40=5 45–2×40=–35 45–3×40=–75 45–4×40=–115
2 0–0=0 45–40=5 2×45–2×40=10 2×45–3×40=–30 2×45–4×40=–70
3 0–0=0 45–40=5 2×45–2×40=10 3×45–3×40=15 3×45–4×40=–25
4 0–0=0 45–40=5 2×45–2×40=10 3×45–3×40=15 4×45–4×40=20
Payoff Matrix
Courses of Action
0 1 2 3 4
States
of
Nature
0 0 –40 –80 –120 –160
1 0 5 –35 –75 –115
2 0 5 10 –30 –70
3 0 5 10 15 –25
4 0 5 10 15 20
Courses of Action
Probability
0 1 2 3 4
States
of
Nature
0 0 –40 –80 –120 –160 0.04
1 0 5 –35 –75 –115 0.06
2 0 5 10 –30 –70 0.20
3 0 5 10 15 –25 0.30
4 0 5 10 15 20 0.40
EXPECTED MONETARY VALUE (EMV)
CALCULATION
0 1 2 3 4 P
0 0 –40 –80 –120 –160 0.04
1 0 5 –35 –75 –115 0.06
2 0 5 10 –30 –70 0.20
3 0 5 10 15 –25 0.30
4 0 5 10 15 20 0.40
𝑬𝑴𝑽 𝟎 = 𝟎 × 𝟎. 𝟎𝟒 + 𝟎 × 𝟎. 𝟎𝟔 + 𝟎 × 𝟎. 𝟐 + 𝟎 × 𝟎. 𝟑 + 𝟎 × 𝟎. 𝟒 = 𝟎
0 1 2 3 4 P
0 0 –40 –80 –120 –160 0.04
1 0 5 –35 –75 –115 0.06
2 0 5 10 –30 –70 0.20
3 0 5 10 15 –25 0.30
4 0 5 10 15 20 0.40
𝑬𝑴𝑽 𝟎 = 𝟎 × 𝟎. 𝟎𝟒 + 𝟎 × 𝟎. 𝟎𝟔 + 𝟎 × 𝟎. 𝟐 + 𝟎 × 𝟎. 𝟑 + 𝟎 × 𝟎. 𝟒 = 𝟎
𝑬𝑴𝑽 𝟏 = −𝟒𝟎 × 𝟎. 𝟎𝟒 + 𝟓 × 𝟎. 𝟎𝟔 + 𝟓 × 𝟎. 𝟐 + 𝟓 × 𝟎. 𝟑 + 𝟓 × 𝟎. 𝟒 = 𝟑. 𝟐
𝑬𝑴𝑽 𝟐 = −𝟖𝟎 × 𝟎. 𝟎𝟒 + −𝟑𝟓 × 𝟎. 𝟎𝟔 + 𝟏𝟎 × 𝟎. 𝟐 + 𝟏𝟎 × 𝟎. 𝟑 + 𝟏𝟎 × 𝟎. 𝟒 = 𝟑. 𝟕
𝑬𝑴𝑽 𝟑 = −𝟏𝟐𝟎 × 𝟎. 𝟎𝟒 + −𝟕𝟓 × 𝟎. 𝟎𝟔 + −𝟑𝟎 × 𝟎. 𝟐 + 𝟏𝟓 × 𝟎. 𝟑 + 𝟏𝟓 × 𝟎. 𝟒 = −𝟒. 𝟖
𝑬𝑴𝑽 𝟒 = −𝟏𝟔𝟎 × 𝟎. 𝟎𝟒 + −𝟏𝟏𝟓 × 𝟎. 𝟎𝟔 + −𝟕𝟎 × 𝟎. 𝟐 + −𝟐𝟓 × 𝟎. 𝟑 + 𝟐𝟎 × 𝟎. 𝟒 = −𝟐𝟔. 𝟖
0 1 2 3 4 P
0 0 –40 –80 –120 –160 0.04
1 0 5 –35 –75 –115 0.06
2 0 5 10 –30 –70 0.20
3 0 5 10 15 –25 0.30
4 0 5 10 15 20 0.40
EMV 0 3.2 3.7 –4.8 –26.8
0 1 2 3 4 P
0 0 –40 –80 –120 –160 0.04
1 0 5 –35 –75 –115 0.06
2 0 5 10 –30 –70 0.20
3 0 5 10 15 –25 0.30
4 0 5 10 15 20 0.40
EMV 0 3.2 3.7 –4.8 –26.8
𝑬𝑴𝑽 𝒎𝒂𝒙 = 𝑬𝑴𝑽 𝟐 = 𝟑. 𝟕
REGRET TABLE
Payoff Table
0 1 2 3 4
0 0 –40 –80 –120 –160
1 0 5 –35 –75 –115
2 0 5 10 –30 –70
3 0 5 10 15 –25
4 0 5 10 15 20
Regret/Opportunity Loss Table
Courses of Action
0 1 2 3 4
States
of
Nature
0 0–0=0
1 5–0=5
2 10–0=10
3 15–0=15
4 20–0=20
Payoff Table
0 1 2 3 4
0 0 –40 –80 –120 –160
1 0 5 –35 –75 –115
2 0 5 10 –30 –70
3 0 5 10 15 –25
4 0 5 10 15 20
Regret/Opportunity Loss Table
Courses of Action
0 1 2 3 4
States
of
Nature
0 0–0=0 0–(–40)=40
1 5–0=5 5–5=0
2 10–0=10 10–5=5
3 15–0=15 15–5=10
4 20–0=20 20–5=15
Payoff Table
0 1 2 3 4
0 0 –40 –80 –120 –160
1 0 5 –35 –75 –115
2 0 5 10 –30 –70
3 0 5 10 15 –25
4 0 5 10 15 20
Regret/Opportunity Loss Table
Courses of Action
0 1 2 3 4
States
of
Nature
0 0–0=0 0–(–40)=40 0–(–80)=80 0–(–120)=120 0–(–160)=160
1 5–0=5 5–5=0 5–(–35)=40 5–(–75)=80 5–(–115)=120
2 10–0=10 10–5=5 10–10=0 10–(–30)=40 10–(–70)=80
3 15–0=15 15–5=10 15–10=5 15–15=0 15–(–25)=40
4 20–0=20 20–5=15 20–10=10 20–15=5 20–20=0
Regret/Opportunity Loss Table
Courses of Action
0 1 2 3 4
States
of
Nature
0 0 40 80 120 160
1 5 0 40 80 120
2 10 5 0 40 80
3 15 10 5 0 40
4 20 15 10 5 0
Regret/Opportunity Loss Table
Courses of Action
P
0 1 2 3 4
States
of
Nature
0 0 40 80 120 160 0.04
1 5 0 40 80 120 0.06
2 10 5 0 40 80 0.20
3 15 10 5 0 40 0.30
4 20 15 10 5 0 0.40
EXPECTED OPPORTUNITY LOSS (EOL)
CALCULATION
Regret/Opportunity Loss Table
Courses of Action
P
0 1 2 3 4
States
of
Nature
0 0 40 80 120 160 0.04
1 5 0 40 80 120 0.06
2 10 5 0 40 80 0.20
3 15 10 5 0 40 0.30
4 20 15 10 5 0 0.40
𝑬𝑶𝑳 𝟎 = 𝟎 × 𝟎. 𝟎𝟒 + 𝟓 × 𝟎. 𝟎𝟔 + 𝟏𝟎 × 𝟎. 𝟐 + 𝟏𝟓 × 𝟎. 𝟑 + 𝟐𝟎 × 𝟎. 𝟒 = 𝟏𝟒. 𝟖
Regret/Opportunity Loss Table
Courses of Action
P
0 1 2 3 4
States
of
Nature
0 0 40 80 120 160 0.04
1 5 0 40 80 120 0.06
2 10 5 0 40 80 0.20
3 15 10 5 0 40 0.30
4 20 15 10 5 0 0.40
𝑬𝑶𝑳 𝟎 = 𝟎 × 𝟎. 𝟎𝟒 + 𝟓 × 𝟎. 𝟎𝟔 + 𝟏𝟎 × 𝟎. 𝟐 + 𝟏𝟓 × 𝟎. 𝟑 + 𝟐𝟎 × 𝟎. 𝟒 = 𝟏𝟒. 𝟖
𝑬𝑶𝑳 𝟏 = 𝟒𝟎 × 𝟎. 𝟎𝟒 + 𝟎 × 𝟎. 𝟎𝟔 + 𝟓 × 𝟎. 𝟐 + 𝟏𝟎 × 𝟎. 𝟑 + 𝟏𝟓 × 𝟎. 𝟒 = 𝟏𝟏. 𝟔
𝑬𝑶𝑳 𝟐 = 𝟖𝟎 × 𝟎. 𝟎𝟒 + 𝟒𝟎 × 𝟎. 𝟎𝟔 + 𝟎 × 𝟎. 𝟐 + 𝟓 × 𝟎. 𝟑 + 𝟏𝟎 × 𝟎. 𝟒 = 𝟏𝟏. 𝟏
𝑬𝑶𝑳 𝟑 = 𝟏𝟐𝟎 × 𝟎. 𝟎𝟒 + 𝟖𝟎 × 𝟎. 𝟎𝟔 + 𝟒𝟎 × 𝟎. 𝟐 + 𝟎 × 𝟎. 𝟑 + 𝟓 × 𝟎. 𝟒 = 𝟏𝟗. 𝟔
𝑬𝑶𝑳 𝟒 = 𝟏𝟔𝟎 × 𝟎. 𝟎𝟒 + 𝟏𝟐𝟎 × 𝟎. 𝟎𝟔 + 𝟖𝟎 × 𝟎. 𝟐 + 𝟒𝟎 × 𝟎. 𝟑 + 𝟎 × 𝟎. 𝟒 = 𝟒𝟏. 𝟔
Regret/Opportunity Loss Table
Courses of Action
P
0 1 2 3 4
States
of
Nature
0 0 40 80 120 160 0.04
1 5 0 40 80 120 0.06
2 10 5 0 40 80 0.20
3 15 10 5 0 40 0.30
4 20 15 10 5 0 0.40
EOL 14.8 11.6 11.1 19.6 41.6
Regret/Opportunity Loss Table
Courses of Action
P
0 1 2 3 4
States
of
Nature
0 0 40 80 120 160 0.04
1 5 0 40 80 120 0.06
2 10 5 0 40 80 0.20
3 15 10 5 0 40 0.30
4 20 15 10 5 0 0.40
EOL 14.8 11.6 11.1 19.6 41.6
𝑬𝑶𝑳 𝒎𝒊𝒏 = 𝑬𝑶𝑳 𝟐 = 𝟏𝟏. 𝟏
EXPECTED VALUE WITH PERFECT
INFORMATION
EV with PI
Payoff Matrix
Courses of Action
P
0 1 2 3 4
States
of
Nature
0 0 –40 –80 –120 –160 0.04
1 0 5 –35 –75 –115 0.06
2 0 5 10 –30 –70 0.20
3 0 5 10 15 –25 0.30
4 0 5 10 15 20 0.40
𝑬𝑽 𝒘𝒊𝒕𝒉 𝑷𝑰 = 𝚺 𝑴𝒂𝒙 𝑬𝑴𝑽 𝒇𝒐𝒓 𝒆𝒂𝒄𝒉 𝑺𝒕𝒂𝒕𝒆 𝒐𝒇 𝑵𝒂𝒕𝒖𝒓𝒆 × 𝑷
Payoff Matrix
Courses of Action
P
0 1 2 3 4
States
of
Nature
0 0 –40 –80 –120 –160 0.04
1 0 5 –35 –75 –115 0.06
2 0 5 10 –30 –70 0.20
3 0 5 10 15 –25 0.30
4 0 5 10 15 20 0.40
𝑬𝑽 𝒘𝒊𝒕𝒉 𝑷𝑰 = 𝚺 𝑴𝒂𝒙 𝑬𝑴𝑽 𝒇𝒐𝒓 𝒆𝒂𝒄𝒉 𝑺𝒕𝒂𝒕𝒆 𝒐𝒇 𝑵𝒂𝒕𝒖𝒓𝒆 × 𝑷
Payoff Matrix
Courses of Action
P
0 1 2 3 4
States
of
Nature
0 0 –40 –80 –120 –160 0.04
1 0 5 –35 –75 –115 0.06
2 0 5 10 –30 –70 0.20
3 0 5 10 15 –25 0.30
4 0 5 10 15 20 0.40
𝑬𝑽 𝒘𝒊𝒕𝒉 𝑷𝑰 = 𝟎 × 𝟎. 𝟎𝟒 + 𝟓 × 𝟎. 𝟎𝟔 + 𝟏𝟎 × 𝟎. 𝟐 + 𝟏𝟓 × 𝟎. 𝟑 + 𝟐𝟎 × 𝟎. 𝟒
∴ 𝑬𝑽 𝒘𝒊𝒕𝒉 𝑷𝑰 = 𝟏𝟒. 𝟖
EXPECTED VALUE OF PERFECT
INFORMATION
EVPI
𝑬𝑽𝑷𝑰 = 𝑬𝑽 𝒘𝒊𝒕𝒉 𝑷𝑰 − 𝑬𝑴𝑽 𝒎𝒂𝒙
𝒐𝒓 𝑬𝑽𝑷𝑰 = 𝟏𝟒. 𝟖 − 𝟑. 𝟕
𝒐𝒓 𝑬𝑽𝑷𝑰 = 𝟏𝟏. 𝟏
𝑬𝑽𝑷𝑰 = 𝑬𝑽 𝒘𝒊𝒕𝒉 𝑷𝑰 − 𝑬𝑴𝑽 𝒎𝒂𝒙
𝒐𝒓 𝑬𝑽𝑷𝑰 = 𝟏𝟒. 𝟖 − 𝟑. 𝟕
𝒐𝒓 𝑬𝑽𝑷𝑰 = 𝟏𝟏. 𝟏
𝑨𝒍𝒕𝒆𝒓𝒏𝒂𝒕𝒆𝒍𝒚,
𝑬𝑽𝑷𝑰 = 𝑬𝑶𝑳 𝒎𝒊𝒏 = 𝑬𝑶𝑳 𝟐 = 𝟏𝟏. 𝟏
DECISION MAKING UNDER
UNCERTAINTY
DECISION CRITERIA UNDER CONDITION OF
UNCERTAINTY
• Maximin.
• Maximax.
• Minimax Regret.
• Hurwicz Criterion.
• Baye’s/Lapalce’s Criterion.
CRITERION OF PESSIMISM (MAXIMIN)
• Also called ‘Waldian Criterion.’
• Determine the lowest outcome for each alternative.
• Choose the alternative associated with the best of these.
CRITERION OF OPTIMISM (MAXIMAX)
• Suggested by Leonid Hurwicz.
• Determine the best outcome for each alternative.
• Select the alternative associated with the best of these.
MINIMAX REGRET CRITERION
• Attributed to Leonard Savage.
• For each state, identify the most attractive alternative.
• Place a zero in those cells.
• Compute opportunity loss for other alternatives.
• Identify the maximum opportunity loss for each alternative.
• Select the alternative associated with the lowest of these.
CRITERION OF REALISM (HURWICZ CRITERION)
• A compromise between maximax and maximin criteria.
• A coefficient of optimism α (0≤α≤1) is selected.
• When α is close to 1, the decision-maker is optimistic about the
future.
• When α is close to 0, the decision-maker is pessimistic about the
future.
HURWICZ CRITERION
• Select the strategy which maximises:
LAPLACE CRITERION
• Assign equal probabilities to each state.
• Compute the expected value for each alternative.
• Select the alternative with the highest alternative.
CRITERION OF PESSIMISM (MAXIMIN)
Payoff Matrix
Courses of Action
0 1 2 3 4
States
of
Nature
0 0 –40 –80 –120 –160
1 0 5 –35 –75 –115
2 0 5 10 –30 –70
3 0 5 10 15 –25
4 0 5 10 15 20
Payoff Matrix: Maximin Criterion
Courses of Action
0 1 2 3 4
States
of
Nature
0 0 –40 –80 –120 –160
1 0 5 –35 –75 –115
2 0 5 10 –30 –70
3 0 5 10 15 –25
4 0 5 10 15 20
𝑴𝒂𝒙𝒊𝒎𝒊𝒏 𝑪𝒓𝒊𝒕𝒆𝒓𝒊𝒐𝒏:
≫ 𝑴𝒂𝒙𝒊𝒎𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒐𝒍𝒖𝒎𝒏
𝒎𝒊𝒏𝒊𝒎𝒂
≫ 𝒊. 𝒆. , 𝒇𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒊𝒏𝒊𝒎𝒖𝒎
𝒑𝒂𝒚𝒐𝒇𝒇 𝒊𝒏 𝒆𝒂𝒄𝒉 𝒄𝒐𝒍𝒖𝒎𝒏
≫ 𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒂𝒙𝒊𝒎𝒖𝒎 𝒐𝒇
𝒕𝒉𝒆𝒔𝒆 𝒑𝒂𝒚𝒐𝒇𝒇𝒔
≫ 𝑻𝒉𝒆 𝒄𝒐𝒖𝒓𝒔𝒆 𝒐𝒇 𝒂𝒄𝒕𝒊𝒐𝒏
𝒄𝒐𝒓𝒓𝒆𝒔𝒑𝒐𝒏𝒅𝒊𝒏𝒈 𝒕𝒐 𝒕𝒉𝒊𝒔
𝒑𝒂𝒚𝒐𝒇𝒇 𝒊𝒔 𝒕𝒉𝒆 𝒂𝒏𝒔𝒘𝒆𝒓
Payoff Matrix: Maximin Criterion
Courses of Action
0 1 2 3 4
States
of
Nature
0 0 –40 –80 –120 –160
1 0 5 –35 –75 –115
2 0 5 10 –30 –70
3 0 5 10 15 –25
4 0 5 10 15 20
Column Minima 0 –40 –80 –120 –160
𝑴𝒂𝒙𝒊𝒎𝒊𝒏 𝑪𝒓𝒊𝒕𝒆𝒓𝒊𝒐𝒏:
≫ 𝑴𝒂𝒙𝒊𝒎𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒐𝒍𝒖𝒎𝒏
𝒎𝒊𝒏𝒊𝒎𝒂
≫ 𝒊. 𝒆. , 𝒇𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒊𝒏𝒊𝒎𝒖𝒎
𝒑𝒂𝒚𝒐𝒇𝒇 𝒊𝒏 𝒆𝒂𝒄𝒉 𝒄𝒐𝒍𝒖𝒎𝒏
≫ 𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒂𝒙𝒊𝒎𝒖𝒎 𝒐𝒇
𝒕𝒉𝒆𝒔𝒆 𝒑𝒂𝒚𝒐𝒇𝒇𝒔
≫ 𝑻𝒉𝒆 𝒄𝒐𝒖𝒓𝒔𝒆 𝒐𝒇 𝒂𝒄𝒕𝒊𝒐𝒏
𝒄𝒐𝒓𝒓𝒆𝒔𝒑𝒐𝒏𝒅𝒊𝒏𝒈 𝒕𝒐 𝒕𝒉𝒊𝒔
𝒑𝒂𝒚𝒐𝒇𝒇 𝒊𝒔 𝒕𝒉𝒆 𝒂𝒏𝒔𝒘𝒆𝒓
Payoff Matrix: Maximin Criterion
Courses of Action
0 1 2 3 4
States
of
Nature
0 0 –40 –80 –120 –160
1 0 5 –35 –75 –115
2 0 5 10 –30 –70
3 0 5 10 15 –25
4 0 5 10 15 20
Column Minima 0 –40 –80 –120 –160
Maximum of the column minima=0
𝑴𝒂𝒙𝒊𝒎𝒊𝒏 𝑪𝒓𝒊𝒕𝒆𝒓𝒊𝒐𝒏:
≫ 𝑴𝒂𝒙𝒊𝒎𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒐𝒍𝒖𝒎𝒏
𝒎𝒊𝒏𝒊𝒎𝒂
≫ 𝒊. 𝒆. , 𝒇𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒊𝒏𝒊𝒎𝒖𝒎
𝒑𝒂𝒚𝒐𝒇𝒇 𝒊𝒏 𝒆𝒂𝒄𝒉 𝒄𝒐𝒍𝒖𝒎𝒏
≫ 𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒂𝒙𝒊𝒎𝒖𝒎 𝒐𝒇
𝒕𝒉𝒆𝒔𝒆 𝒑𝒂𝒚𝒐𝒇𝒇𝒔
≫ 𝑻𝒉𝒆 𝒄𝒐𝒖𝒓𝒔𝒆 𝒐𝒇 𝒂𝒄𝒕𝒊𝒐𝒏
𝒄𝒐𝒓𝒓𝒆𝒔𝒑𝒐𝒏𝒅𝒊𝒏𝒈 𝒕𝒐 𝒕𝒉𝒊𝒔
𝒑𝒂𝒚𝒐𝒇𝒇 𝒊𝒔 𝒕𝒉𝒆 𝒂𝒏𝒔𝒘𝒆𝒓
Payoff Matrix: Maximin Criterion
Courses of Action
0 1 2 3 4
States
of
Nature
0 0 –40 –80 –120 –160
1 0 5 –35 –75 –115
2 0 5 10 –30 –70
3 0 5 10 15 –25
4 0 5 10 15 20
Column Minima 0 –40 –80 –120 –160
Maximum of the column minima=0
𝑴𝒂𝒙𝒊𝒎𝒊𝒏 𝑪𝒓𝒊𝒕𝒆𝒓𝒊𝒐𝒏:
≫ 𝑴𝒂𝒙𝒊𝒎𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒐𝒍𝒖𝒎𝒏
𝒎𝒊𝒏𝒊𝒎𝒂
≫ 𝒊. 𝒆. , 𝒇𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒊𝒏𝒊𝒎𝒖𝒎
𝒑𝒂𝒚𝒐𝒇𝒇 𝒊𝒏 𝒆𝒂𝒄𝒉 𝒄𝒐𝒍𝒖𝒎𝒏
≫ 𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒂𝒙𝒊𝒎𝒖𝒎 𝒐𝒇
𝒕𝒉𝒆𝒔𝒆 𝒑𝒂𝒚𝒐𝒇𝒇𝒔
≫ 𝑻𝒉𝒆 𝒄𝒐𝒖𝒓𝒔𝒆 𝒐𝒇 𝒂𝒄𝒕𝒊𝒐𝒏
𝒄𝒐𝒓𝒓𝒆𝒔𝒑𝒐𝒏𝒅𝒊𝒏𝒈 𝒕𝒐 𝒕𝒉𝒊𝒔
𝒑𝒂𝒚𝒐𝒇𝒇 𝒊𝒔 𝒕𝒉𝒆 𝒂𝒏𝒔𝒘𝒆𝒓
∴ 𝒕𝒉𝒆 𝒃𝒆𝒔𝒕 𝒄𝒐𝒖𝒓𝒔𝒆 𝒐𝒇 𝒂𝒄𝒕𝒊𝒐𝒏 𝒂𝒄𝒄𝒐𝒓𝒅𝒊𝒏𝒈 𝒕𝒐 𝒕𝒉𝒆 𝒎𝒂𝒙𝒊𝒎𝒊𝒏 𝒄𝒓𝒊𝒕𝒆𝒓𝒊𝒐𝒏 = 𝟎
CRITERION OF OPTIMISM (MAXIMAX)
Payoff Matrix
Courses of Action
0 1 2 3 4
States
of
Nature
0 0 –40 –80 –120 –160
1 0 5 –35 –75 –115
2 0 5 10 –30 –70
3 0 5 10 15 –25
4 0 5 10 15 20
Payoff Matrix: Maximax Criterion
Courses of Action
0 1 2 3 4
States
of
Nature
0 0 –40 –80 –120 –160
1 0 5 –35 –75 –115
2 0 5 10 –30 –70
3 0 5 10 15 –25
4 0 5 10 15 20
𝑴𝒂𝒙𝒊𝒎𝒂𝒙 𝑪𝒓𝒊𝒕𝒆𝒓𝒊𝒐𝒏:
≫ 𝑴𝒂𝒙𝒊𝒎𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒐𝒍𝒖𝒎𝒏
𝒎𝒂𝒙𝒊𝒎𝒂
≫ 𝒊. 𝒆. , 𝒇𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒂𝒙𝒊𝒎𝒖𝒎
𝒑𝒂𝒚𝒐𝒇𝒇 𝒊𝒏 𝒆𝒂𝒄𝒉 𝒄𝒐𝒍𝒖𝒎𝒏
≫ 𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒂𝒙𝒊𝒎𝒖𝒎 𝒐𝒇
𝒕𝒉𝒆𝒔𝒆 𝒑𝒂𝒚𝒐𝒇𝒇𝒔
≫ 𝑻𝒉𝒆 𝒄𝒐𝒖𝒓𝒔𝒆 𝒐𝒇 𝒂𝒄𝒕𝒊𝒐𝒏
𝒄𝒐𝒓𝒓𝒆𝒔𝒑𝒐𝒏𝒅𝒊𝒏𝒈 𝒕𝒐 𝒕𝒉𝒊𝒔
𝒑𝒂𝒚𝒐𝒇𝒇 𝒊𝒔 𝒕𝒉𝒆 𝒂𝒏𝒔𝒘𝒆𝒓
Payoff Matrix: Maximax Criterion
Courses of Action
0 1 2 3 4
States
of
Nature
0 0 –40 –80 –120 –160
1 0 5 –35 –75 –115
2 0 5 10 –30 –70
3 0 5 10 15 –25
4 0 5 10 15 20
Column Maxima 0 5 10 15 20
𝑴𝒂𝒙𝒊𝒎𝒂𝒙 𝑪𝒓𝒊𝒕𝒆𝒓𝒊𝒐𝒏:
≫ 𝑴𝒂𝒙𝒊𝒎𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒐𝒍𝒖𝒎𝒏
𝒎𝒂𝒙𝒊𝒎𝒂
≫ 𝒊. 𝒆. , 𝒇𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒂𝒙𝒊𝒎𝒖𝒎
𝒑𝒂𝒚𝒐𝒇𝒇 𝒊𝒏 𝒆𝒂𝒄𝒉 𝒄𝒐𝒍𝒖𝒎𝒏
≫ 𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒂𝒙𝒊𝒎𝒖𝒎 𝒐𝒇
𝒕𝒉𝒆𝒔𝒆 𝒑𝒂𝒚𝒐𝒇𝒇𝒔
≫ 𝑻𝒉𝒆 𝒄𝒐𝒖𝒓𝒔𝒆 𝒐𝒇 𝒂𝒄𝒕𝒊𝒐𝒏
𝒄𝒐𝒓𝒓𝒆𝒔𝒑𝒐𝒏𝒅𝒊𝒏𝒈 𝒕𝒐 𝒕𝒉𝒊𝒔
𝒑𝒂𝒚𝒐𝒇𝒇 𝒊𝒔 𝒕𝒉𝒆 𝒂𝒏𝒔𝒘𝒆𝒓
Payoff Matrix: Maximax Criterion
Courses of Action
0 1 2 3 4
States
of
Nature
0 0 –40 –80 –120 –160
1 0 5 –35 –75 –115
2 0 5 10 –30 –70
3 0 5 10 15 –25
4 0 5 10 15 20
Column Maxima 0 5 10 15 20
Maximum of the column maxima=20
𝑴𝒂𝒙𝒊𝒎𝒂𝒙 𝑪𝒓𝒊𝒕𝒆𝒓𝒊𝒐𝒏:
≫ 𝑴𝒂𝒙𝒊𝒎𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒐𝒍𝒖𝒎𝒏
𝒎𝒂𝒙𝒊𝒎𝒂
≫ 𝒊. 𝒆. , 𝒇𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒂𝒙𝒊𝒎𝒖𝒎
𝒑𝒂𝒚𝒐𝒇𝒇 𝒊𝒏 𝒆𝒂𝒄𝒉 𝒄𝒐𝒍𝒖𝒎𝒏
≫ 𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒂𝒙𝒊𝒎𝒖𝒎 𝒐𝒇
𝒕𝒉𝒆𝒔𝒆 𝒑𝒂𝒚𝒐𝒇𝒇𝒔
≫ 𝑻𝒉𝒆 𝒄𝒐𝒖𝒓𝒔𝒆 𝒐𝒇 𝒂𝒄𝒕𝒊𝒐𝒏
𝒄𝒐𝒓𝒓𝒆𝒔𝒑𝒐𝒏𝒅𝒊𝒏𝒈 𝒕𝒐 𝒕𝒉𝒊𝒔
𝒑𝒂𝒚𝒐𝒇𝒇 𝒊𝒔 𝒕𝒉𝒆 𝒂𝒏𝒔𝒘𝒆𝒓
Payoff Matrix: Maximax Criterion
Courses of Action
0 1 2 3 4
States
of
Nature
0 0 –40 –80 –120 –160
1 0 5 –35 –75 –115
2 0 5 10 –30 –70
3 0 5 10 15 –25
4 0 5 10 15 20
Column Maxima 0 5 10 15 20
Maximum of the column maxima=20
𝑴𝒂𝒙𝒊𝒎𝒂𝒙 𝑪𝒓𝒊𝒕𝒆𝒓𝒊𝒐𝒏:
≫ 𝑴𝒂𝒙𝒊𝒎𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒐𝒍𝒖𝒎𝒏
𝒎𝒂𝒙𝒊𝒎𝒂
≫ 𝒊. 𝒆. , 𝒇𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒂𝒙𝒊𝒎𝒖𝒎
𝒑𝒂𝒚𝒐𝒇𝒇 𝒊𝒏 𝒆𝒂𝒄𝒉 𝒄𝒐𝒍𝒖𝒎𝒏
≫ 𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒂𝒙𝒊𝒎𝒖𝒎 𝒐𝒇
𝒕𝒉𝒆𝒔𝒆 𝒑𝒂𝒚𝒐𝒇𝒇𝒔
≫ 𝑻𝒉𝒆 𝒄𝒐𝒖𝒓𝒔𝒆 𝒐𝒇 𝒂𝒄𝒕𝒊𝒐𝒏
𝒄𝒐𝒓𝒓𝒆𝒔𝒑𝒐𝒏𝒅𝒊𝒏𝒈 𝒕𝒐 𝒕𝒉𝒊𝒔
𝒑𝒂𝒚𝒐𝒇𝒇 𝒊𝒔 𝒕𝒉𝒆 𝒂𝒏𝒔𝒘𝒆𝒓
∴ 𝒕𝒉𝒆 𝒃𝒆𝒔𝒕 𝒄𝒐𝒖𝒓𝒔𝒆 𝒐𝒇 𝒂𝒄𝒕𝒊𝒐𝒏 𝒂𝒄𝒄𝒐𝒓𝒅𝒊𝒏𝒈 𝒕𝒐 𝒕𝒉𝒆 𝒎𝒂𝒙𝒊𝒎𝒂𝒙 𝒄𝒓𝒊𝒕𝒆𝒓𝒊𝒐𝒏 = 𝟒
MINIMAX REGRET CRITERION
Regret/Opportunity Loss Table
Courses of Action
0 1 2 3 4
States
of
Nature
0 0 40 80 120 160
1 5 0 40 80 120
2 10 5 0 40 80
3 15 10 5 0 40
4 20 15 10 5 0
Regret Table (Minimax Regret Criterion)
Courses of Action
0 1 2 3 4
States
of
Nature
0 0 40 80 120 160
1 5 0 40 80 120
2 10 5 0 40 80
3 15 10 5 0 40
4 20 15 10 5 0
𝑴𝒊𝒏𝒊𝒎𝒂𝒙 𝑹𝒆𝒈𝒓𝒆𝒕 𝑪𝒓𝒊𝒕𝒆𝒓𝒊𝒐𝒏:
≫ 𝑴𝒊𝒏𝒊𝒎𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒐𝒍𝒖𝒎𝒏
𝒎𝒂𝒙𝒊𝒎𝒂 (𝒊𝒏 𝒕𝒉𝒆 𝒓𝒆𝒈𝒓𝒆𝒕 𝒕𝒂𝒃𝒍𝒆)
≫ 𝒊. 𝒆. , 𝒇𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒂𝒙𝒊𝒎𝒖𝒎
𝒓𝒆𝒈𝒓𝒆𝒕 𝒊𝒏 𝒆𝒂𝒄𝒉 𝒄𝒐𝒍𝒖𝒎𝒏
≫ 𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒊𝒏𝒊𝒎𝒖𝒎 𝒐𝒇
𝒕𝒉𝒆𝒔𝒆 𝒓𝒆𝒈𝒓𝒆𝒕𝒔
≫ 𝑻𝒉𝒆 𝒄𝒐𝒖𝒓𝒔𝒆 𝒐𝒇 𝒂𝒄𝒕𝒊𝒐𝒏
𝒄𝒐𝒓𝒓𝒆𝒔𝒑𝒐𝒏𝒅𝒊𝒏𝒈 𝒕𝒐 𝒕𝒉𝒊𝒔
𝒓𝒆𝒈𝒓𝒆𝒕 𝒊𝒔 𝒕𝒉𝒆 𝒂𝒏𝒔𝒘𝒆𝒓
Regret Table (Minimax Regret Criterion)
Courses of Action
0 1 2 3 4
States
of
Nature
0 0 40 80 120 160
1 5 0 40 80 120
2 10 5 0 40 80
3 15 10 5 0 40
4 20 15 10 5 0
𝑴𝒊𝒏𝒊𝒎𝒂𝒙 𝑹𝒆𝒈𝒓𝒆𝒕 𝑪𝒓𝒊𝒕𝒆𝒓𝒊𝒐𝒏:
≫ 𝑴𝒊𝒏𝒊𝒎𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒐𝒍𝒖𝒎𝒏
𝒎𝒂𝒙𝒊𝒎𝒂 (𝒊𝒏 𝒕𝒉𝒆 𝒓𝒆𝒈𝒓𝒆𝒕 𝒕𝒂𝒃𝒍𝒆)
≫ 𝒊. 𝒆. , 𝒇𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒂𝒙𝒊𝒎𝒖𝒎
𝒓𝒆𝒈𝒓𝒆𝒕 𝒊𝒏 𝒆𝒂𝒄𝒉 𝒄𝒐𝒍𝒖𝒎𝒏
≫ 𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒊𝒏𝒊𝒎𝒖𝒎 𝒐𝒇
𝒕𝒉𝒆𝒔𝒆 𝒓𝒆𝒈𝒓𝒆𝒕𝒔
≫ 𝑻𝒉𝒆 𝒄𝒐𝒖𝒓𝒔𝒆 𝒐𝒇 𝒂𝒄𝒕𝒊𝒐𝒏
𝒄𝒐𝒓𝒓𝒆𝒔𝒑𝒐𝒏𝒅𝒊𝒏𝒈 𝒕𝒐 𝒕𝒉𝒊𝒔
𝒓𝒆𝒈𝒓𝒆𝒕 𝒊𝒔 𝒕𝒉𝒆 𝒂𝒏𝒔𝒘𝒆𝒓
Regret Table (Minimax Regret Criterion)
Courses of Action
0 1 2 3 4
States
of
Nature
0 0 40 80 120 160
1 5 0 40 80 120
2 10 5 0 40 80
3 15 10 5 0 40
4 20 15 10 5 0
Column maxima 20 40 80 120 160
𝑴𝒊𝒏𝒊𝒎𝒂𝒙 𝑹𝒆𝒈𝒓𝒆𝒕 𝑪𝒓𝒊𝒕𝒆𝒓𝒊𝒐𝒏:
≫ 𝑴𝒊𝒏𝒊𝒎𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒐𝒍𝒖𝒎𝒏
𝒎𝒂𝒙𝒊𝒎𝒂 (𝒊𝒏 𝒕𝒉𝒆 𝒓𝒆𝒈𝒓𝒆𝒕 𝒕𝒂𝒃𝒍𝒆)
≫ 𝒊. 𝒆. , 𝒇𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒂𝒙𝒊𝒎𝒖𝒎
𝒓𝒆𝒈𝒓𝒆𝒕 𝒊𝒏 𝒆𝒂𝒄𝒉 𝒄𝒐𝒍𝒖𝒎𝒏
≫ 𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒊𝒏𝒊𝒎𝒖𝒎 𝒐𝒇
𝒕𝒉𝒆𝒔𝒆 𝒓𝒆𝒈𝒓𝒆𝒕𝒔
≫ 𝑻𝒉𝒆 𝒄𝒐𝒖𝒓𝒔𝒆 𝒐𝒇 𝒂𝒄𝒕𝒊𝒐𝒏
𝒄𝒐𝒓𝒓𝒆𝒔𝒑𝒐𝒏𝒅𝒊𝒏𝒈 𝒕𝒐 𝒕𝒉𝒊𝒔
𝒓𝒆𝒈𝒓𝒆𝒕 𝒊𝒔 𝒕𝒉𝒆 𝒂𝒏𝒔𝒘𝒆𝒓
Regret Table (Minimax Regret Criterion)
Courses of Action
0 1 2 3 4
States
of
Nature
0 0 40 80 120 160
1 5 0 40 80 120
2 10 5 0 40 80
3 15 10 5 0 40
4 20 15 10 5 0
Column maxima 20 40 80 120 160
Minimum of the column maxima=20
𝑴𝒊𝒏𝒊𝒎𝒂𝒙 𝑹𝒆𝒈𝒓𝒆𝒕 𝑪𝒓𝒊𝒕𝒆𝒓𝒊𝒐𝒏:
≫ 𝑴𝒊𝒏𝒊𝒎𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒐𝒍𝒖𝒎𝒏
𝒎𝒂𝒙𝒊𝒎𝒂 (𝒊𝒏 𝒕𝒉𝒆 𝒓𝒆𝒈𝒓𝒆𝒕 𝒕𝒂𝒃𝒍𝒆)
≫ 𝒊. 𝒆. , 𝒇𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒂𝒙𝒊𝒎𝒖𝒎
𝒓𝒆𝒈𝒓𝒆𝒕 𝒊𝒏 𝒆𝒂𝒄𝒉 𝒄𝒐𝒍𝒖𝒎𝒏
≫ 𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒊𝒏𝒊𝒎𝒖𝒎 𝒐𝒇
𝒕𝒉𝒆𝒔𝒆 𝒓𝒆𝒈𝒓𝒆𝒕𝒔
≫ 𝑻𝒉𝒆 𝒄𝒐𝒖𝒓𝒔𝒆 𝒐𝒇 𝒂𝒄𝒕𝒊𝒐𝒏
𝒄𝒐𝒓𝒓𝒆𝒔𝒑𝒐𝒏𝒅𝒊𝒏𝒈 𝒕𝒐 𝒕𝒉𝒊𝒔
𝒓𝒆𝒈𝒓𝒆𝒕 𝒊𝒔 𝒕𝒉𝒆 𝒂𝒏𝒔𝒘𝒆𝒓
Regret Table (Minimax Regret Criterion)
Courses of Action
0 1 2 3 4
States
of
Nature
0 0 40 80 120 160
1 5 0 40 80 120
2 10 5 0 40 80
3 15 10 5 0 40
4 20 15 10 5 0
Column maxima 20 40 80 120 160
Minimum of the column maxima=20
𝑴𝒊𝒏𝒊𝒎𝒂𝒙 𝑹𝒆𝒈𝒓𝒆𝒕 𝑪𝒓𝒊𝒕𝒆𝒓𝒊𝒐𝒏:
≫ 𝑴𝒊𝒏𝒊𝒎𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒐𝒍𝒖𝒎𝒏
𝒎𝒂𝒙𝒊𝒎𝒂 (𝒊𝒏 𝒕𝒉𝒆 𝒓𝒆𝒈𝒓𝒆𝒕 𝒕𝒂𝒃𝒍𝒆)
≫ 𝒊. 𝒆. , 𝒇𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒂𝒙𝒊𝒎𝒖𝒎
𝒓𝒆𝒈𝒓𝒆𝒕 𝒊𝒏 𝒆𝒂𝒄𝒉 𝒄𝒐𝒍𝒖𝒎𝒏
≫ 𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒊𝒏𝒊𝒎𝒖𝒎 𝒐𝒇
𝒕𝒉𝒆𝒔𝒆 𝒓𝒆𝒈𝒓𝒆𝒕𝒔
≫ 𝑻𝒉𝒆 𝒄𝒐𝒖𝒓𝒔𝒆 𝒐𝒇 𝒂𝒄𝒕𝒊𝒐𝒏
𝒄𝒐𝒓𝒓𝒆𝒔𝒑𝒐𝒏𝒅𝒊𝒏𝒈 𝒕𝒐 𝒕𝒉𝒊𝒔
𝒓𝒆𝒈𝒓𝒆𝒕 𝒊𝒔 𝒕𝒉𝒆 𝒂𝒏𝒔𝒘𝒆𝒓
∴ 𝒕𝒉𝒆 𝒃𝒆𝒔𝒕 𝒄𝒐𝒖𝒓𝒔𝒆 𝒐𝒇 𝒂𝒄𝒕𝒊𝒐𝒏 = 𝟎

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Decision theory introductory problem

  • 3. Suppose an electrical goods merchant buys, for resale purposes in a market, electric irons in the range of 0 to 4. His resources permit him to buy nothing or 1 or 2 or 3 or 4 units. These are his alternative courses of action or strategies. The demand for electric irons on any day is something beyond his control and hence is a state of nature. Let us presume that the dealer does not know how many units will be bought from him by the customers. The demand could be anything from 0 to 4. The dealer can buy each unit of electric iron @ Rs.40 and sell it at Rs.45 each, his margin being Rs.5 per unit. Assume the stock on hand is valueless. Portray in a payoff table and opportunity loss table the quantum of total margin (loss), that he gets in relation to various alternative strategies and states of nature.
  • 4. Payoff Matrix Courses of Action 0 1 2 3 4 States of Nature 0 0–0=0 1 0–0=0 2 0–0=0 3 0–0=0 4 0–0=0
  • 5. Payoff Matrix Courses of Action 0 1 2 3 4 States of Nature 0 0–0=0 0–40=–40 1 0–0=0 45–40=5 2 0–0=0 45–40=5 3 0–0=0 45–40=5 4 0–0=0 45–40=5
  • 6. Payoff Matrix Courses of Action 0 1 2 3 4 States of Nature 0 0–0=0 0–40=–40 0–2×40=–80 0–3×40=–120 0–4×40=–160 1 0–0=0 45–40=5 45–2×40=–35 45–3×40=–75 45–4×40=–115 2 0–0=0 45–40=5 2×45–2×40=10 2×45–3×40=–30 2×45–4×40=–70 3 0–0=0 45–40=5 2×45–2×40=10 3×45–3×40=15 3×45–4×40=–25 4 0–0=0 45–40=5 2×45–2×40=10 3×45–3×40=15 4×45–4×40=20
  • 7. Payoff Matrix Courses of Action 0 1 2 3 4 States of Nature 0 0 –40 –80 –120 –160 1 0 5 –35 –75 –115 2 0 5 10 –30 –70 3 0 5 10 15 –25 4 0 5 10 15 20
  • 8. Courses of Action Probability 0 1 2 3 4 States of Nature 0 0 –40 –80 –120 –160 0.04 1 0 5 –35 –75 –115 0.06 2 0 5 10 –30 –70 0.20 3 0 5 10 15 –25 0.30 4 0 5 10 15 20 0.40
  • 9. EXPECTED MONETARY VALUE (EMV) CALCULATION
  • 10. 0 1 2 3 4 P 0 0 –40 –80 –120 –160 0.04 1 0 5 –35 –75 –115 0.06 2 0 5 10 –30 –70 0.20 3 0 5 10 15 –25 0.30 4 0 5 10 15 20 0.40 𝑬𝑴𝑽 𝟎 = 𝟎 × 𝟎. 𝟎𝟒 + 𝟎 × 𝟎. 𝟎𝟔 + 𝟎 × 𝟎. 𝟐 + 𝟎 × 𝟎. 𝟑 + 𝟎 × 𝟎. 𝟒 = 𝟎
  • 11. 0 1 2 3 4 P 0 0 –40 –80 –120 –160 0.04 1 0 5 –35 –75 –115 0.06 2 0 5 10 –30 –70 0.20 3 0 5 10 15 –25 0.30 4 0 5 10 15 20 0.40 𝑬𝑴𝑽 𝟎 = 𝟎 × 𝟎. 𝟎𝟒 + 𝟎 × 𝟎. 𝟎𝟔 + 𝟎 × 𝟎. 𝟐 + 𝟎 × 𝟎. 𝟑 + 𝟎 × 𝟎. 𝟒 = 𝟎 𝑬𝑴𝑽 𝟏 = −𝟒𝟎 × 𝟎. 𝟎𝟒 + 𝟓 × 𝟎. 𝟎𝟔 + 𝟓 × 𝟎. 𝟐 + 𝟓 × 𝟎. 𝟑 + 𝟓 × 𝟎. 𝟒 = 𝟑. 𝟐 𝑬𝑴𝑽 𝟐 = −𝟖𝟎 × 𝟎. 𝟎𝟒 + −𝟑𝟓 × 𝟎. 𝟎𝟔 + 𝟏𝟎 × 𝟎. 𝟐 + 𝟏𝟎 × 𝟎. 𝟑 + 𝟏𝟎 × 𝟎. 𝟒 = 𝟑. 𝟕 𝑬𝑴𝑽 𝟑 = −𝟏𝟐𝟎 × 𝟎. 𝟎𝟒 + −𝟕𝟓 × 𝟎. 𝟎𝟔 + −𝟑𝟎 × 𝟎. 𝟐 + 𝟏𝟓 × 𝟎. 𝟑 + 𝟏𝟓 × 𝟎. 𝟒 = −𝟒. 𝟖 𝑬𝑴𝑽 𝟒 = −𝟏𝟔𝟎 × 𝟎. 𝟎𝟒 + −𝟏𝟏𝟓 × 𝟎. 𝟎𝟔 + −𝟕𝟎 × 𝟎. 𝟐 + −𝟐𝟓 × 𝟎. 𝟑 + 𝟐𝟎 × 𝟎. 𝟒 = −𝟐𝟔. 𝟖
  • 12. 0 1 2 3 4 P 0 0 –40 –80 –120 –160 0.04 1 0 5 –35 –75 –115 0.06 2 0 5 10 –30 –70 0.20 3 0 5 10 15 –25 0.30 4 0 5 10 15 20 0.40 EMV 0 3.2 3.7 –4.8 –26.8
  • 13. 0 1 2 3 4 P 0 0 –40 –80 –120 –160 0.04 1 0 5 –35 –75 –115 0.06 2 0 5 10 –30 –70 0.20 3 0 5 10 15 –25 0.30 4 0 5 10 15 20 0.40 EMV 0 3.2 3.7 –4.8 –26.8 𝑬𝑴𝑽 𝒎𝒂𝒙 = 𝑬𝑴𝑽 𝟐 = 𝟑. 𝟕
  • 15. Payoff Table 0 1 2 3 4 0 0 –40 –80 –120 –160 1 0 5 –35 –75 –115 2 0 5 10 –30 –70 3 0 5 10 15 –25 4 0 5 10 15 20 Regret/Opportunity Loss Table Courses of Action 0 1 2 3 4 States of Nature 0 0–0=0 1 5–0=5 2 10–0=10 3 15–0=15 4 20–0=20
  • 16. Payoff Table 0 1 2 3 4 0 0 –40 –80 –120 –160 1 0 5 –35 –75 –115 2 0 5 10 –30 –70 3 0 5 10 15 –25 4 0 5 10 15 20 Regret/Opportunity Loss Table Courses of Action 0 1 2 3 4 States of Nature 0 0–0=0 0–(–40)=40 1 5–0=5 5–5=0 2 10–0=10 10–5=5 3 15–0=15 15–5=10 4 20–0=20 20–5=15
  • 17. Payoff Table 0 1 2 3 4 0 0 –40 –80 –120 –160 1 0 5 –35 –75 –115 2 0 5 10 –30 –70 3 0 5 10 15 –25 4 0 5 10 15 20 Regret/Opportunity Loss Table Courses of Action 0 1 2 3 4 States of Nature 0 0–0=0 0–(–40)=40 0–(–80)=80 0–(–120)=120 0–(–160)=160 1 5–0=5 5–5=0 5–(–35)=40 5–(–75)=80 5–(–115)=120 2 10–0=10 10–5=5 10–10=0 10–(–30)=40 10–(–70)=80 3 15–0=15 15–5=10 15–10=5 15–15=0 15–(–25)=40 4 20–0=20 20–5=15 20–10=10 20–15=5 20–20=0
  • 18. Regret/Opportunity Loss Table Courses of Action 0 1 2 3 4 States of Nature 0 0 40 80 120 160 1 5 0 40 80 120 2 10 5 0 40 80 3 15 10 5 0 40 4 20 15 10 5 0
  • 19. Regret/Opportunity Loss Table Courses of Action P 0 1 2 3 4 States of Nature 0 0 40 80 120 160 0.04 1 5 0 40 80 120 0.06 2 10 5 0 40 80 0.20 3 15 10 5 0 40 0.30 4 20 15 10 5 0 0.40
  • 20. EXPECTED OPPORTUNITY LOSS (EOL) CALCULATION
  • 21. Regret/Opportunity Loss Table Courses of Action P 0 1 2 3 4 States of Nature 0 0 40 80 120 160 0.04 1 5 0 40 80 120 0.06 2 10 5 0 40 80 0.20 3 15 10 5 0 40 0.30 4 20 15 10 5 0 0.40 𝑬𝑶𝑳 𝟎 = 𝟎 × 𝟎. 𝟎𝟒 + 𝟓 × 𝟎. 𝟎𝟔 + 𝟏𝟎 × 𝟎. 𝟐 + 𝟏𝟓 × 𝟎. 𝟑 + 𝟐𝟎 × 𝟎. 𝟒 = 𝟏𝟒. 𝟖
  • 22. Regret/Opportunity Loss Table Courses of Action P 0 1 2 3 4 States of Nature 0 0 40 80 120 160 0.04 1 5 0 40 80 120 0.06 2 10 5 0 40 80 0.20 3 15 10 5 0 40 0.30 4 20 15 10 5 0 0.40 𝑬𝑶𝑳 𝟎 = 𝟎 × 𝟎. 𝟎𝟒 + 𝟓 × 𝟎. 𝟎𝟔 + 𝟏𝟎 × 𝟎. 𝟐 + 𝟏𝟓 × 𝟎. 𝟑 + 𝟐𝟎 × 𝟎. 𝟒 = 𝟏𝟒. 𝟖 𝑬𝑶𝑳 𝟏 = 𝟒𝟎 × 𝟎. 𝟎𝟒 + 𝟎 × 𝟎. 𝟎𝟔 + 𝟓 × 𝟎. 𝟐 + 𝟏𝟎 × 𝟎. 𝟑 + 𝟏𝟓 × 𝟎. 𝟒 = 𝟏𝟏. 𝟔 𝑬𝑶𝑳 𝟐 = 𝟖𝟎 × 𝟎. 𝟎𝟒 + 𝟒𝟎 × 𝟎. 𝟎𝟔 + 𝟎 × 𝟎. 𝟐 + 𝟓 × 𝟎. 𝟑 + 𝟏𝟎 × 𝟎. 𝟒 = 𝟏𝟏. 𝟏 𝑬𝑶𝑳 𝟑 = 𝟏𝟐𝟎 × 𝟎. 𝟎𝟒 + 𝟖𝟎 × 𝟎. 𝟎𝟔 + 𝟒𝟎 × 𝟎. 𝟐 + 𝟎 × 𝟎. 𝟑 + 𝟓 × 𝟎. 𝟒 = 𝟏𝟗. 𝟔 𝑬𝑶𝑳 𝟒 = 𝟏𝟔𝟎 × 𝟎. 𝟎𝟒 + 𝟏𝟐𝟎 × 𝟎. 𝟎𝟔 + 𝟖𝟎 × 𝟎. 𝟐 + 𝟒𝟎 × 𝟎. 𝟑 + 𝟎 × 𝟎. 𝟒 = 𝟒𝟏. 𝟔
  • 23. Regret/Opportunity Loss Table Courses of Action P 0 1 2 3 4 States of Nature 0 0 40 80 120 160 0.04 1 5 0 40 80 120 0.06 2 10 5 0 40 80 0.20 3 15 10 5 0 40 0.30 4 20 15 10 5 0 0.40 EOL 14.8 11.6 11.1 19.6 41.6
  • 24. Regret/Opportunity Loss Table Courses of Action P 0 1 2 3 4 States of Nature 0 0 40 80 120 160 0.04 1 5 0 40 80 120 0.06 2 10 5 0 40 80 0.20 3 15 10 5 0 40 0.30 4 20 15 10 5 0 0.40 EOL 14.8 11.6 11.1 19.6 41.6 𝑬𝑶𝑳 𝒎𝒊𝒏 = 𝑬𝑶𝑳 𝟐 = 𝟏𝟏. 𝟏
  • 25. EXPECTED VALUE WITH PERFECT INFORMATION EV with PI
  • 26. Payoff Matrix Courses of Action P 0 1 2 3 4 States of Nature 0 0 –40 –80 –120 –160 0.04 1 0 5 –35 –75 –115 0.06 2 0 5 10 –30 –70 0.20 3 0 5 10 15 –25 0.30 4 0 5 10 15 20 0.40 𝑬𝑽 𝒘𝒊𝒕𝒉 𝑷𝑰 = 𝚺 𝑴𝒂𝒙 𝑬𝑴𝑽 𝒇𝒐𝒓 𝒆𝒂𝒄𝒉 𝑺𝒕𝒂𝒕𝒆 𝒐𝒇 𝑵𝒂𝒕𝒖𝒓𝒆 × 𝑷
  • 27. Payoff Matrix Courses of Action P 0 1 2 3 4 States of Nature 0 0 –40 –80 –120 –160 0.04 1 0 5 –35 –75 –115 0.06 2 0 5 10 –30 –70 0.20 3 0 5 10 15 –25 0.30 4 0 5 10 15 20 0.40 𝑬𝑽 𝒘𝒊𝒕𝒉 𝑷𝑰 = 𝚺 𝑴𝒂𝒙 𝑬𝑴𝑽 𝒇𝒐𝒓 𝒆𝒂𝒄𝒉 𝑺𝒕𝒂𝒕𝒆 𝒐𝒇 𝑵𝒂𝒕𝒖𝒓𝒆 × 𝑷
  • 28. Payoff Matrix Courses of Action P 0 1 2 3 4 States of Nature 0 0 –40 –80 –120 –160 0.04 1 0 5 –35 –75 –115 0.06 2 0 5 10 –30 –70 0.20 3 0 5 10 15 –25 0.30 4 0 5 10 15 20 0.40 𝑬𝑽 𝒘𝒊𝒕𝒉 𝑷𝑰 = 𝟎 × 𝟎. 𝟎𝟒 + 𝟓 × 𝟎. 𝟎𝟔 + 𝟏𝟎 × 𝟎. 𝟐 + 𝟏𝟓 × 𝟎. 𝟑 + 𝟐𝟎 × 𝟎. 𝟒 ∴ 𝑬𝑽 𝒘𝒊𝒕𝒉 𝑷𝑰 = 𝟏𝟒. 𝟖
  • 29. EXPECTED VALUE OF PERFECT INFORMATION EVPI
  • 30. 𝑬𝑽𝑷𝑰 = 𝑬𝑽 𝒘𝒊𝒕𝒉 𝑷𝑰 − 𝑬𝑴𝑽 𝒎𝒂𝒙 𝒐𝒓 𝑬𝑽𝑷𝑰 = 𝟏𝟒. 𝟖 − 𝟑. 𝟕 𝒐𝒓 𝑬𝑽𝑷𝑰 = 𝟏𝟏. 𝟏
  • 31. 𝑬𝑽𝑷𝑰 = 𝑬𝑽 𝒘𝒊𝒕𝒉 𝑷𝑰 − 𝑬𝑴𝑽 𝒎𝒂𝒙 𝒐𝒓 𝑬𝑽𝑷𝑰 = 𝟏𝟒. 𝟖 − 𝟑. 𝟕 𝒐𝒓 𝑬𝑽𝑷𝑰 = 𝟏𝟏. 𝟏 𝑨𝒍𝒕𝒆𝒓𝒏𝒂𝒕𝒆𝒍𝒚, 𝑬𝑽𝑷𝑰 = 𝑬𝑶𝑳 𝒎𝒊𝒏 = 𝑬𝑶𝑳 𝟐 = 𝟏𝟏. 𝟏
  • 33. DECISION CRITERIA UNDER CONDITION OF UNCERTAINTY • Maximin. • Maximax. • Minimax Regret. • Hurwicz Criterion. • Baye’s/Lapalce’s Criterion.
  • 34. CRITERION OF PESSIMISM (MAXIMIN) • Also called ‘Waldian Criterion.’ • Determine the lowest outcome for each alternative. • Choose the alternative associated with the best of these.
  • 35. CRITERION OF OPTIMISM (MAXIMAX) • Suggested by Leonid Hurwicz. • Determine the best outcome for each alternative. • Select the alternative associated with the best of these.
  • 36. MINIMAX REGRET CRITERION • Attributed to Leonard Savage. • For each state, identify the most attractive alternative. • Place a zero in those cells. • Compute opportunity loss for other alternatives. • Identify the maximum opportunity loss for each alternative. • Select the alternative associated with the lowest of these.
  • 37. CRITERION OF REALISM (HURWICZ CRITERION) • A compromise between maximax and maximin criteria. • A coefficient of optimism α (0≤α≤1) is selected. • When α is close to 1, the decision-maker is optimistic about the future. • When α is close to 0, the decision-maker is pessimistic about the future.
  • 38. HURWICZ CRITERION • Select the strategy which maximises:
  • 39. LAPLACE CRITERION • Assign equal probabilities to each state. • Compute the expected value for each alternative. • Select the alternative with the highest alternative.
  • 41. Payoff Matrix Courses of Action 0 1 2 3 4 States of Nature 0 0 –40 –80 –120 –160 1 0 5 –35 –75 –115 2 0 5 10 –30 –70 3 0 5 10 15 –25 4 0 5 10 15 20
  • 42. Payoff Matrix: Maximin Criterion Courses of Action 0 1 2 3 4 States of Nature 0 0 –40 –80 –120 –160 1 0 5 –35 –75 –115 2 0 5 10 –30 –70 3 0 5 10 15 –25 4 0 5 10 15 20 𝑴𝒂𝒙𝒊𝒎𝒊𝒏 𝑪𝒓𝒊𝒕𝒆𝒓𝒊𝒐𝒏: ≫ 𝑴𝒂𝒙𝒊𝒎𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒐𝒍𝒖𝒎𝒏 𝒎𝒊𝒏𝒊𝒎𝒂 ≫ 𝒊. 𝒆. , 𝒇𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒊𝒏𝒊𝒎𝒖𝒎 𝒑𝒂𝒚𝒐𝒇𝒇 𝒊𝒏 𝒆𝒂𝒄𝒉 𝒄𝒐𝒍𝒖𝒎𝒏 ≫ 𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒂𝒙𝒊𝒎𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆𝒔𝒆 𝒑𝒂𝒚𝒐𝒇𝒇𝒔 ≫ 𝑻𝒉𝒆 𝒄𝒐𝒖𝒓𝒔𝒆 𝒐𝒇 𝒂𝒄𝒕𝒊𝒐𝒏 𝒄𝒐𝒓𝒓𝒆𝒔𝒑𝒐𝒏𝒅𝒊𝒏𝒈 𝒕𝒐 𝒕𝒉𝒊𝒔 𝒑𝒂𝒚𝒐𝒇𝒇 𝒊𝒔 𝒕𝒉𝒆 𝒂𝒏𝒔𝒘𝒆𝒓
  • 43. Payoff Matrix: Maximin Criterion Courses of Action 0 1 2 3 4 States of Nature 0 0 –40 –80 –120 –160 1 0 5 –35 –75 –115 2 0 5 10 –30 –70 3 0 5 10 15 –25 4 0 5 10 15 20 Column Minima 0 –40 –80 –120 –160 𝑴𝒂𝒙𝒊𝒎𝒊𝒏 𝑪𝒓𝒊𝒕𝒆𝒓𝒊𝒐𝒏: ≫ 𝑴𝒂𝒙𝒊𝒎𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒐𝒍𝒖𝒎𝒏 𝒎𝒊𝒏𝒊𝒎𝒂 ≫ 𝒊. 𝒆. , 𝒇𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒊𝒏𝒊𝒎𝒖𝒎 𝒑𝒂𝒚𝒐𝒇𝒇 𝒊𝒏 𝒆𝒂𝒄𝒉 𝒄𝒐𝒍𝒖𝒎𝒏 ≫ 𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒂𝒙𝒊𝒎𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆𝒔𝒆 𝒑𝒂𝒚𝒐𝒇𝒇𝒔 ≫ 𝑻𝒉𝒆 𝒄𝒐𝒖𝒓𝒔𝒆 𝒐𝒇 𝒂𝒄𝒕𝒊𝒐𝒏 𝒄𝒐𝒓𝒓𝒆𝒔𝒑𝒐𝒏𝒅𝒊𝒏𝒈 𝒕𝒐 𝒕𝒉𝒊𝒔 𝒑𝒂𝒚𝒐𝒇𝒇 𝒊𝒔 𝒕𝒉𝒆 𝒂𝒏𝒔𝒘𝒆𝒓
  • 44. Payoff Matrix: Maximin Criterion Courses of Action 0 1 2 3 4 States of Nature 0 0 –40 –80 –120 –160 1 0 5 –35 –75 –115 2 0 5 10 –30 –70 3 0 5 10 15 –25 4 0 5 10 15 20 Column Minima 0 –40 –80 –120 –160 Maximum of the column minima=0 𝑴𝒂𝒙𝒊𝒎𝒊𝒏 𝑪𝒓𝒊𝒕𝒆𝒓𝒊𝒐𝒏: ≫ 𝑴𝒂𝒙𝒊𝒎𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒐𝒍𝒖𝒎𝒏 𝒎𝒊𝒏𝒊𝒎𝒂 ≫ 𝒊. 𝒆. , 𝒇𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒊𝒏𝒊𝒎𝒖𝒎 𝒑𝒂𝒚𝒐𝒇𝒇 𝒊𝒏 𝒆𝒂𝒄𝒉 𝒄𝒐𝒍𝒖𝒎𝒏 ≫ 𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒂𝒙𝒊𝒎𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆𝒔𝒆 𝒑𝒂𝒚𝒐𝒇𝒇𝒔 ≫ 𝑻𝒉𝒆 𝒄𝒐𝒖𝒓𝒔𝒆 𝒐𝒇 𝒂𝒄𝒕𝒊𝒐𝒏 𝒄𝒐𝒓𝒓𝒆𝒔𝒑𝒐𝒏𝒅𝒊𝒏𝒈 𝒕𝒐 𝒕𝒉𝒊𝒔 𝒑𝒂𝒚𝒐𝒇𝒇 𝒊𝒔 𝒕𝒉𝒆 𝒂𝒏𝒔𝒘𝒆𝒓
  • 45. Payoff Matrix: Maximin Criterion Courses of Action 0 1 2 3 4 States of Nature 0 0 –40 –80 –120 –160 1 0 5 –35 –75 –115 2 0 5 10 –30 –70 3 0 5 10 15 –25 4 0 5 10 15 20 Column Minima 0 –40 –80 –120 –160 Maximum of the column minima=0 𝑴𝒂𝒙𝒊𝒎𝒊𝒏 𝑪𝒓𝒊𝒕𝒆𝒓𝒊𝒐𝒏: ≫ 𝑴𝒂𝒙𝒊𝒎𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒐𝒍𝒖𝒎𝒏 𝒎𝒊𝒏𝒊𝒎𝒂 ≫ 𝒊. 𝒆. , 𝒇𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒊𝒏𝒊𝒎𝒖𝒎 𝒑𝒂𝒚𝒐𝒇𝒇 𝒊𝒏 𝒆𝒂𝒄𝒉 𝒄𝒐𝒍𝒖𝒎𝒏 ≫ 𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒂𝒙𝒊𝒎𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆𝒔𝒆 𝒑𝒂𝒚𝒐𝒇𝒇𝒔 ≫ 𝑻𝒉𝒆 𝒄𝒐𝒖𝒓𝒔𝒆 𝒐𝒇 𝒂𝒄𝒕𝒊𝒐𝒏 𝒄𝒐𝒓𝒓𝒆𝒔𝒑𝒐𝒏𝒅𝒊𝒏𝒈 𝒕𝒐 𝒕𝒉𝒊𝒔 𝒑𝒂𝒚𝒐𝒇𝒇 𝒊𝒔 𝒕𝒉𝒆 𝒂𝒏𝒔𝒘𝒆𝒓 ∴ 𝒕𝒉𝒆 𝒃𝒆𝒔𝒕 𝒄𝒐𝒖𝒓𝒔𝒆 𝒐𝒇 𝒂𝒄𝒕𝒊𝒐𝒏 𝒂𝒄𝒄𝒐𝒓𝒅𝒊𝒏𝒈 𝒕𝒐 𝒕𝒉𝒆 𝒎𝒂𝒙𝒊𝒎𝒊𝒏 𝒄𝒓𝒊𝒕𝒆𝒓𝒊𝒐𝒏 = 𝟎
  • 47. Payoff Matrix Courses of Action 0 1 2 3 4 States of Nature 0 0 –40 –80 –120 –160 1 0 5 –35 –75 –115 2 0 5 10 –30 –70 3 0 5 10 15 –25 4 0 5 10 15 20
  • 48. Payoff Matrix: Maximax Criterion Courses of Action 0 1 2 3 4 States of Nature 0 0 –40 –80 –120 –160 1 0 5 –35 –75 –115 2 0 5 10 –30 –70 3 0 5 10 15 –25 4 0 5 10 15 20 𝑴𝒂𝒙𝒊𝒎𝒂𝒙 𝑪𝒓𝒊𝒕𝒆𝒓𝒊𝒐𝒏: ≫ 𝑴𝒂𝒙𝒊𝒎𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒐𝒍𝒖𝒎𝒏 𝒎𝒂𝒙𝒊𝒎𝒂 ≫ 𝒊. 𝒆. , 𝒇𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒂𝒙𝒊𝒎𝒖𝒎 𝒑𝒂𝒚𝒐𝒇𝒇 𝒊𝒏 𝒆𝒂𝒄𝒉 𝒄𝒐𝒍𝒖𝒎𝒏 ≫ 𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒂𝒙𝒊𝒎𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆𝒔𝒆 𝒑𝒂𝒚𝒐𝒇𝒇𝒔 ≫ 𝑻𝒉𝒆 𝒄𝒐𝒖𝒓𝒔𝒆 𝒐𝒇 𝒂𝒄𝒕𝒊𝒐𝒏 𝒄𝒐𝒓𝒓𝒆𝒔𝒑𝒐𝒏𝒅𝒊𝒏𝒈 𝒕𝒐 𝒕𝒉𝒊𝒔 𝒑𝒂𝒚𝒐𝒇𝒇 𝒊𝒔 𝒕𝒉𝒆 𝒂𝒏𝒔𝒘𝒆𝒓
  • 49. Payoff Matrix: Maximax Criterion Courses of Action 0 1 2 3 4 States of Nature 0 0 –40 –80 –120 –160 1 0 5 –35 –75 –115 2 0 5 10 –30 –70 3 0 5 10 15 –25 4 0 5 10 15 20 Column Maxima 0 5 10 15 20 𝑴𝒂𝒙𝒊𝒎𝒂𝒙 𝑪𝒓𝒊𝒕𝒆𝒓𝒊𝒐𝒏: ≫ 𝑴𝒂𝒙𝒊𝒎𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒐𝒍𝒖𝒎𝒏 𝒎𝒂𝒙𝒊𝒎𝒂 ≫ 𝒊. 𝒆. , 𝒇𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒂𝒙𝒊𝒎𝒖𝒎 𝒑𝒂𝒚𝒐𝒇𝒇 𝒊𝒏 𝒆𝒂𝒄𝒉 𝒄𝒐𝒍𝒖𝒎𝒏 ≫ 𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒂𝒙𝒊𝒎𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆𝒔𝒆 𝒑𝒂𝒚𝒐𝒇𝒇𝒔 ≫ 𝑻𝒉𝒆 𝒄𝒐𝒖𝒓𝒔𝒆 𝒐𝒇 𝒂𝒄𝒕𝒊𝒐𝒏 𝒄𝒐𝒓𝒓𝒆𝒔𝒑𝒐𝒏𝒅𝒊𝒏𝒈 𝒕𝒐 𝒕𝒉𝒊𝒔 𝒑𝒂𝒚𝒐𝒇𝒇 𝒊𝒔 𝒕𝒉𝒆 𝒂𝒏𝒔𝒘𝒆𝒓
  • 50. Payoff Matrix: Maximax Criterion Courses of Action 0 1 2 3 4 States of Nature 0 0 –40 –80 –120 –160 1 0 5 –35 –75 –115 2 0 5 10 –30 –70 3 0 5 10 15 –25 4 0 5 10 15 20 Column Maxima 0 5 10 15 20 Maximum of the column maxima=20 𝑴𝒂𝒙𝒊𝒎𝒂𝒙 𝑪𝒓𝒊𝒕𝒆𝒓𝒊𝒐𝒏: ≫ 𝑴𝒂𝒙𝒊𝒎𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒐𝒍𝒖𝒎𝒏 𝒎𝒂𝒙𝒊𝒎𝒂 ≫ 𝒊. 𝒆. , 𝒇𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒂𝒙𝒊𝒎𝒖𝒎 𝒑𝒂𝒚𝒐𝒇𝒇 𝒊𝒏 𝒆𝒂𝒄𝒉 𝒄𝒐𝒍𝒖𝒎𝒏 ≫ 𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒂𝒙𝒊𝒎𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆𝒔𝒆 𝒑𝒂𝒚𝒐𝒇𝒇𝒔 ≫ 𝑻𝒉𝒆 𝒄𝒐𝒖𝒓𝒔𝒆 𝒐𝒇 𝒂𝒄𝒕𝒊𝒐𝒏 𝒄𝒐𝒓𝒓𝒆𝒔𝒑𝒐𝒏𝒅𝒊𝒏𝒈 𝒕𝒐 𝒕𝒉𝒊𝒔 𝒑𝒂𝒚𝒐𝒇𝒇 𝒊𝒔 𝒕𝒉𝒆 𝒂𝒏𝒔𝒘𝒆𝒓
  • 51. Payoff Matrix: Maximax Criterion Courses of Action 0 1 2 3 4 States of Nature 0 0 –40 –80 –120 –160 1 0 5 –35 –75 –115 2 0 5 10 –30 –70 3 0 5 10 15 –25 4 0 5 10 15 20 Column Maxima 0 5 10 15 20 Maximum of the column maxima=20 𝑴𝒂𝒙𝒊𝒎𝒂𝒙 𝑪𝒓𝒊𝒕𝒆𝒓𝒊𝒐𝒏: ≫ 𝑴𝒂𝒙𝒊𝒎𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒐𝒍𝒖𝒎𝒏 𝒎𝒂𝒙𝒊𝒎𝒂 ≫ 𝒊. 𝒆. , 𝒇𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒂𝒙𝒊𝒎𝒖𝒎 𝒑𝒂𝒚𝒐𝒇𝒇 𝒊𝒏 𝒆𝒂𝒄𝒉 𝒄𝒐𝒍𝒖𝒎𝒏 ≫ 𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒂𝒙𝒊𝒎𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆𝒔𝒆 𝒑𝒂𝒚𝒐𝒇𝒇𝒔 ≫ 𝑻𝒉𝒆 𝒄𝒐𝒖𝒓𝒔𝒆 𝒐𝒇 𝒂𝒄𝒕𝒊𝒐𝒏 𝒄𝒐𝒓𝒓𝒆𝒔𝒑𝒐𝒏𝒅𝒊𝒏𝒈 𝒕𝒐 𝒕𝒉𝒊𝒔 𝒑𝒂𝒚𝒐𝒇𝒇 𝒊𝒔 𝒕𝒉𝒆 𝒂𝒏𝒔𝒘𝒆𝒓 ∴ 𝒕𝒉𝒆 𝒃𝒆𝒔𝒕 𝒄𝒐𝒖𝒓𝒔𝒆 𝒐𝒇 𝒂𝒄𝒕𝒊𝒐𝒏 𝒂𝒄𝒄𝒐𝒓𝒅𝒊𝒏𝒈 𝒕𝒐 𝒕𝒉𝒆 𝒎𝒂𝒙𝒊𝒎𝒂𝒙 𝒄𝒓𝒊𝒕𝒆𝒓𝒊𝒐𝒏 = 𝟒
  • 53. Regret/Opportunity Loss Table Courses of Action 0 1 2 3 4 States of Nature 0 0 40 80 120 160 1 5 0 40 80 120 2 10 5 0 40 80 3 15 10 5 0 40 4 20 15 10 5 0
  • 54. Regret Table (Minimax Regret Criterion) Courses of Action 0 1 2 3 4 States of Nature 0 0 40 80 120 160 1 5 0 40 80 120 2 10 5 0 40 80 3 15 10 5 0 40 4 20 15 10 5 0 𝑴𝒊𝒏𝒊𝒎𝒂𝒙 𝑹𝒆𝒈𝒓𝒆𝒕 𝑪𝒓𝒊𝒕𝒆𝒓𝒊𝒐𝒏: ≫ 𝑴𝒊𝒏𝒊𝒎𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒐𝒍𝒖𝒎𝒏 𝒎𝒂𝒙𝒊𝒎𝒂 (𝒊𝒏 𝒕𝒉𝒆 𝒓𝒆𝒈𝒓𝒆𝒕 𝒕𝒂𝒃𝒍𝒆) ≫ 𝒊. 𝒆. , 𝒇𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒂𝒙𝒊𝒎𝒖𝒎 𝒓𝒆𝒈𝒓𝒆𝒕 𝒊𝒏 𝒆𝒂𝒄𝒉 𝒄𝒐𝒍𝒖𝒎𝒏 ≫ 𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒊𝒏𝒊𝒎𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆𝒔𝒆 𝒓𝒆𝒈𝒓𝒆𝒕𝒔 ≫ 𝑻𝒉𝒆 𝒄𝒐𝒖𝒓𝒔𝒆 𝒐𝒇 𝒂𝒄𝒕𝒊𝒐𝒏 𝒄𝒐𝒓𝒓𝒆𝒔𝒑𝒐𝒏𝒅𝒊𝒏𝒈 𝒕𝒐 𝒕𝒉𝒊𝒔 𝒓𝒆𝒈𝒓𝒆𝒕 𝒊𝒔 𝒕𝒉𝒆 𝒂𝒏𝒔𝒘𝒆𝒓
  • 55. Regret Table (Minimax Regret Criterion) Courses of Action 0 1 2 3 4 States of Nature 0 0 40 80 120 160 1 5 0 40 80 120 2 10 5 0 40 80 3 15 10 5 0 40 4 20 15 10 5 0 𝑴𝒊𝒏𝒊𝒎𝒂𝒙 𝑹𝒆𝒈𝒓𝒆𝒕 𝑪𝒓𝒊𝒕𝒆𝒓𝒊𝒐𝒏: ≫ 𝑴𝒊𝒏𝒊𝒎𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒐𝒍𝒖𝒎𝒏 𝒎𝒂𝒙𝒊𝒎𝒂 (𝒊𝒏 𝒕𝒉𝒆 𝒓𝒆𝒈𝒓𝒆𝒕 𝒕𝒂𝒃𝒍𝒆) ≫ 𝒊. 𝒆. , 𝒇𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒂𝒙𝒊𝒎𝒖𝒎 𝒓𝒆𝒈𝒓𝒆𝒕 𝒊𝒏 𝒆𝒂𝒄𝒉 𝒄𝒐𝒍𝒖𝒎𝒏 ≫ 𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒊𝒏𝒊𝒎𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆𝒔𝒆 𝒓𝒆𝒈𝒓𝒆𝒕𝒔 ≫ 𝑻𝒉𝒆 𝒄𝒐𝒖𝒓𝒔𝒆 𝒐𝒇 𝒂𝒄𝒕𝒊𝒐𝒏 𝒄𝒐𝒓𝒓𝒆𝒔𝒑𝒐𝒏𝒅𝒊𝒏𝒈 𝒕𝒐 𝒕𝒉𝒊𝒔 𝒓𝒆𝒈𝒓𝒆𝒕 𝒊𝒔 𝒕𝒉𝒆 𝒂𝒏𝒔𝒘𝒆𝒓
  • 56. Regret Table (Minimax Regret Criterion) Courses of Action 0 1 2 3 4 States of Nature 0 0 40 80 120 160 1 5 0 40 80 120 2 10 5 0 40 80 3 15 10 5 0 40 4 20 15 10 5 0 Column maxima 20 40 80 120 160 𝑴𝒊𝒏𝒊𝒎𝒂𝒙 𝑹𝒆𝒈𝒓𝒆𝒕 𝑪𝒓𝒊𝒕𝒆𝒓𝒊𝒐𝒏: ≫ 𝑴𝒊𝒏𝒊𝒎𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒐𝒍𝒖𝒎𝒏 𝒎𝒂𝒙𝒊𝒎𝒂 (𝒊𝒏 𝒕𝒉𝒆 𝒓𝒆𝒈𝒓𝒆𝒕 𝒕𝒂𝒃𝒍𝒆) ≫ 𝒊. 𝒆. , 𝒇𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒂𝒙𝒊𝒎𝒖𝒎 𝒓𝒆𝒈𝒓𝒆𝒕 𝒊𝒏 𝒆𝒂𝒄𝒉 𝒄𝒐𝒍𝒖𝒎𝒏 ≫ 𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒊𝒏𝒊𝒎𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆𝒔𝒆 𝒓𝒆𝒈𝒓𝒆𝒕𝒔 ≫ 𝑻𝒉𝒆 𝒄𝒐𝒖𝒓𝒔𝒆 𝒐𝒇 𝒂𝒄𝒕𝒊𝒐𝒏 𝒄𝒐𝒓𝒓𝒆𝒔𝒑𝒐𝒏𝒅𝒊𝒏𝒈 𝒕𝒐 𝒕𝒉𝒊𝒔 𝒓𝒆𝒈𝒓𝒆𝒕 𝒊𝒔 𝒕𝒉𝒆 𝒂𝒏𝒔𝒘𝒆𝒓
  • 57. Regret Table (Minimax Regret Criterion) Courses of Action 0 1 2 3 4 States of Nature 0 0 40 80 120 160 1 5 0 40 80 120 2 10 5 0 40 80 3 15 10 5 0 40 4 20 15 10 5 0 Column maxima 20 40 80 120 160 Minimum of the column maxima=20 𝑴𝒊𝒏𝒊𝒎𝒂𝒙 𝑹𝒆𝒈𝒓𝒆𝒕 𝑪𝒓𝒊𝒕𝒆𝒓𝒊𝒐𝒏: ≫ 𝑴𝒊𝒏𝒊𝒎𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒐𝒍𝒖𝒎𝒏 𝒎𝒂𝒙𝒊𝒎𝒂 (𝒊𝒏 𝒕𝒉𝒆 𝒓𝒆𝒈𝒓𝒆𝒕 𝒕𝒂𝒃𝒍𝒆) ≫ 𝒊. 𝒆. , 𝒇𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒂𝒙𝒊𝒎𝒖𝒎 𝒓𝒆𝒈𝒓𝒆𝒕 𝒊𝒏 𝒆𝒂𝒄𝒉 𝒄𝒐𝒍𝒖𝒎𝒏 ≫ 𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒊𝒏𝒊𝒎𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆𝒔𝒆 𝒓𝒆𝒈𝒓𝒆𝒕𝒔 ≫ 𝑻𝒉𝒆 𝒄𝒐𝒖𝒓𝒔𝒆 𝒐𝒇 𝒂𝒄𝒕𝒊𝒐𝒏 𝒄𝒐𝒓𝒓𝒆𝒔𝒑𝒐𝒏𝒅𝒊𝒏𝒈 𝒕𝒐 𝒕𝒉𝒊𝒔 𝒓𝒆𝒈𝒓𝒆𝒕 𝒊𝒔 𝒕𝒉𝒆 𝒂𝒏𝒔𝒘𝒆𝒓
  • 58. Regret Table (Minimax Regret Criterion) Courses of Action 0 1 2 3 4 States of Nature 0 0 40 80 120 160 1 5 0 40 80 120 2 10 5 0 40 80 3 15 10 5 0 40 4 20 15 10 5 0 Column maxima 20 40 80 120 160 Minimum of the column maxima=20 𝑴𝒊𝒏𝒊𝒎𝒂𝒙 𝑹𝒆𝒈𝒓𝒆𝒕 𝑪𝒓𝒊𝒕𝒆𝒓𝒊𝒐𝒏: ≫ 𝑴𝒊𝒏𝒊𝒎𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒐𝒍𝒖𝒎𝒏 𝒎𝒂𝒙𝒊𝒎𝒂 (𝒊𝒏 𝒕𝒉𝒆 𝒓𝒆𝒈𝒓𝒆𝒕 𝒕𝒂𝒃𝒍𝒆) ≫ 𝒊. 𝒆. , 𝒇𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒂𝒙𝒊𝒎𝒖𝒎 𝒓𝒆𝒈𝒓𝒆𝒕 𝒊𝒏 𝒆𝒂𝒄𝒉 𝒄𝒐𝒍𝒖𝒎𝒏 ≫ 𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝒎𝒊𝒏𝒊𝒎𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆𝒔𝒆 𝒓𝒆𝒈𝒓𝒆𝒕𝒔 ≫ 𝑻𝒉𝒆 𝒄𝒐𝒖𝒓𝒔𝒆 𝒐𝒇 𝒂𝒄𝒕𝒊𝒐𝒏 𝒄𝒐𝒓𝒓𝒆𝒔𝒑𝒐𝒏𝒅𝒊𝒏𝒈 𝒕𝒐 𝒕𝒉𝒊𝒔 𝒓𝒆𝒈𝒓𝒆𝒕 𝒊𝒔 𝒕𝒉𝒆 𝒂𝒏𝒔𝒘𝒆𝒓 ∴ 𝒕𝒉𝒆 𝒃𝒆𝒔𝒕 𝒄𝒐𝒖𝒓𝒔𝒆 𝒐𝒇 𝒂𝒄𝒕𝒊𝒐𝒏 = 𝟎