decision making

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ComSci: Renas R. Rekany
Artificial Intelligence
Decision Making
Renas R. Rekany
2018

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Decision making is the study of identifying and choosing
alternatives based on the values and preferences of the
decision maker. Making a decision implies that there are
alternative choices to be considered, this indicates that
alternatives must be considered and a decision must be made
to fit out needs.
Decision making is trivial in AI, as robots should possess the
necessary knowledge to make a decision when required. as
such is the aim of AI. Usually a problem might not have an
exact solution. thus, a solution must be found for a problem
at hand based on problem analysis and requirements.

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Decision making is based on:
1. Alternative.
2. Criterion.
3. Probability of event occurrence.
4. Outcome evaluation.
There two major decision making categories:
1. Simple decision making.
2. Complex decision making.

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Problem 1:
Suppose a company has 10 computers and should be
served by 5 staff [ so each staff should be responsive
for serving of 2 computers].
So, decision making is a choice of two alternatives.

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1. Identify the Problem
2. Gather Information
3. Analyze the Situation
4. Configure Goals
5. Evaluate Alternatives
6. Select a Preferred (Best) Alternative
7. Act on the Decision

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Simple Decision Making Concept:
A simple decision refers to the process of decision making in an
situation where one of the alternatives and it’s outcome significantly
overweigh the other alternatives. to make such a decision can be
simple as the results are obviously as desired or are closest to the
solution.
Concepts of simple decision making
1. Expected Monetary Value (EMV)
2. Posterior Probability Function.
3. Quantitative Approach
4. Qualitative Approach

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The EMV utility function tries to capture the average between a best
case and worst case scenario. it also tries to include the probability of
that outcome occurring.
Example1:
An author wants to publish his book; the following contracts are
available.
Hint) To find the best contract, EMV is applied. to calculate the best
possible solution, EMV suggest the sum of multiplying each value by
its chance of occurrence, the value should indicate the better contract.
Degree of success Probability A B C
High 0.3 60,000 73,000 56,000
Medium 0.5 33,000 32,000 31,000
Low 0.2 15,000 13,000 20,000

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Sol) Expected monetary value (EMV) [expected profit]
EA = 0.3 * 60,000 + 0.5 * 33,000 + 0.2 * 15,000 = 37,000
EB = 0.3 * 73,000 + 0.5 * 32,000 + 0.2 * 13,000 = 40,000
EC = 0.3 * 56,000 + 0.5 * 31,000 + 0.2 * 20,000 = 36,000
Since the EMV is maximum for the second contract, author should
choose the second contract.

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Posterior probability is part of Bayesian Statics,
which indicates the truth of a statement by the degree
of belief.
[Posterior probability can be calculated from the
multiplication of prior and conditional probabilities
over the sum of multiplying prior and conditional
probabilities for all possibilities]

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Example2:
A business man want to open a restaurant, the construction
company offered three contracts to open the restaurant, which
contract should he choose?
Should the restaurant be open? and for how many clients?
Prior Probabilities Conditional Probabilities
100 clients 60% 100 clients 70%

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Sol) Since the joint Probability is more than 50%, the
restaurant should be opened.
Based on the posterior probability, the restaurant must be
opened for 100 clients.
State Prior
Prob.
Conditiona
l Prob.
Joint Prob. Posterior Prob.
100 60% 70% 0.6*07=0.42 0.42/0.51=0.8
200 10% 0% 0.1*0=0 0/0.51=0
300 30% 30% 0.3*0.3=0.09 0.09/0.51=0.2
Total Joint Prob. 0.51 1.0

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Example3:
An auctioneer has developed a simple mathematical
model for deciding the starting bid he will require when
auctioning a used automobile. Essentially, he sets the starting
bid at 70% percent of what he predicts the final winning bid
will (or should) be. He predicts the winning bid by starting
with the car's original selling price and making two
deductions, one based on the car's age and the other based on
the car's mileage. The age deduction is $800 per year and the
mileage deduction is $0.025 per mile.
Hint: Suppose a four-year old car with 60,000 miles on the
odometer is up for auction. If its original price was $12,500,
what starting bid should the auctioneer require?

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Original Price – age deduction (per year) – mileage deduction (per
miles)
Bid = 0.7 (expected winning Bid).
Bid = 0.7 (P – 800 (A) – 0.025 (M)).
Bid = 0.7 * P – 560 (4) – 0.0175 (60,000).
Bid = 0.7 * 12,500 – 560 * 4 – 0.0175 * 60,000 = $5460

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-Expected Profit without Perfect Information (EP w/o PI)
-Expected Profit with Perfect Information (EP w PI)
-Expected Profit of Perfect Information = EP w PI – EP w/o PI
Example4: Make Decision to select the alternative
according to:
Alternative First Second Third Fourth
1 0 -20 -50 -100
2 -50 40 -20 -60
3 -70 -50 40 100
Prob. 0.2 0.4 0.1 0.3

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1. EP w/o PI:
Altr.1→ 0.2*0 + 0.4*(-20) + 0.1*(-50) + 0.3*(-100) = -43
Altr.2→ 0.2*(-50) + 0.4*40 + 0.1*(-20) + 0.3*(-60) = -14
Altr.3→ 0.2*(-70) + 0.4(-50) + 0.1*40 + 0.3*100 = 0
Decision is Altr.3 (best decision)
2. EP w PI:
= 0.2*0 + 0.4*40 + 0.1*40 + 0.3*100 = 50
3. EP of PI → EP w PI – EP w/o PI → 50 - 0 = 50

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Example5: The probability for having a head in coin is 0.5,
and for having a tail is 0.5, make a decision if Ali should start
a bid or not? according to the following:
1. Win 3,000,000$ if he not betting.
2. Win nothing if he got a head.
3. Win 8,000,000$ if he got a tail.
Sol:
4m > 3m
The decision
is Yes.

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1. Expected Profit
2. Standard Deviation
Example6: John is trying to make a choice on a bet: John is
paid 2000$ when Tiger shoots less than 65, John is sure of
that 80%, John is paid 1500$ if Tiger shoots more than 65.
however, John is paid 3000$ if Sergio shoots less than 65 and
is sure 70% of this event and pays 200$ if he shoots more
than 65.

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1. Expected Profit
2. Standard Deviation
Solution:
1) Expected Profit:
Expected Profit (for Tiger): -
= 2000*0.8 + 1500*0.2=1900$
Expected Profit (for Sergio): -
=3000*0.7+200*0.3=2160$
John should invest Sergio, because 2160$>1900$.

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Standard Deviation: Tiger
1)Subtract each possible outcome from expected profit.
2000-1900=100
1500-1900= -400
2)Square the results
100 ^2=10000
-400 ^2=160000
3)Multiply each result by corresponding probability.
0.8*10000=8000
0.2*160000=32000
4)Sum the results 8000+32000=40000$
5)Find square root of 40000 = 200

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Standard Deviation: Sergio
1)Subtract each possible outcome from expected profit.
3000-2160=840
200-2160= -1960
2)Square the results.
840^2=705600
-1960^2=3841000
3)Multiply each result by corresponding probability.
0.7*705600=493920
0.3*3841000=1152480
4)Sum the result 493920+1152480=1,646,400
5)Find square root of 1,646,400= 1283

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Standard deviation for Sergio 1283 is much
more than for Tiger 200, so invest for Tiger.

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1. Maximax Pay off (Optimistic approach).
2. Maximin Pay off (Pessimistic approach).
3. Laplace (Likely equal hood method).
4. Hurwicz Criterion (Criterion Realism).
5. Minimax Criterion (Regret Criterion).

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Example7: Illustration: Considering a manufacturing
company that is thinking of various alternatives to
increase its production to meet the increasing market
demand.
State O1 O2 O3 Maximax (1) Maximin (2)
A 1000 1000 1000 1000 1000
B 10,000 -7000 500 10,000 -7000
C 5000 0 800 5000 0
D 8000 -2000 700 8000 -2000

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1. Maximax: The best decision is
B > D > C > A → 10,000 > 8000 > 5000 > 1000
2. Maximin: The best decision is
A > C > D > B → 1000 > 0 > -2000 > -7000
3. Laplace:
For A: (1000+1000+1000)/3= 1000
For B: (10,000+(-7000) +500)/3= 1166
For C: (5000+0+800)/3 = 1933
For D: (8000+(-2000) +700)/3 = 2233
So, the best decision is D > C > B > A

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4. Hurwicz: α = 0.6
For each alternative: (The best * α + (1- α) * The worst)
For A: 1000*0.6 + (1- 0.6) * 1000 = 1000
For B: 10,000 * 0.6 + 0.4 * (-7000) = 3200
For C: 5000*0.6 + 0.4 * 0 = 3000
For D: 8000*0.6 + 0.4 * (-2000) = 4000
So, the best decision is D > B > C > A

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5. Minimax:
Best one in column – alternative
A: 10,000 – 1000 = 9000 1000 – 1000 = 0 1000 – 1000 = 0
B: 10,000 – 10,000 = 0 1000 – (-7000) =8000 1000 – 500 = 500
C: 10,000 – 5000 = 5000 1000 – 0 = 1000 1000 – 800 = 200
D: 10,000 – 8000 = 2000 1000 – (-2000) =3000 1000 – 700 = 300
Max of first row is 9000, max of second row is 8000, max of third row is
5000, max of fourth row is 3000.
So, the decision is from Min D > C > B > A

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Example8:
Suppose Ali has $100 to invest:
1. For stock share, will get $105 for certain.
2. Invest $100 for something else, $150 will
returned back to him with 60% or $80 with
probability 40%.
Find: Risk Free Return, Expected Risk, Risk
Premium.

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Sol:
1. Risk free return:
$105 - $100 = $5
2. Expected risk return:
0.6 * $150 + 0.4 * $80 = $122
3. Risk premium return:
Expected risk – Invested money
$122 – $100 = $22

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Example9:
Suppose a business man want to extend his
company. He is sure in returning the
investment 50% (Certain). According to the
following incomes find risk premium return
using utility function (UR=3𝑹 𝟐 + 5R + 5).

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Sol:
Utility for sure return: 3 * (0.5)^2 + 5 * (0.5) + 5 = 8.25
Utility for risky return: 3 * (0.15)^2 + 5 * (0.15) + 5 = 6
: 3 * (0.25)^2 + 5 * (0.25) + 5 = 6.4
: 3 * (0.55)^2 + 5 * (0.55) + 5 = 9
: 3 * (0.05)^2 + 5 * (0.05) + 5 = 5.3
Income Probability
15% 5%
25% 30%
55% 40%
5% 25%

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6 * 0.05 = 0.3
6.4 * 0.3 = 1.92
9 * 0.4 = 3.6
5.3 * 0.25 = 1.3
Total risk premium = 0.3 + 1.92 + 3.6 + 1.3 = 7.12
The utility for sure return greater than risky
return. So, the business man mustn’t take risk
investment.

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Example 10:
EOLi = σ 𝑷𝒊 * OLij
P(s1) = 0.5 , P(s2) = 0.3 , P(s3) = 0.2
Alternative S1 S2 S3
A1 50 70 100
A2 40 80 90
A3 90 70 60

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Sol:
A1 90 - 50 = 40 80 – 70 = 10 100 – 100 = 0
A2 90 – 40 = 50 80 – 80 = 0 100 – 90 = 10
A3 90 – 90 = 0 80 – 70 = 10 100 – 60 = 40
Eol(A1) = 0.5 * 40 + 0.3 * 10 + 0.2 * 0 = 23
Eol(A2) = 0.5 * 50 + 0.3 * 0 + 0.2 * 10 = 27
Eol(A3) = 0.5 * 0 + 0.3 * 10 + 0.2 * 40 = 11
Min Eol is 11, so the decision make for alternative A3.

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Example 11: Make Decision in which city to live according to:
Cities
Climate
5
Arts
4
Entertainment
3
Cost of living
2
Crime
1
Atlanta 4 4 2 1 0
Boston 3 5 3 0 1
Chicago 1 5 3 1 0
Denver 1 3 3 1 1
Eugene 4 2 5 2 2

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1) Weighted Additive Model (WADD)
Atlanta→ 5*4 + 4*4 + 3*2 + 2*1 + 1*0 = 44
Boston→ 5*3 + 4*5 + 3*3 + 2*0 + 1*1 = 45
Chicago→ 5*1 + 4*5 + 3*3 + 2*1 + 1*0 = 36
Denver→ 5*1 + 4*3 + 3*5 + 2*1 + 1*1 = 29
Eugene→ 5*4 + 4*2 +3*5 + 2*2 + 1*2 = 49
So, the decision is for Eugene.

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2) Satisficing (SAT) Heuristic
Set a minimum threshold for each attribute (for example 1)
Choose first item that exceeds threshold for all attributes
-Minimum Threshold is 1:
Atlanta rejected because of cost of living is 1.
Boston rejected because of cost of living is 0.
Chicago and Denver are rejected because of Climate is 1.
Eugene is the best decision because all attributes are more
than threshold.

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3) Lexicographic (LEX)
Choose the first important attribute and it’s (Climate) then we
keep Atlanta and Eugene because both of them equal to 4,
since the next important attribute is (Arts), So, between Atlanta
and Eugene we decide Atlanta because is more than Eugene
(4>2).

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4) Elimination_By_Aspect (EBA)
Set a minimum threshold for each attribute (for example 2)
-Choose the items that equal and greater than threshold for all
attributes minimum threshold is 2.
Chicago and Denver are rejected because of Climate is 1.
We keep choosing the Atlanta, Boston and Eugene till we
compare with attribute (Cost of living), Atlanta and Boston will
be rejected because both are less than threshold 2, So the best
decision is Eugene.

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5)Majority Conforming Dimensions (MCD)
Is used for pair of alternatives choose one that has the highest
number of winning.
Atlanta → 1 winning
Pair1 So, the decision is Eugene.
Eugene → 3 winning
Atlanta→ 2 winning
Pair2 So, the decision is Boston
Boston→ 3 winning

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6)Frequency of good and bad features (FRQ)
Minimum value of all attributes is zero, and max value is five.
So, 0+5/2=2.5
-Determine reference point for each attribute to separate “good” from “bad”
(2.5). Choose the alternative with maximum “good”.
Cities
Climate
5
Arts
4
Entertainment
3
Cost of living 2
Crime
1
Atlanta +4 +4 -2 -1 -0
Boston +3 +5 +3 -0 -1
Chicago -1 +5 +3 -1 -0
Denver -1 +3 +3 -1 -1
Eugene +4 -2 +5 -2 -2
So, the decision is Boston because has maximum good+ equal to 3 pluses.

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7) Equal Weight (EQW)
Assign equal weights to all attributes and select the best alternative.
Atlanta→ 4+4+2+1+0= 11
Boston→ 3+5+3+0+1= 12
Chicago→ 1+5+3+1+0= 10
Denver→ 1+3+3+1+1= 9
Eugene→ 4+2+5+2+2= 15
So, the decision is for Eugene.

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