ISE 562; Dr. Smith
EVPI and EVSI Part I
Decision Theory
ISE 562; Dr. -Part I
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• T erminology
• EVPI and EVSI
Today-Part II
• More on utility and how to find the certainty equivalent
• How to identify risk attitude
• Other methods for calculating EV of
information; ENGS
10/8/2022 2
ISE 562; Dr. glossary of terms
• a*: optimal decision
• CE: certainty equivalent
• CS: cost of sampling
• DM: decision maker
• E[]: expectation operator
• EV: expected value
• EMV: expected monetary value
• ENGS: expected net gain of sampling
• EL: expected loss
• EOL: expected opportunity loss
• ERPI; EVPI: expected return/payoff under perfect information
• EU: expected utility
• R(a): book notation for return (payoff) of alternative a.
• Utility: functional assignment of numerical preference value to levels
of an attribute (U(x): x[0, 1])
• RP: risk premium
• VPI(): value of perfect information
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Trade Bill
P(pass)=.60
P(fails)=0.40
ISE 562; Dr. Value
Expected value decision = plant corn
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=.6(35000)+(.4)(8000)=24,200 =.6(18000)+(.4)(12000)=15,600
=.6(22000)+(.4)(20000)=21,200
Trade Bill
P(pass)=.60
P(fails)=0.40
ISE 562; Dr. analysis and value of information
Decisions often have opportunities to gather data (at some cost) to enable a better decision
Classic example: exploratory oil wells Sending precursor missions to testing prior to release of new product
Surveys to assess customer reaction Studies to define/refine decision components Hiring experts/consultants
ISE 562; Dr. Opportunity Loss (choose option with minimum EOL)
=.6(0)+(.4)(12000)=4800 =.6(17000)+(.4)(8000)=13,400
=.6(13000)+(.4)(0)=7,800
EOL decision = plant corn
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Additional information to make a better
decision is usually not free.
– Cost of sampling
– Cost of testing
– Cost of survey
– Cost of studies, forecasts
– Cost of consultants
Key question: Does the value of the
information justify the cost of obtaining
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ISE 562; Dr. SE 562; Dr. Smith
• There are three information value concepts
described:
1. Expected Value of Perfect Information called EVPI: Theoretical maximum value of information to obtain the optimal decision.
2. Expected Value of Sample Information called EVSI: Difference in expected value with and without sample information.
3. Expected Net Gain of Sampling ENGS. The EVSI with the cost of sampling included.
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ISE 562; Dr. Smith
1. Expected Value of Perfect Information
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ISE 562; Dr. value of perfect information
• If we knew the best outcome would yield $X, we should only be willing to pay an amount difference between best outcome and expected value outcome. EVPI=[(EV given PI) – EV(a*)]
• To calculate EV (of) PI we need
Expected value of optimal decision, expected value= EV(a*)
Expected value of decision given perfect info (the best we could do for each state)
Trade Bill
P(pass)=.60
P(fails)=0.40
Trade Bill
P(pass)=.60
P(fails)=0.40
ISE 562; Dr. calculate EVPI we need
1. Expected value of the decision Expected value
ISE 562; Dr. calculate EVPI we need
2. Expected value of decision given perfect info (the best we could do for each state)
Expected value given PI=.6(35000)+(.4)(20000)=29,000
=[(EV given PI) – EV] =29000 – 24200 =$4,800
… seen this before?
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*Expected value decision = plant corn
10/8/2022 11
=.6(35000)+(.4)(8000)=24,200 =.6(18000)+(.4)(12000)=15,600
=.6(22000)+(.4)(20000)=21,200
Trade Bill
P(pass)=.60
P(fails)=0.40
ISE 562; Dr. Smith
• EVPI = $4,800 and EOL = $4,800
• EVPI(a*) = EOL(a*)!
• Always true: regret measures the difference between the best decision under a state of nature and the decision actually made.
• The EVPI is what we should be willing
to pay to avoid the regret of not getting
the optimal decision
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ISE 562; Dr. : Expected Opportunity Loss (choose option with minimum EOL)
=.6(0)+(.4)(12000)=4800 =.6(17000)+(.4)(8000)=13,400
=.6(13000)+(.4)(0)=7,800
EOL decision = plant corn
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ISE 562; Dr.
EV(A) = .4(-4) + .4(5) + .2(10)
EV(T) = .4(-8) + .4(6) + .2(12) =1.6 EV(C) = .4(-2) + .4(3) + .2(8) =2.0
EVgivenPI = .4(-2) + .4(6) + .2(12) = 4.0 So
EVPI = 4.0 – 2.4 = 1.6
Market Acceptance
P(Med)=0.4 P(High)=0.20 Medium High
5 10 6 12 38
P(Low)=0.4 Alternatives Low
AutoPlanner -4 ThermalView -8 CalcForce -2
=2.4 = EV(a*)
ISE 562; Dr. example
Payoff Table
Alternative Route A: 105-110-10 B: 105-710-10 C: 105-605-10
States of Nature, estimated minutes of travel time
.60 Normal traffic 52
Rush hour traffic
132 75 105 68
=67.1 = EV(a*)
EV(A) = .6(52) + .3(125) + .1(98) EV(B) = .6(62) + .3(132) + .1(75) EV(C) = .6(48) + .3(105) + .1(68)
EVgivenPI = .6(48) + .3(105) + .1(68) = 67.1 So
EVPI = 67.1 – 67.1 = 0. Why? Because C dominates the other alternatives. No information would add any benefit.
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ISE 562; Dr. Smith
2. Expected Value of Sample Information
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ISE 562; Dr. Smith
• There is another (common) situation where the value of information can be computed—the expected value of sample information, EVSI
• It is the difference in expected value with and without additional information
• EVSI = EVwith info – EVwithout info
• Because we have gathered new
information we need to update our
probabilities … but how?
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Reverend Bayes is back!
Need to compute posterior probabilities after sampling
Let’s work a problem.
posterior probability ( prior prob.)(likelihood ) ( prior prob.)(likelihood )
ISE 562; Dr. Smith
ISE 562; Dr. problem
Decision to invest in apartments, office Buiding, or warehouse. Payoffs in thousands
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States of nature
Apartments
Office Bldg
ISE 562; Dr. basic calculations:
States of nature
Apartments
Office Bldg
The basic calculations: max EV=44
EV given PI = .60(100)+.40(30)=72
EVPI = EV given PI – EV = 72-44=28
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=.6(50)+.4(30)=42 =.6(100)+.4(-40)=44 =.6(30)+.4(10)=22
ISE 562; Dr. decision tree with payoffs (and no added
information): 0.6 Good
0 42 0.4 Poor
44 0 44 0.4
0 22 0.4 Poor
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ISE 562; Dr. the decision maker decides to obtain sample information by hiring an economic analyst to forecast future economic conditions. A report will be provided indicating a positive result for good economic conditions or negative for poor economic conditions.
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ISE 562; Dr. Smith
• Based on consultant’s past record of forecasting, the decision maker has estimated the conditional probabilities for the following events:
– g=event good economic conditions
– p=poor economic conditions
– P=positive economic report
– N=negative economic report
ISE 562; Dr. Smith
• The consultant’s record: – P(P|g)=.80
– P(N|g)=.20 – P(P|p)=.10 – P(N|p)=.90
• These values represent the likelihood part of Baye’s rule
posterior probability ( prior prob.)(likelihood ) ( prior prob.)(likelihood )
ISE 562; Dr. Smith
• From other sources (e.g., published forecasts of indices), the decision maker also has estimates of the prior probabilities for the states of nature:
– P(g)=.60
– P(p)=.40
• Given the prior and likelihood values, we can compute the posterior probabilities for the states of nature using Bayes rule.
posterior probability ( prior prob.)(likelihood ) ( prior prob.)(likelihood )
ISE 562; Dr. Smith
• Posterior probabilities (from Bayes rule)
posterior probability ( prior prob.)(likelihood ) ( prior prob.)(likelihood )
ISE 562; Dr. Smith
•Bayesrule:P(X |Y) P(Y|Xr)P(Xr)
r P(Y |Xi)P(Xi)
• For this example:
P(g | P)
.80(.60)
.80(.60) .10(.40) 10/8/2022
.48 0.923 .52
P(P | g) P(g) P(P|g)P(g)P(P| p)P(p)
ISE 562; Dr. the other states:
P(g | N) P( p | P) P( p | N )
P(N | g) P(g) .20(.60) .12 0.25 P(N|g)P(g)P(N|p)P(p) .20(.60).90(.40) .48
P(P | p) P( p) .10(.40) .04 0.077 P(P|p)P(p)P(P|g)P(g) .10(.40).80(.60) .52
P(N | p) P( p) .90(.40) .36 0.75 P(N|p)P(p)P(N|g)P(g) .90(.40).20(.60) .48
Left hand side values are in P(A|B) form (conditional)
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ISE 562; Dr. the posterior probabilities we can calculate the probabilities on the branches of the decision tree.
Starting with conditional probability equation: P(AB)=P(A|B)P(B)
We want P(P) and P(N) so
• P(P)=denominator of bayes calculation for branch P • P(P)=P(Pg)+P(Pp)=P(g)P(P|g) + P(p)P(P|p) =0.52
• P(N)=denominator of bayes calculation for branch N • P(N)=P(Ng)+P(Np)=P(g)P(N|g) + P(p)P(N|p) =0.48
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ISE 562; Dr. branch probabilities using table:
posterior probability ( prior prob.)(likelihood ) ( prior prob.)(likelihood )
ISE 562; Dr.
States of nature
Good Conditions
Poor Conditions
Good Conditions
Poor Conditions
Prior prob.
Conditional (likelihoods)
P(P|g) =.80
P(P|p) =.10
P(N|g) =.20
P(N|p) =.90
Prior prob x likelihood
P(Pg) =.48
P(Pp) =.04
P(Ng) =.12
P(Np) =.36
Posterior prob.
=.48/.52 =.923
=.04/.52 =.077
=.12/.48 =.25
=.36/.48 =.75
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ISE 562; Dr. (P)=denominator of bayes calculation for branch “P” = .52
P(N)=denominator of bayes calculation for branch “N” = .48
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ISE 562; Dr. we can revise the decision tree to show the effect of (new) sample information.
Display tree
ISE 562; Dr. (finally) the EVSI can be calculated:
• EVSI = EVwith info – EVwithout info
• From example, the expected value of the
decision is $63.194 with sample info
• Without the sample information we use the
original value of 44
See tree A
• So EVSI = 63.194 – 44 = 19.194
• This implies the decision maker should be willing to pay the economic analyst up to $19,194 for the forecast.
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ISE 562; Dr. Smith
• Note that the cost of consultant was zero in this example.
• What if hiring consultant cost $5k?
• Where does it go in the tree? See tree B • There are two ways
– Calculate net payoff by subtracting cost at end of tree branches
– Wait until rollback of tree calculations and subtract the test cost when test is performed
• Like this…
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ISE 562; Dr. EVSI does NOT include cost of sampling, CS
EVSI = [58.194+5*] – 44 = 19.194 EMV cost EMV cost
with SI without SI
EVSI = [63.194**] – 44 = 19.194
Net payoff approach (Tree C)
Cost at point of expense (Tree D)
Same result
*need to add the CS back to get correct EVSI if net payoffs used **CS not included
ISE 562; Dr. Smith
• To determine if sampling is cost effective compute expected net gain of sampling, ENGS:
– If positive, sample
– If negative, not worth the sampling cost
• Let CS = “cost of sampling”
ENGS = EVSI – CS
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ISE 562; Dr. payoff approach (Tree C)
ENGS = [58.194*] – 44 = 14.194 CS
Cost at point of expense (Tree D)
ENGS = [63.194**] – 44 – 5 = 14.194 Same result
CS included
*included in net payoffs
**not included in gross payoffs so subtract here
ISE 562; Dr. Smith
EVPI and EVSI Part II
Decision Theory
ISE 562; Dr. Smith
• More on utility—how to find the certainty equivalent
• How to identify risk attitude
• Other methods for calculating EV of information; ENGS
ISE 562; Dr. shop problem:
A machine shop owner is attempting to decide whether to purchase a new drill press, a lathe, or a grinder. The return from each will be determined by whether the company succeeds in getting a government military contract. The profit or loss from each purchase and the probabilities is in the following table.
State of nature Polisher
P = 0.40 Contract $40000 20000 12000
P=0.60 No Contract $-8000 4000 10000
The machine shop owner is considering hiring a military consultant to ascertain whether the shop will get the government contract. The consultant is a former military officer who uses various personal contacts to find out such information. By talking to other shop owners who have hired the consultant, the owner has estimated a .70 probability that the consultant would present a favorable report, given that the contract is awarded to the shop, and a .80 probability that the consultant would present an unfavorable report, given that the contract is not awarded. Using decision tree analysis, determine the decision strategy the owner should follow, the expected value of this strategy, and the maximum fee the owner should pay the consultant.
ISE 562; Dr. the machine shop problem
$40000 20000 12000
No Contract
$-8000 4000 10000
State of nature
Polisher er
EVPI = EV given PI – EV(a*)
EV given PI = .40(40) + .60(10) = 22 EV(a*)= .4(40) – .60(-8) = 11.2
So EVPI = 22 – 11.2 = 10.8
10/8/2022 43
ISE 562; Dr. have
Fav [16.48]
Unfav [10.4]
[16.48] .6
Don’t hire
ISE 562; Dr. Smith
Don’t hire
ISE 562; Dr. Case
Unfavorable Case
Bayes Rule Calculation With Imperfect Information
P(F|) P()
No Contract
P(U|) P()
No Contract
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ISE 562; Dr. Case
Unfavorable Case
Bayes Rule Calculation With Perfect Information
P(F|) P()
No Contract
P(U|) P()
No Contract
10/8/2022 47
ISE 562; Dr. VPI:
EVgivenPI Fav [22.0]
C40 No C -8
[22.0]hire
Don’t hire
Unfav [10.0]
NoC 10 C 40
NoC -8 0 C20
EV(a*) without PI
Difference = EVPI = 22 – 11.2 = 10.8
ISE 562; Dr. about EVSI?
EVSI for the original problem= EV with info – EV without info
= 16.48 – 11.20 [25.6] = 5.28
Fav [16.48]
[16.48] .6
Don’t hire
Unfav [10.4]
ISE 562; Dr. Smith
EVSI with perfect information = EV with info – EV without info = 22 – 11.20 = 10.8 = EV0PI for the NoorigCinal problem—we
have maxed out thePoEl VSI because the prediction is perfect;
[40.0] Lathe 1 C 20
the EVSI is lower for the original problem because of the
No C uncertainty in prediction (the value is lower).4
NoC 10 C 40
NoC -8 0 C20
EV given PI
[22.0]hire
Don’t hire
Unfav [10.0]
Difference = EVSI = 22 – 11.2 = 10.8 = EVPI 50
EV(a*) without PI
ISE 562; Dr. to find the certainty equivalent • What is it?
– The point where the decision maker is indifferent between a lottery (gamble) versus certainty of receiving some payoff value.
ISE 562; Dr. xample: CE
• Consider the positive increasing utility
function over the range -4000 to
Curve is concave
ISE 562; Dr. Smith
Now suppose we have the lottery:
.90 .70 .50
-4000 -1000 CE
The EU =1/2(.90)+1/2(.50)=0.70
The value of .70 corresponds to x= 1000
So DM is indifferent between lottery and 1000 for certain
The CE = -1000 53
ISE 562; Dr. Smith
Note that if CE
Curve is convex
ISE 562; Dr. exam:
Solve (x+8)2/400 = 0.50 for x=6.14
EMV= 1⁄2 (-8) + 1⁄2(12)= 2
EU = 1⁄2 (0)+ 1⁄2 (1) =0.50
The value of .50 corresponds to +6.14 so CE=+6.14
In this case CE > EMV so DM is a risk taker 56
Curve is convex
ISE 562; Dr. , suppose utility function looks like
.80 .60 .40
EMV= 1⁄2 (4000) + 1⁄2(-4000)=0
EU = 1⁄2 (.8)+ 1⁄2 (.4) =0.60
The value of .60 corresponds to 0 so CE=0
In this case CE = EMV so DM is risk neutral 57
Curve is linear
ISE 562; Dr. in book for CE U(x) 1
G= gain in utility if lottery is won L= loss in utility if lottery is lost
.90 .70 .50
G =U(4000) – U(0)= .9 -.8 = .10
L =U(0)–U(-4000)=.8-.5=.30
If G-L < 0 then risk averse If G-L > 0 then risk taker
If G-L = 0 then risk neutral
Since G-L<0, risk averse
0 +4000 Curve is concave
-4000 -1000
ISE 562; Dr. in book for CE
G =U(4000) – U(0)= .60 -.38 = .22 .30 L =U(0)–U(-4000)=.38-.30=.08 0
If G-L < 0 then risk averse If G-L > 0 then risk taker
If G-L = 0 then risk neutral
Since G-L>0, risk taker 10/8/2022
Curve is convex
1000 4000 Curve is convex
ISE 562; Dr. in book for CE .80
Curve is linear
G =U(4000) – U(0)= .8 -.60 = .20
L =U(0) – U(-4000) = .60 – .40 = .20
If G-L < 0 then risk averse If G-L > 0 then risk taker
If G-L = 0 then risk neutral
Since G-L=0, risk neutral 10/8/2022
ISE 562; Dr. we have two alternatives:
Question 2
1⁄410 1⁄47
Andutilityfunction: U(x) 1 (x4)2 36
Find the CE’s 10/8/2022
4x10 and x6 U(x)4
18.00 16.00 14.00 12.00 10.00 8.00 6.00 4.00 2.00 0.00
1 2 3 4 5 6 7 8 9 10 11 12 13
ISE 562; Dr. calculate EV’s 36
1⁄410 1⁄47
Question 2
EV(A1)= 1⁄4 (10)+ 3⁄4 (4) = 5.5 EV(A2)= 1⁄4 (7)+ 3⁄4 (5) = 5.5
EU(A1)= 1⁄4 (1)+ 3⁄4 (0)= +.25 EU(A2)= 1⁄4 (.25)+ 3⁄4 (.0278)= +0.083 Now find the CE’s
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U(x) 1 (x4)2 4x10 and x6 U(x)4
18.00 16.00 14.00 12.00 10.00 8.00 6.00 4.00 2.00 0.00
1 2 3 4 5 6 7 8 9 10 11 12 13
ISE 562; Dr. 2 Find CE(A1) by computing the value of x for which U(x)=0.25
x6 U(x)4 x6 .254
x6 U(x)4 x6 .0834
16.00 14.00 12.00 10.00 8.00 6.00 4.00 2.00 0.00
U(4)= 0.0625 – 0 = 0.0625; G-L>0: risk take
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