Business Decision Models BU/EC275
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Transcript of Business Decision Models BU/EC275
Business Decision ModelsBU/EC275
Created by: Greg Overholt
Agenda
• Waiting Lines– Economic Analysis
• Simulation– Crystal Ball / Arena
• Decision Analysis– Payouts/Trees– EVPI (Perfect Information) – EVSI (Sample Info)
• Utility/Scoring
High-Level
• Last term was all about deterministic models – when you have the same constraints and coefficients so you will get the same results.
• This term is all about stochastic models – include some element of randomness
Waiting Lines
• 4 Types• Line Characteristics• Metrics• Calculations!
Study of waiting line = Queuing Theory
Overall Purpose
4 TypesAutograph line (only 1 server)
Old-School Cafeteria, where you went down the food line
Eg: Customs (tons of servers and stations!)
Bank Tellers(1 line, many servers)
Characteristics!• Line structure: A/B/s - This format is called Kendall’s
notation.• A: Arrival Distribution • B: Service Distribution • s: Number of Servers
• A/B are typically either:– M: Markov’s dists: Poisson/Exponential Distribution– D: Constant (deterministic)– G: General Normal Distribution (mean / stdev)
• Terms– Balking: when they come in and see the line so they don’t
enter the queue– Reneging: when people leave the queue
Poisson• Poisson is to describe the
distribution of the probability that ‘x’ occurrences in an interval (number of successes). Discrete distribution = aka, the probability of each number is > 0 (vs continuous that is 0 for each individual possibility).
The rate / per hour (typically)
=POISSON(x,λ,cumulative)If ‘cumulative’ = TRUE less thenIf ‘cumul…’ = FALSE equal to
Poisson Questions
Students buy Tim Horton’s coffee at a rate of 100/hr (λ)
Q1. What is the average time between arrivals? - 100 in an hour, so 60 mins/100 students
= .6 mins per student
Q2. What is prob that exactly 100 students come in the next hour?
f(100) = 100100*e-100 / 100!f(100) = .0399 or 3.99% probability.
Exponential• Exponential is the probability
that the person will arrive/served within the first ‘x’ mins. – P(x < 2) = the probability that
a person will arrive in 2 mins or less. (CUMULATIVE)
=EXPONDIST(x,λ,cum)cumulative =TRUE less then
IF P > X, then take away the ‘1-’ (right tail test!)
Exponential Question
At the Tim Hortons, they take, on average, 1.5 mins to serve a customer.
Q1. What is the service rate (per hour)?
- We want this in hours, so 1.5 mins = .025 hours.
- λ = 1/u = 1/.025 = 40 customers / hour. Q2. What is the prob that the service will take exactly 1.5
mins?- 0 (continuous probability, and the exact prob = 0)
Q3. What is prob that service will take more then 3 mins? P(x > 3) = e -40(3/60) = .13533 = 13.53%
Relationship between P and E
• The relationship between the Poisson and exponential is very important. E.g. if the distribution for the time between arrivals is exponential with mean = B , then the distribution of the number of arrivals per unit of time is Poisson with mean = 1 / B– INVERSE RELATIONSHIP!!
Waiting Line Inputs
λ / μ = ρ (or U) = Utilization factor (probability that server is busy / prob that a customer has to wait)
I = 1 – U: This is the probability that the server is idle == the probability that there are no customers in the queuing system (P0)
If ρ = 60% (so busy for 60% of the time), I (idle %) would be 40% = the chance that the server is not serving a customer, and no one is in the queue.
Terms / Things to Calculate
Other Operating characteristics of the queuing system:
• Po = probability the service facility is idle (pronounced “P-naught”)
• Pn = probability of n units in system• Pw = probability an arriving unit must wait• Lq = average number of units in queue awaiting
service• L = average number of units in system• Wq = average time a unit spends in queue awaiting
service• W = average time a unit spends in system
Example!
Joe Ferris is a stock trader for the firm ofSmith, Jones, Johnson, and Thomas, Inc.Stock transactions arrive at a mean rate
of20 per hour. Each order received by Joerequires an average of two minutes toprocess. (Assuming a M/M/1 system)
Example!
Arrival Rate DistributionQuestion: What is the probability that no orders are
received within a 15-minute period?
What is the mean number of orders in the 15 mins? (20 per hours = 1 every 3 mins, so average = 5 every 15 mins = λ
Answer: P(x) = (λxe-λ)/x!P(0) = (50e-5)/0!
= e-5 = .0067 = .67% chance
=POISSON(0,5,false)
Example!
Arrival Rate DistributionQuestion: What is the probability that more than 6 orders
arrive within a 15-minute period?Answer:P(x > 6) = 1 - [P(0) + P(1) + P(2) + P(3) + P(4) + P(5) +
P(6)]= 1 - 0.762 = 0.238 = 23.8% chance that more then 6 orders arrive.
=1-POISSON(6,5,true)
Example!
Service Rate DistributionQuestion: What is the mean service rate per hour?
Answer: Since Joe Ferris can process an order in an average time of 2 minutes (.033 hours) then the mean service rate, μ, is μ = 1/(mean service time), or 60/2 = 30 / hr
Example!
Service Time DistributionQuestion: What % of orders will take less than one minute
to process?
Answer: Since units are expressed in hrs, P(T ≤ 1 minute) = P(T ≤ 1/60 hour) (μ = 30 per hour – from previous slide)
Using exponential distribution,P(T ≤ t) = 1 - e-μt P(T<1/60) = 1 - e-30(1/60)
= 1 - .6065 = .3935 = 39.35% = EXPONDIST(1/60, 30,1)
The value gotten from e-μt is the area to the right, but here we want the area to the left (so 1 – value)
Calculate System variables
NOTE: L > Lq and W > Wq. If you calculate something otherwise, re-check your work!
Average Time in SystemQuestion: What is average time an order must wait from
time Joe receives order until it is finished being processed (i.e. its
turnaround time)?
Answer: This is an M/M/1 queue with λ = 20 per hour & μ = 30 per hour. Average time an order waits in the system is:
W = 1/(μ - λ)= 1/(30 - 20)= 1/10 hour or 6 minutes
Example!
Average Length of QueueQuestion: What is the average # of orders Joe has waiting
to be processed?Answer: Average # of orders waiting in queue is:
Lq = λ2/[μ(μ - λ)]= (20)2/[(30)(30-20)]= 400/300 = 4/3 = on average there are 1.33 orders waiting
in the queue.
Example!
Utilization FactorQuestion: What percentage of time is Joe processing
orders?
Answer: % of time Joe is processing orders is equivalent to utilization factor, λ/μ. Thus, % of time he is processing orders is:
λ/μ = 20/30= 2/3 or 66.67% of the time.
Example 2 (M/M/2)Smith, Jones, Johnson, & Thomas, Inc. has begun a major
advertising campaign which it believes will increase its business 50%. To handle the increased volume, the company has hired an additional floor trader, Sue Hanson, who works at the same speed as Joe Ferris (2 minute or 30 per hour).
Recall λ = 20 per hour & μ = 30 per hour with just Joe (M/M/1)
> Note that the new arrival rate of orders, λ, is 50% higherthan that of problem (A). Thus, λ = 1.5(20) = 30 per hour.
Example 2 (M/M/2)Sufficient Service RateQuestion: Why will Joe Ferris alone not be able to handle
the increase in orders?
Answer: Since Joe Ferris processes orders at a mean rate of μ = 30 per hour, then λ = μ = 30 and utilization factor, λ/μ
= 1.
This implies queue of orders will grow infinitely large. Hence, Joe alone cannot handle this increase in demand.
CANNOT HAVE λ being = or > then μ - can’t have people arriving a rate faster then you can serve them – queue will grow infinitely!
Example 2 (M/M/2)Probability of no Units in SystemQuestion: What is probability that neither Joe nor Sue
will be working on an order at any point in time?
Answer: Given λ = 30, μ = 30, k = 2 & [λ/μ] = 1, probability that neither Joe nor Sue will be working is:
A = 33% chance that neither Joe nor Sue will be working.
Summation only applies to term directly to right
Example 2 (M/M/2)Average Time in System
Question: What is average time in the system (turnaround time) for an order with both Joe & Sue working?
Answer: Average turnaround time is average waiting time in system (not just the line!!) = W.
Big Problem!!
• What is the ideal number of servers??– Balance speed and cost (both service cost and
waiting cost!!)• Need to compare TOTAL costs across the
different scenarios.
How?
• Change Queue Priority• Change Queue Style (1 line reduces wait
time vs multiple lines)• Improve Service Rate
– More channels / servers– Faster channels (technology / self-serve)
• EXAM: Will ask you to do an ‘economic analysis’ of 2 scenarios that have different number of server! (compare total costs!)
Economic Analysis
•The advertising campaign of SJJT, Inc. was so successful that business actually doubled. The mean rate of stock orders arriving is now 40/hr and the company must decide how many traders to employ. Each trader hired can process an order in an average time of 2 min.•Based on a number of factors the brokerage firm has determined the average waiting cost/minute for an order to be $0.50. Traders hired will earn $20/hr in wages & benefits.
Using this information compare the total hourly cost of having 2 traders with that of having 3 traders.
Economic Analysis of Waiting Lines
Total Hourly Cost = Cost of Service + Cost of Waiting= (Total salary cost per hour) + (Total hourly cost for orders in the system) = ($20 per hour) x (Number of traders) + ($30 waiting cost per hour) x (Number of Orders per hour) x (Average wait per order)
= 20k + 30 x W but L = W Little’s Flow Equations
= 20k + 30L
Economic Analysis
Economic AnalysisThus, L must be determined for k = 2 traders and for k = 3
traders with = 40/hr and = 30/hr (since the average service time is 2 minutes (1/30 hr).
02!1
Pkk
Lk
q
qLL WE HAVE WE DON’TTOTAL COST = 20k + 30L
WE HAVE
WE DON’T
Economic AnalysisCost of Two Servers
Probability of no Units in System
5
1
3
8
3
41
1
40302
302
!2
3040
!1
3040
!0
3040
1
k
k
!k
/
!n
/
1P
210
1k
0n
kn0
20% chance that there will be no units in the system.
Cost of System
hrTC
LL
Pkk
L
q
k
q
/112$51230220
51203401516
15
16
5
1
40302!1
30403040
!1 2
2
02
Economic AnalysisCost of Two Servers
Probability of no Units in System
59
15
45
32
9
8
3
41
1
40303
303
!3
3040
!2
3040
!1
3040
!0
3040
1
k
k
!k
/
!n
/
1P
3210
1k
0n
kn0
25% chance that there will be no units in the system.
Economic AnalysisCost of Three Servers
Cost of System
hr/34.104$885130830320TC
885130834885128LL
885
128
59
15
40303!2
30403040P
k!1kL
q
2
3
02
k
q
Economic AnalysisCost of Three Servers
System Cost Comparison
WagesCost/Hr
WaitingCost/Hr
TotalCost/Hr
2 Traders 40.00 72.00 112.00
3 Traders 60.00 44.34 104.34
Simulation
The process of designing a mathematical or logical model of a real system and then conducting computer-based experiments with the model to describe, explain, and predict the behavior of the real system.
Advantages:- Leads to a better understanding of the real system- Simulation is far more general then mathematical
models- Simulation answers ‘what-if’ questionsDisadvantages:- No guarantee that it will provide good answers- Time consuming- The simulation technique still lacks a standardized
approach- Unlike analytical techniques, it is NOT an optimizing
technique.
SIMULATION
Simulation is the most widely used Decision
Analysis technique
Simulation is NOT an optimizing technique, but a descriptive
tool of the system under various conditions
Types of Simulations
Static vs Dynamic• Does time have a role in model?Continuous change vs Discrete change• Can system change continuously or only at
discrete points in timeDeterministic vs Probabilistic• Is everything certain, or is there uncertainty?
– Use Monte Carlo technique for uncertainty
Most operational models are: • Dynamic, Discrete change, Probabilistic
Equation for profit model
• Without simulation, you can get worst / avg / best case results.. But need a formula to figure out profit:
Let P = the unit price. Marketing has set this already: P = $40:00.
Let Q = the quantity to produce (i.e. the answer to the question we’ve posed).
Let D = the quantity sold in the market (i.e. the Demand).
Let FC = the fixed costs (depending on which machine tool we end up buying), and
Let UC = the unit cost (the planning numbers above). • PROFIT = P * [min(Q,D)] - (UC * Q) - FC
What Program to use!!• Excel: when time isn’t a factor, and probabilities
are known! • Crystal Ball: Crystal Ball automatically
calculates thousands of different "what if" cases, saving the inputs and results of each calculation as individual scenarios. Analysis of these scenarios reveals to you the range of possible outcomes, their probability of occurring, which input has the most effect on your model and where you should focus your efforts.
• Arena: Arena is most effective when modeling and analyzing business, service, or manufacturing processes or flows. Use when things are very dynamic, changes with time, and multiple steps
Probabilistic ‘Excel’ Style question:
Exam Example
• A consultant knows that the monthly costs incurred is uniformly distributed in the range ($3000, $5000) if he has less than 4 clients per month. If he has 4 to 6 clients that month, his expenditure is either $5000 or $6000 [both are equally likely]. He never has more than 6 clients a month and 30% of the time he has less than 4 clients a month. His monthly revenue from clients is either $4000 or $7,000. The probability that his revenue is $4000 is twice the probability that his revenue is $7000.
Client Probability: 30% = less then 4,70% will have 6 clientsSo, RN’s 0 0.3 = less then 40.3 1 = 6 clients
Expense Probability: If has less then 4 clients, then take RN*$2000, and add that to $3,000.
If has 6 clients, then 50% will be $5K, 50% will be 6K, so if 0.0 0.5, then take $5K, if above 0.5 then use $6K
Revenue Probability: $4K is twice as probable.. So 66% chance of being 4K, 33% chance of being 7K.
So if 0.0 0.66 go with $4K, if above .66 go with $7K.
Client Probability: 30% = less then 4,70% will have 6 clientsSo, RN’s 0 0.3 = less then 40.3 1 = 6 clients
Expense Probability: If has less then 4 clients, then take RN*$2000, and add that to $3,000.
If has 6 clients, then 50% will be $5K, 50% will be 6K, so if 0.0 0.5, then take $5K, if above 0.5 then use $6KRevenue Probability:
$4K is twice as probable.. So 66% chance of being 4K, 33% chance of being 7K.
So if 0.0 0.66 go with $4K, if above .66 go with $7K.
no
yes
no
yes
$6K
.35*2K + 3K= 3,700
.27*2K + 3K= 3,540
$5K
$4K
$7K
$4K
$4K
-$2K
$3.4K
$0.46K
$1K
Crystal Ball
• The Monte Carlo technique is defined as a technique for selecting numbers randomly from a probability distribution for use in a trial (computer run) of a simulation model.
• Crystal Ball is a risk analysis and forecasting program that uses Monte Carlo simulation to provide a statistical range of results.
Crystal Ball
Previous Final Exam Question: What is initial seed value? - the starting random number generated that you want to use in your sequence. So if you have a seed value of 100, then your test will use the 100th, 101st, 102nd random number generated until you don’t need anymore.
Crystal Ball
Crystal Ball
• Overall: Spreadsheets do not account for the variability of the real world.
• CB allows you to you range of inputs to explore the range of possible outcomes AND the probability of their occurrence.
Simulation
• Purpose of simulation is to estimate population parameters, such as the mean and variance, and to gain knowledge about distribution of outcome variables in order to assess risk
Process!
• Need to first know the appropriate input probability distribution:
1. Plot points on a histogram2. Eye-ball to see if looks like
normal/uniform/exponential/lognormal (normal, but skewed, no values below 0)
3. Use Goodness-of-Fit tests to see which fits best using crystal ball!
Testing Simulation Results Goodness of Fit Methods – MC!
• Chi-Square– most common
• p-value < 0.5 indicates no fit, i.e., Reject HO• p-value > 0.5 indicates that it is a good fit!
– (formal method to compare two distributions)
• Kolmogorov-Smirnov– largest vertical distance between cumulative distributions,
value < 0.03 indicates good fit • (compare the difference of cumulative prob.)
• Anderson-Darling– Modification of K-S, placing greater weight on the tails of the
distribution, value < 1.5 indicates good fit
Crystal Ball – Choosing Distribution
• Look at p-value / KS / Anderson.– Look if it says insufficient data
for chi-square test (needs expected value > 5)
Model Verification and Validation
• Challenge: build an accurate model & convince of its validity and usefulness. Needs to be:– Realistic– Simple
• Verification: does it perform as intended?
• Validation: does conceptual model accurately predict real system?
ARENA
Arena
• Delay– Entity just sits; no resource involved; multiple
entities can be simultaneously delayed– Example: traffic light
• Seize Delay– Entity seizes specified qty of resource, then has
delay; another module releases (see Delay Release)
• Delay Release– Entity seized resources at upstream module; now
delays & releases• Seize Delay Release
– Like above, but entity released after delay – Typical queuing situations
Arena
Arena Q’s
Arena Q’s
Arena Q
DECISION ANALYSIS
Payout and Trees
• Payoffs– Vij Value of Decision di, in State sj
– Used to evaluate alternatives– Form a ‘payout table’ (matrix of actions vs
outcomes)
s1 s2 s3
d1 V11 V12 V13
d2 V21 V22 V23
d3 V31 V32 V33Dec
isio
n
States of Nature
Payout and Trees
• Decision Trees– chronological representation of problem– nodes and branches– 2 types of nodes: square = decision, round = outcome– payoffs at end of branches
s1
s2
s1
s2
V23
s3
s3
d1
d2
V11
V12
V13
V21
V22
s1
s2
s1
s2
V23
s3
s3
d1
d2
V11
V12
V13
V21
V22
Decision Nodes are denoted with boxes
Random Event Nodes are denoted with circles
Payoffs are at the end of the branches
Types of Prob’s• CERTAINTY - know which outcome (state of
nature) will occur– select best alternative
• UNCERTAINTY - no probability information available– Use methods described shortly
• RISK - probabilities of outcomes (states of nature) are available– Find the ‘Expected Values’ (multiple outcome
by probability) and take the highest one
Decisions Under Uncertainty
• When decision maker doesn’t know which outcome will occur & has no probability estimates
• Commonly used criteria:– Assumed equally likely (Laplace)– the optimistic approach (MAXIMAX or MINIMIN)– the conservative approach (MAXIMIN or MINIMAX)– the MINIMAX REGRET approach
s1 s2 s3
d1 10,000 6,500 -4,000
d2 8,000 6,000 1,000
d3 5,000 5,000 5,000
d4 8,000 5,900 500
Decis
ionStates of NaturePayoff
Table
• Would you drop any of these?
Decisions Under Uncertainty
• d4 is dominated by d2, so we can drop d4 from further consideration
Dominance
• A decision dominates another if it is better than or equal to another decision in every possible state of nature
• Always look for dominance it will save you work.– Reduce the size of your payoff table, or– Prune your decision tree
s1 s2 s3
d1 10,000 6,500 -4,000
d2 8,000 6,000 1,000
d3 5,000 5,000 5,000 Dec
isio
nStates of NaturePayoff
Table
• What decision would you choose?• Why?
Decisions Under Uncertainty
Laplace Criterion (assumes equal probability)
• Assume all states are equally likely• Can use either the average or sum of the payoffs
S1 S2 S3
D1 10,000 6,500 -4,000 4,167 12,500
D2 8,000 6,000 1,000 5,000 15,000
D3 5,000 5,000 5,000 5,000 15,000
TotalPayoff
De
cisi
on
States of NaturePayoffTable
AveragePayoff
Optimistic Approach(MAXIMAX)
• Used by optimistic (aggressive?) decision maker• Decision with the largest possible payoff is chosen
MAXIMAX: Maximize the maximum payoff (take the max of the maximums)
s1 s2 s3
d1 10,000 6,500 -4,000 10,000
d2 8,000 6,000 1,000 8,000
d3 5,000 5,000 5,000 5,000 Dec
isio
n
States of NaturePayoffTable
MaximumPayoff
Conservative/Pessimistic Approach (MAXIMIN)
MAXIMIN (Maximize the minimum payoff)
• For each decision identify the minimum payoff, select the maximum of these minimum payoffs
s1 s2 s3
d1 10,000 6,500 -4,000 -4,000
d2 8,000 6,000 1,000 1,000
d3 5,000 5,000 5,000 5,000 Dec
isio
n
States of NaturePayoffTable
MinimumPayoff
Minimax Regret Approach
• Construct a regret (opportunity loss) table • For each state of nature calculate the difference
between each payoff and the largest payoff for that state of nature.
• Using regret table, the maximum regret for each possible decision is listed
• Chose decision which minimizes the maximum regrets
• Minimize the maximum regret
s1 s2 s3
d1 10,000 6,500 -4,000
d2 8,000 6,000 1,000
d3 5,000 5,000 5,000
s1 s2 s3
d1 0 0 9,000 9,000
d2 2,000 500 4,000 4,000
d3 5,000 1,500 0 5,000
MaximumRegret
Dec
isio
nD
ecis
ion
States of NaturePayoffTable
RegretTable
States of Nature
Minimization Problems
• These problems have been maximization problems (seeking the best payoff)
• What if these had been minimization problems? (seeking the best/lowest cost)
• Minimin (Optimistic)– choose the option that has the absolute lowest
cost• Minimax (Conservative)
– identify maximum costs for each decision, then select the minimum of these
• Minimax Regret– minimal of the maximum regrets
Minimization Objective
s1 s2 s3
d1 10,000 6,500 -4,000 -4,000 10,000
d2 8,000 6,000 1,000 1,000 8,000
d3 5,000 5,000 5,000 5,000 5,000
Minimin Minimax
s1 s2 s3
d1 5,000 1,500 0 5,000
d2 3,000 1,000 5,000 5,000
d3 0 0 9,000 9,000
MaximumCost
MinimaxRegret
MaximumRegret
Dec
isio
nD
ecis
ion
States of NaturePayoffTable
RegretTable
States of Nature
MinimumCost
Decisions Under Risk
• 3 options re risk– Avoid (safe decisions)– Manage (insure, share risk, prioritize, …)– Embrace (casinos)
• Probabilities known for Outcomes (States of Nature)– Use Expected Value Approach
• The decision yielding the best expected return is chosen = sum of all outcomes * their probability
Expected Value Approach
s1 s2 s3
0.6 0.2 0.2
10,000 6,500 -4,000 EV6,000 1,300 -800 6,500
P(sj)
Payoff
State
EV(Payoff)
s1 s2 s3 EV(di)
0.6 0.2 0.2
d1 10,000 6,500 -4,000 6,500
d2 8,000 6,000 1,000 6,200
d3 5,000 5,000 5,000 5,000 Dec
isio
n
States of NaturePayoffTable
P(sj)
Expected Value of Perfect Information (EVPI)
• The expected value of perfect information (EVPI) is the maximum amount a decision maker would pay for additional information.
• Approach 1: EVPI equals the expected value given perfect information minus the expected value without perfect information.
EVPI = EVwPI - EVwoPI • Provides an upper bound on the expected value of any sample or
survey information
• Approach 2: EVPI equals the expected opportunity loss (EOL) for the best decision.
EVPI
s1 s2 s3 EV(Di)
0.6 0.2 0.2
d1 10,000 6,500 -4,000 6,500
d2 8,000 6,000 1,000 6,200
d3 5,000 5,000 5,000 5,000
10,000 6,500 5,000 8,300 -6,500
EVPI 1,800
Best
Dec
isio
n
States of NaturePayoffTable
P(sj)
Example: Oil Firm’s Dilemma
• An Alberta oil company routinely seeks new sites for oil drilling. With no other information, there is a 50-50 chance of striking oil. If a well is found, a profit of $150,000 is realized. If the site is dry, a loss of $100,000 is incurred. Find the optimal strategy and the EVPI.
Example
Dry (50%) Oil (50%) EV
Drill -100 150 25K
Don’t Drill 0 0 0K
EVwoPI: Expected Value with provided probability and without perfect information
EVwPI: Expected Value with perfect info (so you’d know if there was oil or not)
You would drill if there was oil, and you would not drill if it was dry:150K(.5) + 0(.5) = 75K
WHAT IS THE VALUE OF HAVING PERFECT INFORMATION:
50K (75K – 25K)
Perfect vs Sample Info
• In the real world, there isn’t ‘perfect information’, so the best we can get is sample information (past results, trends, expert analysis).
• Using the sample information, we will use that to best determine the optimal choice.
Bayes’ Theorem
BUT… what if you could get sample information (pay for a survey to be done for $20K), do you do it / how does that affect your decision making?
For the Alberta oil company, we may conduct a geological survey for $20,000 providing strong evidence of there being oil or not. Past history indicates that when there really is oil, the survey is correct 90% of the time; when the site is dry, the survey is correct 80% of the time.
• Prior Probabilities (current information)• Prior to obtaining new information, P(states of
nature) are called prior probabilities. This is stuff we already know from the first page. Let S1 = OIL and S2 = DRY (our two states of nature). The probabilities associated with those are 50% each, hence:
– P(S1) = 0.50 and P(S2) = 0.50
New Information• We can spend $20K and get a survey done.
The survey will have two outcomes (again, states of nature):– IOil survey indicates oil– IDry survey indicates dry
• The survey info tell us:– P(IOil | S1) = 0.90 – P(IDry | S2) = 0.80
• We can easily calculate the complements (since we’ll need them as well):– P(IDry | S1) = 0.10 – P(IOil | S2) = 0.20
.45
.10
.55
.45/.55 = .81
.10/.55 = .19
All the instances when it will have oil (when survey is right with oil present, and when survey is wrong with survey saying dry)
.05
.40
.45
.05/.45 = .11
.40/.45 = .89
All the instances when it will be dry (when survey is wrong with oil present, and when survey is actually dry with survey saying dry)
Sum = 1.00
CHECK: Marginal P(Idry) + P(Ioil) = 1.00 .45 + .55 = 1.00
No survey
survey
L1 – Survey says OIL!
D2 (don’t drill it)
D2 (don’t drill it)
D1 (drill it)
D1 (drill it)
P(S | I )
L2 – Survey says DRY!
S2 – DRY!
S1 – OIL!
S2 – DRY!
S1 – OIL!
.82
.18
.11
.89.45
.55
$130
-$120
-$120
$130
-$20
-$20-$20
$85
$85
$37.75
$25-$20
-$20
-$92
$37.75
• Expected Value with Sample Info = $37.75
• Expected Value without Sample Info = $25.0
EVSI = 37.75 - 25.0 = 12.75K
Is the survey worthwhile? – YES it is > 0! How efficient is this sample information in
helping us make the best decision?
PERFECT INFORMATION
Dry (50%) Oil (50%) EV
Drill -100 150 25K
Don’t Drill 0 0 0K
EVwoPI: Expected Value with provided probability and without perfect information
EVwPI: Expected Value with perfect info (so you’d know if there was oil or not)
You would drill if there was oil, and you would not drill if it was dry:150K(.5) + 0(.5) = 75K
WHAT IS THE VALUE OF HAVING PERFECT INFORMATION:
50K (75K – 25K)
Efficiency of the Forecast
• The Efficiency (E) of the Sample Information (Forecast) is:
• In our example, this is:
%100EVPI
EVSIE
E = 12.75 / 50 = 26%
UTILITY
Utility
• Utility measures the total worth of a particular outcome, reflecting the decision maker’s attitude towards a collection of factors. – These factors may be profit, loss, and risk.– Captures personal preference!!
• The amount of happiness you experience for that outcome– Done on a 100 point scale
Deriving Utility
• Assign top profit to utility of 100• Assign lowest amount (loss) to utility of 0.• Each indiv person has to assign the amount of ‘good’
they get for each value in between the extremes– If Risk Averse (don’t like risk): the amount of utility you
get from even a little profit is a lot!– If Risk Lover: then the only amounts that you really enjoy
are the really big profits! So the lower profit levels have a low utility, since they don’t do much for you in regards to happiness.
– If Risk Neutral: your utility function is a straight line, you enjoy each additional amount of profit equally.
Risk Diagram
Payoffs
Util
ity
Risk Averse
Risk Neutral
Risk Lover
Utility Exam Q
• Will give you 2 utility functions, you need to decide which of a few alternatives will they choose.– Need to calculate the EU: Expected Utility.
• With the utility table, assign the 0-100 according to their personal preference (usually given).
• Multiply the utilities with the probability of achieving the level to get the expected utility.
• For Risk Neutral, just do the expected value, multiplying the $ outcomes with the probability.
Example
s1 s2 s3
P(sj) 0.1 0.3 0.6
d1 100,000 40,000 -60,000
d2 50,000 20,000 -30,000
d3 20,000 20,000 -10,000
d4 40,000 20,000 -60,000
• Consider the following three-state, four-decision problem with the following payoff table in dollars:
Example
Risk-Neutral Decision Maker• If decision maker is risk neutral, expected
value approach is applicable:
s1 s2 s3
P(sj) 0.1 0.3 0.6 EV
d1 100,000 40,000 -60,000 -14,000
d2 50,000 20,000 -30,000 -7,000
d3 20,000 20,000 -10,000 2,000
d3 has the highest expected value
Example
Decision Makers with Different Utilities
• Suppose that two decision makers have the utility values shown on the right
Decision DecisionAmount Maker I Maker II$100,000 100 100 $50,000 94 58 $40,000 90 50 $20,000 80 35
-$10,000 60 18 -$30,000 40 10 -$60,000 0 0
Example - Utility Functions
0
10
20
30
40
50
60
70
80
90
100
-$80,000 -$40,000 $0 $40,000 $80,000 $120,000
Amounts
Uti
lity
DM I
DM II
Example
Expected Utility: Decision Maker I
s1 s2 s3
P(sj) 0.1 0.3 0.6 EU
d1 100 90 0 37
d2 94 80 40 57
d3 80 80 60 68
Example
Expected Utility: Decision Maker II
s1 s2 s3
P(sj) 0.1 0.3 0.6 EU
d1 100 50 0 25.0
d2 58 35 10 22.3
d3 35 35 18 24.8
Which values this problem more?
• Decision Maker 1 (risk adverse) EU = 68– He assigned a utility of:– 60 for -$10,000– 80 for $20,000
• Each utility point between those is valued at $1,500 ($30/(80-60)).
• So Value of this opportunity = -10,000 + 8*1,500 = $2,000
• Decision Maker 2 (risk taker) EU = 25– He assigned a utility of:– 18 for -$10,000– 35 for $20,000
• Each utility point between those is valued at 1,667 ($30/17)
• So value of this opportunity = -10,000 + 1,667*7 = $1,667
Result
• Given the expected payouts, and the utility of each person, the risk adverse Dec Maker 1 values this opportunity more then the risk taking Dec Maker 2 since $2,000 > $1,667.
Scoring
• Used to make a decision about multiple options, that have various factors that you need to consider.
• This approach gives each option’s factors a score. Each factor needs to be weighted by its importance in your decision and then multiplied to get a total score for each alternative.
Example
Score CriterionCriterion Company A Company B Weights
Starting Salary 0.80 0.90 0.30Career Potential 0.95 0.65 0.40Job Security 0.60 0.95 0.20Location 0.95 0.45 0.10
Weighted Avg. Score: 0.84 0.77 1.00
Goal: Choose your first job
Scoring models are. . .• Versatile: They can use with any decision
choice. They handle both qualitative & quantitative variables.
• Transparent: They make assumptions explicit. Everything is captured in the set of alternatives, criteria, and ratings.
• Explanatory: They help explain our reasoning to others. – But be forewarned! GIGO – Garbage In, Garbage
Out – is one potential downside. They can be too subjective! You should treat the results as approximate, not precise.
WRAP-UP
• Thank you on behalf of our partner communities in Guatemala (May) and Nicaragua (August)
• GET INVOLVED: Here in the summer? SOS high school! Next year? Help others that will be in your spot next year!
THANK YOU FOR YOUR SUPPORT
SECOND BAYES EXAMPLE
Exam Q
Given the existing payout table, with the states being growth of:
• Greater then 2.5% (s1)• between .5% and 2.5% (s2) • less then .5% (s3)
s1 (gro > 2.5) s2 (gro > 0.5) s3 (gro < 0.5) EV(di)
0.6 0.2 0.2d1 10,000 6,500 -4,000 6,500
d2 8,000 6,000 1,000 6,200
d3 5,000 5,000 5,000 5,000
PayoffTable
States of Nature
P(sj)
Dec
isio
nWith the same example, you have acquired sample information (Interest rate probabilities), that will improve the accuracy of your expected value. How much is this new sample information worth??
Exam Q
• Now, also our sample info, interest rates, can be in 3 states:f1: Interest Rates remain the same or fall
f2: Interest Rates rise between 0.1-1.5%
f3: Interest Rates rise more than 1.5%
• You are also given the probabilities of the interest rates given various growth rates P(fi|si) in a table.
• GOAL: We want to make 3 different expected value tables based on the 3 interest rate levels, so want to find all 9 P(si|fi) – 3 for each table.
Exam Q
• Based on past experience, you know (given!!):P(f1|s1) = 0.70, P(f2|s1) = 0.18, P(f3|s1) = 0.12
P(f1|s2) = 0.27, P(f2|s2) = 0.28, P(f3|s2) = 0.45
P(f1|s3) = 0.13, P(f2|s3) = 0.18, P(f3|s3) = 0.69
NEXT: You know the probability of each of s1/s2/s3 as .6/.2/.2, so multiplying each of the above 9 cells, by the probability of its ‘S’ state.
P (fi | si): GIVEN in table forms1 s2 s3
f1 0.7 0.27 0.13f2 0.18 0.28 0.18f3 0.12 0.45 0.69
Sum 1 1 1
Exam Q
• Table below after having multiplied each interest probability by the chance of its growth state (.6/.2/.2)
Probabilty of P (fi | si) multiplied by the S probabilty.
s1 s2 s3Probabilty of each Interest level
f1 0.42 0.054 0.026 0.5f2 0.108 0.056 0.036 0.2f3 0.072 0.09 0.138 0.3
Probably of each State 0.6 0.2 0.2
You now know the probability of each interest rate happening!! (.5 / .2 / .3), and the marginal probability (the 9 cells). You want to find the probability of s1/s2/s3 given a interest level (f1-f3) to make your three tables – you have all of it!!
Exam Q
You want P(si|fi) / P(fi):- .42/.5 = .84- .054/.5 = .108- .026 / .5 = .052- Etc..
P(si | fi ) s1 s2 s3 sum f1 0.84 0.108 0.052 1f2 0.54 0.28 0.18 1f3 0.24 0.3 0.46 1
Probabilty of P( si | fi ) - the probabilty a given growth state will happen given an interest level.
3 resulting tables!
Using these probability, create 3 tables to find the optimal decision:
P(si | fi ) s1 s2 s3 sum f1 0.84 0.108 0.052 1f2 0.54 0.28 0.18 1f3 0.24 0.3 0.46 1
Probabilty of P( si | fi ) - the probabilty a given growth state will happen given an interest level.
s1 s2 s3 EV(di)
0.84 0.108 0.052
d1 10,000 6,500 -4,000 8,894
d2 8,000 6,000 1,000 7,420
d3 5,000 5,000 5,000 5,000 Dec
isio
n
States of NaturePayoffTable
P(sj|f1)
IF THE FORECAST IS F1 (interest rate stays the same or falls)
BEST FOR THIS INTEREST RATE
s1 s2 s3 EV(di)
0.54 0.28 0.18
d1 10,000 6,500 -4,000 6,500
d2 8,000 6,000 1,000 6,180
d3 5,000 5,000 5,000 5,000
PayoffTable
States of Nature
P(sj|f2)
Dec
isio
n3 resulting tables!
Using these probability, create 3 tables to find the optimal decision:
P(si | fi ) s1 s2 s3 sum f1 0.84 0.108 0.052 1f2 0.54 0.28 0.18 1f3 0.24 0.3 0.46 1
Probabilty of P( si | fi ) - the probabilty a given growth state will happen given an interest level.
IF THE FORECAST IS F2 (Interest Rates rise between 0.1-1.5%)
BEST FOR THIS INTEREST RATE
s1 s2 s3 EV(di)
0.24 0.3 0.46
d1 10,000 6,500 -4,000 2,510
d2 8,000 6,000 1,000 4,180
d3 5,000 5,000 5,000 5,000
PayoffTable
States of Nature
P(sj|f3)
Dec
isio
n3 resulting tables!
Using these probability, create 3 tables to find the optimal decision:
P(si | fi ) s1 s2 s3 sum f1 0.84 0.108 0.052 1f2 0.54 0.28 0.18 1f3 0.24 0.3 0.46 1
Probabilty of P( si | fi ) - the probabilty a given growth state will happen given an interest level.
IF THE FORECAST IS F3 (Interest Rates rise more then 1.5%)
BEST FOR THIS INTEREST RATE
Almost there!
Forecast Decision EV|di P(fk) (EV|di)xP(fk)
f1 d1 8894 0.5 4447
f2 d1 6500 0.2 1300
f3 d3 5000 0.3 1500
7247 6500 747
EVwoSIEVSI
Now, you have your 3 optimal decisions (d1,d1,d3) from the 3 interest rates possibilities, so now combine them given their probabilities of each interest rate scenario happening to get a total expected value with this sample information!!
SUM of the 3 expected values
The original chosen decision (d1) with just the initial .6 / .2 / .2 probabilities
Initial State Table
s1 (gro > 2.5) s2 (gro > 0.5) s3 (gro < 0.5) EV(di)
0.6 0.2 0.2d1 10,000 6,500 -4,000 6,500
d2 8,000 6,000 1,000 6,200
d3 5,000 5,000 5,000 5,000
PayoffTable
States of Nature
P(sj)
Dec
isio
n
Almost there!
Forecast Decision EV|di P(fk) (EV|di)xP(fk)
f1 d1 8894 0.5 4447
f2 d1 6500 0.2 1300
f3 d3 5000 0.3 1500
7247 6500 747
EVwoSIEVSI
EVwSI
SUM of the 3 expected values
The original chosen decision (d1) with just the initial .6 / .2 / .2 probabilities
Expected Value with Sample Information gave an expected value of 7247, whereas the EV without the sample info gave a value of 6500 a difference of $747. SO, if the cost to gather the sample info is less then 747, then do it!! If not, then it is not worth the extra cost!
Efficiency of the Forecast
• The Efficiency (E) of the Sample Information (Forecast) is:
• In our example, this is:
%5.41
%1001800
747E
%100EVPI
EVSIE