business decision models bu/ec275

Business Decision Models BU/EC275Created by: Greg Overholt

Presented by: Stephen Duarte [email protected]

Agenda

• Waiting Lines– Economic Analysis

• Simulation– Monte Carlo / Arena

• Decision Analysis– Payouts/Trees– EVPI (Perfect Information) – EVSI (Sample Info)

• Utility/Scoring

• Remember to ask questions!! Thumbs up/down?

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: Departures(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

• Terms– Balking: when they come in and see the line so they

don’t enter the queue– Reneging: when people leave the queue

Poisson• the distribution of the probability

that ‘x’ occurrences in an interval (number of successes).

• Discrete distribution = aka, the probability of each number is > 0

• PROBABILITY FOR A SPECIFIC VALUE

The rate / per hour (typically)

Poisson

• Excel=POISSON(x,λ,cumulative)If ‘cumulative’ = TRUE less thenIf ‘cumul…’ = FALSE equal to

Poisson Questions

Students arrive at Tim Horton’s at a rate of 10/hr (λ)

Q1. What is the average time between arrivals? (What is µ?)- 10 in an hour, so 60 mins/10 students

= 6 mins per student

Inverse Relationship! Example

Q2. What is prob that exactly 5 students come in the next hour?

f(5) = 105*e-10 / 5!f(5) = .0378 or 3.78% 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)

IF P > X, then take away the ‘1-’ (right tail test!)

• Excel

=EXPONDIST(x,λ,cum)cumulative =TRUE less then

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%

Poisson vs Exponential

Exponential

Poisson

P(x=xo)=POISSON(x,λ,false)P(x≤xo)=POISSON(x,λ,true) P(x>xo)=1 - POISSON(x,λ,true)

P(x=xo)= EXPONDIST(x,λ,false) = 0P(x≤xo)= EXPONDIST(x,λ,true)P(x>xo)= 1 - EXPONDIST(x,λ,true)cumulative =TRUE less then

Waiting Line Inputs

NOW WE SEPARATE λ AND µ

• λ / μ = U = Utilization factor – probability that server is busy– probability that a customer has to wait

• I = 1 – U : Idle time– the probability that the server is idle – the probability that there are no customers in

the queuing system (P0)• If U = 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 Sub Maker at MR Sub in the terrace.

Sub orders arrive at a mean rate of20 per hour. Each order received by Joerequires an average of two minutes toMake the sub. (Assuming a M/M/1

system)

Example!

Arrival Rate DistributionQuestion: What is the probability that no orders are

received within a 15-minute period?

Poisson because you are looking for a specific value

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 average service rate per hour? (What

is μ?)

Answer: Since Joe Ferris can process an order in an average time of 2 minutes (.033 hours)

We know that: Average Service Time = 1/ μ, and we are looking for average service rate (which is just μ).

So:Average Service Time = 0.033 hours = 1/ μ0.033 = 1/ μ -> μ = 1/0.033 Therefore, μ = 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 making subs?

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)Mr Sub 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 sub maker, 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 that Joe is working all the time, he is unable to take breaks, in order to comply with labour standards, we cannot do this

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 Mr Sub. was so successful that business actually doubled. The mean rate of sub orders arriving is now 40/hr and the company must decide how many sub makers to employ. Each sub maker hired can process an order in an average time of 2 min.•Based on a number of factors the sub shop has determined the average waiting cost/minute for an order to be $0.50. Sub makers hired will earn $20/hr in wages & benefits.

Using this information compare the total hourly cost of having 2 sub makers with that of having 3 sub makers.

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 (sub makers) + ($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 sub makers and for k =

3 2 sub makers 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

Wages Cost/Hr

Waiting Cost/Hr

Total Cost/Hr

2 Traders 40.00 72.00 112.00

3 Traders 60.00 44.34 104.34

Sub Makers

Sub Makers

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

What Program to use!!

• Excel: when time isn’t a factor, and probabilities are known!

• 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 4 to 6 clientsSo, RN’s 0 0.3 = less then 40.3 1 = 4 to 6 clients

Expense Probability: If has less then 4 clients, then take RN*$2000, and add that to $3,000.

If has 4 to 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.

Client Probability: 30% = less then 4,70% will have 6 clientsSo, RN’s 0 0.3 = less then 40.3 1 = 4 to 6 clients

Expense Probability: If has less then 4 clients, then take RN*$2000, and add that to $3,000.

If has 4 to 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.

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

Monte Carlo

• 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.

• Initial seed value is the number you begin with

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?

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

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 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?


• 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?


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


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


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


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


RegretTable

States of Nature

MinimumCost

Cost

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


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


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.

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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


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


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!


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


IF THE FORECAST IS F2 (Interest Rates rise between 0.1-1.5%)


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!


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


IF THE FORECAST IS F3 (Interest Rates rise more then 1.5%)


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

business decision models bu/ec275

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