OIDD 101 Midterm Flashcards
0
Q
Process scope
A
Can be defined at the micro level it’s multiple sub processes and also at an aggregate level which tracks the process as a whole
1
Q
Process
A
Set of activities that accepts inputs and produces outputs.
2
Q
Flow unit
A
The flow unit is what is tracked thru the process and generally defines the process output of interest
3
Q
Little’s Law
A
I = R * T
4
Q
I in Little’s law
A
Inventory
The average flow units in the process (units)
5
Q
R in Little’s law
A
Flow rate
Average rate at which flow units enter or leave the process
(Units / time)
6
Q
T in Little’s law
A
Flow time.
Average time a flow unit is in the process. (Time)
7
Q
Make sure to do a Little’s Law sample calculation
For example, incoming calls —> call center —> completed calls
What is I, R, and T?
A
I will be the average number of callers on the phone with the call center
R is the average number of incoming or completed calls per min (what goes in must go out)
T is the average time a caller spends with the call center
8
Q
Why do we need inventory?
A
Flow time
Seasonality
Batching
Buffers —> a buffer is inventory between processes. Allows one process to work while another doesn’t, possibly preventing a reduction in the overall flow rate
Uncertain demand —> produce in advance to reduce chance of running out of inventory
Pricing —> buy when cheap and sell later at higher price
9
Q
4 diff ways to count inventory
A
In terms of: 1. Flow units 2 dollars 3. Days of supply 4. Turns
10
Q
Counting inventory in terms of flow units
A
The I in I=RT
Number of wetsuits, patients, etc.
Useful when the focus is on one particular flow unit
Just measuring in terms of the flow unit
11
Q
Counting inventory in terms of dollars
A
The I in I=RT
The dollar value of inventory
Intuitive measure of firms total inventory
Problem is this isn’t relative
Based on cost to purchase good, not sell it
12
Q
Measuring inventory in terms of days of supply
A
The average number of days a unit spends in the system
Number of days the average amount of inventory would last at the average R if no replenishments
The T in I = RT
Can also be weeks, months, years of supply
T = I/R = inventory /flow rate
13
Q
Measuring inventory in terms of turns
A
The number of times the average amount of inventory exits the system
= 1/T or R/I
For example, say T =2 months
Then annual turns = 6
So, average inventory lasts 2 months. Avg inventory will exit the system once every 2 months or 6 times per year. So annual turns = 6
14
Q
Cost of goods sold
A
This is the flow rate
Flow rate isn’t sales, as inventory is measured in the cost to purchase goods, not in the sales revenue.
R =COGS
15
Q
Problem with having too much inventory
A
Opportunity cost of capital —> money in inventory could be invested in some other asset
Storage, insurance, maintanance costs
Obsolescence costs—> inventory may be worth less tomorrow than it is worth today
Inflation provides benefit to hold inventory, as with inflation cheaper to buy today than tomorrow.
16
Q
Two ways to measure inventory holding costs
A
- Dollars per flow unit per unit time.
For example, 25 cents per customer per minute
$114 per ton of milk powder per year - % of cost of goods
17
Q
Measuring inventory holding costs as percent of cost of goods sold
A
For example, cost of holding inventory is 40%per year.
If a component cost $120, then it costs .4(120) =$48 to hold it for one year or $4 per month
This percent captures all inventory costs
We can also write the annual holding cost as a percentage of COGs as:
Annual inventory holding cost percent / yearly turns
18
Q
Example of inventory holding cost
Suppose Walmart annual inventory holding cost is 25%
I = 44,469
R = COGS = 360,984
A
Annual inventory holding cost = .25(I) = 11,175
Annual inventory holding cost as percent of COGs = this divided by R
Skip a step and just do annual inventory holding cost percent /yearly turns = .25/8.12 = 3.1%
19
Q
Gross margin percent
Example:
Sell drill for $50,purchase it for $30
A
(Price - cost) / price
(50-30)/50 = 40%
20
Q
Capacity
A
The maximum flow rate through a resource.
Be sure to use same units (unit / time)
21
Q
If 6 mixers, 22,000 hotdogs per batch, 20 min to mix, what is capacity?
A
22000 (6)/ 20 = 6600 dogs / min
22
Q
Capacity of the entire process
A
The minimum capacity among the resources
Resource that constrains the entire process is called the bottleneck
While at any minute could produce more than the average, the average capacity is the bottleneck.
23
Q
Processing time, activity time, service time
A
Duration that a flow unit has to spend at a resource, not including any waiting time.
24
Capacity relation to processing time
If one worker, capacity = 1 / processing time
If multiple workers, capacity = m/ processing time, where m is the number of workers or machines for a given part of the process.
25
Demand
Rate at which flow units are requested from the process. (People / hr, widgets / min, etc.)
26
Input
Rate at which flow units can begin the process (unit / time)
27
Flow rate for a process
The rate at which flow units enter or leave the process.
R.
Min (process capacity, demand, input)
Usually input isn’t the min so min of proc cap and demand.
28
Utilization
Fraction of time working
| Flow rate / capacity
29
Supply constrained
Process Capacity < demand
This means that the flow rate will equal the process capacity, the capacity of the bottleneck
So, utilization will be 100% for one of the steps in the process.
30
Demand constrained
Process capacity > demand
Flow rate = demand
Now utilization isn’t 100% for any step
31
Flow time
Time a unit spends in the process.
| This is just T = I/R using Little’s law.
32
Cycle time
Time between when units exit the process.
(Min/unit)
1/R
33
Time to go through an empty worker paced process (1 unit)
Sum of the processing times.
34
Time thru an empty machine paced process (conveyer belt) (1 unit)
Number of resources in sequence * processing time of the bottleneck step
35
Time to produce N units starting with a full system (machine or worker paced)
N * cycle time
| N is the number of flow units you want to produce.
36
Time to produce N units starting with empty worker paced system
Sum of processing times + (N-1) * cycle time
Takes longer with empty system
First unit goes thru for whole processing times, rest go thru every (cycle time)min
37
Time to produce N units starting with empty machine paced system
Number of stations * cycle time + (N-1)(cycle time)
| Conveyer belt. Everyone works at the bottleneck speed.
38
Cost of direct labor
Labor cost per flow unit
How much labor cost per unit we produce
($/unit)
Wages per unit of Time / R
Note that wagers per unit of Time = wages per worker per unit of Time * number of workers
39
Labor content
Total LABOR time to produce a flow unit.
Sum of processing times involving labor
Machine processing times aren’t included
(Time / unit)
40
Labor utilization
Average utilization across workers
R / labor capacity
Labor capaocy = N/ labor content
So labor útil = labor content * (R/N)
Or labor content / (N* cycle time)
Remember cycle time = 1/R
If everything in a process is labor then labor útil =process útil
41
Takt Time
Time between when flow units are demanded (min/unit)
1/demand rate
42
Target manpower
Minimum number of workers needed to satisfy demand (assumes 100% útil)
Labor content / Takt time
Minimum number of workers to get the demand rate that we need.
43
Process improvement
Replicate process, add 1 worker, add 2 workers, balance tasks, integrate work
44
Adding 1 worker
Could move the bottleneck to a different step. New min capacity
45
Balancing tasks
People split two tasks, evenly divides the total processing times of these tasks between the two people.
Capacity goes up
46
Integrating work
Each person does every task. Naturally balanced lines. Never waiting for someone else.
Add up total processing time, that is processing time per worker now.
Then 1/this is the capacity per worker and then multiply by workers to get total capacity
This will be highest capacity, as no idle time and each worker has 100 percent útil.
Problem is this requires more training and different skills. Heart surgeon won’t perform knee surgery and answer phones
47
Line balancing
Attempting to achieve even útil across resources in a process and minimize idle time.
48
Why we line balance
To improve capacity of total process without adding resources.
Integrating work is best way to improve line balancing.
49
When not every flow unit goes thru each resource, what do we do?
Use yields to determine demand.
50
Yield of a resource
Flow rate of good output / flow rate of input
| Say a yield is .336. This means 33.6% of the inputs make it thru the process while the rest are rejected.
51
Yield of a process
Product of resources’ yields.
52
How to get demand per resource when we have attrition loss— some units rejected after each resource
Use yields
For example, 110 applications are demanded and the first step, underwriting, has a yield of .336
So, 110(1-.336) = 73 are rejected after underwriting and 110(.336) = 37 make it thru to the next step, Q and C
This step has a yield of .784
So, 110(.336)(.784)=29mane it thru to get funded and 110(.336)(1-.784)=8 get rejected after second step.
To get demand on resource, add these numbers up.
So underwriting demand is 110and Q and C is 37 loans / day
53
Implied utilization
Demand /capacity of a resource
Can be bigger than 100%
54
Bottleneck in cases with attrition loss
Resource with highest implied utilization is the bottleneck.
55
Big error with this
Make sure that when calculating resources demand you evaluate it as if it is the only resource in the process.
56
How to achieve a target flow rate
Get the whole process yield. Do this by multiplying the resources yields.
Input needed =target output /process yield
Ex. Say we want a target output of 30loans per day
Proces yueld is .263
So 30/.263= 114 loans / day
So, we need 114 loans / day on the underwriting resource and multiply by that resource’s yield to get capaocty for Q and C
57
Process with rework
If a flow unit needs to go thru a process multiple times.
Make sure to use processing times not capacity—>
Say 80%of travelers go thru it once and have a capaocty of 3 travelers /min while 20% go thru twice and cap = 1.5/min
Processing times are 1/3and 1/1.5=2/3
So do average processing time = (,8)(1/3)+(.2)(2/3) =.4 min / traveler
Then invert this to get capaocty = 2.5 travelers /min
Be sure to not take weighted average of capacities!!! Do it with processing times!!!
58
Lessons from airport screening with bottleneck
Don’t send bad units thru bottleneck
| Don’t starve or block the bottleneck. Allow the bottleneck to work at its highest capaocty.
59
When to use minutes of work as flow unit
When each flow unit doesn’t visit each resource.
When the processing times depend on the application type
For example, we have consulting, staff, and internship applications and they all go thru different processes.
So, applications can’t be flow unit, as capaocty of each resource depends on types of applications.
Need a common flow unit that allows us to express all demands and capacities in terms of that flow unit.
So, use minute of work as the flow unit
60
Capacity Minute of work as the flow unit
Each resource can do 60 min of work /hr times the number of employees at the resource
61
Demand in minutes of work
If 5 jobs arrive per hour and each job takes 10 min, then demand is 5jobs /hr * 10 min of work = 50 min of work / hr
Demand on a resource depends on which applications flow thru that resource and the processing time for each application
Make a table with the resources on top and the different flow units on the left side
Multiply processing time in minutes /app * demand in app/ hr to get the demand in min of work / hr
Then add up the demand for each resource with the different units.
62
How to get bottleneck with minutes of work as flow unit
Do implied utilization.
| Demand /capacity
63
Economic order quantity (EOQ) model
Order the right amount to minimize cost.
64
Q
Quantity of units in each order (what we need to choose)
65
I in EOQ model
The average inventory level.
| Q/2
66
R
Flow rate of demand.
| In units /time
67
Number of orders per unit of Time
R/Q
| This shows how many orders we must make per unit of Time
68
Time between shipments
Q/R
| How often we must order new shipments.
69
Inventory holding cost per unit time
h
We get this by multiplying the annual holding cost by the cost per unit and then dividing by 52to get the inventory holding cost per week.
How much it costs to hold one unit for one week.
70
Average inventory cost per unit time
h * Q/2
| How much it costs to hold one unit per week times the average inventory
71
Setup cost
K
Cost per order, independent of the amount ordered.
Fixed cost.
72
Setup cost per unit time
K/ (Q/R) as the time between orders is Q/R
73
Purchase costs
How much it costs per unit * R =purchase costs per week. Not based on Q!
74
EOQ
economic order quantity.
Want to minimize the average setup and holding costs per unit time, C(Q), function
= SQRT (2KR/h)
75
Setup and inventory holding costs per unit
C(Q)/R
C(Q) is costs per unit of time
R is units sold per unit of Time
So, C(Q)/R is setup and inventory holding cost incurred by each unit, not divided by Q, as that wouldn’t make sense.
Note that per unit costs decrease as R goes up.
76
C(Q)
Average setup and holding Costs per unit of Time
| KR/Q + .5 hQ
77
Quantity discount for EOQ
if you order more, can get a discount.
| Use equation for costs and if get a discount on h, see that you should buy in bulk.
78
Setup time
The amount of time needed to get ready for production during which no production actually occurs.
Time preparing
Doesn’t depend on the number of units produced. Just in minutes(or sec, days, etc.)
79
Capacity of a process with setups
Capacity = number of units produced / time to produce those units
Time to produce a batch = setup time + batch size * processing time
So capacity = batch size / (setup time + batch size * proc time)
80
Production cycle
Repeating sequence of identical production runs
For example, suppose they produce 100 steer supports, 200 ribs, then 100 ss, ...
Then one production cycle would be 100 ss and 200 robs
A batch is a set of flow units produced in a production cycle.
The logical flow unit above is a component set, 1ss and 2ribs
Production cycle is a batch of 100 component sets.
81
Batch
Set of flow units produced in a production cycle.
82
Component set
Within the production cycle.
Component set is The flow unit (1 ss, 2 ribs)
Batch (100 component sets) is the amount of flow units in a production cycle (100 ss and 200ribs)
83
Processing time
Time to produce one component set
Time to produce 1ss and 2 ribs
Add up
Min / comp set
84
Capacity given a production cycle
Batch size / (setup time + batch size * proc time)
85
As batch size increases, what happens to capacity?
Capacity goes up
Batch size increases, setup time is amortized over more units, so process capacity goes up
Process capacity maxes out at 1/processing time
Capacity with an infinite batch size.
86
Flow rate
Units produced per unit of Time
| Larger batch increases flow rate thru process, only up to a point
87
Utilization
Percent of time producing (not idle, not setting up)
| Larger batch increases u. Only up to a point
88
Inventory
Average number of units in a process
| Larger batch increases I
89
Problem with large batches
Tradeoff between flow rate (capacity) and inventory
| Both increase, but having too much inventory is costly.
90
Batch size and location of the bottleneck
Say we have two steps, one with a setup time and one without a setup time.
For the resource with setup times, the capacity will be dependent on the batch size while for the resource without setup times will have a constant capacity.
With larger batch sizes, the resource with the setup times will have a higher capacity and with lower batch sizes, the resource without setup times will have higher capaocty
We need to find the smallest batch size such that they have equal capacities.
91
Choosing right batch size
We want to have equal capacities among the resources.
If supply constrained, want equal batch sizes.
Otherwise, demand rate would be target capacity
Rearrange capacity equation to be batch size = (capacity* setup time) /(1-capacity * proc time)
Capacity in this equation is typically set to target flow rate (R), which is usually demand
Setup time is the total setup time for the flow unit in the production cycle
Processing time is the total processing time for the flow unit.
If we have multiple resources with setups in a process and are targeting the same capacity, we choose the batch size = maximum batch size calculated among resources with setups.
92
Utilization in a batch process
Utilization is the fraction of time the resource is producing output.
Without setups is flow rate / capaocty
Capacity is output rate when producing.
When not being utilized, is idle.
For resources with setups, útil = flow rate / output rate when producing
This is flow rate / (1/proc time) =flow rate * proc time
When not being útil, either idle or being setup.
93
Utilization with a small or big batch
If batch size is too small and supply constrained, then both step utilization will be too small. Process with setup time will be bottleneck as will have smallest capaocty
If batch size is too big and supply constrained, then the step that doesn’t have setup time will have 100 percent útil. But the other one will have idle time. Process without setup time will be bottleneck as smaller capaocty
Note that utilization’s don’t show bottleneck. Based on capaocities here
We want the batch with the just right size. Have equal capacities.
94
Utilization with just right batch
Equal capaocities, same flow rate.
Neither have idle time.
Utilization for process with setup time can’t get to 100% as has setup time.
95
Batch size and utilization’s
Increasing batch size increases utilization of process with setup times but only up to a point
Because large batches need idle time to match assembly, can’t get 100%util because constrained by demand or assembly.
96
Maximum inventory
```
Production time * rate of increase
B (1-Rp)
B is batch size
p is processing time
R flow rate
```
97
Average inventory
1/2 maximum inventory =.5B(1-Rp)
Note that this equation is only applicable to one type of inventory (Allie dolls)at a time in process with multiple types of inventory (A, B, C, D)dolls.
So we do rib pairs instead of component sets.
98
Avg inventory example
Batch size is 200rib pairs
Processing time is 1min per rib pair
Flow rate is 1/3 rib pairs per min
So then we do AI= .5 B(1-Rp) = 67.7 rib pairs.
99
Batching and product variety (chicken and tomato soup)
Define flow unit as 1 gallon of soup
Assume we iterate between chicken and tomato production
Production cycle includes both chicken and tomato soup
Batch is total quantity in gallons of chicken and tomato soup.
How to minimize inventory and satisfy demand?
Target capacity adds up the demand for each kind of soup
Each batch involves making both soups, so add up the setup time.
Processing time is gallons per hour. Same for both soup. Just use that number.
Then use batch size equation. Get it and then produce in proportion to demand.
If we get three kinds of soup, Same thing. Batch size increases.
100
Variety and batch size
Adding variety makes Lot bigger batch size necessary.
Consequence from adding variety.
With lot of variety, takes longer and need to hold more inventory.
101
What generates queues
Arrival rate that predictably exceeds service rate (toll booth congestion during thanksgiving)
Variation in arrival and service rates in a process where average service rate more than adequate to process average arrival rate. For example, calls to a brokerage are higher during certain times by random chance.
102
Predictable queue setting
Demand will be bigger than capaocty for a certain time.
103
Predictable queue growth rate, length of queue at time T, time to serve average customer arriving at time T, average time to serve customer in the queue
queue growth rate = demand - capacity
Length of queue at time T =T(demand -capacity)
This gives you the length in customers.
Time to serve Qth person in queue = Q/capacity
Time to serve customer arriving at time T = T (Demand/ capacity -1)
Average time to serve customers in the queue is 1/2 of the time to serve customer arriving at time T
104
How to address predictable queue setting
Peak load pricing (congestion pricing): charge more during peaks demand
Tempeorary capocty for peak period (hire more workers during peak)
Pre processing strategies: do more of the work off peak, ahead of time.
105
Average inter arrival time
a.
| Average time between arrivals
106
Standard deviation of inter arrival times
How spread out of variable inter arrival times are.
107
Coefficient of variation of arrival process
Coefficient of variation is a measure of the relative variability of a process. The standard deviation divided by the average.
Allows for standard deviation and mean to be on the same scale.
Standard deviation of interarrival time / average interarrival time.
108
Service time variability
CVp = standard deviation of proc time / average proc time
Measures how variable the service time is.
Higher it is, more variable. More spread out.
109
Multi server queue assumptions
All servers equally skilled (all take p time to process each customer)
Each customer is served by only one server.
Customers wait until their service is completed
There is sufficient capacity to serve all demand.
110
Demand in queue process
where a is the average interarrival time (time per customer),
Demand =1/a
111
Capaocty of queue process
m (1/p) where p is the average processing time, the capacity of each server is 1/p and there are m servers.
112
Multi server queue implied util and util
Implied util =demand / capacity = p/ (am)
Utilization = flow rate /cap
If demand is less than capaocty, which must happen to have a stable queue, then flow rate = demand and util = implied util.
113
Stable and unstable queue
If implied util is less than 100%, then system is stable. Size of the queue doesn’t keep growing. Average Capacity is bigger than demand
If implied util is bigger than or equal to 100%, unstable. Size of the queue will keep growing.
Only analyze stable queue (avg demand < avg cap)
114
How to interpret utilization
Say util = 86%
At any given moment, there is a 86% probability a server is busy serving a customer and 14% chance the server is idle.
Or, at any given moment, on average 86% of servers are busy serving customers and 14% are idle.
115
Time spent in the system
Flow time = time in queue + time in the process (p)
Only works for stable system.
Time in queue equation on eq sheet. Add this to p to get flow time.
This says how long average customer will spend in the system (waiting for service plus time in service)
116
Average inventory in the system
Since demand is less than cap, flow rate = demand = 1/a
Avg Inventory in queue = (1/a)(Tq)= Tq/a
Avg inventory in the process = (1/a)(p) = p/a
Note, doesn’t depend on m because as long as stable system, m>p/a, the flow rate is constant for any value of m (1/a)
117
Utilization and system performance (time in queue)
As utilization approaches 100%, time in queue increases dramatically.
Servers are so busy, time waiting skyrockets. Customers have to wait a while, trade off between how busy servers are and customer wait time.
118
What happens with utilization and time in queue when we increase amount of servers?
Increase m, the tradeoff curve between útil and time in queue is less steep.
Can have more utilization with less time in queue.
Curve shifts right. Increases heavily at a higher util.
Big scale is less sensitive to util.
Economies of scale.
Can keep útil constant but decrease time in queue
Can keep time in q constant, but increase util
Or can reduce time in queue and increase util.
Better service and lower labor costs with Econ of scale.
119
Pooled system vs partially pooled system vs separate queue system
Variability, total demand, processing time, utilization, probability a server is busy (util) is the same.
Pooled system is one queue with all the servers (say m=4)
This will be the most effective, have regular a and m. Lowest time in queue
Partially pooled system will have 2 queues with 2 servers per queue.
a is double because half of the people arriving go to each queue and m is half. Time in queue is bigger.
Separate queue system has 4 queues with one server each. A is four times as big, m is 1. Time in queue much bigger.
Pooling dramatically reduces time in queue because with separate queue system can have idle customers and servers at the same time. That is not good.
Note that inventory in queue is different for each. Double for partially pooled system and quadruple for separate queue system.
Same inventory in the process.
120
Limitations to pooling
May require workers to have broader set of skills, more training and higher wages.
May disrupt customer server relationship.
Pooling could increase time in queue for one customer class at expense of other (removing first class line)
121
Potty parity laws
Women’s restrooms have double flushing capocty of men’s restrooms. m increases for women, decreasing capaocty and util.
Women have larger p, fewer servers (lower m), go in batches (higher CVa), take longer and have bigger variety (higher CVp) resulting in longer time to wait for bathroom.
122
Two things to take into account when deciding if to invest in drug
Efficacy — is it an effective treatment
| Toxicity— how harmful are the side effects
123
Prior probability
Your initial probability assessment.
Your probability estimate before learning new information about it.
For example, 25% chance the drug will pass before we learn new info.
124
Simple decision tree
Squares represent action or decision nodes, meaning we get to control. Either invest or don’t invest
Circles are event nodes. We don’t control. Nature makes decision (either passes or fails)
Action description above each line, outcome description and probability, payoff outcomes on the right.
125
Rules about events and probabilities
Each Event node must have an outbound arrow for all possible outcome of that event
Each possible outcome must have a probability associated with it
The sum of probabilities for a branch must equal 1
126
Three different models for choosing what to do
Maximin
Máximax
Expected value decision
127
Maximin
For each action, evaluate the minimum payoff that could occur.
Choose the action that yields the maximum minimum payoff.
So for example, with invest we could get 280 or -80 and with abandon we get 0. We then look at -80 vs 0 and see that since 0 is bigger than -80, we go with the 0 branch and abandon.
Doesn’t rely on probabilities
Conservative.
Will use this in high uncertainty and high stake situations.
Looks at the worst that could happen with each tree and see which is less worse.
128
Maximax
Evaluate the maximum payoff that could occur for each action and choose the action that yields the maximum maximum payoff.
For example, we have 280 or -80 on one branch (280) and 0 on the other beach. Choose the 280 branch.
Doesn’t rely on prob. Aggressive. Not used often
129
Maximize expected value
Start at the right and successively evaluate the expected value at each node as you move left.
Say there is a 25% chance it passes (make $280) if you invest and 75% it fails (lose $80). Then expected value = .25(280)+.75(-80)=10
This compared with the other branch, abandon, which has an expected value of 0
Choose the branch with the higher expected value.
130
Expected value of perfect information (EVPI)
The difference between the expected value of the decision with perfect information and the expected value of the decision without perfect info (original scenario)
Suppose we could choose our actions after learning which event outcome occurs, meaning we could choose actions with perfect info.
Expected value couldn’t decrease and prob will increase because choosing the same action before or after learning info should lead to same outcome, and we can’t be worse off but could be better off.
131
How to evaluate EVPI
Draw the decision tree with all actions happening AFTER unknown events.
So circle of events we can’t control first and then the events we can control (square)
Get the new expected value.
Subtract this by our prior probability.
This gives us the EVPI. Don’t forget to subtract by prior prob!!!
132
Posterior prob
Probability assessment after learning some info.
Probability of an action given some other action.
Prob (pass | good) means the probability it passes given it had a good signal.
133
Joint probability
Probability that two events occur.
Prob (A & B)
Means that both A andB occur.
Probability before we do the test that we get a good signal and the compound passes.
134
Linking prior and joint prob
For example, probability of pass = prob (pass and good) + prob (pass and bad)where it must either be good or bad if it passes.
135
Bayes Rule
P(A|B) = P(A&B)/P(B)
Can rearrange:
P (A and B) = P(A|B)*P(B)
136
Putting prior and posterior prob together
P(A) = P(A|B)*P(B) + P(A|C) * P(C)
| Where P A given B and A given C are only posterior probabilities.
137
Expected value of sample information
Difference between expected value with the test and expected value without the test
The maximum you are willing to pay for the sample—wouldn’t pay more than this to conduct the test
EVSI less than or equal to EVPI. Sample info can’t be more valuable than perfect info
138
When is EVSI 0?
If your subsequent actions don’t depend on the outcome of the test.
Test doesn’t influence our decision.
139
When does EVSI = EVPI
If the test is perfect diagnostic, meaning it never makes a mistake.
P(Pass| good) = 1 and P(pass|bad) =0
140
Characteristics of a good test
Quick to implement, cheap, informative (diagnostic—> more diagnostic, higher EVSI)
Diagnostic test is as likely as possible to give a good signal and when it does, it gives correct signal. Only says good when product will pass
Want tiles in quadrants I and IV
141
Completely diagnostic test
If the signal is good, then product will surely pass
If signal is bad, will surely fail
Learned as much as we can from the test
Everything is top left or bottom right quadrant
142
Completely undiagnostic (uninformative)test
In all cases P(pass) = P(pass given good) = P (pass given bad)
This means that whether the signal is good or bad, odds the product will pass is the same
Don’t learn anything from these tests.
Of the goods, 4/16 pass and of the bads, 1/4 pass
Same 25 percent.
143
Weak test— too conservative
When a movie critic likes a movie (good signal), surely wins Best Picturue Award, but he likes at most one movie per year and sometimes doesn’t like any. Very picky, gives a bad signal to many movies that win.
P(pass given good) = 1/1
P(pass given bad) = 4/19 = 21%
P(good) = 1/20=5%
P(pass) =25%
A good signal is great news (will surely pass), but test misses 4 of 5 good products. Too picky.
144
Weak test—too liberal
Sue, a movie critic, almost always likes (good signal)the movie that wins the best picture award,but she also likes most movies and most of those aren’t even nominated
P(pass given good)= 5/17=29%
P(pass given bad) =0
P(good)= 80%
P(pass) =25%
Bad signal surely is bad news (will fail),but the test labels most products as good, even those that will fail
145
Rules for good decision making
Information can be valuable even if not perfect. Imperfect info can be useful when it will change future actions
Info useless unless change future actions.
“How does the test change how we manage the condition?” should be the main question asked
Don’t judge decision based on info we don’t know till after. Don’t be q MMQB
judge based on what you knew when you made the decision.
