Dormitory
On-Campus Apartment
Off-Campus Apartment
At Home
Total
No Regular Exercise
32
74
110
39
255
Sporadic Exercise
30
64
25
6
125
Regular Exercise
28
42
15
5
90
Total
90
180
150
50
470
Chi-Test
p-value =
Assignment 5 – 20 Points
Deliverable 1 (15 points):
Part A: The General Ford Motors Corporation (GFMC) is planning the introduction of a brand new
SUV—the Vector. There are two options for production. One is to build the Vector at the company’s
existing plant in Indiana, sharing production time with its line of minivans that are currently being
produced there. If sales of the Vector are just moderate, this will work out well as there is sufficient
capacity to produce both types of vehicles at the same plant. However, if sales of the Vector are strong,
this option would require the operation of a third shift, which would lead to significantly higher costs.
A second option is to open a new plant in Georgia. This plant would have sufficient capacity to
meet even the largest projections for sales of the Vector. However, if sales are only moderate, the plant
would be underutilized and therefore less efficient.
This is a new design, so sales are hard to predict. However, GFMC predicts that there would be
about a 60% chance of strong sales (annual sales of 100,000), and a 40% chance of moderate sales (annual
sales of 50,000). The average revenue per Vector sold is $30,000.
Production costs per vehicle for the two production options depend upon sales, as indicated in the
table below:
Moderate Sales
Strong Sales
Shared Plant in Indiana
$16,000
$24,000
Dedicated Plant in Georgia
$22,000
$20,000
The fixed costs for adding Vector production to the plant in Indiana would total $200 million per
year (regardless of sales volume). The fixed costs for opening a plant in Georgia would total $400
million per year (regardless of sales volume).
(i)
Construct a decision tree to determine which production option maximizes the expected
annual profit, considering fixed costs, production costs, and sales revenues. What risk
exists in this choice?
Part B: Due to the uncertainty in expected sales for the Vector, GFMC is considering conducting a
marketing survey to determine customer attitudes toward the Vector and better predict the
likelihood of strong sales. The marketing survey would give one of two results—a positive attitude
or a negative attitude toward the design. GFMC has used this marketing survey for other vehicles.
In the past, the marketing survey indicated positive attitudes toward the design 70% of the time and
negative attitudes 30% of the time. When the marketing survey indicates positive results the
probability of having strong sales is 0.84 and the probability of having moderate sales is 0.16.
When the marketing survey indicates negative results the probability of having strong sales is 0.36
and the probability of having moderate sales is 0.64.
(i)
Construct a decision tree to determine the expected profit from using the survey. Which plant
should GFMC use if the survey indicates a positive attitude? Which plant should GFMC use if the
survey indicates a negative attitude?
(ii)
What is the expected value of the sample information in part b? What does this say about how
large the cost of the marketing survey can be before it would no longer be worthwhile to conduct
the survey?
Deliverable 2 (5 points):
1
The Assignment 5 Data excel file has a Chi Square worksheet that displays data on how often a college
senior exercises and where they live. Based on the data, is there a relationship between exercise and
student's living arrangement? Do you think where a person lives affect their exercise status? (Use alpha =
0.01)
Instructions
Submit your assignment using the Assignment Folder. The file name should follow
the format:
Your-last-name Assgn5.xlxs
2
Decision Analysis given Uncertainty
Expected value E(X) - the average of many trials of a discrete probability distribution
What you expect is not necessarily what you get
Discrete probability distribution
All outcomes are listed
A probability of occurrence is assigned to each
The probabilities must sum to one
Calculation of expected value:
Multiply each outcome times the probability of that outcome occurr
Sum up all of the products
Outcome (X)
3
5
8
P(X)
X * P(X)
0.2
0.5
0.3
1 E(X) =
The expected value is NOT what you expect to get!
How can expected value be used in decision making?
Calculate the expected value of each of the possible decisions and select the decision
Bonds
Stocks
%
P(%)
%
P(%)
High Growth
10
0.2
15
0.2
Growth
7
0.25
10
0.25
No Change
5
0.5
5
0.5
Loss
3
0.05
-5
0.05
1 E(X) =
1
Personality Types and Decision Making
Conservative or Pessimism (MaxMin)
Choose the minimum for each and choose the maximum
Bonds
Stocks
Deposit
Aggressive or Optimism (MaxMax)
Choose the maximum for each and choose the maximum
Bonds
Stocks
Deposit
Expected Opportunity Loss (EOL)
Determine the largest number for each growth and subtract each entry in the row fro
Multiply each difference by the probability and sum to calculate the EOL for each act
Choose the action with the smallest EOL
Bonds
Stocks
%
P(%)
%
P(%)
High Growth
10
0.2
15
0.2
Growth
7
0.25
10
0.25
No Change
5
0.5
5
0.5
Loss
3
0.05
-5
0.05
EOL =
We have looked at a variety of methods of choosing an investment and have hinted a
I would like to quantify how risky each investment is (quantify the variation)
Coefficient of Variation = Standard deviation divided by mean expressed in percentag
How do we calculate the standard deviation of a discrete probability distribution?
Take each outcome and subtract the expected value, square each difference, multipl
all products and take the square root
Bonds
%
10
7
P(%)
0.2
0.25
5
3
Stocks
%
15
10
5
-5
0.5
0.05
E(X) =
σ=
CV =
σ=
CV =
P(%)
0.2
0.25
0.5
0.05
E(X) =
obability distribution
that outcome occurring
What would 100 trials look like?
Calculate standard deviation
nd select the decision with the highest expected value
Deposit
%
P(%)
6.5
0.2
6.5
0.25
6.5
0.5
6.5
0.05
E(X) =
1 E(X) =
ch entry in the row from the largest
e the EOL for each action
Deposit
%
P(%)
6.5
0.2
6.5
0.25
6.5
0.5
6.5
0.05
EOL =
EOL =
ment and have hinted at variations in returns
the variation)
expressed in percentage terms
bility distribution?
ch difference, multiply each squared difference by the probability, sum up
Chi-Square test is used to determine if there is a relationship between two categorical variables
Actual Data
Week 1
Week 2
Week 3
Total
Sand
97
120
82
299
Misrun
8
15
4
27
Shift
18
12
0
30
Drop
8
13
12
33
Corebreak
23
21
38
82
Broken
21
17
25
63
Other
5
15
19
39
Sand
Misrun
Shift
Drop
Corebreak
Broken
Other
Sand
Misrun
Shift
Drop
Corebreak
Broken
Other
Percentages
Week 1
Week 2
Week 3
Total
Expected
Week 1
Week 2
Week 3
Total
wo categorical variables
Total
180
213
180
573
α = 0.05
If p-value is less than α, then conclude there is a relationship
between week and distribution of rejects
CHITEST
p-value =
Total
Total
A friend proposes a wager: You will pay her $9.00, and then a
fair die will be rolled. If the die comes up a 3, 4, 5, or 6, then your friend will pay
you $15.00. If the die comes up 1 or 2, she will pay you nothing. Furthermore,
your friend agrees to repeat this game as many times as you wish to play.
Success Payout to You
(2/3)
You Pay $9
Accept Offer
Failure Payout to You
(1/3)
Reject Offer
You Pay $0
Net Profit
To absorb some short-term excess production capacity
at its Arizona plant, Special Instrument Products is considering a short manufacturing
run for either of two new products, a temperature sensor or a pressure
sensor. The market for each product is known if the products can be successfully
developed. However, there is some chance that it will not be possible to
successfully develop them.
Revenue of $1,000,000 would be realized from selling the temperature sensor
and revenue of $400,000 would be realized from selling the pressure sensor. Both
of these amounts are net of production cost but do not include development cost.
If development is unsuccessful for a product, then there will be no sales, and the
development cost will be totally lost. Development cost would be $100,000 for
the temperature sensor and $10,000 for the pressure sensor.
Success Revenue = $1,
0.5
Cost $100,000
Temperate Sensor
Failure Revenue = $0
0.5
Success Revenue = $40
0.8
Pressure Sensor
Cost $10,000
0.2
Failure Revenue = $0
Neither
Net Profit
ABC
Bid
EV =
0.333
$9,500
0.667
EV =
0.667
$8,500
0.333
EV =
$7,500
No Bid
Complex
Bid
Manufacturing
Process
EV =
New
$10,000
Current
$9000/$8000
EV =
New
$10,000/$9,000
Current
$8,000
EV =
New
$10,000/$9,000/$8,000
Current
Cost per
Computer
0.25
$8,500
0.5
$7,500
0.25
$5,000
$8,000
0.25
$8,500
0.5
$7,500
0.25
$5,000
$8,000
Net
Profit
0.25
$8,500
0.5
$7,500
0.25
$5,000
$8,000
Bid Cost =
Computers =
$1,000,000
10000
ABC Computer Company is considering submission
of a bid for a government contract to provide 10,000 specialized computers
for use in computer-aided design. There is only one other potential bidder for
this contract, Complex Computers, Inc., and the low bidder will receive the contract.
ABC's bidding decision is complicated by the fact that ABC is currently
working on a new process to manufacture the computers. If this process works as
hoped, then it may substantially lower the cost of making the computers. However,
there is some chance that the new process will actually be more expensive
than the current manufacturing process. Unfortunately, ABC will not be able to
determine the cost of the new process without actually using it to manufacture
the computers.
If ABC decides to bid, it will make one of three bids: $9,500 per computer,
$8,500 per computer, or $7,500 per computer. Complex Computers is certain
to bid, and it is equally likely that Complex will bid $10,000, $9,000, or $8,000
per computer. If ABC decides to bid, then it will cost $1,000,000 to prepare the
bid due to the requirement that a prototype computer be included with the bid.
This $1,000,000 will be totally lost regardless of whether ABC wins or loses the
bidding competition.
With ABC's current manufacturing process, it is certain to cost $8,000 per
computer to make each computer. With the proposed new manufacturing process,
there is a 0.25 probability that the manufacturing cost will be $5,000 per computer
and a 0.50 probability that the cost will be $7,500 per computer. Unfortunately,
there is also a 0.25 probability that the cost will be $8,500 per computer.
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