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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