...Task 1 Simulation Trials QUANTITATIVE ANALYSIS Competency 309.3.1: Decision-Making Models - The graduate uses optimizing models and other models as aids for making more informed decisions. ________________________________________ Introduction: In this task, you are asked to conduct a small number of simulation trials. This task should be completed with manual calculations. Please be aware that the simulation results of such a small number of trials would not be sufficient for one to draw valid conclusions in a real situation. A large number of trials run on a computer would be necessary in order to arrive at a valid conclusion. Random numbers are provided in the attached template; these numbers are discrete, uniform, and between 1 and 100 inclusive. Scenario: Management has asked you to help estimate the average cost per unit to produce a new product so that they can project profits at different selling prices. You will conduct a Monte Carlo simulation for cost behavior using provided probabilistic data for the following cost estimates, which are provided in the attached worksheet: cost per unit for materials, labor, and utilities and their probability distributions. In your calculations, round to the nearest cent. Requirements: Note: When the files opens, enter your name and student ID as they appear in Taskstream into the respective boxes so the task questions and problems will populate the worksheet. If the attached template does not open or is missing...
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...QAT1 Task 5 Revised Task A. Develop New Product 1.) Develop Thoroughly= $210,200 2.) Develop Rapidly= $55,700 Consolidate Existing Product 3.) Strengthen Products= $64,900 4.) Reap Without Investing= $6,400 Task B. The decision alternative is to develop new products thoroughly, or decision alternative 1, with the expected value (EV) of $210, 200. 1. Decision alternative 1 has the highest expected value ($210,200) of all four branches, thus making it the most favorable decision. To find expected values for each decision alternative, first you multiply the Predicted Gains (or, Payoffs) by the Probability for each States of Nature (in this case, Market Reactions: Good, Moderate, and Poor). Then, you add the totals of each Market Reaction and that becomes the expected value for each decision alternative. The highest expected value from all the decisions is generally the most favorable decision. The following are the calculations for the expected values of each of the four decision alternatives: The formula is: EV= Predicted Gains(Probability) Decision alternative 1: Good: 500,000(.4) = 200,000 Moderate: 25,000(.4) = 10,000 Poor: 1,000(.2) = 200 Total: $210,200 Decision alternative 2: Good: 500,000(.1) = 50,000 Moderate: 25,000(.2) = 5,000 Poor: 1,000(.7) = 700 Total: $55,700 Decision alternative 3: Good: 200,000(.3) = 60,000 Moderate: 10,000(.4) = 4,000 Poor: 3,000(.3) = 900 Total: $64,900 Decision...
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...QAT1 Task 5 #258453 In the given scenario, a company is considering alternatives for improving profits by either developing new products, or consolidating existing products. There are 4 separate branches that split from the 2 main branches. Develop new product: 1). Develop thoroughly: a) Good demand .47 $500,000 b) Moderate demand .38 $25,000 c) Poor demand .15 $1000 So, with the above given variables, to calculate the expected value, you multiply each probability times the corresponding payoff. Then add the results for each decision outcome. Calculations: a) .47 (500,000) = $235,000 b) .38 (25000) = $9500 c) .15 (1000) = $150 Branch 1 expected value = $244,650. 2). Develop rapidly: a) Good demand .06 $500,000 b) Moderate demand .16 $25000 c) Poor demand .78 $1000 Calculations: a) .06 (500,000) = $30000 b) .16 (25,000) = $4000 c) .78 (1,000) = $780 Branch 2 expected value = $34780 Consolidate existing product: 3). Strengthen products: a) Good demand .69 $2,000 b) Moderate demand .27 $10,000 c) Poor demand .04 $3,000 Calculations: a) .69 (2,000) = $1380 b) .27 (10,000) = $2700 c) .04 (3,000) = $120 Branch 3 expected value = $4200 4). Reap without investing: a) Good demand .32 $10,000 b) Poor demand ...
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...Quantitative Analysis for Business (QAT1) Submitted 05/05/2015 Assignment 309.3.3-04 Version LMF5-28 Student: Richard McClanahan Student ID: 000343792 TASK #5 Answer Task 5A Calculate the expected value for EACH of the four decision branches. 1. Develop Thoroughly: GOOD) $500,000 (0.45) = $225,000 MOD.) $25,000 (0.10) = $2,500 POOR) $1,000 (0.45) = $450 TOTAL EXPECTED VALUE: $227,950 2. Develop Rapidly: GOOD) $500,000 (0.52) = $260,000 MOD) $25,000 (0.23) =$5,700 POOR) $1,000 (0.25) =$250 TOTAL EXPECTED VALUE: $265,950 3. Strengthen Products GOOD) $2,000 (0.33) = $660 MOD) $10,000 (0.52) = $5,200 POOR) $3,000 (0.15) = $450 TOTAL EXPECTED VALUE: $6,310 4. Reap without investing GOOD) $10,000 (0.33) = $3,300 POOR) $1,000 (0.67) = $670 TOTAL EXPECTED VALUE: $3,970 EXPLINATION: We take the projected payoff and multiply that payoff by the probability factor. So if the good payoff to develop a product rapidly is $500,000, we then multiply that by the probability factor of 52%, or 0.52. That gives us a probable payoff of $260,000. Following this simply process, we extrapolate these results as listed above. ANSWER TASK 5B After calculating the total expected value for each decision alternative, the most profitable decision would be to RAPIDLY DEVELOP new products for a probable...
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...Jane Doe 02/10/2014 QAT1 309.3.2-01-06 A. See table 1.1 attachment B. 1. Expected time to complete: * a= the optimistic completion time estimate * b= the pessimistic completion time estimate * m= the most likely completion time estimate The activities mean completion time is: t= (a+4m+b)/6 * Task A. (2+12+4)/6=3 * Task B. (5+24+13)/6=7 * Task C. (3+16+8)/6=4.5 * Task D. (10+44+15)/6=11.5 * Task E. (4+20+6)/6=5 * Task F. (8+40+12)/6=10 * Task G. (4+24+11)/6=6.5 * Task H. (8+40+18)/6=11 * Task I. (3+24+12)/6=6.5 * Task J. (2+12+7)/6=3.5 a. The activities completion time variance is: ²= ((b-a)/6)² * Task A. ((4-2)/6)²=0.10 * Task B. ((13-5)/6)²=1.76 * Task C. ((8-3)/6)²=0.68 * Task D. ((15-10)/6)²=0.68 * Task E. ((6-4)/6)²=0.10 * Task F. ((12-8)/6)²=0.43 * Task G. ((11-4)/6)²=1.34 * Task H. ((18-8)/6)²=2.75 * Task I. ((12-3)/6)²=2.25 * Task J. ((7-2)/6)²=0.68 2. See Network Diagram attachment 3. a. Expected duration of the entire project: 33.5 weeks b. Slack for project task A: LS-ES=slack time = 6.5-0= 6 c. Slack for project task H: LS-ES=slack time= 19-18.5= 0.5 weeks d. The week project task F. is scheduled to start: Task F begins when Task B ends, which= week 7 e. The week project task I. is scheduled to finish: Task J. LS= task I. LF time=week 30 4. Probability that the project will be done in 34 weeks: ²=²b+²d+²h+²j= ²=1.76+.43+1...
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