...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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...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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...espondence concerning this article should be addressed xxxxxx, care of Western Governors University. E-mail: xxxxx@wgu.edu QAT1 Task 2: Competency 309.3.1-03, 11, 12 Determine the equations for each of the three constraints. Nutrient C : 4x + 4y ≤ 30 Flavor A : 12x + 6y ≤ 72 Color : 6x + 15y ≤ 90 Identify each constraint as Minimum or Maximum. Nutrient C constraint is a Maximum Flavor A constraint is a Maximum Color is a Maximum Determine the total contribution to profit that lies on the Object Function as plotted on the graph. By visual observation the objective function line indicates shows Brand X = 3 and Brand Y = 4. Using these values and the objective function of 30y + 40x = 30*4 +40*3 = 240 Determine how many cases of each type should be produced to generate the greatest profit. Solving for the intersection of Color (4x + 4y = 30) and Nutrient (6x + 15y = 90) yields the values of Y = 5 case and X = 2.5 cases for a total of 7.5 Cases. Entering these values into the objective function: 30y + 40x = 30*5 + 40*2.5 = 250 Solving for the intersection of Flavor (12x + 6y = 72) and Color (4x + 4y = 30) yields the values of Y = 3 and X = 4.5 cases for a total of 7.5 cases. Entering these values into the objective function: 30y + 40x = 30*3 + 40*4.5 = 270 Solving for the intersection Flavor (12x + 6y = 72) and Nutrient (6x + 15y = 90) yields the values of Y = 4.5 and X =3.75 for a total of 8.25 cases. Entering these values into the objective function: ...
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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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