...If a business wants to determine the optimale replenishment lot size, the use of the Economic Production Lot Model can be used. This model is a variation of the EOQ Model. This model procides the most optimied approach of ordering as it considers, demand, available production, ordering cost, set up costs, and holding costs in order to develop the inventory to be ordered to maintain a minimum annual cost (Rendoer 2012). Company B Details Holding Rate: 6% HR: Holding Rate Demand Units Per Year: 15,470,000 D: Demand Expected Production: 910,000 EP: Expected Production Production Cost Per Unit: $400.00 UPC: Unit Production Cost Inventory Cost Per Unit: $200.00 ICPU: Inventory Cost Per Unit *Holding Cost: $12.00 HC: Holding Cost * The holding cost is the per unit cost of inventory multiplied by the holding cost rate $200(.06) = $12.00 Process Description: The optimal lot size per order is equal to the square root of the equation of the annual demand (D). It is multiplied by 2, then multiplied by the production set up cost (UPC). Which is then divided by the holding cost (HC). Then demand (D) divided by epected production (EP), then subtracted by 1. The end result of the equation provides the optimal lot size order. The Equation: You must follow the mathmatical order of operations; simplify and solve - numerator divided by the demoniator...
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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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...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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