Quantitative Analysis of Business - Coursework Example

? Editing work for Quantitative Analysis for Business Affiliation with more information about affiliation, research grants, conflict of interest and how to contact Editing Coursework for Quantitative Analysis of Business This paper is based on the quantitative analysis of business. Four tasks have been taken into consideration. Quantitative analysis of business is very important for making decisions related to management. “Using quantitative analysis, managers often build statistical or mathematical models for entering data relating to business opportunities and goal setting/attaining” (Limitations of Using Quantitative Methods in Business Decision Making, 2011). TASK 1 Material costs will range between $33.00 and $39.00 per unit, labor costs between $22.00 and $28.00 per unit and utility costs between $3.00 and $6.00 per unit. According to the probabilistic distributions given, the earlier mentioned three costs have been calculated below along with the total cost and the average cost: Materials Labor Utilities Total cost Trial Random no. Cost p.u. Random no. Cost p.u Random no. Cost p.u 1 28 35 21 25 85 4 64 2 14 38 50 24 64 4 66 3 44 35 90 25 15 4 64 4 51 38 71 28 66 6 72 5 91 34 10 23 45 4 61 6 38 38 11 22 74 6 66 7 79 35 70 24 62 3 62 8 70 38 16 27 18 4 69 9 98 38 88 22 74 3 63 10 11 38 67 25 35 6 69 Total Avg. 367 245 44 656 Avg. 36.7 24.5 4.4 65.6 A1. Average materials cost per unit is $36.7 (Highest material cost is $39.00 and the lowest cost is $33.00.)  10 random numbers are picked in the range 33-38. Then it is checked if there are any equal terms. 34 - 1 35 – 3 38 – 6 The numbers are repeated 10 times. The fraction of trails matching point out avg. material cost is 36.7 A 2. Average labor cost per unit is $24.50 (Highest cost is found to be $26.00 and the lowest cost $22.00.) A 3. Average utilities cost per unit is $4.40. (Highest cost is $6.00 while the lowest cost is $3.00) A 4. Average total cost per unit is $65.60 B. For the company to realize an average markup of $20 on each unit implies that the difference between production cost and selling price must be $20. This shows that using the simulation results, the selling price is found to be $85.60.) Task: 2 A. Determine the equations for each of the three constraints that are plotted on the attached “graph 1”, showing all work necessary to arrive at the equations. 1. Identify each constraint as a minimum or maximum constraint. The objective function is Z= 30X+72Y+90 Subject to 7.5X + 7.5Y ? 30 (equation for Nutrient C) ------- 1 6X + 12Y ? 72 (equation for Flavour Additive) ------- 2 15X + 6Y ? 90 (Equation for Color Additive) ------- 3 X ? 0, Y ? 0 Since the feasible region is below the constraints the constraints are minimum constraints. B. Determine the total contribution to profit, if the company produces a combination of cases of brand X and brand Y that lies on the purple objective function (profit line) as it is plotted on the attached “graph 1”. If the company chooses to produce a combination of brand X and Y as given in graph then the different combinations would be (0, 8), (1, 6.6), (2, 5.4), (3, 4), (4, 2.6), (5, 1.3), (6, 0). The contribution to profit at various combinations can be obtained using the objective function where profit= 30X+72Y+90 and substituting the value of X and Y for each set in this function we get profit for each combination. Combination of Brand X & Y Profit = 30X+72Y+90 0,8 666 1, 6.6 595.2 2, 5.4 538.8 3, 4 468 4, 2.6 397.2 5, 1.3 333.6 6,0 270 When the company produces 8 units of Brand Y and no Brand X the profit function is maximized (666). But this combination is outside the feasible region. So, the combination that gives maximum profit (468) to the producer within the optimal region is 3 units of X and 4 units of Y. The Total Contribution of Profit: The total contribution to profit which can be obtained by producing 3 units of brand X and 4 units of brand Y is 468 which is obtained by putting the values of X and Y in the profit function. So the total contribution of profit = 468. C. Determine, with necessary calculations, how many cases each of Brand X and Brand Y you recommend should be produced during each production period for optimum production, if company A wants to generate the greatest amount of profit. In my opinion company should produce a combination of 6 units of Brand X and 6 units of Brand Y. Any combination between this will yield a profit between $ 270 and $ 522 which lies within the feasible region and this is best suited for any producer since any producer will choose a combination within the feasible region. The combination can vary in between this which can be as follows: (0, 6), (1, 5), (2, 4), (3, 3), (4, 2), (5, 1) and (6, 0). This combination is chosen by using the end points of feasible region. Constraints Combination of X & Y Value of Z= 30X+72Y+90 7.5X + 7.5Y ? 30 (0,7.5), (7.5, 0) 630 & 315 6X + 12Y ? 72 (0, 12), (6, 0) 954 & 270 15X + 6Y ? 90 (0, 6), (15, 0) 522 & 540 6X + 6Y ? 36 (0, 6) (6, 0) 270 & 522 The optimum production of brand X and Y is obtained by joining the two ultimate points of the feasible region. Thus a new constraint is formed which gives the optimal production function with two extreme values of value 270 & 522 which is obtained by a combination of (0, 6), (6, 0). D. Determine the total contribution to profit that would be generated by the production level you recommend in part C and justify your answer. The total contribution to profit that is generated by the production of 6 units of Brand X and 6 units of Brand Y is given in the following table: Combination of Brand X & Y Value of Z= 30X+72Y+90 0,6 522 1, 5 480 2, 4 438 3, 3 396 4, 2 354 5, 1 312 6,0 270 The total contribution to profit using the optimum production function 6X+6Y? 36 ranges between 270 and 522. Task Four 1. PERT/CPM Analysis - Task Detail Table 1.1 Task Preceding Activity Optimistic Time to Complete (weeks) Probable Time to Complete (weeks) Pessimistic Time to Complete (weeks) Expected Time to Complete (weeks) Variance (weeks) START           A START 2 3 4 3 0.11 B START 5 6 13 7 1.77 C A 3 4 8 4.5 0.69 D B 10 11 15 11.5 0.69 E C 4 5 6 5 0.11 F B 8 10 12 10 0.44 G F 4 6 11 6.5 1.36 H D,E 8 10 18 11 2.77 I G 3 6 12 6.5 2.25 J H,I 2 3 7 3.5 0.69 END         68.5   2. PERT chart 3. a) The expected duration of the entire project is 33.5 weeks. This is obtained by summing up the duration of critical activities of the company. Expected duration = 7+10+6.5+6.5+3.5 = 33.5 weeks b) Slack of an event is calculated by finding out the difference between the earliest event time and the latest event time. Therefore, slack for project task A = L2- E2 = 9.5-3 = 6.5 c) Slack for project task H = L9 –E9 = 30-29.5 = 0.5 d) The week project task F is scheduled to start in week 7, which is the earliest start time for task F. It should also be noted that the earliest start time of a task would be equal to the earliest event time associated with the tail event of that task. e) The week project task I am scheduled to finish in week 30, which is the earliest finish time for task I. The earliest finish time is equal to earliest start time added with the activity duration. 4. Probability of completing the project is seen to be 34 weeks. The probability can be calculated by using the formula Z = t-te/? The variance of the critical path= 1.77+0.44+1.36+2.25+0.69 = 6.51 = ?6.51 = 2.551 Z = t-te/? Z = 34-33.5/2.551 = 0.5/2.551 = 0.196 When Z is positive, probability = 0.5+table value When Z= 0.196, the table value is 0.07535 So, probability = 0.5+0.07535 = 0.57535 Therefore probability of completing the project in 34 weeks is 0.57535. Therefore there is a 57.53% probability to complete the project in 34 weeks. B. 1. Maximum reduction in time = expected time – crash time 2. Crash cost per week = crash cost – normal cost /normal time – crash time Task Expected Complete Time   Normal Cost Crash Cost Maximum Reduction in Time Crash Cost per week Normal (weeks) Crash (weeks) START             A 3 2 $ 8,400 $ 11,200 1 $ 2,800 B 7 5 $ 28,000 $ 40,000 2 $ 6,000 C 4.5 3 $ 18,000 $ 24,000 1.5 $ 4,000 D 11.5 7 $ 36,800 $ 44,800 4.5 $ 1,777.7 E 5 3 $ 14,000 $ 16,800 2 $ 1,400 F 10 6 $ 20,000 $ 24,000 4 $ 1,000 G 6.5 5 $ 18,200 $ 28,000 1.5 $ 6,533.3 H 11 8 $ 44,000 $ 64,000 3 $ 6,666.6 I 6.5 4.5 $ 26,000 $ 36,000 2 $ 5,000 J 3.5 3 $ 21,000 $ 36,000 0.5 $ 30,000 END 68.5           C. 1. It is always better to crash earlier activities on the critical path. So the activity to be crashed in order to complete the project in 30 weeks is F, which is on the critical path. 2. F should be crashed by 3.5 weeks as the crash cost of task F is the lowest. 3. Additional crashing cost of activity F is $3500. Task 5: A. Expected Value for Each Decision: Expected Value is the method to join probabilities and payoffs for both nodes. EV is a way to calculate the comparative qualities of decision options. One needs to compute the expected values after every payoff and, therefore, probabilities values are recognized. Following is the calculation of Expected Value: Developed thoroughly (Good) 0.4?500000 = 200000 Developed thoroughly (Moderate) 0.4?25,000 = 10000 Developed thoroughly (Poor) 0.2 ?1000 = 200 SO, EV (Developed thoroughly) = 0.4?500000+0.4?25,000+0.2 ?1000 = 210200 Developed rapidly (Good) 0.1? 500000 = 50000 Developed rapidly (Moderate) 0.2 ? 25000 = 5000 Developed rapidly (Poor) 0.7?1000 = 700 SO, EV (Developed rapidly) = 0.1? 500000+0.2 ? 25000+0.7?1000 = 55700. Strengthen product (good) 0.3?200000 = 60000 Strengthen product (Moderate) 0.4?10000 = 4000 Strengthen product (Poor) 0.3?3000 = 900 SO, EV (Strengthen product) = 0.3?200000+0.4?10000+0.3?3000 = 64900. Reap without investing (Moderate) 0.6?10000 = 6000 Reap without investing (Poor) 0.4?1000 = 400 SO, EV (Reap without investing) = 0.6?10000+0.4?1000= 6400. The values are obtained by considering probability in the decision tree diagram and multiplying it with the values in accordance with the decision option. This value indicates the expected value for the decision branch. In each decision branch, there are various options and adding all the given values gives one overall expected value. The highest EV is for the strategy developed thoroughly and it is 210200. B. The decision alternative that has the most favorable total expected value is through developing new products thoroughly. This is obtained by taking all expected values and adding them. They give the highest value as compared to the rest. The expected value obtained for developing new products is $265,900. This value is calculated by adding up the expected values obtained for the two decisions of developing thoroughly and developing rapidly. The expected value obtained for consolidation of existing products is $71,300 This value is derived by adding the expected values obtained for the options of strengthening and reaping without investing. The value for development of new products was higher than the consolidation of existing products and, therefore, the decision to take the option is arrived at. Additionally, the decisions for the consolidation of the existing products seemed not to be viable due to the given options, which are not suitable considering a business arena. One needs to invest what they get to gain more value. When introducing a new product in the market, it is very difficult to determine the success of product, cost of product etc. There are a lot of challenges that companies face nowadays in the development of new products. These include competition from already existing products, insufficient time to make new designs, better value and the need for stable price reductions so as to keep up with the demands of consumers. When the product is freshly brought into the market, the company has to incur a lot of costs for its promotion process. It is a difficult task for a new product to have market share. In the case of existing products, it will have good demand and market share. So it need not undertake so many advertisement or other promotional programs. Due to these reasons, it is a better choice for the company to upgrade existing products rather than innovating and taking risks. Reference List Limitations of Using Quantitative Methods in Business Decision Making, (2011). Walden University. Retrieved Sep. 28, 2011, from thinkup.waldenu.edu/...decision-making/.../11851-limitation-of-using-quantitative-methods-business-decision-making Read More