The Expected Useful Life of Nerverdeadi Batteries - Assignment Example

Name: Lecturer: Course name: Course code: Date: Q1a).the expected useful life of nerverdeadi batteries Mean life= 19.8 days Standard deviation = 5days X is the useful life of the batteries. Upper limit 5 batteries, lower limit -5 batteries Therefore, it takes 23 days for neverdeadi batteries to exist before it is replaced with another one under normal distribution. This is illustrated by the normal distribution curve above at the point of intersection. Standard deviation shows the variance or divergence of the data from the mean value. The above data shows a standard deviation of 5 days signifying that the accuracy of the Neverdeadi battery useful life is better and thus the results of the above normal distribution would give a 95% significance level on the accuracy of the expected useful life. Q1B) Purcell battery is exponentially distributed with a mean life of 47.7 days. Therefore the useful life of the battery would be. Let T be the time of the battery X, y, Z be the failure time of the three batteries. Hence the flash light fails at time T T = min(X, Y, Z). Anytime Purcell battery is working, its remaining useful life is the same as the minimum life time of the two batteries. By the deficiency in reminiscence property of the exponential distribution, the lifetime of the battery is equal to the minimum of two exponential arbitrary variables not in connection with the past life of the batteries. Hence, the outstanding lifetime of the battery is an exponential random variable having equal rate to the total of the rates of the two batteries. Mean life time of the battery is 47.7 days, then the rate would be (1/47.7*2batteries) = 20/477.therefore, the reciprocal value would be the mean of exponential random variable Mean random variable =47.7/2=23.85 days The succeeding time before the battery fails would be E (T) = 1*23.85days= 23.85 days 1c) for exponential distribution, Graph the function From the above graph analysis, it can be depicted that puracells battery have a useful life of 27 days as compared to neverdeadi battery which is normally distribute and has a useful of 23 days. This because parcels battery is exponentially distributed with a mean of 47.7 days as unlike the neverdeadi cells which is normal distributed with a mean of 19.8 days and a standard deviation of 5 days. 1c).if the number of batteries is increased to five days, then the useful life for puracells would be If the company increases the number of battery from 3 to 5 days, the number of useful life of the battery would increase to 125 days as compared to the previous 23 days. For neverdeadi cells, if the number of batteries are increased to 5 batteries. The number of days increases from 24 days to 25 days but quite below the puracells. Exponential distribution is a constant version of statistical distribution exclusive of recollection. The purpose of this is that, it is used to replica the instance for a progression to take place at a steady state. Normal distribution is a bell shape curve which is symmetric about the mean unlike graph of exponential distribution. Its range by standard deviation above and below the mean of the data set. Standard deviation determines how extreme the trial divergence from the mean is. While F(x) is the frequency with which a definite significance of X transpires in a population. Q2) In a spreadsheet, construct an input table for your analysis. input analysis. outcomes(X) Dry well Gas only Gas and oil Oil only Total probabilities(X) 0.4 0.4 0.3 0.1 1 A. returns 0 400 800 1600 2,800 E. returns 0 160 240 160 560 b) Construct a decision tree for this problem. -400 0.2NO oil/Gas Gas only 0.4 0 Oil 0.5 Gas and oil 0.3 -400 Oil only 800 1600 0.5 Drill no oil 0.5 No drill 0.5 The expected monetary value (EMV) for the four possible outcome would be. 1).No gases no oil. (Total revenue* total cost) joint probability =EMV 0- 400(0.5*0.2*0)-=$-40 million 2) Gas only EMV= 400-400(0.5*0.4*) =$0 (break even) 3) Gas and oil EMV=$800-400(0.5*0.3) = $60 million 4), oil only EMV= 1600-400(0.5*0.1=$60 million From the above analysis of the expected monetary values.WWW limited have an option of either investing on oil only or gas and only as the both outcomes gives the same monetary value expected of $60 million c).For any probability node(s) in your decision tree, show the details of the calculation of EMV at that node. At the probability node of 0.4 (gas only), the joint probability is (0.5*0.4=0.2); it is expected that revenue is 400, therefore. The expected monetary value is derived as follows. EMV= 400-400(0.5*0.4*) =$0 (break even). From the above analysis, a break even point is derived. Break even point is a position where the company will not earn profit or loss but it will be in a position to recover its drilling cost only. Therefore at this stage, the company should not consider this option due to the fact the expected monetary value on the project will be equal to zero. D1) a sensitivity analysis to investigate how variation in the fee for leasing the land to the competitor affects the optimal decision Sensitivity analysis is a practice which aids in establishing how a dissimilar principle of an autonomous variable affects specific dependent variable under a specified set of hypothesis. This performance is used within detailed a limitation which relies on greater than one input and thus Sensitivity analysis is used to ascertain whether a particular probability is vital in establishing an ideal decision. Lowest possible value =$60 million (Drill option) Highest possible value =$150milliion (lease option) Therefore, expected lowest value = (0.5*0.3) + (0.5*.1)*60 million =$12 million Highest value = (0.5*0.3) + (0.5*.1)*150,000 million = $30 million D 2) From the above sensitivity analysis test, the decision option with the highest value gives rise to a difference of ($30million – 12 million) = $8, million. This implies that the decisions to either invest on drilling for oil or leasing the land are sensitive to the value of the probability. E1). Perform a sensitivity analysis to investigate how variation in the estimated payoff values affects the optimal decision. The optimal decision is to invest on oil only which gives a highest monetary value of $60 million. Therefore the effects on estimated pay off would be Outcome Estimated Payoff Joint Probability Dry well (no oil or gas) $0 0.2*0.5=0.1 Gas only $400 million 0.4*0.5=0.2 Gas and oil $800 million 0.3*0.5=0.15 Oil only $1,600 million 0.1*0.5=0.05 A).Dry well (0.2*0.5)*60 million= $ 6 million B).Gas only (0.4*0.5)*60 million = $ 12 million C. Gas and oil (0.3*0.5)*60 million = $ 9 million D). Oil only (0.1*0.5)*60 million = 3 million. E2) from the above sensitivity analysis, the optimal decision is altered by the variation of the estimated pay-offs due to the effect of probabilities on different state of occurrence on the possibility node. It can be observed that the optimal expected monetary value (EMV) would be $12 million in which the company is suppose to drill for gas only due to the highest value generated as well as the outcome as the highest chances of occurrence depicted by the joint probability value. Therefore, sensitivity analysis is an important tool used in trying to understand the impact of the actual outcome of a particular variable. an analyst uses this tool in determining how changes in one variable will affect the targeted variable any necessary recommendation should be made in order to guarantee company sustenance as well as attaining maximum benefit with lowest cost. A good model of sensitivity analysis helps managers as well as decision makers to come up with a long-term strategy by evaluating sustainability for diverse circumstances of the expansion. Projects with high monetary values as projected should be considered based on the impact off the sensitivity analysis. Q3A.1) Q3A.2) Q3A2)from the above analysis, the waiting time is in control because all points is falling between the upper and lower control limit and thus the variation observed is only emanating from the sources frequent to the progression. This implies that the product is within the desired pricing strategies as well as performance. From the management point of view is that a good qualitative analysis on the product is put in place and a superior estimation and forecasting is enhanced. The effect is that the company will operate into a foreseeable future because the risk is well taken care of. (b) For each “other out-of-control-indication” (if there are any),state which “other out-of-control-indication” has occurred, A characteristic of out of control is depicted in the above data analysis. It can be envisaged that there is presence of 4 out of 5 points in zone B under the same side of the centre line or beyond. This therefore means that managers ought to pay close supervision on the product before the trend analysis becomes out of control.Also, from the above data the subsamples that are involved in the “other out-of-control-indication”. Are 21, 25, 31 and 33.this can be depicted from the above trend analysis in which their tendency are above the centre line under the same zone B. Q3c) what are the important aspects of the control charts that should be brought to the attention of management The management should consider control chart because some features such as under control and out of controls are key aspects of control limit that is used to observe the trend performance and any deviation that occurs. For example if there is existence of under control, then no modifications or alteration to the progression managed limits are preferred. Making close supervision on trend analysis on the product performance helps managers to understand product’s performance and the market and any deviation can be taken care off immediately hence making the company a price leader in the market as well as the leading company. control charts can be used by managers to forecast future trend performance of the progression in that the variation origin of the process can be depicted the by the chart control and thus management can help retain the performance in the desired echelon and not to degrade thus making the company not to run into insolvency but be in a position to analyze the future needs of the customers and the product. The control chart is an instrument of quality control that is used for time-series analysis a, or making a comparison of samples which was taken on the same time or performance of specific individuals. This analysis tool will aid managers in making critical decision on a project with similar useful life and hence managers are freed from making poor decision on a project that might cause the company a huge loose. Also quality and superior decision making is achieved with the aids of a chart controls because figures displayed clear shows the reasons for project deviation either below or upper the expected limit and gives a recommendation. This measure will guarantee the company a successive performance and growth as well. Q3.Wd) Discuss any further points which can be made about the waiting time data in Steak Out.xlsx Audits analysis. For example, discuss any features, limitations or further information. Data of waiting time shows a wide range of time between tables. This implies that some customers will be oppressed with the service times hence to reduce this effect; WWW limited should consider reducing the time length of waiting by considering the effect of the standard deviations. Capability ratio appraises the expectations of the consumer with the practice of normal deviation performances. This is assessed using the standard deviation with either +3 or -3 off the process mean. Reference 1. Gopalakrishnan, N. In Simplified Six Sigma : Methodology, Tools and Implementation (p. PPG 287). 2. Kim Heldman, V. M. (2009). In PMP Project Management Professional Exam Review Guide (p. PPG 139). 3. M. H. Hamza, I. A. (1994). In Applied Identification, Modelling, and Simulation: AIMS '84 (p. PPG 82). Read More