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Questions:
Suppose you own an expensive car and purchase auto insurance. This insurance has a \$1000 deductible, so that if you have an accident and the damage is less than \$1000, you pay for it out of your pocket. However, if the damage is greater than \$1000, you pay the first \$1000 and the insurance pays the rest. In the current year there is probability 0.025 that you will have an accident. If you have an accident, the damage amount is normally distributed with mean \$3000 and standard deviation \$750.

1.Use Excel to simulate the amount you have to pay for damages to your car. This should be a one-line simulation, so run 5000 iterations by copying it down. Then find the average amount you pay, the standard deviation of the amounts you pay, and a 95% confidence interval for the average amount you pay. (Note that many of the amounts you pay will be 0 because you have no accidents).

2.Continue, the simulation by creating a two-way data table, where the row input is the deductible amount, varied from \$500 to \$2000 in multiples of \$500. Now find the average amount you pay, the standard deviation of the amounts you pay, and a 95% confidence internal for the average amount you pay for each deductible amount.

3.Do you think it is reasonable to assume that damage amounts are normally distributed? What would you criticize about this assumption? What might you suggest instead?

4.Suppose instead that the damage amount is triangularly distributed with parameters 500,1500, and 7000. That is, the damage in an accident can be as low as \$500, and there is definite skewness to the right. (It turns out, as you an verify in @RISK, that the mean of this distribution is \$3000). Use @RISK to simulate the amount you pay for damage. Run 5000 iterations, then answer the following questions. In each case, explain how the indicated event would occur.

a.What is the probability that you pay a positive amount but less than \$750?
b.What is the probability that you pay more than \$600?
c.What is the probability that you pay exactly \$1000(the deductible)?

According to the given information, The Pulsometer Pump Company has three alternative options and can anticipate three different market situations. For each of these combinations, they have already identified expected profit. Hence, the payoff matrix criteria has been applied to decide which strategy needs to be applied by them.

The payoff calculations have been done as mentioned below:

 Maximax Criteria Market Factors Strategy 15% Stable -10% Max S1 240 130 0 240 S2 210 150 70 210 S3 170 150 70 170 Maximax 240 Decision S1 strategy needs to be applied

 Maximin Criteria Market Factors Strategy 15% Stable -10% Min S1 240 130 0 0 S2 210 150 70 70 S3 170 150 70 70 Maximin 70 Decision Either S1 or S2 needs to be applied

 Minimax Criteria Market Factors Strategy 15% Stable -10% Max S1 0 20 70 70 S2 30 0 0 30 S3 70 0 0 70 Minimax 30 Decision S2 needs to be applied
 15% 0.6 £ 2,40,000.00 S1 Strategy Stable EMV £ 1,83,000.00 0.3 £ 1,30,000.00 -10% 0.1 0 15% 0.6 £ 2,10,000.00 Pumping Equipment S2 Strategy Stable Production EMV £ 1,78,000.00 0.3 £ 1,50,000.00 -10% 0.1 £    70,000.00 15% 0.6 £ 1,70,000.00 S3 Strategy Stable EMV £ 1,54,000.00 0.3 £ 1,50,000.00 -10% 0.1 £    70,000.00

As per this decision tree and EMV calculations, it can be said that strategy 1 will be the feasible option to attain maximum profit. This method has given confidence to the management, however, they wants further investigation and thus a third party consultation service option has been assessed though below calculations.

 EMV as per consultation Report Strategy EMV Consultancy Fees Net Value S1 £ 1,08,600.00 £               5,000.00 £ 1,03,600.00 S2 £    53,600.00 £               5,000.00 £    48,600.00 S3 £    24,100.00 £               5,000.00 £    19,100.00

The above table has shown that even after considering third party consultation service, the option will remain the same, that is, they should go with strategy 1. Hence, it is not necessary to go for consultation service.

Finally, they wants to apply bays theorem option to ensure the decision making process. Accordingly posterior probability of each of the strategy has been calculated and basis these probabilities, EMV has been calculated:

 EMV as per bays theorem Strategy EMV Consultancy Fees Net Value S1 £          2,54,464.29 £               5,000.00 £ 2,49,464.29 S2 £          1,32,194.55 £               5,000.00 £ 1,27,194.55 S3 £             38,633.86 £               5,000.00 £    33,633.86

This also confirms the above decision and thus it can be concluded that strategy 1 should be chosen as final decision.

According to the given information, the insurance will be applicable if the car experienced and accident and the damages crossed \$1000 amount. In such case beyond the \$1000, everything will be covered by the insurance authority. Now, there is a probability of 0.025 that accident will happen.

Here, first of all excel simulation has been performed considering 5000 iterations and the average amount you pay, the standard deviation of the amounts you pay, and a 95% confidence interval for the average amount you pay are calculated as mentioned below:

 Details Average Amount Pay \$   48.28 Standard Deviation of Pay \$329.64 Confidence Interval 0.95 Standard Error \$     0.07 Lower Limit \$   39.15 Upper Limit \$   57.42

The above table has shown that the average amount pay has to incur is \$48.28 and it will remain in between \$39.15 and \$57.42

Now, as per requirement, further study has been done considering changes in deductible amount, varied from \$500 to \$2000 in multiples of \$500. A two way table has been shown the implication of such changes on the average amount you pay, the standard deviation of the amounts needs to be paid, and a 95% confidence internal for the average amount needs to be paid for each deductible amount. It is shown as mentioned below:

 Average Amount Pay Standard Deviation of Pay Lower Limit Upper Limit \$48.28 \$329.64 \$39.15 \$57.42 \$500.00 \$63.69 \$416.13 \$52.15 \$75.22 \$1,000.00 \$50.38 \$331.49 \$41.19 \$59.56 \$1,500.00 \$40.18 \$268.98 \$32.73 \$47.64 \$2,000.00 \$28.38 \$207.98 \$22.62 \$34.15

Considering damages amount is normally distributed is not always holds true. The reason is that there is a significant amount of probabilities involved behind the extent to which the accident can take place.

Now, the information has shown that instead of normal distribution, if damages due accident follows triangular distribution with value, \$500, \$1500 and \$7000, then using @RISK software, following result can be found:

 Insurance Deductible Amount \$1,000.00 Probability of Accident 0.025 Average damages \$3,000.00 Stdev of damages \$    750.00 Damages [Triangular Distribution] \$2,908.23 Amount payable \$1,908.23 P(probability<\$750) 0.200 P(probability>\$600) 0.852 P(probability=\$1000) 0.699

From the above payoff calculations, it has found that three different options are highlighted applying three different payoff matrix techniques. Hence, the organisation further applies probability aspects associated with each three strategies and performed EMV calculations (Schwalbe, 2015). Further these EMVs have been shown using decision tree as mentioned bellow.

As per this table, it can be said that the probability of damages less than \$750 is 0.200. Similarly, the probability that the damages will be more than \$600 is 0.852 and the probability that damages amount will be exactly \$1000 is 0.699.

1. Calculation of Expected Time and Mean Time
 Activity No Task Name Optimistic Duration Most Likely Duration Pessimistic Duration Immediate Predecessor Expected Time Variance 1 Team meeting 0.5 1 1.5 1 0.027778 2 Hire Contractors 6 7 8 1 7 0.111111 3 Network Design 12 14 16 1 14 0.444444 4 Order Ventilation system 18 21 30 1 22 4 5 Install Ventilation system 5 7 9 4 7 0.444444 6 Order new racks 13 14 21 1 15 1.777778 7 Install racks 17 21 25 6 21 1.777778 8 Order power supplies and cables 6 7 8 1 7 0.111111 9 Install power supplies 5 5 11 8,12 6 1 10 Install cables 6 8 10 8,12 8 0.444444 11 Renovation of data centre 19 20 27 2,3 21 1.777778 12 City inspection 1 2 3 2,5,7 2 0.111111 13 Facilities 7 8 9 10 8 0.111111 14 Operations/System 5 7 9 10 7 0.444444 15 Operations/Telecommunications 6 7 8 10 7 0.111111 16 System & applications 7 7 13 10 8 1 17 Customer service 5 6 13 10 7 1.777778 18 Power check 0.5 1 1.5 9,10,11 1 0.027778 19 Install test servers 5 7 9 1,2,13,14,15,16 7 0.444444 20 Management safety check 1 2 3 5,18,19 2 0.111111 21 Primary systems check 1.5 2 2.5 20 2 0.027778 22 Set date for move 1 1 1 21 1 0 23 Complete move 1 2 3 22 2 0.111111

Network Diagram

Calculation of Variance and SD

 Activity No Optimistic Duration Most Likely Duration Pessimistic Duration Var SD (a) (b) (c) V=((c-a)/6)^2 Sqrt(V) 1 0.5 1 1.5 0.028 0.167 2 6 7 8 0.111 0.333 3 12 14 16 0.444 0.667 4 18 21 30 4.000 2.000 5 5 7 9 0.444 0.667 6 13 14 21 1.778 1.333 7 17 21 25 1.778 1.333 8 6 7 8 0.111 0.333 9 5 5 11 1.000 1.000 10 6 8 10 0.444 0.667 11 19 20 27 1.778 1.333 12 1 2 3 0.111 0.333 13 7 8 9 0.111 0.333 14 5 7 9 0.444 0.667 15 6 7 8 0.111 0.333 16 7 7 13 1.000 1.000 17 5 6 13 1.778 1.333 18 0.5 1 1.5 0.028 0.167 19 5 7 9 0.444 0.667 20 1 2 3 0.111 0.333 21 1.5 2 2.5 0.028 0.167 22 1 1 1 0.000 0.000 23 1 2 3 0.111 0.333

Probability Calculation

Expected time for project completion is T (mean) = 69 (from critical path analysis)

Variance (V) = 5.833

SD = 2.415

Z= -1/2.415

Since, Z= -0.4140

Therefore, applying normal distribution function for Z,

Probability= 33.94%

1. Measurement of Project Risk Exposure

Risk assessment is a crucial method of identification and analysis of the factors of risk in a project (Kerzner and Kerzner, 2017). The risk monitoring tools have been helpful for forming the expectation of the risk factors in project.

Project Risk Exposure can be mitigated by the use of the high level of risk mitigation and planning. The risk management and analysis is done for taking care of the impact of risk in projects for the deployment of the effective mitigation strategies in project (Harrison and Lock, 2017).

1. Risk Identification can be implied for identifying the factors of risk favouring the alignment of successful risk mitigation (Nicholas and Steyn, 2017)
2. Risk Impact Assessment is helpful for analysing the consequences of cost, schedule and technical performance of project activities
3. Risk prioritization is done for analysing the impact of risk in the project for developing effective implication of the operations
Risk methodology

In order to replicate the models designed in research paper 2, here @RISK software is used. First of all, the network diagram is designed by excel and presented in the above section. Subsequently, to find out the risk factor, triangular and uniform distribution has been used (Salvatore and Brooker, 2015).

Compare results with research paper 2:

The below mentioned table has shown that the risk factors for each of the activity is greater than what mentioned in the requirement file.

 Name Weather Soil Productivity Equipment Delay of materials Description Output Output Output Output Output Iteration / Cell \$D\$2 \$D\$3 \$D\$4 \$D\$5 \$D\$6 Act 2 0.429720627 0.73817623 0.473555815 0.452330268 0.612962158 Act 3 0.654646714 0.67283892 0.637431347 0.891686636 0.49652601 Act 4 0.799126109 0.604221584 0.162661848 0.058870516 0.581955774 Act 5 0.731424743 0.937093848 0.521483078 0.585388118 0.47251929 Act 6 0.557872187 0.430008462 0.32176467 0.222236455 0.738026223 Act 7 0.610947454 0.203394483 0.730326802 0.912389271 0.410686392 Act 8 0.833723846 0.877287282 0.401538833 0.732767695 0.826364793 Act 9 0.927478614 0.342713985 0.589362036 0.642357412 0.136475043 Act 10 0.316579455 0.574692097 0.666757944 0.35063763 0.663537465 Act 11 0.123153871 0.806614414 0.95355877 0.12110178 0.250617032 Mean 0.598467362 0.618704131 0.545844114 0.496976578 0.518967018 SD 0.236718107 0.225201513 0.211860946 0.290692737 0.202208091

Change in distribution of weather

In case of change in distribution for weather to uniform distribution from triangular distribution, the risk factor will remain identical irrespective of iteration (Donegan, 2016).

Sensitivity analysis:

The sensitivity analysis has shown that all risk factors are significantly positively associated. It means, it will influence the overall duration by same rate by which all aspects are increased (K?ivan and Cressman, 2017).

References:

Donegan, H.A., 2016. Decision analysis. In SFPE handbook of fire protection engineering (pp. 3048-3072). Springer, New York, NY.

Harrison, F. and Lock, D., 2017. Advanced project management: a structured approach. Routledge.

Kerzner, H. and Kerzner, H.R., 2017. Project management: a systems approach to planning, scheduling, and controlling. John Wiley & Sons.

K?ivan, V. and Cressman, R., 2017. Interaction times change evolutionary outcomes: Two-player matrix games. Journal of theoretical biology, 416, pp.199-207.

Nicholas, J.M. and Steyn, H., 2017. Project management for engineering, business and technology. Taylor & Francis.

Salvatore, D. and Brooker, R.F., 2015. Managerial economics in a global economy. Oxford University Press.

Schwalbe, K., 2015. Information technology project management. Cengage Learning.

Cite This Work

My Assignment Help. (2020). Optimizing Auto Insurance Deductible: A Simulation Study. Retrieved from https://myassignmenthelp.com/free-samples/cen4017-risk-management-in-projects/standard-deviation.html.

"Optimizing Auto Insurance Deductible: A Simulation Study." My Assignment Help, 2020, https://myassignmenthelp.com/free-samples/cen4017-risk-management-in-projects/standard-deviation.html.

My Assignment Help (2020) Optimizing Auto Insurance Deductible: A Simulation Study [Online]. Available from: https://myassignmenthelp.com/free-samples/cen4017-risk-management-in-projects/standard-deviation.html
[Accessed 07 August 2024].

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My Assignment Help. Optimizing Auto Insurance Deductible: A Simulation Study [Internet]. My Assignment Help. 2020 [cited 07 August 2024]. Available from: https://myassignmenthelp.com/free-samples/cen4017-risk-management-in-projects/standard-deviation.html.

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