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Cen4017 Risk Management In Projects Assessment Answers

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)?

Answer:

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.


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