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.
 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=((ca)/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%
 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).
 Risk Identification can be implied for identifying the factors of risk favouring the alignment of successful risk mitigation (Nicholas and Steyn, 2017)
 Risk Impact Assessment is helpful for analysing the consequences of cost, schedule and technical performance of project activities
 Risk prioritization is done for analysing the impact of risk in the project for developing effective implication of the operations
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 
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).
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).
Donegan, H.A., 2016. Decision analysis. In SFPE handbook of fire protection engineering (pp. 30483072). 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: Twoplayer matrix games. Journal of theoretical biology, 416, pp.199207.
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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