Cen4017 Risk Management In Projects Assessment Answers
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.
- 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%
- 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
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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