exam paper in (tpk5115) project risk...

Department of Mechanical and Industrial Engineering

Examination paper for TPK5115 - Project Risk Management

Academic contact during examination: Jørn VatnPhone: 414 73 730

Examination date: December, 3rd 2019Examination time (from-to): 1500 - 1900Permitted examination support material: C: “Formulas in Project Risk” and a specific

basic calculator is allowed

Other information:All students bring with them “Formulas in Project Risk” which contains 16 pages and shall be printed with double sided print. The students are allowed to make notes in the “Formulas in Project Risk” and on its cover. For each problem the answers should be clearly separated from the intermediate result required to derive the answers. Always use double underlining for every numeric answer asked for. When it is not obvious you should always write down your assumptions as part of your answer. Each problem specified with a letter has the same weight (i.e., 1a has the same weight as 2b and so on). Students are free to answer in Norwegian or English. Written exam counts 70 %, and works 30 %, of the final grade awarded.

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NTNU – Department of Mechanical and Industrial EngineeringExam paper in (TPK5115) Project Risk Management)

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Problem 1As part of a highway project section is under construction. This section is based on a historical project, but we make so many simplifications that the relation to the actual project is very vague. Figure 1 shows the entrance to a typical double passage tunnel and Figure 2 shows a bridge being part of the new section.

Figure 1 Double passage tunnel

Figure 2 BridgeIn the first part of this problem we assume that the work in the two tunnel passages could be done in parallel, and that the work on the bridge starts when the two tunnel passages are completed. Preparation work starts prior to the tunnel work, and finalization includes road surfacing, road marking, safety measures and so on. Table 1 shows model parameters to be used in the modelling.

Table 1 Activity durations in months, assuming PERT distributionsActivity L M H 2

Preparation 1 2 4 2.17 0.31Tunnel A, bursting work 8 10 18 11.00 3.00Tunnel A, completion (sealing, lightning fixture etc) 5 6 9 6.33 0.51Tunnel B, bursting work 7.5 9.5 17 10.4

22.74

Tunnel B, completion (sealing, lightning fixture etc) 5 6 9 6.33 0.51Bridge 9 11 16 11.50 1.61Finalization 2 3 5 3.17 0.31

The due date for the project is DD = 36 months after project start-up.

In order to use the SSP method, we need the eMax() and varMax() functions giving expected value and variance of the maximum of two independent and normally distributed stochastic variables with parameters (1, 1

2) and (2, 22) respectively. Relevant values are given in Table 2.

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Table 2 eMax(), varMax(), eMax() and varMax() values for different set of input parameters1 1

2 2 12 eMax(1,1

2,2,22) varMax(1,1

2,2,22) eMin(1,1

2,2,22) varMin(1,1

2,2,22)

19.5 3.8218.9

2 3.56 20.32 2.57 18.1 2.52

19.5 3.5118.9

2 3.25 20.27 2.36 18.15 2.31

19.5 3.5119.4

2 3.25 20.50 2.31 18.42 2.30

19.5 3.5119.9

2 3.25 20.76 2.30 18.66 2.34

Table 2 also shows results for eMin() and varMin(). These functions give the expected value and variance for the minimum of two independent and normally distributed stochastic variables with parameters (1, 1

2) and (2, 22) respectively. Note that eMin() and varMin() are functioning in the

same way as eMax() and varMax().

a) Draw a flow network diagram for the project. Let T denote the duration of the project. Use the CPM method to find a first assessment of the duration of the project. Show by a calculation example that the PERT distribution is a more plausible distribution for the durations than the triangular distribu-tion.

b) Use the PERT method to obtain E(T), SD(T), Pr(T>DD) and expected number of months exceeding the due date.

c) Use the SSP method to obtain the same quantities as in problem 1b). Compare with the PERT solu-tion.

d) Explain how to obtain the same quantities as in problem 1b) with Monte Carlo simulation.

e) It is possible to start the work with the bridge when the work in one of the tunnels is completed. If such a strategy is used it is expected that the duration of the work with the bridge is increased with ¾ months (increase in the expected value, no change in the variance). Modify the SSP to investigate if such a strategy pays off compared to the original strategy. Hint: Use the result for eMin() and varMin() in Table 2. Explain your approach and indicate how to solve this problem with Monte Carlo simulation.

f) There is some uncertainty regarding the rock structure for tunnel B. The investigation from the geo-logists shows that there is some risk of problems which will cause delays. More precisely it con-sidered 20% probable that problems will arise. If problems are encountered, it is assumed that it takes one extra month with the bursting work if no measures are implemented. With measures im-plemented this extra duration is reduced to half a month. Such a measure could be to make an agree-ment with another contractor to assist in case of problems. We do not consider uncertainty in the ex-tra work. Explain what is meant by a risk register and how such a register is used in the risk manage-ment of projects. Give an example of such a risk register where the rock structure problem is spe-cified in the risk register.

g) Find Pr(T>DD) and expected number of months exceeding the due date both if the risk mitigating measure is implemented, and if it is not implemented. In this problem you shall stick to the original plan, i.e., the start-up of the bridge work will not start before the work in both tunnels are completed.

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Problem 2We proceed with the situation described in Problem 1, but we will now have focus on the Bridge activity. Historical data for previous projects has been collected and are shown in Table 3. The leftmost column lists the length of the bridges, the middle column is the corresponding durations, and the rightmost column is residuals discussed later on. The data will be used to establish an empirical model for the duration of the bridge work.

Table 3 Duration of bridge projects (in months) for various lengths in metresLenght Duration Residual

100 15.3 0.75350 10.6 0.1175 11.1 -0.765 11 -0.10390 14.2 0.924

110 15.4 -0.648115 16.4 -0.485150 24.5 0.148

The length of the new bridge is 120 meters. Some of our experienced project managers claim that the expected duration of the work would be on the form 0 + 1x + 2x2, where x is the length of the bridge. The data has been analysed with Excel, and Appendix A shows a print with the relevant results. The residual column in Table 3 is the difference between the observed durations and the predicted durations by the model proposed by our experts. The standard deviation of the residuals is found to be SD = 0.611, and the mean value of the residuals is as expected found to be zero.

a) Describe the regression model used to produce the results in Appendix A, and write down the estim-ates. Verify that the first residual in Table 3 is correctly calculated.

b) Assume that the residuals are PERT distributed with parameters L, M and H. Estimate the paramet-ers by the method of moments.

c) Now, for the new bridge with x = 120, obtain estimates for L, M and H still assuming that the dura-tions are PERT distributed. Describe how you could obtain uncertainty estimates for these paramet-ers.

Table 4 Result from bootstrapping to be used later onParameter to estimate SD-Bootstrap

L 0.42M 0.30H 0.55

Problem 3We have an experienced project engineer in our group which will be used as an expert to complement the limited amount of statistical data. We are a bit suspicious regarding his unbiasedness (he is optimistic in nature) and would therefore ask him some control questions regarding durations of some

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project activities in completed projects where he was only involved in the planning, and not in the final execution of the projects. The results are given in Table 5.

Table 5 True values and estimates of selected activities (in months)Activity True value (x) Assessment by expert (Y)Activity 1 6 4Activity 2 12 10Activity 2 8 6Activity 4 9 7Activity 5 15 10Activity 6 20 19

Running Excel regression with x = dependent and Y = independent variable, we find x =0 + 1Y 2.85 + 0.945 Y and “Standard Error” = 1.49. Further, running Excel regression with Y = dependent and x = independent variable, we find Y =0 + 1x - 2.19 + 0.99 x and “Standard Error” = 1.53.

a) Explain what is meant by calibration, and decide whether it is required to calibrate our expert or not.

b) In the remaining part of this problem, we calibrate the expert independent of the statistical “results” from the result in 3a). We now ask the expert to assess the most likely duration of the bridge activ-ity. The assessment is 12 months. What will be the calibrated value to use in the further analysis, and what is the uncertainty (standard deviation).

c) Combine the result from the statistical analysis in Problem 2 and the expert judgement statement to find an updated value of the M-value to use for the bridge activity.

Problem 4In Problem 1 the challenge with the rock structure was identified as an important threat to the project. The economical consequences shall now be investigated. In order to make an agreement with the other contractor (RockSavior), we have to pay 2 MNOKs so he can allocate resources and tools. If the agreement is made, we have to pay this amount independent of whether we need assistance or not. If rock structure problems arise, the impact depends on the progress in the project. If we are ahead of the schedule the consequences are rather limited, whereas if we are delayed the situation is much worse. In the modelling we simplify and assume that at the point when we eventually will have problems with the rock structure, we are either ahead of schedule, or delayed. The probability that the project is delayed at the critical time is considered to be Pr(Delay) = 40%. Table 6 shows delay penalty costs (penalty of default) for various situations. B is here the extra duration related to rock structure problems. B = 0 means that there is no rock structure problems, B = 1 means that there are rock structure problems and we do not use the RockSavior contractor. B = ½ means that we use the RockSavior contractor in case of rock structure problems.

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Table 6 Delay penalty costs (MNOKs) for various situationsSituation B = 0 B = ½ B = 1Delayed at the critical time 10 20 50Ahead of schedule at the critical time 0 10 25

If problems occur and we decide to use the RockSavior contractor the cost is 20 MNOKs.

a) Establish a decision tree for the situation. State your assumptions. Hint: When deciding whether to use the services of the RockSaviour or not we assume that it is known if there are rock structure problems and the progress of the project (delayed or not delayed).

b) Perform the necessary quantitative calculations to determine the optimal strategy.

Appendix APrint from Excel for problem 2a)

SUMMARY OUTPUT

Regression Statistics

Multiple R 0.990853

R Square 0.981789Adjusted R Square 0.974505

Standard Error 0.722669

Observations 8

ANOVA

df SS MS FSignificanc

e F

Regression 2 140.777570.3887

5134.779

6 4.48E-05

Residual 5 2.6112550.52225

1Total 7 143.3888

Coefficie

ntsStandard

Error t Stat P-value Lower 95%Upper 95%

Lower 95.0%

Upper 95.0%

Intercept 12.18112 2.4533824.96503

1 0.00423 5.874518.4877

4 5.8745 18.48774

Lenght -0.09131 0.051892 -1.759590.13879

7 -0.22470.04208

4 -0.2247 0.042084

Lenght_sq 0.00115 0.0002584.44797

70.00671

5 0.0004850.00181

4 0.000485 0.001814

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Tentative solution, TPK5115 20191a)

PERT verification:=2.167 =(1+4*2+4)/6

2=0.306 =(2.17-1)*(4-2.17)/7

Triangular:=2.2 =(1+3*2+4)/5

2=0.3889 =(1*1+2*2+4*4-1*2-1*4-2*4)/18

I.e., PERT is the best one

Critical path = {Prep, A-Burst,A-Final,Bridge,Finalization}, CPM = 32

b) PERT-MethodSum expectations on the critical path: E( T ) = 34.17 Sum variances on the critical path: 5.74 i.e., SD( T ) = √5.74 2.4

Pr( T > 36) = 1 – Pr(T 36) = 1 - 1−Φ ( 36−34.172.4 )=1−Φ (0.76 )=1−0.78 = 0.22

E(#mnth>36) = B (36 , μ , σ )=( μ−36 )[1−Φ( 36−μσ )]+σϕ (36−μ

σ )=¿0.31

where μ is the expected value and is the standard deviationc) SSP - The standard setup is:Activity i i

2 EiF Vi

F

Preparation 2.17 0.31 2.17 0.31Tunnel A, bursting work 11 3 13.17 3.31Tunnel A, completion (sealing..) 6.33 0.51 19.5 3.82Tunnel B, bursting work 10.42 2.74 12.59 3.05Tunnel B, completion (sealing..) 6.33 0.51 18.92 3.56V 20.27 2.67Bridge 11.5 1.61 31.77 4.28Finalization 3.17 0.31 34.94 4.59

For the virtual node we need an “eMax()” function with arguments 1 = 19.5,12 = 3.82-0.31=3.51, 2 =

7.18.92,12 = 3.56-0.31=3.25, i.e., we use the second data row in Table 1. For the “variance” of V we

add 0.31.

E(T)= 34.94SD(T)= 2.14

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u=(36-E(T))/SD(T) 0.49(u) 0.69Pr(T>36) = 1 -(u) 0.31E(#mnth>36) 0.43

Where we here also use that E(#mnth>36) is given by the backorder function B (36 , μ , σ ) .Expected duration, probability of delay, and expected number of months exceeding 36 are all higher than for the PERT because we include all paths, not only the critical path.d) MCSWe use Excel with built in functions for generating pseudo random numbers. Here we need =rndPERT(r,L,M,H) to generate PERT distributed data. r is generated by the RAND() function in Excel. Other tools could have different names of these functions. Then we need cells (or alternatively variables in a programming language) to hold S = Start, D = Duration, and F = Finalization of each activity. S is found by the maximum of the finalizations of the preceding activities in the flow diagram, or just the finalization of the preceding activity if there is only one activity. Here we only use the =MAX() function to obtain the starting point of the Bridge. D if found by =rndPERT(), and F is found by adding D to the S-value. For each “realization” of the model, we collect the F-value of the last activity, and use this value to provide statistics, typically expected value and standard deviation. To find Pr(T > 36) we typically run statistics on the expression:

=IF(F_last > 36, 1, 0)and to find expected number of months exceeding 36 by running statistics on the expression:

=MAX(F_last-36,0)e) We now use the same model as in 1c). But we now replace “eMax” with “eMin”, and “varMax” with “varMin”, still using the second data row in Table 1. This gives when adding 0.75 to the bridge duration:

E(T)= 33.57SD(T)= 2.13u=(36-E(T))/SD(T) 1.14(u) 0.87Pr(T>36) = 1 -(u) 0.13E(#mnth>36) 0.13

This is better than in 1b)!To obtain the same result with MCS we replace the =MAX() function with the =MIN() function. And add 0.75 to the duration of the bridge work. Note that we might also for the “F_Last” put the maximum of the finalization of the tunnels and the finalization of the “project” to ensure that both tunnels are competed before we end the project. But the likelihood for that this extra max-function is relevant is very low……

f) Risk register: identify, document, analyse, and follow up of important risk elements. Se course compendium. List events (welding apparatus, and earth failure), probabilities, impact, and measures as part of the risk register.Need to list the “Rock structure problem”. Give probability, and consequences with and without the measure.

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g) We rerun the model, but now adding to the duration of B-Bursting: 1/2 if problem and measure implemented, and If problem and no measure. We then use the eMax and varMax with data from row 3 and 4 in Table 1. This gives:

Quantity

1/2

E(T)= 34.94 35.17 35.43SD(T)= 2.14 2.131 2.128u=(36-E(T))/SD(T) 0.49 0.39 0.27(u) 0.69 0.65 0.61Pr(T>36) = 1 -(u) 0.31 0.35 0.39E(#mnth>36) 0.43 0.5 0.59

For Pr(T>36) we use the law of total probability, and for E(#mnth>36) we use the double expectation rule yielding:Quantity With measure Without measurePr(T>36) 0.318 0.326E(#mnth>36) 0.444 0.462

PROBLEM 2a) We use that Y = 0 + 1x + 2x2 + , where Y is the duration of the project, and is an error term with zero mean value and the same standard deviation for all observations. The estimates are:Beta_Intercept = 0 = 12.18Beta_Lenght = 1 = -0.091 Beta_LengthSquared = 2= 0.00115

To verify the first residual we calculate 15.3 - 12.18 - 0.0921×100 - 0.00115×100×100 = 0.75

b) Residuals can be used for estimation L, M and H corresponding to with the MoM in the standard way and are shown in column 2 in the table below. c) Formulas for the final estimates (x=120) are given in column 3 in the table below, and the final estimates are given in column 4: Parameter to estimates MoM -Estimates

(b)Formulas for a given x (length):

x=120

L -1.40 L ( x)= L+¿0 + 1x + 2x2 16.37M -0.11 M (x )=M+¿0 + 1x + 2x2 17.67H 1.86 H ( x )=H+¿0 + 1x + 2x2 19.64

Uncertainty in the estimates are found by Bootstrapping, see course compendium.PROBLEM 3a) The expert is systematically optimistic, i.e., he always assesses a lower value (Z=-5). Since n > 5,

we then calibrate.b) Using the Excel regression results with x= dependent, Y = independent (x =+¿0 + 1Y 2.85 +

0.945 Y) we insert for Y = 12 and get x = 14.19

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c) Using the Excel regression with Y = dependent, x = independent, we only need “Standard Error” = 1.53 = Std. Dev for the expert. Weighted estimate (Data & Expert) =

x= (x_Data/SD_Data^2+x_Expert/SD_Expert)/(1/SD_Data^2+1/SD_Expert) = 17.47

, where SD_Data = 0.30 found from the Bootstrapping (for M). We see that the result is much closer to the data estimate than the calibrated expert estimate due to the less variance in the data es-timate compared to the expert estimate.

PROBLEM 4First decision node: Make an agreement with RockSaviour (Y/N)If we do not make an agreement, we can not do anythingIf we make an agreement, and not get problem, there is nothing to do, since RockSaviour cannot help us. If we have problems, whether to use RocSavior depends on whether we are delayed or not.

Decision diagram follows: If we made the agreement, we should ask RockSaviour to help us only if we are delayed. If not,

it is too expensive. In total, it is better not to make an agreement with RockSaviour, the minimum expected cost is

10.2 MNOKs. Making the agreement gives expected cost equal to 11.4.

Red numbers are expected values, black numbers are probabilities (%).

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