An Integrated Efficiency-Risk Approach in Sustainable Project Control

Read this paper, which describes the most common project management tools and then presents a hybrid model combing different elements from each. It then uses the model in a case study analysis. Think about how the hybrid simultaneously controls for model parameters. How does this increase project sustainability and efficiency in the case study?

5. Case Study

Mobarakeh Steel Co. is a leading Iranian company that produces steel sheets. This company is committed to playing a central role in the industrial, economic, and social development of the country, to improving its technologies in the steel industry, and to acting as an international organization that produces about 50% of the country's steel for use in industries, such as automobiles, energy, heavy steel, fluid transfer tubes, packaging, electricity, profile, and tube. This company comprises seven industrial complexes distributed across the country and has more than 20,000 employees in its different departments. One of the projects related to this plant is the development of a custom product, which is used as a case study for applying the proposed approach of integrated risk–efficiency methodology. The Primavera Risk Analysis software version 8.7 (Oracle Corporation, Redwood City, CA, USA) is used for the application. The project is composed of three parts, with each part built separately and then assembled. The project activities, activity duration, and costs required are extracted, as shown in Table 3. Cut and paste, error root square, and Monte Carlo simulation are used to calculate the schedule and cost buffer of the projects. The duration and costs are estimated as a probability (triangular probability distribution).

Table 3. Activity list, duration, and budget cost of project.

ID Activity Name Expected Duration Expected Cost
Description Min Duration (days) Most Likely (days) Max Duration (days) Min Cost ($) Most Likely Cost ($) Max Cost ($)
0010 Project of assembled product 116,250 143,000 233,060
0020 Part No. 1 52,000 65,000 104,000
0025 Start project Milestone -
0030 Design 8 10 16 8000 10,000 16,000
0040 Civil 8 10 16 40,000 50,000 80,000
0050 Test 8 10 16 4000 5000 8000
0060 Part No. 2 20,250 27,000 47,250
0070 Design 6 8 14 3000 4000 7000
0080 Civil 6 8 14 15,000 20,000 35,000
0090 Test 6 8 14 2250 3000 5250
0100 Part No. 3 20,000 24,000 39,860
0110 Design 5 6 10 2500 3000 5000
0120 Civil 5 6 10 12,500 15,000 24,900
0130 Test 5 6 10 5000 6000 9960
0140 Product Assembly 24,000 27,000 41,950
0150 Assembly 8 9 14 16,000 18,000 28,000
0160 Product test 5 6 10 8000 9000 13,950
0170 Project completion Milestone -


After scheduling the project, the baseline planned duration (BPD0) is estimated to be equal to 48 days, and the budget cost at completion regardless of the buffer cost (BAC0) is calculated using Equation (25):

\begin{aligned} &B C A C_{0}=\sum\nolimits_{i=1}^{N} B P V_{i}=\$ 143000 \\ &\text { where } \\ &N=\text { quantity of activity } \end{aligned}            (25)

The results of the deterministic scheduling are illustrated in Figure 5.

Figure 5. Deterministic scheduling.


Based on the equations in Section 2, the schedule and cost buffer are calculated by cut and paste, error root square, and Monte Carlo simulation as Equations (26) and (27). Based on the calculations, the results of the error root square are obtained:

 \begin{aligned} &For \, Schedule \, Buffer \\ & B_{C-P} =.5 \times \sum\nolimits_{i=1}^{N}\left(T_{p}-T_{m}\right)=14 \text { days } \\ & B_{r s e} =\sqrt{\sum\nolimits_{i=1}^{N}\left(T_{p}-T_{m}\right)^{2}}=12.6 \text { days } \\ & B_{m c s} =B P D_{\% 90}-B P D_{d}=12 \text { days } \end{aligned}          (26)

 \begin{aligned} & For \, Cost \, Buffer \\  &B_{C-P}=.5 \times \sum\nolimits_{i=1}^{N}\left(C_{p}-C_{m}\right)=\$ 90060 \\ &B_{r s e}=\sqrt{\sum\nolimits_{i=1}^{N}\left(C_{p}-C_{m}\right)^{2}}=\$ 37766 \\ &B_{m c s}=B C A C_{\%} 90-B C A C_{d}=\$ 37750 \\ &\text { where } \\ &N=\text { equal quantity of activity } \\ &C_{p}=\text { pessimestic cost of activity } \\ &C_{m}=\text { most likely cost of activity } \\ &B C A C_{d}=\text { deterministic Budget Cost At Completion of project } \\ &B C A C_{\%} 90=\text { Budget Cost At Completion of project with a probability of 90%} \end{aligned}          (27)

The project in two control periods of 15 and 30 days is updated and monitored. In the mentioned periods all activities, except no. 130 (related to the test activity of part 3), are conducted based on a scheduling plan. Using the proposed methodology of integrated risk–efficiency methodology, the duration progress, cost progress, usage percentage of the schedule, and cost buffer are calculated using Equations (28) and (29):

 \begin{align} &C V=B C W P-A C W P=E V-A C W P= \, \$ 5975  \, for  \, period  \, 1  \, and  \, 1  \, \$ 4667  \, for  \, period  \, 2\\ &P_{C}=\frac{C V}{C B}=16\%  \, for  \, period  \, 1  \, and  \, 39 \%  \, for  \, period  \, 2 \\ &W_{C}=\frac{E V}{B C A C_{0}}=48\%  \, for  \,  period  \, 1  \, and  \, 67\%  \, for  \, period  \, 2 \end{align}           (28)

\begin{align} &E S_{T}=T+\frac{E V-P V_{T}}{P V_{T+1}-P V_{T}}=15 \, days  \, for  \, period  \, 1  \, and  \, 21  \, days  \, for \,  period  \, 2 \\  & S V_{T}=E S_{T}-A T=0  \, days  \, for  \, period  \, 1  \, and  \, 9  \, days  \, for  \, period  \, 2 \\ &P_{T}=\frac{S V_{T}}{S B}=0 \%  \, for \,  period  \, 1  \, and  \, 71  \, \%  \, for  \, period  \, 2 \\ &W_{T}=\frac{E S_{T}}{B P D_{0}}=31\%  \, for  \, period  \, 1  \, and  \, 48  \%  \, for  \, period  \, 2 \end{align}       (29)

The duration and cost at completion of the project in the two control periods of 15 and 30 days are also estimated using the proposed methodology.

  • BCAC: Budget cost at completion with regard to the cost buffer (combination of performance and risk parameters), which is calculated using Equation (30):

B C A C=B C A C_{0}+C B=143000+37766=\$ 180766              (30)

  • BC: Cost buffer percentage of BCAC0, which is calculated using Equation (31):

 B_{C}=\frac{C B}{B C A C_{0}}=\frac{37766}{143000}=26.4 \%               (31)

  • ECAC0: During the phases of project control, estimated cost at completion, regardless of the cost buffer, and only after control of project buffer usage is done. In other words, with the help of cost buffer usage, the adjusted BCAC0is estimated, which is the same as ECAC0. Equation (32) calculates this value:

E C A C_{0}=B C A C_{0}+B C A C_{0} * P_{C}=B C A C_{0}\left(1+P_{C}\right)=\text{\$16588 for period 1 and \$198770 for period 2}            (32)

  • ECAC: Estimated cost at completion with regard to cost buffers (combination of performance and risk parameters), which is calculated using Equation (33);

E C A C=B C A C_{0}\left(1+P_{C}\right)+B_{C} * B C A C_{0} * P_{C}+B_{C} * B C A C_{0} *\left(1-P_{C}\right) *\left(1-\frac{P_{C} * B_{C}}{W_{C}}\right)==\text{\$200841 for period 1 and \$232983 for period 2}              (33)

  • BPD: Total baseline planned duration, with regard to the schedule buffer (combination of performance and risk parameters), which is calculated using Equation (34):

B P D=B P D_{0}+S B=48+12.6=60.6 \text { days }               (34)

  • BT: Schedule buffer percentage of BPD0which is calculated using Equation (35):

\mathrm{B}_{\mathrm{T}}=\frac{\mathrm{SB}}{\mathrm{BPD}_{0}}=26.2 \%            (35)

  • EDAC0: Within the phases of project control, the estimated duration at completion, regardless of the buffer, and only after control of project buffer usage is done. In other words, with the help of schedule buffer usage, the adjusted BPD0is estimated, which is the same as EDAC0. Equation (36) calculates this value:

\begin{align}E D A C_{0}=B P D_{0}+B P D_{0} * P_{T}=B P D_{0}\left(1+P_{T}\right)=48 \text { days for period } 1 \text { and } 82 \text { days for period } 2 \end{align}             (36)

  • EDAC: Estimated duration at completion with regard to schedule buffers (the combination of performance and risk parameters), which is calculated using Equation (37):

\begin{align} E D A C=B P D_{0}\left(1+P_{T}\right)+B_{T} * B P D_{0} * P_{T}+B_{T} * B P D_{0} *\left(1-P_{T}\right) *\left(1-\frac{P_{T} * B_{T}}{W_{T}}\right)= 61 \text { days for period } 1 \text { and } 93 \text { days for period } 2 \end{align}         (37)