SOLUTION TO PARK AND RIDE PROJECT COURSEWORK - Assignment Example

?SOLUTION TO PARK AND RIDE PROJECT WORK Part A Based on the Project Logic Evaluation presented in Appendix B, complete the Network Analysis forthe Park and Ride Project to determine the total duration of the project based on the ‘normal’ activity durations and to identify the activities that lie on the critical path(s). [20 marks] Solution: The network analysis for the Park and Ride project was analysed based on the project logic evaluation shown in Appendix B of the coursework. The events, which are represented by circles, were numbered to indicate a milestone such as the start and completion of an activity (Kerzner, 2009; De Marco, 2011). The basic rule for numbering events is that the starting point of an activity is lower than the completion point. Meanwhile, the activities or tasks which need to be completed are represented by arrows. In Figure 1, the tasks indicated in Appendix B were replaced by activity codes using letters. Task duration in days is shown beside their respective codes as numbers in parenthesis. Table 1 presents the activity data for the project with the activity codes, description of each task, normal task duration, the early start (ES), early finish (EF), late start (LS) and late finish (LF) times based on normal duration. The ES and EF times were computed based on the analysis of the network using a forward pass. In a forward pass, computation is made from left to right. The earliest starting time of an activity is the earliest finish time of its predecessor. When an activity has no predecessor, such as for initial activity (or activities), the ES of this activity is 0. The earliest finish time is the sum of the early start time and the duration of the activity (Kerzner, 2009; De Marco, 2011). Meanwhile, the LS and LF times were calculated using a backward pass or a right to left computation. The late start of the final activity is taken as the late finish of this activity and from here the late start of the final activity is computed by subtracting the activity duration from the LS time. If there are two or more terminal activities, the highest LS time of these activities should be adopted as the LS time of the rest of the terminating activities (Kerzner, 2009; Demarco, 2011). Table 1 is presented below and the network diagram is shown as Figure 1 on page 4. Table 1. Activity Data for the Park and Ride Project Using ‘Normal’ Task Durations – Float Times Task Description Duration (in days) Early Start (ES) Early Finish (EF) Late Start (LS) Late Finish (LF) Total Float (TF) Free Float (FF) A Excavate Site 10 0 10 0 10 0 0 B Install Ground Drainage 5 10 15 15 20 5 5 C Install Piled Foundations 10 10 20 10 20 0 0 D Erect Steel Frame 10 20 30 20 30 0 0 E Pour In-situ Concrete Floors 9 30 39 30 39 0 0 F Install Electricity, Lighting and IT Cabling 10 39 49 39 49 0 0 G Electrical and Lighting Fit Out 8 49 57 51 59 2 2 H Fix IT Hardware and Screens 10 49 59 49 59 0 0 I Tar-macadam to Access Ramps and Parking 4 49 53 55 59 6 0 J Fix Automated Entrance Barriers 3 49 52 56 59 7 7 K Commission Services and IT Equipment 10 59 69 59 69 0 0 L Landscaping 10 53 63 59 69 6 6 1. The critical path based on the ‘normal’ activity durations There are three ways of determining if an activity is critical. First, from the tabulation of the ES, EF, LS and LF times, if the ES and EF times of an activity are identical to the LS and LF times, then this activity is critical. An examination of Table 1 revealed that seven activities are critical and these are highlighted in blue and bold font in the table. These are activities are: A, C, D, E, F, H, and K. Another way of determining which activities are critical is by computing the float or slack. Float or slack is the difference between the early schedule (ES, EF) and the late schedule (LS, LF). Tasks with zero (0) float are critical (Kendrick, 2010). As reflected in Table 1, the critical activities have 0 total float and 0 free float. To differentiate, total float is the amount of time (i.e. days, in this project) that an activity can be delayed without delaying the targeted completion schedule; while, free float is the amount of time that a project task can be delayed or extended, without delaying the start of the following activity (Kay, 2010). In this project, four of the five non-critical activities (i.e., B, G, J and L) have the same number of total float times. Only Activity I has a total float of 6, but zero free float. Finally, the critical path may be traced from the project network diagram shown in Figure 1. There are 8 paths from start (event 1) to completion (event 12), namely: (1) A – C – D – E – F – H – K; (2) A – C – D – E – F – G – K; (2) A – C – D – E – F – J- K; (4) A – C – D – E – F – I – L; (5) A – B – D – E – F – H – K; (6) A – B – D – E – F – G – K; (7) A – B – D – E – F – J- K; and (8) A – B – D – E – F – I – L. Summing up the individual task duration for each path gives 69, 67, 62, 63, 62, 64, 57 and 58, respectively. It is apparent that the longest path is 69 days. The path with the longest duration is the critical path (Urdhwareshe, 2011). Hence, the critical path is A – C – D – E – F – H - K. [Answer] 2. Total duration of the project based on the ‘normal’ activity durations The total duration of the project is computed by summing up the duration of all activities in the critical path (Urdhwareshe, 2011; Halpin & Senior, 2011). Total Project Duration = 10 + 10 + 10 + 9 + 10 + 10 + 10 = 69 days. [Answer] Part B Determine the total cost of the project using the ‘normal’ durations and the costs for the construction activities. [10 marks] Solution: Table 2 presents the task codes and their respective duration and construction cost, as well as the indirect cost and the total cost of the project using normal durations. Table 2. Activity Data for the Park and Ride Project Using ‘Normal’ Task Durations and Project Cost (1) (2) (3) (4) (5) (6) (7) (8) Task Description Duration (in days) Cost1 Early Start (ES) Early Finish (EF) Late Start (LS) Late Finish (LF) A Excavate Site 10 50,000 0 10 0 10 B Install Ground Drainage 5 7,500 10 15 15 20 C Install Piled Foundations 10 70,000 10 20 10 20 D Erect Steel Frame 10 50,000 20 30 20 30 E Pour In-situ Concrete Floors 9 54,000 30 39 30 39 F Install Electricity, Lighting and IT Cabling 10 35,000 39 49 39 49 G Electrical and Lighting Fit Out 8 28,000 49 57 51 59 H Fix IT Hardware and Screens 10 35,000 49 59 49 59 I Tar-macadam to Access Ramps and Parking 4 20,000 49 53 55 59 J Fix Automated Entrance Barriers 3 10,500 49 52 56 59 K Commission Services and IT Equipment 10 35,000 59 69 59 69 L Landscaping 10 25,000 53 63 59 69 Indirect cost2 693 ?3,000 ?207,000.004 Cost of all tasks ?420,000.00 Total Normal Project Cost ?627,000.005 1The cost is given in sterling pounds (?) • 2Indirect costs associated with the project is given at ?3,000 per day • 3The total project duration based on the network analysis is 69 days • 4The total indirect cost incurred by the project is calculated as the product of 69 and ?3,000 • 5The total normal project cost incurred is calculated as the sum of the cost of each task and the total indirect cost. The cost of all tasks is computed by adding the cost of each task indicated in column (4) for activities A to L, which equals ?420,000.00. The indirect cost is calculated by multiplying the given indirect cost per day associated with the project (?3,000.00), and the total duration of the project solved in Part A (69 days), the product of which is ?207,000.00. The total project cost is computed as the sum of the cost of all tasks and the total indirect cost. Total Project Cost = Cost of all Tasks + Total Indirect Cost Total Project Cost = ?420,000.00 + ?207,000.00 Hence, Total Project Cost = ?627,000.00 [Answer] Part C Produce a Linked Barchart that presents a summary of the ‘normal’ project programme, and which indicates the available float for any non-critical activities. [15 marks]. Solution: The Linked Barchart is exhibited next page. It summarizes the ‘normal’ project programme of the Park and Ride Project showing each activity as a bar against a time scale where the length of the bar is proportional to the task duration and the start and completion schedule aligned to a calendar on top (or at times, at the bottom) of the figure (The Chartered Institute of Building, 2011; Uher, 2003). Each task is shown on a separate line. Activities are listed vertically at the edge if the paper (on the left, in this coursework). Arrows are used to link the activities in order to highlight the interrelationship among the activity bars (Lester, 2007; Lock, 2007). Critical activities, non-critical activities and available float were distinguished using a colour scheme – tan, light turquoise and light yellow, respectively. The available float in days is indicated in the light yellow bar bordered by broken lines by a number. FIGURE 2. LINKED BAR CHART: PARK AND RIDE PROJECT USING “NORMAL” ACTIVITY DURATIONS Part D Assume the project is presently underway and the normal project programme (you have determined above) has been followed so far. The project has now reached the end of day 39. At this point, all of the tasks programmed so far have been completed successfully except for the task to pour the insitu concrete floors. The task has been delayed and will now not be completed for a further 5 days. The cost of the particular task will not increase per se but the delay will have a knock on effect for the project and will mean that unless action is taken, the overall project completion will be delayed. The penalty costs that will be incurred as a consequence of a late completion of the project beyond the originally scheduled completion date will be ?7,500 per additional day. In the light of this information, and in view of the fact that the project incurs indirect costs at a rate of ?3,000 per day, determine the most economical solution (time and total cost) that is now available to the client as far as the completion of this project is concerned. (Important: You must present the complete procedure that you follow in determining the economical solution as well as reporting the solution). [40 marks]. Solution: The following methodology was adopted in the determination of the most economical solution for the problem at hand: 1. To have a proper basis from which to compare the most economical solution, a network diagram1 was prepared depicting the problem scenario without crashing any activity. The cost of the project was computed grounded on the additional cost of a 5-day delay. Since no additional cost will be incurred even if pouring of insitu concrete floors (Activity E) will be delayed, the additional cost for the 5-day delay is calculated by adding the product of the ?7,500 penalty per additional day and the number of days that the project will be delayed (5 days) to the total project cost based on ‘normal’ activity durations. The benchmark cost will thus, be ?37,500 .00 plus ?627,000.00 or ?664,000.002. This entails an additional ?37,500 to the total project cost or 5.98% of the total project cost at normal duration. 2. The most economical solution would, therefore, be one which is less than an additional 5.98% of the total project cost at normal duration. Since there will be a number of options to solve the delay problem, the solution with the least additional cost in terms of the total project cost and is less than 5.98% will be selected as the most economical solution. 3. All possible options were tried but only three of these options were exhibited in this paper. It should also be made clear at this point that the first four tasks of the project which were already completed when the delay occurred in the pouring of insitu concrete floors (Activity E) can not anymore be crashed since these are assumed to be completed. These activities are: excavate site (Activity A), install ground drainage (Activity B), install piled foundations (Activity C), and erect steel frame (Activity D). 4. Selection of the tasks to be crashed was based on the crash cost and if it is a critical activity. 5. A network diagram for each option was rendered in order to determine the total project duration with one or more activities crashed. These diagrams are shown as Figures 3 -6 on pages 15-18. 6. Total project cost for each option was computed based on the total project duration. Calculation was similar to the computation of the total project cost using ‘normal’ activity durations, except that the penalty for delay will be included in the equation. These tables are presented as Tables 3 - 5 on pages 11 - 13. 7. A summary table to report the solution showing the benchmark costs and additional costs3 with no crashing and the results when one or more activities are crashed is presented as Table 6 on page 14 for visual comparison. The lowest additional cost as a percentage of the ‘normal’ project cost was recommended as the most economical solution. Option 1: Crash Activity F (install electricity, lighting and IT cabling) to 8 days and Activity L (landscaping) to 7 days. Activity F is a critical activity and it is succeeded by 4 other activities, hence crashing F is a potential solution. Meanwhile as for Activity L, although it is not a critical activity, it is a terminating activity, and it has one of the lowest crash cost among the activities in the project. Table 3 presents the data and results regarding the project cost with activities F and L crashed. All costs are in pounds (?). Project duration was based on the network diagram in Figure 4. Table 3. Total Project Cost for Option 1 Task Code Task Description Original Duration (days) Revised Duration (days) Normal Cost Crash Cost Total Project Cost A Excavate Site 10 10 50000 8000 50000 B Install Ground Drainage 5 5 7500 3000 7500 C Install Piled Foundations 10 10 70000 10000 70000 D Erect Steel Frame 10 10 50000 7500 50000 E Pour Insitu Concrete Floors 9 14 54000 54000 F Install Electricity, Lighting and IT Cabling 10 8 35000 30001 38000 G Electrical and Lighting Fit Out 8 8 28000 4500 28000 H Fix IT Hardware and Screens 10 10 35000 4500 35000 I Tarmacadam to access ramps and parking 4 4 20000 20000 J Fix automated Entrance Barriers 3 3 10500 10500 K Commission Services and IT Equipment 10 10 35000 35000 L Landscaping 10 7 25000 3500 28500 Construction Cost for Option 1 426500 Indirect costs 69 3,000 2070002 Penalty cost 3 7500 225003 Total Project Cost for Option 1 6560005 Difference with ‘normal’ total cost 6270004 290006 % Difference 4.62% COLOR CODE: Red =activities which can not be crashed; Blue (bold) = crashed activities; Black = activities in their ‘normal’ duration. LEGEND: 1 Crash cost of 4500 for three days means 1500 per day, so 3000 for crashing F to 8 days • 2Total Indirect Cost = ‘normal’ project duration (69) x indirect cost per day (3000) = 207,000 • 3Penalty cost = number of days exceeding ‘normal’ project duration (3) x penalty per day = 7500 • 4 ‘Normal’ project cost (no delay, no crashing) • 5 Total project cost for Option 1 • 6Difference between crash cost for Option 1 and ‘normal’ project cost Option 2: Crash Activity F (install electricity, lighting and IT cabling) to 7 days and Activity H (fix IT and hardware screens) to 8 days. The rationale for selecting Activity F to crash is already explained in Option 1. Lesson learned from Option 1 showed that in spite of the lower cost of crushing Activity L (landscaping), since it is not a critical activity, the 69-day normal limit of the activity duration was not met. On the other hand, Activity H is the only critical activity following F. This has potential of facilitating project completion within the 69-day ‘normal’ project duration so that penalty cost will not have to affect the total project cost. Table 4 presents the data and results regarding the project cost with activities F and H crashed. All costs are in pounds (?). Table 4. Total Project Cost for Option 2 Task Code Task Description Original Duration (days) Revised Duration (days) Normal Cost Crash Cost Total Project Cost A Excavate Site 10 10 50000 8000 50000 B Install Ground Drainage 5 5 7500 3000 7500 C Install Piled Foundations 10 10 70000 10000 70000 D Erect Steel Frame 10 10 50000 7500 50000 E Pour Insitu Concrete Floors 9 14 54000 54000 F Install Electricity, Lighting and IT Cabling 10 7 35000 4500 39500 G Electrical and Lighting Fit Out 8 8 28000 4500 28000 H Fix IT Hardware and Screens 10 8 35000 30001 38000 I Tarmacadam to access ramps and parking 4 4 20000 20000 J Fix automated Entrance Barriers 3 3 10500 10500 K Commission Services and IT Equipment 10 10 35000 35000 L Landscaping 10 10 25000 3500 25000 Construction Cost for Option 2 427500 Indirect costs 69 3,000 2070002 Penalty cost 0 7500 03 Total Project Cost for Option 2 6345005 Difference with ‘normal’ total cost 6270004 75006 % Difference 1.20% COLOR CODE: Red =activities which can not be crashed; Blue (bold) = crashed activities; Black = activities in their ‘normal’ duration. LEGEND: 1 Crash cost of 4500 for three days means 1500 per day, so 3000 for crashing H to 8 days • 2Total Indirect Cost = ‘normal’ project duration (69) x indirect cost per day (3000) = 207,000 • 3Penalty cost = number of days exceeding ‘normal’ project duration (0) x penalty per day = 0 • 4 ‘Normal’ project cost (no delay, no crashing) • 5 Total project cost for Option 2 • 6Difference between crash cost for Option 2 and ‘normal’ project cost Option 3: Crash Activity F (install electricity, lighting and IT cabling) to 8 days and Activity H (fix IT and hardware screens) to 7 days. The rationale for selecting Activity F and Activity H had already been discussed in Option 2. In this option, however, F is crashed to only 8 days while H is crushed to the maximum allowable under the circumstances of Part D. Since Option H allowed for project completion within the 69-day ‘normal’ project duration and penalty cost was not incurred, it would be interesting to find out if exchanging the crash times for the activities will increase or decrease the total project cost computed for Option 2. Table 5 presents the data and results regarding the project cost with activities F and H crashed for 7 and 8 days, respectively. All costs are in pounds (?). Table 5. Total Project Cost for Option 3 Task Code Task Description Original Duration (days) Revised Duration (days) Normal Cost Crash Cost Total Project Cost A Excavate Site 10 10 50000 8000 50000 B Install Ground Drainage 5 5 7500 3000 7500 C Install Piled Foundations 10 10 70000 10000 70000 D Erect Steel Frame 10 10 50000 7500 50000 E Pour Insitu Concrete Floors 9 14 54000 54000 F Install Electricity, Lighting and IT Cabling 10 8 35000 30001 38000 G Electrical and Lighting Fit Out 8 8 28000 4500 28000 H Fix IT Hardware and Screens 10 7 35000 4500 39500 I Tarmacadam to access ramps and parking 4 4 20000 20000 J Fix automated Entrance Barriers 3 3 10500 10500 K Commission Services and IT Equipment 10 10 35000 35000 L Landscaping 10 10 25000 3500 25000 Construction Cost for Option 2 427500 Indirect costs 69 3,000 2070002 Penalty cost 1 7500 75003 Total Project Cost for Option 2 6420005 Difference with ‘normal’ total cost 6270004 150006 % Difference 2.39% COLOR CODE: Red =activities which can not be crashed; Blue (bold) = crashed activities; Black = activities in their ‘normal’ duration. LEGEND: 1 Crash cost of 4500 for three days means 1500 per day, so 3000 for crashing F to 8 days • 2Total Indirect Cost = ‘normal’ project duration (69) x indirect cost per day (3000) = 207,000 • 3Penalty cost = number of days exceeding ‘normal’ project duration (1) x penalty per day = 0 • 4 ‘Normal’ project cost (no delay, no crashing) • 5 Total project cost for Option 3 • 6Difference between crash cost for Option 3 and ‘normal’ project cost Based on the analysis of the various options to solve the delay in project schedule caused by the 5-day delay of Activity E (the pouring of insitu concrete floors), findings revealed that the most economical solution for the problem at hand is Option 2 – crashing Activity F (installation of electricity, lighting and IT cabling) to 7 days and crashing Activity H (fixing of IT hardware and screen to 8 days). Option 2 will provide the lowest total project cost since the project can be finished within the ‘normal’ project schedule of 69 days with the least additional outlay of ?7,500 or only 1.20% in excess of the ‘normal project’ cost. Hence, no penalty cost will be incurred. With this scheme, however, extreme care should be exercised since two critical paths were identified in the network diagram shown in Figure 5: A – C – D – E – F – H – K and A – C – D – E – F – G – K. Table 6. Summary Table of Findings Assumptions Made: Parameters Involved NO CRASHING F crashed to 8 days L crashed to 7 days F crashed to 7 days H crashed to 8 days F crashed to 8 days L crashed to 7 days CRASH OPTION 1 CRASH OPTION 2 CRASH OPTION 3 Assumption 1: The indirect cost of ?3000 per day does not apply after the normal project duration (69 days) is exceeded. No. of days exceed-ing normal project duration 5 3 0 1 Construction cost ?420,000 ?426,500 ?427,500 ?427,500 Indirect cost incurred ?207,000 ?207,000 ?207,000 ?207,000 Penalty cost incurred ?37,500 ?22,500 0 ?7,500 Total Project cost ?664,500 ?656,000 ?634,500 ?642,000 Cost in excess of normal project duration ?37,500 ?29,000 ?7,500 ?15,000 Percentage of additional cost to normal project cost 5.98% 4.62% 1.20% 2.39% Assumption 2: The indirect cost of ?3000 per day applies even after the normal project duration (69 days) is exceeded. No. of days exceed-ing normal project duration 5 3 0 1 Construction cost ?420,000 ?426,500 ?427,500 ?427,500 Indirect cost incurred ?222,000 ?216,000 ?207,000 ?210,000 Penalty cost incurred ?37,500 ?22,500 0 ?7,500 Total Project cost ?679,500 ?665,000 ?634,500 ?645,000 Cost in excess of normal project duration ?52,500 ?38,000 ?7,500 ?17,500 Percentage of additional cost to normal project cost 8.37% 6.06% 1.20% 2.79% The next four figures present the network diagram used to determine the total project duration during for the scheme with no crashing employed and the three crashing options presented. As shown in the solution in this section of the coursework, crashing is a process of schedule compression to decrease the project duration or selected activities for corrective action, usually when the project schedule is delayed (Sherrer, 2010; Grisham, 2010; Steiss, 2005). Crashing the project schedule usually involves throwing additional resource, such as manpower and equipment to tasks which are delayed (Grisham, 2010; Kuehn, 2011; McCarthy, 2011). Part E Produce a Linked Barchart that presents the revised project programme for the most economical solution that is now available. [15 marks]. Solution: The most economical solution the for the delay in Task E as discussed in Part D is Option 2 – crashing Activity F to 7 days and crashing Activity H to 8 days. Figure 7 presents the Linked Barchart for the revised programme depicting the most economical solution. FIGURE 7. LINKED BAR CHART: PARK AND RIDE PROJECT OF THE REVISED PROGRAMME REFERENCES De Marco, A., 2011. Project management for facility constructions: A guide for engineers and architects. New York: Springer. Grisham, T.W., 2010. International project management: Leadership in complete environments. Hoboken: John Wiley & Sons. Halpin, D.W. and Senior, B.A., 2011. Construction management, 4th ed. Hoboken: John Wiley and Sons. Kay, R., 2010. An APMP primer, PRINCE 2 ed. Buckinghamshire: Association for Project Management / LuLu. Kendrick, T., 2010. The project management tool kit: 100 tips and techniques for getting the job. New York: American Management Association. Kerzner, H., 2009. Project Management: A systems approach to planning, scheduling and controlling, 10th ed. Hoboken: John Wiley & Sons. Kuehn, U., 2011. Integrated cost and schedule control in project management, 2nd ed. Vienna: Management Concepts. Lester, A., 2007. Project management: management planning and control - managing engineering, construction and manufacturing projects to PMI, ASP & BSI standards, 5th ed. Oxford: Butterworth-Heinemann. Lock, D., 2007. The essentials of project management (3rd ed.). Hampshire, GBR: Bower Publishing. McCarthy, J.F., 2011. Construction project management. Westchester: Pareto – Building Improvement. Sherrer, J.A, 2010. Project management road trip: For the project management professional. Newtown Square: Author / Philippine Management Institute. Steiss, A.W., 2005. Strategic facilities planning: capital budgeting and debt administration. Lanham: Lexington Books / Rowan & Littlefield Publishing. The Chartered Institute of Building, 2011. Guide to good practice in the management of time in complex projects. West Sussex: Author. Urdhwareshe, H., 2011. Six sigma for business excellence: Approach, tools and applications. New Delhi: Dorley Kindersley / Pearson Education. Uher, T.E., 2003. Programming & scheduling techniques. Sydney: University of New South Wales Press. Read More