Assignment 3

FACULTY OF CIVIL ENGINEERINGSKAA 4113 CONSTRUCTION AND PROJECT MANAGEMENT SEMESTER 1 – 2017 / 2018 ASSIGNMENT 3 Analyse resource requirement for manpower, machinery, material and money in construction project LECTURER : MR. ABDUL RAHIM B. ABDUL HAMID SESSION 2016/2017 Q15. . Which of the followings are the appropriate actions to control wastage on site? I. wastage on the site can be controlled. Resource levelling will generally: A.. I.to be obtained solely from the performing organization. Subcontract the works A.SEMESTER 1. II. Employ skill workers III. All of the above Reason: By employing skill workers and using precast materials. Q16.. An output of the process Estimate activity resources is a document called "activity resource requirements". It identifies the types and quantities of resources. Use precast materials IV. Q17. C.. Reason: Resource levelling is a technique in which start and finish dates are adjusted based on resource constraints with the goal of balancing demand for resources with the available supply.while the term resources is limited to equipment and materials. Double handling of materials II. . A.. Reduce the overutilization of resources. Reason: Activity resource requirements determines the resources for each work package and each work period... Reduce the time needed to do the project. Increase the total time necessary to do all the tasks. . On the other hand. II and III only C. . II and IV only B.required for each activity in a work package. Reduce resources to the lowest skill that is possible. double handling of materials and subcontract the works cause the wastage occurred on site. D.. III and IV only D.while the term resources is limited to human resources.. D. B. B. C... . Reason: Poor machineries and equipment output are caused by the problem of under capacity. Crashing works only for activities on critical path where additional resources will shorten the activity’s duration. Calculate Total Cost of this project: Assume six (6) working days a week except Sunday. Under capacity B.00. unskilled operators and lack of maintenance culture. C. Question 20 to 22 will refer to the following table: Table Q20-Q22 Activity Normal Time Crash Time Normal Cost Crash Cost (week) (week) (RM) (RM) 1-2 5 2 21000 27000 1-3 3 1 5000 8000 2-4 2 1 10000 12000 2-5 4 1 7500 12000 3-4 5 2 24000 36000 4-5 7 2 8000 22000 Q20. To bring the project on schedule. Resource planning C. Lack of maintenance culture.Q18. D. Timely replacement of parts.500 . The project is running late. Q19. You are the project manager responsible for building a new project. Fast-tracking B. Schedule Management Reason: Crashing is a technique used to shorten the schedule duration for the least increment cost by adding resources. Unskilled operators. you decide to add additional resources to the critical path. The problems related to the poor machineries and equipment output during construction are attributed to the followings. Given indirect cost per day is RM300. RM 75. Crashing D. except: A. A. This is an example of… A. 33 2-5 4 1 7500 12000 250 3-4 5 2 24000 36000 666. RM 102.67 Indirect Cost = 300 (RM/day) Total Cost = 102500 (RM) Crash Activity Direct Indirect Duration Total (RM) at Cost (RM) Cost (RM) (day) Normal 75500 27000 102500 90 1-2 81500 27000 108500 90 1-3 78500 25200 103700 84 3-4 87500 25200 112700 84 4-5 89500 18000 107500 60 Activity 3-4 shows the highest increment if time compression needs to be done. RM 117. B. Activity 4-5 Reason: Normal Crash Time Normal Cost Crash Cost Cost slope Activity Time (RM) (RM) (RM/day) (week) (week) 1-2 5 2 21000 27000 333. it takes (3+5+7) weeks  6 days/week = 90 days Total Cost = RM (21000 + 5000 + 10000 + 7500 + 24000 + 8000) + (90 days  RM 300 / day) = RM 102.250 D.800 Reason: Critical Path is 1-3-4-5. Activity 1-2 B. Activity 1-3 C.500 Q21. .33 1-3 3 1 5000 8000 250 2-4 2 1 10000 12000 333. Thus.67 4-5 7 2 8000 22000 466.000 C. RM 13. Activity 3-4 D. Which activity indicates the highest increment if time compression needs to be done? A. 0. What is the cost performance index? A. The deviation from the original plan then have to be rectify.75 B. Instead.700 C. the person assigned to the task was 75% complete. and take two weeks to complete. The project manager is making sure that the product of the project has been completed according to the project management plan. Executing C. A decision has been made which activities 1-3 and 4-5 needed to be crashed 2 weeks each due to financial reason. the project manager was only able to assign one person to this task. Planning B.Q22. At the end of two weeks. After planning.50 . 1. Q24. RM 117. full time. What part of the project management process is he in? A. 0.15 Reason: In two weeks. BCWP = 100  75% = 75. ACWP = 50 CPI = BCWP/ACWP = 75/50 = 1. 1. RM 105. The implementation will have to be closely monitor to see how much it varies from the planning. Monitoring and controlling D.50 C. Calculate the new Total Cost of this project? A.000 Reason: Crash Direct Cost Indirect Total (RM) Duration Activity (RM) Cost (RM) (day) Normal 75500 27000 102500 90 1-3 78500 25200 103700 84 4-5 84100 21600 105700 72 Q23. RM 114. Closing Reason: Monitoring and Controlling are similar to check and act. the implementation will take place. A task was scheduled to employ two persons.700 B. let say two persons have BCWS = 100. RM 102.25 D.500 D. Indirect cost include cost of hiring and purchasing plant and machinery C. indirect cost and profit D. Profit is also known as contractor’s mark up Reason: Indirect cost is known as overhead cost. Total cost is the summation of direct cost and indirect cost only. Which of the following statement about project cost is true? A. Total cost is the summation of direct cost.Q25. . Direct cost is also known as overhead cost B. Direct cost include cost of hiring and purchasing plant and machinery. 90 = RM 11. To simplify the calculation. Table Q26-Q29 Activity Duration Predecessor Total Value of (month) Work (RM  1000) Start 0 - A 2 Start 800 B 2 A (FS .000 B. III.400. RM 11.900. Progress payment is on monthly basis and measurement will be made at the end of each month and the contractor receives the payment one (1) month later IV. The gross profit margin is 10 % of the value.000 at the start of the project. V.3) 5200 E 2 C (FS .1) 1600 C 3 B 1200 D 4 C (FS .900. II. G 500 Finish 0 H Based on the above information’s.000  0. F.000 D. The retention money is repaid 3 (three) months after the practical completion.000 . The contractor is given advance payment of RM 700.610. answer the following questions. Q26.000 Total Cost = RM 12.610. you may assume that all costs must be met at the instant they are incurred.000 Reason: Total Value = RM 12. C 1800 (SS+3) H 1 D.160.900.1) 600 F 2 E 1200 G 1 B (FS+4). RM 12. Retention money is 5 %. What is the total cost of the project? A.000 C. The information of the project is as the followings: I. You may also assume that the monthly value and cost for each activity are equally distributed. RM 12.Questions Q26 – Q29 are referring to the Table Q26 Table Q26 shows the value of works and the program for the construction of a building. RM 11. the advance payment will be returned to the client through progress payment deduction starting from the end of fourth month to the end tenth month equally. 000 C. End of month 5 B. RM 470.550.000 B. What is the expected monthly payment received of all works that are scheduled to be completed in month 9 only? A. RM 2.000 Q28.000 C.000 .625. RM 556.000 D.180.340. RM 2. the maximum cash is required? A. End of month 7 D. RM 475. RM 2. RM 2. Calculate the maximum cash required for this project.000 D. At the end of which month. A. End of month 8 Q29.910.Q27. RM 2.000 B. End of month 6 C. Month Activity 1 2 3 4 5 6 7 8 9 10 11 A 400 400 B 800 800 C 400 400 400 D 1300 1300 1300 1300 E 300 300 F 600 600 G 1800 H 500 Monthly Value 400 1200 800 1700 1700 2000 1600 2400 600 500 Cumulative Monthly Value 400 1600 2400 4100 5800 7800 9400 11800 12400 12900 Monthly Cost 360 1080 720 1530 1530 1800 1440 2160 540 450 Cumulative Monthly Cost 360 1440 2160 3690 5220 7020 8460 10620 11160 11610 Payment 700 380 1140 660 1515 1515 1800 1420 2180 470 475 Cumulative Payment 700 1080 2220 2880 4395 5910 7710 9130 11310 11780 12255 Required Cash 740 1080 1470 2340 2625 2550 2910 2030 300 * RM  1000 .Reason: For Q27-29. III and IV C. Capital cost III. Spare part availability A. Use in accordance to drawing and specification advised by the manufacturer IV. spare part availability. IV. Different method of construction will require different types of machinery. SESSION 2016/2017 Q15. Q16. Annual maintenance cost II. Acquire schedule of service for the engine II. . The followings are the factors that influence the selection of a particular model of in ownership of a machine: I. II and III D. I and II B. Operator cost. All the above Reason: In order to achieve the required productivity. Inspection before running the machine every day III. III and IV C. Reject usage of overage machine A. Which of the following are the processes for machinery maintenance monitoring in order to achieve the required productivity? I. capital cost. Method of construction C. market usage and productivity. The selection of machineries for work execution is based on A. economic life span. Popularity of machine Reason: Machineries are selected based on method of construction. II and III D. I. Availability of a skilled operator D. I and II B. Model of machine B. I. All the above Reason: The factors need to be considered is maintenance cost. operator cost. Q17. all the above processes must be performed.SEMESTER 2. Q18. RM12. 2 B 4 A 4 C 3 B (FS+1) 3 D 3 B 3 E 1 A 1 F 3 C (SS+2). what is the labour resource allocation for day 9 of the project? A. Using a priority listing of activities based on early start-total float sort and late start-total float sort. Reduce number of machineries B.800 in 5 days. The contractor planned to shorten the work due to weather condition which affects the crash cost increase to RM120. Subsequently. Appropriate working capital for the project Reason: Reduce number of machineries will delay the duration of project. RM5. D. Based on the early start resource loading diagram. Given a normal cost and time for a roof finishing work is RM95. RM24. Minimized material wastage D. the cost of the project will increase. In order to increase profitability of a project the following strategies with regard to managing resources have to be carried out. RM19. Q19.080/day 𝐂𝐫𝐚𝐬𝐡 𝐂𝐨𝐬𝐭−𝐍𝐨𝐫𝐦𝐚𝐥 𝐂𝐨𝐬𝐭 Reason: Cost Slope = 𝐍𝐨𝐫𝐦𝐚𝐥 𝐃𝐮𝐫𝐚𝐭𝐢𝐨𝐧−𝐂𝐫𝐚𝐬𝐡 𝐃𝐮𝐫𝐚𝐭𝐢𝐨𝐧 𝟏𝟐𝟎𝟖𝟎𝟎−𝟗𝟓𝟒𝟎𝟎 = 𝟏𝟎−𝟓 = RM5080/day Question 20 to 22 will refer to the following table: Table Q20-22 shows the scheduling data for a small project. What is the expected cost slope of this activity? A. 4 . Optimized workers used for all activities C. prepare resource loading diagrams. except: A. F 2 Q20. E 3 G 2 C.080/day C.400 and 10 days. Table Q20-Q22 Activity Duration Predecessor No.080/day B.160/day D. of Labour A 2 . 6 D. 5 C. 9 Reason: 10 9 8 7 Total Labour 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Time (days) . 6 D. Based on the late start resource loading diagram. 5 C. 3 B. 2 Reason: 7 6 5 Total Labour 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Time (days) Q21. what is the labour resource allocation for day 11 of the project? A. B. III and IV D. . 18 C. I. 10 Reason: In Early Start. I. Q23. III. Which of the following are the reasons for the contractor to shorten the project time and conduct time-cost trade-off analysis? I. III and IV C. To reduce overhead cost of the project since the duration of project decrease. ii. The delay incurred for the earlier activities. II. The client requested the project to be completed ahead of the schedule. 16 B. iii.Q22. 14 D. If the labour resources for the entire project duration are limited to 6 units only. II. Contractor needs to rationalise the unit rate of the activity. IV. the maximum number of resources is 6 units. A. All the above Reason: The reason for the contractor to shorten the project time and conduct time-cost trade-off analysis is: i. II and III B. what is the duration of this project? A. To catch up with the lost time due to the delay of earlier activities. Instruction by the client for the earlier hand over the project. Thus. To reduce the overhead cost of the project. the project duration will not change. determine the S-Curve. A. forecast profit margin.Q24. C. Overview report B. It is an effort to ensure adequate cash is available. S-curves C. A. Cash flow forecasting is an effort to ……………………………of the project. forecast the resource requirements. Milestone report Reason: S-curve is a project management tool that display cumulative values in terms of number of activities or costs over time. Q25. Histograms D. The performance report format that best shows the cumulative values over time for a project is called. B. . Reason: Cash flow forecasting is the projection of income and expenses during the life of the project. determine cash requirements. D. You may also assume that the monthly value and cost for each activity are equally distributed.Questions Q26 – Q29 are referring to the Table Q26 Table Q26 shows the value of works and the program for the construction of a building. you may assume that all costs must be met at the instant they are incurred. II.000 B.000 Total Cost = RM 12. What is the total cost of the project? A.800. 1800 C(SS+3) H 1 D. B(SS+1) 900 D 5 B (FS .520.000  0. RM 12. III.000 D. The gross profit margin is 10% of the value and retention money is 5 %. Activity A starts on 1 February 2017.800.070. RM 11. To simplify the calculation.90 = RM 11. The contractor is given advance payment of RM 700.000 . the advance payment will be returned to the client through progress payment deduction starting from the end of May to the end November equally V. The retention money is repaid in full 3 (three) months after the practical completion.900. Table Q26 Activity Duration Predecessor Total Value of (Month) Work (RM  1000) Start 0 - A 3 Start 1200 B 2 A (SS + 1) 1400 C 3 A. Progress payment is on a monthly basis and measurement will be made at the end of each month and the contractor receives the payment one (1) month later IV.1) 5000 E 2 D (FS . The information’s of the project are as the followings: I.000 Reason: Total Value = RM 12.000 C.2) 800 F 2 E 1200 G 1 B (FS+4). F. RM 11.000 at the start of the project. G 500 Finish 0 H Q26. RM 12.800.520. 180.725.135.000 C.000 D. A.000 . RM 2.230.000 B. At the end of which month. RM 2. RM 1. End of October Q29. RM 2.000 D. RM 2. End of August C.515. RM 1.080. Calculate the maximum cash required for this project. RM 1. End of September D. the maximum cash is required? A.035.000 Q28. What is the expected monthly payment received of all works that are scheduled to be completed in month August only? A.910. RM 2.Q27. End of July B.000 C.000 B. Reason: Month Activity 2 3 4 5 6 7 8 9 10 11 12 A 400 400 400 B 700 700 C 300 300 300 D 1000 1000 1000 1000 1000 E 400 400 F 600 600 G 1800 H 500 Monthly Value 400 1100 2100 1300 1300 1700 1400 2400 600 500 0 Cumulative Monthly Value 400 1500 3600 4900 6200 7900 9300 11700 12300 12800 12800 Monthly Cost 360 990 1890 1170 1170 1530 1260 2160 540 450 0 Cumulative Monthly Cost 360 1350 3240 4410 5580 7110 8370 10530 11070 11520 11520 Payment 700 380 1045 1895 1135 1135 1515 1230 2180 470 475 Cumulative Payment 700 1080 2125 4020 5155 6290 7805 9035 11215 11685 12160 Required Cash 650 2160 2285 1560 1955 2080 2725 2035 305 -165 .