INF3708 May/Jun 2011 exam paper — questions

Free sample

  1. Question 1.1 · Nature of software projects · 1 mark

    Regarding the differences between General Project Management and Software Project Management, in terms of the inherent characteristics of software and the software environment, which of the following statements is/are true: (i) Invisibility, (ii) Complexity, (iii) Conformity, (iv) Flexibility?Show the full question

  2. Question 1.2 · Software development models · 1 mark

    When producing a system in a project, different models can be chosen. The following are advantages of one of the process models: (i) large projects may benefit from the limited iteration process allowed; (ii) logical flow aids in understanding; (iii) sequential project processes are easier to plan and implement; (iv) allows project completion times to be forecast with a relative degree of accuracy; (v) it is relatively simple and easy to understand; (vi) enables allocation of tasks within a phase; (vii) the progress can be evaluated at the end of each phase. Which process model has the above advantages?Show the full question

  3. Question 1.3 · Software development models · 1 mark

    Prototypes can be used to eliminate risk and facilitate communication by the following means: (i) specific assumptions, dependencies or concepts are tested, resulting in a better understanding of the system; (ii) end-user participation is encouraged during all stages of development, thereby largely reducing product uncertainty; (iii) the systems development process becomes clear to all stakeholders and tangible deliverables are produced on a continuing basis, allowing for regular end-user assessment and testing; (iv) the iterative approach may identify possible risk areas early in the life cycle, alerting the project manager to apply risk management criteria to reduce the possible influence thereof on the project. Which of the above statements is/are true?Show the full question

  4. Question 1.4 · Evaluation and investment appraisal · 1 mark

    Which of the following factors is taken into consideration by the Net Present Value calculation?Show the full question

  5. Question 1.5 · Nature of software projects · 1 mark

    Software Project Management scope normally comprises the following elements: (i) Project Feasibility, (ii) Project Initiation, (iii) Project Planning, (iv) Project Execution, (v) Project Control, (vi) Project Termination. Which combination of these elements is correct for the Project Management scope?Show the full question

  6. Question 2 · Monitoring, control and earned value · 4 marks

    Name and identify any two (2) tools that can be used to visualise the progress of a project during project monitoring and control, and give a brief description of each of the two tools you have named.Show the full question

  7. Question 3.1 · Evaluation and investment appraisal · 3 marks

    A project scenario presents estimated cash flows (in rands) for three different projects over six years, as follows. Project 1: Year 0 = -R170 000; Year 1 = -R10 000; Year 2 = +R20 000; Year 3 = +R50 000; Year 4 = +R50 000; Year 5 = +R60 000; Year 6 = +R60 000. Project 2: Year 0 = -R150 000; Year 1 = +R5 000; Year 2 = +R20 000; Year 3 = +R30 000; Year 4 = +R80 000; Year 5 = +R90 000; Year 6 = -R10 000. Project 3: Year 0 = -R280 000; Year 1 = +R10 000; Year 2 = +R30 000; Year 3 = +R50 000; Year 4 = +R120 000; Year 5 = +R120 000; Year 6 = +R120 000. Based on this table of estimated cash flows, answer the questions below. Using the cash flow figures given for Project 1, Project 2 and Project 3 (Year 0 through Year 6 as listed in the scenario), calculate the net profit of each of the three projects.Show the full question

  8. Question 3.2 · Evaluation and investment appraisal · 1 mark

    A project scenario presents estimated cash flows (in rands) for three different projects over six years, as follows. Project 1: Year 0 = -R170 000; Year 1 = -R10 000; Year 2 = +R20 000; Year 3 = +R50 000; Year 4 = +R50 000; Year 5 = +R60 000; Year 6 = +R60 000. Project 2: Year 0 = -R150 000; Year 1 = +R5 000; Year 2 = +R20 000; Year 3 = +R30 000; Year 4 = +R80 000; Year 5 = +R90 000; Year 6 = -R10 000. Project 3: Year 0 = -R280 000; Year 1 = +R10 000; Year 2 = +R30 000; Year 3 = +R50 000; Year 4 = +R120 000; Year 5 = +R120 000; Year 6 = +R120 000. Based on this table of estimated cash flows, answer the questions below. Based on the net profit figures you calculated in question 3.1 for Project 1, Project 2 and Project 3, state which project you would select to develop.Show the full question

  9. Question 3.3 · Evaluation and investment appraisal · 4 marks

    A project scenario presents estimated cash flows (in rands) for three different projects over six years, as follows. Project 1: Year 0 = -R170 000; Year 1 = -R10 000; Year 2 = +R20 000; Year 3 = +R50 000; Year 4 = +R50 000; Year 5 = +R60 000; Year 6 = +R60 000. Project 2: Year 0 = -R150 000; Year 1 = +R5 000; Year 2 = +R20 000; Year 3 = +R30 000; Year 4 = +R80 000; Year 5 = +R90 000; Year 6 = -R10 000. Project 3: Year 0 = -R280 000; Year 1 = +R10 000; Year 2 = +R30 000; Year 3 = +R50 000; Year 4 = +R120 000; Year 5 = +R120 000; Year 6 = +R120 000. Based on this table of estimated cash flows, answer the questions below. Applying the shortest payback method, as discussed in Hughes and Cotterell, to the cash flow figures of Project 1, Project 2 and Project 3, determine which project you would now select for development, and explain why.Show the full question

  10. Question 3.4 · Evaluation and investment appraisal · 6 marks

    A project scenario presents estimated cash flows (in rands) for three different projects over six years, as follows. Project 1: Year 0 = -R170 000; Year 1 = -R10 000; Year 2 = +R20 000; Year 3 = +R50 000; Year 4 = +R50 000; Year 5 = +R60 000; Year 6 = +R60 000. Project 2: Year 0 = -R150 000; Year 1 = +R5 000; Year 2 = +R20 000; Year 3 = +R30 000; Year 4 = +R80 000; Year 5 = +R90 000; Year 6 = -R10 000. Project 3: Year 0 = -R280 000; Year 1 = +R10 000; Year 2 = +R30 000; Year 3 = +R50 000; Year 4 = +R120 000; Year 5 = +R120 000; Year 6 = +R120 000. Based on this table of estimated cash flows, answer the questions below. Using the cash flow data for Project 1, Project 2 and Project 3, calculate the Return on Investment (ROI) of each of the three projects.Show the full question

  11. Question 3.5 · Evaluation and investment appraisal · 1 mark

    A project scenario presents estimated cash flows (in rands) for three different projects over six years, as follows. Project 1: Year 0 = -R170 000; Year 1 = -R10 000; Year 2 = +R20 000; Year 3 = +R50 000; Year 4 = +R50 000; Year 5 = +R60 000; Year 6 = +R60 000. Project 2: Year 0 = -R150 000; Year 1 = +R5 000; Year 2 = +R20 000; Year 3 = +R30 000; Year 4 = +R80 000; Year 5 = +R90 000; Year 6 = -R10 000. Project 3: Year 0 = -R280 000; Year 1 = +R10 000; Year 2 = +R30 000; Year 3 = +R50 000; Year 4 = +R120 000; Year 5 = +R120 000; Year 6 = +R120 000. Based on this table of estimated cash flows, answer the questions below. Based on the ROI values you calculated in question 3.4 for Project 1, Project 2 and Project 3, state which project you would select to develop.Show the full question

  12. Question 3.6 · Evaluation and investment appraisal · 6 marks

    A project scenario presents estimated cash flows (in rands) for three different projects over six years, as follows. Project 1: Year 0 = -R170 000; Year 1 = -R10 000; Year 2 = +R20 000; Year 3 = +R50 000; Year 4 = +R50 000; Year 5 = +R60 000; Year 6 = +R60 000. Project 2: Year 0 = -R150 000; Year 1 = +R5 000; Year 2 = +R20 000; Year 3 = +R30 000; Year 4 = +R80 000; Year 5 = +R90 000; Year 6 = -R10 000. Project 3: Year 0 = -R280 000; Year 1 = +R10 000; Year 2 = +R30 000; Year 3 = +R50 000; Year 4 = +R120 000; Year 5 = +R120 000; Year 6 = +R120 000. Based on this table of estimated cash flows, answer the questions below. Assuming a discount rate of 12%, calculate the Net Present Value (NPV) of each of Project 1, Project 2 and Project 3, using the cash flow figures from the scenario. The discount factors at 12% for each year are given as: Year 0 = 1.000; Year 1 = 0.8929; Year 2 = 0.7972; Year 3 = 0.7118; Year 4 = 0.6355; Year 5 = 0.5674; Year 6 = 0.5066.Show the full question

  13. Question 3.7 · Evaluation and investment appraisal · 1 mark

    A project scenario presents estimated cash flows (in rands) for three different projects over six years, as follows. Project 1: Year 0 = -R170 000; Year 1 = -R10 000; Year 2 = +R20 000; Year 3 = +R50 000; Year 4 = +R50 000; Year 5 = +R60 000; Year 6 = +R60 000. Project 2: Year 0 = -R150 000; Year 1 = +R5 000; Year 2 = +R20 000; Year 3 = +R30 000; Year 4 = +R80 000; Year 5 = +R90 000; Year 6 = -R10 000. Project 3: Year 0 = -R280 000; Year 1 = +R10 000; Year 2 = +R30 000; Year 3 = +R50 000; Year 4 = +R120 000; Year 5 = +R120 000; Year 6 = +R120 000. Based on this table of estimated cash flows, answer the questions below. Based on the NPV values you calculated for Project 1, Project 2 and Project 3 in question 3.6, state which project you would now select for development, and give your general conclusion regarding the viability of these projects, basing your answer on the NPVs of each project.Show the full question

  14. Question 4.1 · Activity planning and critical path · 5 marks

    A Critical Path Method (CMP) diagram is provided and must be converted into a Precedence network (Activity-on-node) diagram, using the node-naming convention adopted by Hughes and Cotterell, which is based on British Standard BS 4335 (illustrated by a key box showing Early Start, Duration and Early Finish along the top row of a node, the Task Name in the middle row, and Late Start, Slack and Late Finish along the bottom row). The CMP diagram to be converted shows: node 1 with early start, late start and slack all equal to 0, linked by activity A (duration 3) to node 2, and by activity B (duration 5) to node 3; node 2 shows values of 3 and 6 (split by a diagonal line) with a 3 below; node 3 shows values of 5 and 5 (split by a diagonal line) with a 0 below; node 2 is linked by activity C (duration 6) to node 4, and node 3 is linked by activity D (duration 7) to node 4; node 4 shows values of 12 and 12 (split by a diagonal line) with a 0 below. Convert this CMP diagram into the equivalent Precedence network (Activity-on-node) diagram using the BS 4335 node layout described above.Show the full question

  15. Question 4.2 · Activity planning and critical path · 18 marks

    Consider a set of activities together with their precedents and durations, listed as follows: Task A has no precedents and a duration of 6; Task B has no precedents and a duration of 7; Task C has no precedents and a duration of 28; Task D has precedent B and a duration of 7; Task E has precedent A and a duration of 6; Task F has precedent A and a duration of 9; Task G has precedents D and E and a duration of 5; Task H has precedents F and G and a duration of 8. Using this information, draw a complete Precedence network (Activity-on-node) diagram, and carry out both a forward pass and a backward pass to determine the total project duration and identify the critical path.Show the full question

  16. Question 4.3 · Activity planning and critical path · 4 marks

    Based on the complete Precedence network (Activity-on-node) diagram drawn for the activities A to H in question 4.2 above, list all the possible paths through the network together with each path's total duration.Show the full question

  17. Question 5.1 · PERT and probabilistic scheduling · 3 marks

    A PERT network table lists seven activities (A to G) with their optimistic (a), most likely (m) and pessimistic (b) duration estimates, together with the already-computed expected time (te) and standard deviation (s) for each: Activity A – optimistic 2, most likely 3, pessimistic 4, expected time 3, standard deviation 0.3; Activity B – optimistic 5, most likely 6, pessimistic 8, expected time 6.2, standard deviation 0.5; Activity C – optimistic 3, most likely 4, pessimistic 5, expected time 4, standard deviation 0.3; Activity D – optimistic 1, most likely 3, pessimistic 4, expected time 2.8, standard deviation 0.5; Activity E – optimistic 2, most likely 2, pessimistic 3, expected time 2.2, standard deviation 0.2; Activity F – optimistic 1, most likely 3, pessimistic 5, expected time 3, standard deviation 0.7; Activity G – optimistic 2, most likely 4, pessimistic 5, expected time 3.8, standard deviation 0.5. The target completion date for the project is day 17. An accompanying network diagram shows node 1 as the start event; activity A runs from node 1 to node 2 (where the cumulative expected time is shown as 3 and cumulative standard deviation as 0.3); activity B runs from node 1 to node 3 (cumulative te = 6.2, s = 0.5); activity F also feeds into node 3; activity C runs from node 2 to node 4 (cumulative te = 9, s = 0.7); activity D runs from node 3 to node 4; activity E runs from node 4 to node 5 (cumulative te = 11.2, s = 0.7); and activity G runs from node 5 to node 6, the final event, where the cumulative te is 17 and s is 0.9. Using the PERT network diagram and the expected time/standard deviation data supplied, calculate the standard deviation (s) value associated with activity/node 4 in the network.Show the full question

  18. Question 5.2 · PERT and probabilistic scheduling · 2 marks

    A PERT network table lists seven activities (A to G) with their optimistic (a), most likely (m) and pessimistic (b) duration estimates, together with the already-computed expected time (te) and standard deviation (s) for each: Activity A – optimistic 2, most likely 3, pessimistic 4, expected time 3, standard deviation 0.3; Activity B – optimistic 5, most likely 6, pessimistic 8, expected time 6.2, standard deviation 0.5; Activity C – optimistic 3, most likely 4, pessimistic 5, expected time 4, standard deviation 0.3; Activity D – optimistic 1, most likely 3, pessimistic 4, expected time 2.8, standard deviation 0.5; Activity E – optimistic 2, most likely 2, pessimistic 3, expected time 2.2, standard deviation 0.2; Activity F – optimistic 1, most likely 3, pessimistic 5, expected time 3, standard deviation 0.7; Activity G – optimistic 2, most likely 4, pessimistic 5, expected time 3.8, standard deviation 0.5. The target completion date for the project is day 17. An accompanying network diagram shows node 1 as the start event; activity A runs from node 1 to node 2 (where the cumulative expected time is shown as 3 and cumulative standard deviation as 0.3); activity B runs from node 1 to node 3 (cumulative te = 6.2, s = 0.5); activity F also feeds into node 3; activity C runs from node 2 to node 4 (cumulative te = 9, s = 0.7); activity D runs from node 3 to node 4; activity E runs from node 4 to node 5 (cumulative te = 11.2, s = 0.7); and activity G runs from node 5 to node 6, the final event, where the cumulative te is 17 and s is 0.9. Using the same PERT network diagram and data, calculate the standard deviation (s) value associated with activity/node 5 in the network.Show the full question

  19. Question 5.3 · PERT and probabilistic scheduling · 2 marks

    A PERT network table lists seven activities (A to G) with their optimistic (a), most likely (m) and pessimistic (b) duration estimates, together with the already-computed expected time (te) and standard deviation (s) for each: Activity A – optimistic 2, most likely 3, pessimistic 4, expected time 3, standard deviation 0.3; Activity B – optimistic 5, most likely 6, pessimistic 8, expected time 6.2, standard deviation 0.5; Activity C – optimistic 3, most likely 4, pessimistic 5, expected time 4, standard deviation 0.3; Activity D – optimistic 1, most likely 3, pessimistic 4, expected time 2.8, standard deviation 0.5; Activity E – optimistic 2, most likely 2, pessimistic 3, expected time 2.2, standard deviation 0.2; Activity F – optimistic 1, most likely 3, pessimistic 5, expected time 3, standard deviation 0.7; Activity G – optimistic 2, most likely 4, pessimistic 5, expected time 3.8, standard deviation 0.5. The target completion date for the project is day 17. An accompanying network diagram shows node 1 as the start event; activity A runs from node 1 to node 2 (where the cumulative expected time is shown as 3 and cumulative standard deviation as 0.3); activity B runs from node 1 to node 3 (cumulative te = 6.2, s = 0.5); activity F also feeds into node 3; activity C runs from node 2 to node 4 (cumulative te = 9, s = 0.7); activity D runs from node 3 to node 4; activity E runs from node 4 to node 5 (cumulative te = 11.2, s = 0.7); and activity G runs from node 5 to node 6, the final event, where the cumulative te is 17 and s is 0.9. Using the cumulative expected time and standard deviation at the end of the network together with the project target date of 17, calculate the z value for the project.Show the full question

  20. Question 6.1 · Effort and cost estimation (COCOMO) · 4 marks

    State the equation used in Boehm's formula for calculating effort when applying the COCOMO model, and identify what each of the variables in the equation represents.Show the full question

  21. Question 6.2 · Effort and cost estimation (COCOMO) · 3 marks

    Three systems have been identified, each with its estimated lines of code and system type: System A has 6749 lines of code and is classified as semi-detached mode; System B has 8556 lines of code and is classified as embedded mode; System C has 10485 lines of code and is classified as organic mode. The COCOMO constants for each system type are as follows: Organic mode has c = 2.8 and k = 1.06; Semi-detached mode has c = 3.0 and k = 1.12; Embedded mode has c = 3.4 and k = 1.24. Using this information, determine whether System A can be completed within three years.Show the full question

  22. Question 7 · Effort and cost estimation (COCOMO) · 8 marks

    The SOC-Programming Project has a labour cost table and is scheduled for completion in 50 days. The programming team charges R300 per day in overhead costs for the days scheduled. Phillip and Albert are both project leaders and will each spend an additional week on the project to plan and conduct the post-project review. Phillip will also spend an extra 4 days on the marketing strategy. Daphney works on the project every day of the 50-day schedule. Mpho and Faith work mornings only, which amount to 5 hours per day. Donald works for only 1 week on the project. You may assume a workday consists of 8 hours and a workweek consists of 5 days, giving 8*5=40 hours per week. The hourly costs of the staff members are: Phillip R450; Albert R500; Mpho R380; Daphney R200; Faith R250; Donald R300. Using this information, calculate the total cost of the SOC-Programming Project.Show the full question