BEE Exam Prep › Paper-1 › Chapter 8

BEE Paper-1 — Chapter 8: Project Management

104 questions — 43 objective (1 mark), 45 short (5 marks), 16 long (10 marks). Every answer is checked against the 2014 BEE guidebook and carries its book section reference plus an explanation.
▶ Practice this chapter interactively (timer, read-aloud, progress saving).

Objective questions (1 mark) — 43

📖 §8.3 PERT — expected time formula

1. An activity in a project is having an optimistic time of 8 days, a most likely time of 15 days and a pessimistic time of 16 days. Its expected time of completion is

  1. 14 days
  2. 13 days
  3. 12 days
  4. none of the above
Answer: A) 14 days
Confirmed vs Book-1 §8.3 — PERT expected time T_E = (T_O + 4T_M + T_P)/6. Here (8 + 4×15 + 16)/6 = (8 + 60 + 16)/6 = 84/6 = 14 days. The most-likely time carries a weight of 4, so the answer is not the plain average (13 days). Option (a).
Source: Sep 2021
📖 §8.3 Float or Slack — float = LS−ES = LF−EF

2. From an activity in a project, latest start time is 8 weeks; latest finish time is 12 weeks. The slack time for the activity is ____.

  1. 1 week
  2. 5 weeks
  3. 4 weeks
  4. none of the above
Answer: D) none of the above
Confirmed vs Book-1 §8.3 — Book-1 defines float only as LS − ES or LF − EF. Here only LS = 8 and LF = 12 are given. LF − LS = 12 − 8 = 4 weeks is the activity DURATION t (since LS = LF − t), not the slack. With no ES or EF supplied the float cannot be computed, so (d) none of the above.
Source: Sep 2021
📖 §8.3 CPM — benefits

3. PERT/CPM provides which of the following benefits

  1. predicts the time required to complete the project
  2. shows activities which are critical to maintaining the schedule
  3. graphical view of the project
  4. all the above
Answer: D) all the above
Confirmed vs Book-1 §8.3 — Book-1 lists the CPM/PERT benefits as: gives a graphical view of the project; predicts the time required to complete the project; shows which activities are critical to maintaining the schedule and which are not. All three options are therefore correct → (d) all the above.
Source: Apr 2010
📖 §8.3 Gantt chart

4. The simplest technique for scheduling of tasks and tracking progress is

  1. CPM
  2. Gantt chart
  3. PERT
  4. CUSUM
Answer: B) Gantt chart
Confirmed vs Book-1 §8.3 — Book-1: Gantt's scheduling tool is a horizontal bar graph 'commonly used for scheduling the tasks and tracking the progress of energy management projects'. It is the simplest such tool; CPM/PERT are the more advanced network methods and CUSUM is a monitoring & targeting technique. Option (b).
Source: Apr 2010
📖 §8.3 CPM — critical path (zero slack)

5. All the activities falling in the critical path of the PERT network will have

  1. ES = LS and EF = LF
  2. LF = LS and ES=EF
  3. only ES=LS
  4. only EF = LF
Answer: A) ES = LS and EF = LF
Confirmed vs Book-1 §8.3 — Book-1: 'The critical path is the path through the project network in which none of the activities have slack, that is, the path for which ES = LS and EF = LF for all activities in the path.' Zero float therefore means ES = LS and EF = LF simultaneously → option (a).
Source: Nov 2009
📖 §8.3 (general management — Pareto 80/20; not defined in Book-1 Ch-8 text)

6. The 80/20 Rule in management means

  1. few (20%) are vital and many (80%) are trivial
  2. many (80%) are vital and few (20%) are trivial
  3. 80% of work is outsourced
  4. 20% of work is outsourced
Answer: A) few (20%) are vital and many (80%) are trivial
Confirmed vs Book-1 §8.3 — The 80/20 (Pareto) rule states that a vital few causes (about 20 %) account for the bulk (about 80 %) of the effect, so effort must be concentrated on that vital few. It is a general management/prioritisation principle used when screening and ranking project opportunities. Option (a).
Source: Nov 2009
📖 §8.3 PERT — three time estimates

7. Project management technique which uses three time estimates.

  1. PERT
  2. CUSUM
  3. CPM
  4. none of the above
Answer: A) PERT
Confirmed vs Book-1 §8.3 — Book-1: 'Unlike CPM where times can be estimated with relative certainty, PERT uses 3 time estimates' — T_O (optimistic), T_M (most likely), T_P (pessimistic). CPM is deterministic (one fixed time) and CUSUM is a monitoring technique → option (a) PERT.
Source: Nov 2009
📖 §8.3 Float or Slack — numerical

8. For an activity in a project, Latest start time is 8 weeks and Latest finish time is 12 weeks. If the earliest finish time is 9 weeks, Slack time for the activity is ____.

  1. 3 weeks
  2. 4 weeks
  3. 1 week
  4. none of the above
Answer: A) 3 weeks
Confirmed vs Book-1 §8.3 — Duration t = LF − LS = 12 − 8 = 4 weeks. Given EF = 9, ES = EF − t = 9 − 4 = 5 weeks. Float = LS − ES = 8 − 5 = 3 weeks, and the cross-check LF − EF = 12 − 9 = 3 weeks agrees. Option (a) 3 weeks.
Source: 2018
📖 §8.3 Float or Slack — definition (Book EOC Objective Q9)

9. Which of the following statements is correct regarding ‘float’ for an activity?

  1. Time between its earliest start time and earliest finish time
  2. Time between its latest start time and latest finish time
  3. Time between latest start time and earliest finish time
  4. Time between earliest finish time and latest finish time
Answer: D) Time between earliest finish time and latest finish time
Confirmed vs Book-1 §8.3 — Book-1: total float is 'the time between its earliest and latest start time, or between its earliest and latest finish time', i.e. Float = LS − ES = LF − EF. Of the four choices only (d), the time between earliest finish and latest finish (LF − EF), is one of these two valid expressions. Options (a) and (b) give the activity duration, and (c) is meaningless.
Source: 2017
📖 §8.3 Float or Slack — definition (Book EOC Objective Q5)

10. The time between its earliest and latest start time, or between its earliest and latest finish time of an activity is

  1. delay time
  2. slack time
  3. critical path
  4. start time
Answer: B) slack time
Confirmed vs Book-1 §8.3 — This is the verbatim Book-1 definition of the total float: 'The total float (slack time) for an activity is the time between its earliest and latest start time, or between its earliest and latest finish time.' Slack (float) is therefore option (b).
Source: 2012
📖 §8.3 CPM — critical path (Book EOC Objective Q7)

11. In project management, the critical path in the network is

  1. the path where activates have slack
  2. the shortest path
  3. the path where no activities have slack
  4. none of the above
Answer: C) the path where no activities have slack
Confirmed vs Book-1 §8.3 — Book-1: 'Critical path is a sequence of activities from start to finish with zero slack' — the path in which none of the activities have slack. It is also the longest-duration path, so option (b) 'shortest path' is wrong → option (c).
Source: 2012
📖 §8.3 PERT — expected time (Book EOC Objective Q1)

12. An activity in a project has an optimistic time of 10 days, a most likely time of 15 days and a pessimistic time of 20 days. Its expected time of completion is

  1. 10 days
  2. 15 days
  3. 30 days
  4. 35 days
Answer: B) 15 days
Confirmed vs Book-1 §8.3 — T_E = (T_O + 4T_M + T_P)/6 = (10 + 4×15 + 20)/6 = (10 + 60 + 20)/6 = 90/6 = 15 days. For a symmetric spread the expected time equals the most likely time. Option (b).
Source: Guidebook
📖 §8.3 Limitation of Gantt chart (Book EOC Objective Q2)

13. Network diagrams show logic clearly but do not have ___ like Gantt chart.

  1. nodes
  2. arrows
  3. time scale
  4. events
Answer: C) time scale
Confirmed vs Book-1 §8.3 — Book-1: 'Such requirements are best served by the network diagram, which shows logic clearly but does not have a time scale axis like the Gantt chart.' The missing feature is therefore the time scale → option (c).
Source: Guidebook
📖 §8.2 Financing / project screening (Book EOC Objective Q3)

14. To judge the attractiveness of any investment, the project manager must consider:

  1. Initial capital cost
  2. Net operating cash inflows
  3. salvage value
  4. all the above
Answer: D) all the above
Confirmed vs Book-1 §8.2 — Investment attractiveness is judged on the whole cash-flow picture: the initial capital cost, the net operating cash inflows over the life of the measure, and the salvage value at the end. Book-1 screens projects on economic feasibility (IRR, NPV, cash flow, payback), all of which need these three inputs → (d) all the above.
Source: Guidebook
📖 §8.3 CPM/PERT — critical path identifies minimum project time (Book EOC Objective Q4)

15. The Critical Path in PERT indicates

  1. minimum time required for the completion of the project
  2. delays in the project
  3. maximum time required for the completion of the project
  4. none of the above
Answer: A) minimum time required for the completion of the project
Corrected (was c) — Book-1 §8.3: Book-1 states explicitly: 'Critical path identifies the minimum time to complete project.' Although the critical path is the LONGEST path through the network, its length is the MINIMUM time in which the project can be completed — no shorter completion is possible because those activities have zero float. Hence option (a); option (c) 'maximum time' is the standard distractor (see the Sep-2024 paper, where 'maximum time required' is keyed as the FALSE statement).
Source: Guidebook
📖 §8.3 Work Breakdown Structure (Book EOC Objective Q6)

16. In project management, work breakdown structure defines

  1. temporary endeavour undertaken to create unique product or service
  2. the activities to be completed in the projects
  3. how realistic were the assumptions underlying the project
  4. none of the above
Answer: B) the activities to be completed in the projects
Confirmed vs Book-1 §8.3 — Book-1: 'Work Breakdown Structure (WBS) is the process of dividing complex projects to simpler and manageable tasks... The WBS can be used to identify the tasks before constructing Gantt chart and networks.' WBS therefore defines the activities to be completed in the project → option (b). Option (a) is the definition of a project itself.
Source: Guidebook
📖 §8.3 Gantt chart (Book EOC Objective Q8)

17. The technique used for scheduling the tasks and tracking the progress of energy management projects through a bar chart is called

  1. CPM
  2. Gantt chart
  3. CUSUM
  4. PERT
Answer: B) Gantt chart
Confirmed vs Book-1 §8.3 — Book-1: 'Gantt's scheduling tool takes the form of a horizontal bar graph... commonly used for scheduling the tasks and tracking the progress of energy management projects.' The bar-chart technique is therefore the Gantt chart → option (b).
Source: Guidebook
📖 §8.3 PERT — expected time (Book EOC Objective Q10)

18. An activity has an optimistic time of 15 days, a most likely time of 18 days and a pessimistic time of 27 days. What is the expected time?

  1. 60 days
  2. 20 days
  3. 19 days
  4. 18 days
Answer: C) 19 days
Confirmed vs Book-1 §8.3 — T_E = (T_O + 4T_M + T_P)/6 = (15 + 4×18 + 27)/6 = (15 + 72 + 27)/6 = 114/6 = 19 days. Option (c). Note it exceeds the most-likely 18 days because the pessimistic tail is longer.
Source: Jul 2022
📖 §8.3 PERT — expected time formula

19. What is the expected time, when the optimistic time, most likely time and pessimistic time are 10, 20 and 30 respectively.

  1. 10
  2. 20
  3. 22
  4. 21
Answer: B) 20
Confirmed vs Book-1 §8.3 — T_E = (T_O + 4T_M + T_P)/6 = (10 + 4×20 + 30)/6 = (10 + 80 + 30)/6 = 120/6 = 20. The distribution is symmetric here, so T_E = T_M = 20 → option (b).
Source: Mar 2023
📖 §8.3 CPM/PERT — benefits

20. PERT/CPM provides which of the following:

  1. Predicts the time required to complete the project
  2. Shows activities which are critical for completing the project as per the schedule
  3. Graphical view of the project
  4. All of the above
Answer: D) All of the above
Confirmed vs Book-1 §8.3 — Book-1 lists the CPM benefits: 'Provides a graphical view of the project. Predicts the time required to complete the project. Shows which activities are critical to maintaining the schedule and which are not.' All three statements are true → option (d) All of the above.
Source: Mar 2023
📖 §8.3 Gantt chart — features

21. Which of the following is correct?

  1. Gantt chart is commonly used for scheduling the tasks and tracking the progress.
  2. Gantt charts are developed using bars.
  3. The length of the Gantt chart shows how long the task is expected to be completed.
  4. All of the above
Answer: D) All of the above
Confirmed vs Book-1 §8.3 — Book-1: Gantt charts are 'commonly used for scheduling the tasks and tracking the progress'; they 'are developed using bars to represent each task'; and 'the length of the bar shows how long the task is expected to take to complete'. All three statements are taken from the text → option (d).
Source: Mar 2023
📖 §8.3 Project planning techniques vs CUSUM

22. The technique not used for scheduling the tasks and tracking of the progress of energy management projects is called ____.

  1. CPM
  2. PERT
  3. Gantt chart
  4. CUSUM
Answer: D) CUSUM
Confirmed vs Book-1 §8.3 — Book-1 Ch-8 lists Gantt chart, CPM and PERT as the project scheduling / progress-tracking techniques. CUSUM (cumulative sum of differences) belongs to energy monitoring & targeting (Ch-9), not to project scheduling → option (d).
Source: Jul 2022
📖 §8.3 CPM — critical path is the longest path

23. Which of the following statements about critical path analysis is true?

  1. The critical path is the longest path through the network
  2. The critical path is the shortest path through the network
  3. Tasks with float can never be a task on critical path
  4. none of the above
Answer: A) The critical path is the longest path through the network
Confirmed vs Book-1 §8.3 — Book-1: 'Identify the critical path (longest path through the network)' and 'The critical path is the longest-duration path through the network.' Option (a) is therefore true. Option (c) is false only in wording sense — by definition critical activities have ZERO float, so a task WITH float cannot lie on the critical path, but option (a) is the direct book statement asked for.
Source: Jul 2022
📖 §8.3 Work Breakdown Structure

24. Work Breakdown Structure (WBS) is mainly used for:

  1. Combining small tasks into one large project
  2. Dividing complex projects into simpler, manageable tasks
  3. Preparing cost estimation only
  4. Eliminating tasks from the project
Answer: B) Dividing complex projects into simpler, manageable tasks
Confirmed vs Book-1 §8.3 — Book-1: 'WBS is the process of dividing complex projects to simpler and manageable tasks... much larger tasks are broken down to manageable chunks of work' that can be easily supervised and estimated. Hence option (b).
Source: Sep 2025
📖 §8.3 Limitation of Gantt chart

25. What is a major limitation of the Gantt chart in project management?

  1. It does not show the duration of activities
  2. It does not clearly show logical dependencies between activities
  3. It cannot be used for construction projects
  4. It requires advanced statistical methods for preparation
Answer: B) It does not clearly show logical dependencies between activities
Confirmed vs Book-1 §8.3 — Book-1, 'Limitation of Gantt chart': 'The Gantt chart does not normally show the logical interdependencies between the predecessor and successor activities very well.' Duration IS shown by the bar length, so (a) is wrong → option (b).
Source: Sep 2025
📖 §8.3 Dummy activity

26. In a project network diagram, why is a dummy activity used?

  1. To represent an activity with very small duration
  2. To show logical dependency between activities with the same start and end nodes
  3. To reduce the total project duration
  4. To allocate additional resources to critical activities
Answer: B) To show logical dependency between activities with the same start and end nodes
Confirmed vs Book-1 §8.3 — Book-1: 'Dummy activity is required if two or more activities having identical starting and ending events... shown as dotted line to ensure that activity C starts only after activity B and activity D are completed.' A dummy has ZERO duration and consumes ZERO resources; it only preserves logic → option (b).
Source: Sep 2025
📖 §8.3 CPM — deterministic model

27. Select the correct statement about the Critical Path Method (CPM):

  1. CPM is a deterministic model that does not take into account variation in completion time
  2. CPM is a probabilistic model that takes into account variation in completion time
  3. CPM is a probabilistic model that does not take into account variation in completion time
  4. CPM is a deterministic model that takes into account variation in completion time
Answer: A) CPM is a deterministic model that does not take into account variation in completion time
Confirmed vs Book-1 §8.3 — Book-1: 'CPM is a deterministic model that does not take into account variation in the completion time, so one fixed time is used for an activity.' PERT, by contrast, is the probabilistic model using three time estimates → option (a).
Source: Sep 2025
📖 §8.3 Float or Slack — float = LS−ES = LF−EF

28. Acceptable delay time (slack time/float) is equal to:

  1. Time between Earliest Finish and Latest Finish
  2. Time between Earliest Start and Latest Start
  3. Both a and b
  4. None of the above
Answer: C) Both a and b
Confirmed vs Book-1 §8.3 — Book-1: total float is 'the time between its earliest and latest start time, OR between its earliest and latest finish time', i.e. Float = LS − ES = LF − EF. The worked example confirms it: for activity C, LS(8) − ES(5) = LF(12) − EF(9) = 3 weeks. Both expressions are valid and equal → option (c) Both a and b.
Source: Sep 2025
📖 §8.3 Float or Slack — critical activity test

29. Is the activity critical, given ES = 8 days and LS = 10 days?

  1. Yes
  2. No
  3. More details required
  4. Next activity details required
Answer: B) No
Confirmed vs Book-1 §8.3 — Float = LS − ES = 10 − 8 = 2 days. A critical activity must have ZERO float (ES = LS and EF = LF). Since the float is 2 days (> 0), the activity is not on the critical path → option (b) No.
Source: Sep 2025
📖 §8.3 CPM — critical path properties

30. Which statement is false regarding Critical path?

  1. CP is longest duration path
  2. It identifies minimum time to Complete the project
  3. Activities lies on it cannot be delay
  4. It is maximum time required to complete the project
Answer: D) It is maximum time required to complete the project
Confirmed vs Book-1 §8.3 — Book-1: the critical path is 'the longest-duration path through the network' and 'Critical path identifies the minimum time to complete project'; 'the activities that lie on it cannot be delayed without delaying the project'. So (a), (b) and (c) are all true statements; describing it as the MAXIMUM time required to complete the project is false → option (d).
Source: Sep 2024
📖 §8.3 Float or Slack — numerical

31. For activity in project, Latest start time is 8 weeks and Latest Finish time is 12 weeks. If the earliest finish time is 9 weeks, Slack time for the activity is

  1. 1 Week
  2. 3 Weeks
  3. 4 Weeks
  4. 7 Weeks
Answer: B) 3 Weeks
Confirmed vs Book-1 §8.3 — Duration t = LF − LS = 12 − 8 = 4 weeks; ES = EF − t = 9 − 4 = 5 weeks. Float = LS − ES = 8 − 5 = 3 weeks (check: LF − EF = 12 − 9 = 3 weeks) → option (b).
Source: Sep 2024
📖 Book-1 §8.3 PERT — expected time formula

32. In PERT, the expected time (tE) of an activity is given by:

  1. tE = (to + tm + tp) / 3
  2. tE = (to + 4·tm + tp) / 6
  3. tE = (to + 4·tp + tm) / 6
  4. tE = (tp − to) / 6
Answer: B) tE = (to + 4·tm + tp) / 6
Confirmed vs Book-1 §8.3 — "Expected time = (Optimistic + 4 × Most likely + Pessimistic) / 6", i.e. Tₑ = (Tₒ + 4Tₘ + Tₚ)/6, a weighted average from the assumed beta distribution. The most-likely time carries the weight of 4; the total divisor is 6. (Tₚ − Tₒ)/6 in option (d) is the standard deviation, not the expected time.
Source: AI practice
📖 Book-1 §8.3 PERT; Ch-8 Objective Q1 (identical figures)

33. For an activity with to = 10, tm = 15, tp = 20 days, the expected time tE is:

  1. 15 days
  2. 16.5 days
  3. 14 days
  4. 45 days
Answer: A) 15 days
Confirmed vs Book-1 §8.3 — this is guidebook Ch-8 Objective Q1, answer 15 days. Working: Tₑ = (10 + 4×15 + 20)/6 = (10 + 60 + 20)/6 = 90/6 = 15 days. The estimates here are symmetric, so the plain average coincides with Tₑ — do not let that habit carry over to skewed data (see the 15/18/27 item).
Source: AI practice
📖 Book-1 §8.3 PERT; Ch-8 Objective Q10 (identical figures)

34. For an activity with to = 15, tm = 18, tp = 27 days, the expected time tE is:

  1. 20 days
  2. 19 days
  3. 18 days
  4. 21 days
Answer: B) 19 days
Confirmed vs Book-1 §8.3 — this is guidebook Ch-8 Objective Q10, answer 19 days. Working: Tₑ = (15 + 4×18 + 27)/6 = (15 + 72 + 27)/6 = 114/6 = 19 days. The plain average (15+18+27)/3 = 20 days is the planted distractor — the ×4 weighting on Tₘ genuinely changes the answer.
Source: AI practice
📖 Book-1 §8.3 PERT — standard deviation / variance (Example 8.2 and solved example)

35. In PERT, an activity has an optimistic time of 4 days and a pessimistic time of 22 days. Its standard deviation is:

  1. 3 days
  2. 9 days
  3. 6 days
  4. 18 days
Answer: A) 3
Confirmed vs Book-1 §8.3 — the book’s worked tables use V = ((Tₚ − Tₒ)/6)², so σ = (Tₚ − Tₒ)/6. Working: σ = (22 − 4)/6 = 18/6 = 3 days; the variance would be σ² = 9. Options repaired to carry units. Note the book sums the variances of the CRITICAL-PATH activities only to get project variance.
Source: AI practice
📖 Book-1 §8.3 CPM — Float or Slack or Free Time; Ch-8 Objective Q9

36. The float (slack) of an activity is correctly given by:

  1. Float = ES − LS
  2. Float = LS − ES = LF − EF
  3. Float = EF − ES
  4. Float = LF + EF
Answer: B) Float = LS − ES = LF − EF
Confirmed vs Book-1 §8.3 — "The total float (slack time) for an activity is the time between its earliest and latest start time, or between its earliest and latest finish time", i.e. Float = LS − ES = LF − EF. The book demonstrates it on activity C: LS(8) − ES(5) = LF(12) − EF(9) = 3 weeks. "Critical path is a sequence of activities from start to finish with zero slack."
Source: AI practice
📖 Book-1 §8.3 CPM — Example 8.1, activity C in Table 8.3

37. An activity has ES = 5, EF = 9, LS = 8 and LF = 12 weeks. Its float (slack) is:

  1. 0 weeks
  2. 3 weeks
  3. 4 weeks
  4. 7 weeks
Answer: B) 3
Confirmed vs Book-1 §8.3 — these are exactly the Table 8.3 values for activity C of Example 8.1, whose float the book gives as 3 weeks. Working: Float = LS − ES = 8 − 5 = 3 weeks = LF − EF = 12 − 9 = 3 weeks. Non-zero float means the activity is NOT critical and may be delayed up to 3 weeks without delaying the project. Options repaired to carry units.
Source: AI practice
📖 Book-1 §8.3 CPM (step 4) and PERT — introduction; Ch-8 Short question S-5

38. The essential difference between CPM and PERT is that:

  1. CPM uses three time estimates; PERT uses one
  2. CPM is deterministic (one time estimate); PERT is probabilistic (three time estimates)
  3. Both are probabilistic
  4. PERT considers cost while CPM does not
Answer: B) CPM is deterministic (one time estimate); PERT is probabilistic (three time estimates)
Confirmed vs Book-1 §8.3 — "CPM is a deterministic model that does not take into account variation in the completion time, so one fixed time is used for an activity." PERT "is a probabilistic network model that allows for randomness in activity completion times" and "uses 3 time estimates" (Tₒ, Tₘ, Tₚ). CPM was developed by DuPont in 1957; PERT in the late 1950s for the U.S. Navy’s Polaris project.
Source: AI practice
📖 Book-1 §8.3 CPM — Identify the Critical Path; Ch-8 Objective Q4

39. The critical path in a network is:

  1. The shortest-duration path, equal to the maximum project time
  2. The longest-duration path, equal to the minimum project completion time
  3. The path with the most activities
  4. The path with maximum total float
Answer: B) The longest-duration path, equal to the minimum project completion time
Confirmed vs Book-1 §8.3 — "The critical path is the longest-duration path through the network" and "Critical path identifies the minimum time to complete project" (guidebook Objective Q4 answers: minimum time required). Its activities have zero slack (ES = LS and EF = LF) and cannot be delayed without delaying the project. Longest path but minimum project duration — the two ideas must be held together.
Source: AI practice
📖 Book-1 §8.3 CPM — Identify the Critical Path (method of the solved example, p.209)

40. A network has paths A-B-F = 9 weeks, A-C-D = 7 weeks, and E = 2 weeks. The critical path and project duration are:

  1. A-C-D, 7 weeks
  2. A-B-F, 9 weeks
  3. E, 2 weeks
  4. A-B-F, 18 weeks
Answer: B) A-B-F, 9 weeks
Confirmed vs Book-1 §8.3 — the book’s solved example lists every start-to-finish path, adds the durations and takes the longest as critical (there, A-C-E-H-I = 39 weeks). Working here: A-B-F = 9, A-C-D = 7, E = 2 → longest = A-B-F = 9 weeks, which is also the minimum project duration. The other paths carry float: A-C-D has 9 − 7 = 2 weeks and E has 9 − 2 = 7 weeks of slack.
Source: AI practice
📖 Book-1 §8.3 Limitation of Gantt chart; Ch-8 Objective Q2

41. The main limitation of a Gantt chart is that it:

  1. Does not show task durations
  2. Has no time axis
  3. Does not clearly show the logical interdependencies between activities
  4. Cannot show more than five tasks
Answer: C) Does not clearly show the logical interdependencies between activities
Confirmed vs Book-1 §8.3 — "The Gantt chart does not normally show the logical interdependencies between the predecessor and successor activities very well." "Such requirements are best served by the network diagram, which shows logic clearly but does not have a time scale axis like the Gantt chart" — which is guidebook Objective Q2. The Gantt chart does have a time axis and does show durations, so (a) and (b) are false.
Source: AI practice
📖 Book-1 §8.3 Project Networking Techniques — Dummy activity (Figures 8.2/8.3)

42. A dummy activity in a network diagram:

  1. Consumes resources but no time
  2. Has zero duration and no resources, showing only a logical dependency
  3. Is always on the critical path
  4. Represents the longest activity
Answer: B) Has zero duration and no resources, showing only a logical dependency
Confirmed vs Book-1 §8.3 — "Dummy activity is required if two or more activities having identical starting and ending events ... shown as dotted line to ensure that activity C starts only after activity B and activity D are completed." It exists purely to express that logical dependency, so it carries no duration and consumes no resources — unlike an "activity", which the book defines as a portion of the project that "uses up resource and consumes time". A dummy also appears in the chapter’s solved-example network.
Source: AI practice
📖 Book-1 §8.3 Work Breakdown Structure (WBS); Ch-8 Objective Q6

43. Breaking down a complex project into smaller, manageable tasks is known as:

  1. Crashing
  2. Work Breakdown Structure (WBS)
  3. Float analysis
  4. Critical path method
Answer: B) Work Breakdown Structure (WBS)
Confirmed vs Book-1 §8.3 — "Work Breakdown Structure (WBS) is the process of dividing complex projects to simpler and manageable tasks", broken into chunks that can be easily supervised and estimated. WBS is developed BEFORE dependencies are identified and durations estimated, and is used to identify the tasks before constructing the Gantt chart and the CPM/PERT networks (and later to assign costs in §8.5). Crashing, by contrast, is shortening critical activities by adding resources.
Source: AI practice

Short questions (5 marks) — 45

📖 §8.3 Float or Slack — importance of slack (Book EOC S-1)

1. What is the importance of slack time in a project network?

Model answer: Slack (float) is the length of time an activity can be delayed without affecting the project completion date. Its importance: it identifies critical activities (zero slack) versus non-critical activities (positive slack); it gives scheduling flexibility, allowing resources to be shifted from non-critical to critical activities; it tells the project manager where to focus monitoring and control; and it helps optimise resource use and absorb delays without pushing out the project deadline. Float = LS - ES = LF - EF.
Direct book Short Question S-1. Slack defined in section on Float/Slack/Free Time.
Source: Book end-of-chapter (S-1)
📖 §8.3 PERT — Limitations of PERT (Book EOC S-2)

2. What are the limitations of PERT?

Model answer: Per the guidebook: (1) The activity time estimates are somewhat subjective and depend on judgment. (2) Even if times are well estimated, PERT assumes a beta distribution for the time estimates, but the actual distribution may differ. (3) Even if the beta assumption holds, PERT assumes the project-completion-time distribution is the same as that of the critical path; because other paths can become critical if their activities are delayed, PERT consistently underestimates the expected project completion time.
Book Short Question S-2. Answer taken verbatim from the 'Limitations of PERT' subsection.
Source: Book end-of-chapter (S-2)
📖 §8.2 Contracting — performance contracts (Book EOC S-3)

3. Explain the following: (a) Guaranteed Savings Performance Contract (b) Shared Savings Performance Contract.

Model answer: (a) Guaranteed Savings Performance Contract: all or part of the savings is guaranteed by the contractor, and all or part of the cost of the equipment and/or services is paid down out of the savings as they are achieved. (b) Shared Savings Performance Contract: the contractor provides the financing and is paid an agreed fraction of the actual savings as they are achieved; this payment is used to pay down the debt cost of the equipment and/or services.
Book Short Question S-3. Definitions taken directly from the Contracting subsection.
Source: Book end-of-chapter (S-3)
📖 §8.3 Network Definitions — activity, event, network (Book EOC S-4)

4. Define activity, event and network.

Model answer: Activity: any portion of a project (a task) that is required by the project, uses up resources and consumes time; in an AOA network it is shown as an arrow. Event: the beginning or ending point of one or more activities, called a 'node'; it consumes no time or resources and is shown as a circle. Network: the combination of all the project activities and the events, shown graphically to display the logical dependency relationships between tasks.
Book Short Question S-4. Definitions taken from the 'Network Definitions' subsection.
Source: Book end-of-chapter (S-4)
📖 §8.3 CPM vs PERT (Book EOC S-5)

5. What is the essential difference between CPM and PERT?

Model answer: The essential difference is in the handling of activity time. CPM is a deterministic model that uses a single fixed time estimate per activity (times known with relative certainty), suited to repetitive/known work, and it considers both time and cost. PERT is a probabilistic model that uses three time estimates - optimistic (To), most likely (Tm) and pessimistic (Tp) - to allow for randomness/uncertainty in completion times, and is suited to research/novel projects. In short: CPM = one deterministic time; PERT = three estimates accounting for uncertainty.
Book Short Question S-5. CPM described as deterministic (one fixed time), PERT probabilistic (3 estimates).
Source: Book end-of-chapter (S-5)
📖 §8.1 What is a Project? — definition

6. Define a 'project' and state why projects are described as temporary and unique.

Model answer: A project is a 'temporary endeavour undertaken to create a unique product or service'. Projects are temporary because they have a definite beginning and a definite end. They are unique because the product or service they create differs in some distinguishing way from similar products or services. Example: the design, installation and commissioning of a cogeneration system in an industry. A project manager is responsible for the project, overseeing contractors and managing the schedule and budget.
Verbatim definition from §8.1 What is a Project.
Source: Book §8.1
📖 §8.1 Project management — definition & objective

7. What is project management, and what is its objective?

Model answer: Project management is a set of principles, methods and techniques that people use to effectively plan and control project work. These principles and techniques help complete projects on schedule, within budget and in full conformance to specifications, while also helping achieve organisational goals such as productivity, quality and cost effectiveness. The objective of project management is to optimise project cost, time and quality.
From §8.1. The cost-time-quality balance is the stated objective.
Source: Book §8.1
📖 §8.2 Project Development Cycle (PDC) — six steps

8. List the six steps of the Project Development Cycle (PDC).

Model answer: The Project Development Cycle has six steps: (1) Project Identification and Screening; (2) Technical Design; (3) Financing; (4) Contracting; (5) Implementation; (6) Performance Monitoring. Projects may be identified internally (by the facility manager during day-to-day energy management or detailed audits) or externally (through systematic audits by a consultant, ESCO or industry organisation).
From §8.2. The six PDC stages listed verbatim.
Source: Book §8.2
📖 §8.2 Project identification & screening criteria

9. What criteria are used to rank/screen project opportunities?

Model answer: Screening criteria to rank project opportunities are: economic feasibility of the energy-saving measures (internal rate of return, net present value, cash flow, average payback); sustainability of the savings over the life of the equipment; ease of quantifying, monitoring and verifying the energy savings; availability of technology and ease of adapting it to Indian conditions; and other environmental and social cost benefits (such as reduction in GHG emissions and local pollutants like SO2).
From §8.2 Project Identification and Screening.
Source: Book §8.2
📖 §8.2 Financing — internal & external sources of funds

10. What are the internal and external sources of funds for an energy management project?

Model answer: Internal sources: direct cash provision from company reserves; revenue budget (if payback is less than one year); new share capital. External sources: bank loans; leasing arrangement; payment by savings (a deal arranged with the equipment supplier); energy services contract; private finance initiative. Before applying for funds, all funding options should be discussed with the finance managers; energy savings often improve the viability of non-energy projects.
From §8.2 Financing subsection - both lists given verbatim.
Source: Book §8.2
📖 §8.2 Contracting — types of contract

11. List and briefly describe the types of contract used to implement a project through an outside contractor.

Model answer: (1) Traditional Contract: all specifications are given to a contractor who buys and installs equipment at a fixed price or cost-plus-markup (also called fixed-price/lump-sum); the contractor takes the risk of unforeseen problems for a larger profit - suited to unknown vendors or risky work. (2) Extended Technical Guarantee/Service: extended guarantees on selected equipment and/or service-maintenance agreements. (3) Extended Financing Terms: contractor offers a lease or financing vehicle with payments based on expected savings. (4) Guaranteed Savings Performance Contract: contractor guarantees all/part of the savings, paid down from savings. (5) Shared Savings Performance Contract: contractor finances and is paid an agreed fraction of actual savings.
From §8.2 Contracting. Five contract types listed in the book.
Source: Book §8.2
📖 §8.2 Contracting — management of contracts

12. What points should be taken care of in the management of project contracts?

Model answer: (1) The competence and capability of all contractors must be ensured - one weak link can affect timely performance. (2) Discipline must be inculcated by insisting contractors/suppliers develop realistic, detailed resource and time plans matching the project plan. (3) Penalties (which may be graduated) must be imposed for failure to meet obligations, while incentives may be offered for good performance. (4) Help should be extended to contractors/suppliers with genuine problems - treat them as partners. (5) Project authorities must retain independence to off-load contracts (partly or wholly) to others in time where delays are anticipated.
From §8.2 Contracting subsection - the five management points.
Source: Book §8.2
📖 §8.2 Performance Monitoring & Project Review

13. Why is performance monitoring needed, and what points must be borne in mind when setting up a monitoring system?

Model answer: A monitoring system keeps a tab on project progress; it helps in anticipating deviations from the implementation plan, analysing emerging problems, and taking corrective action. Points to bear in mind: it should focus sharply on the critical aspects of project implementation; it must lay more emphasis on physical milestones and not on financial targets; and monitoring must be kept simple. After commissioning, a periodic project review compares actual with projected performance - it tests the realism of project assumptions, provides a documented log for future decisions, suggests corrective action, and helps uncover judgmental biases.
From §8.2 Performance Monitoring and Project Review subsections.
Source: Book §8.2
📖 §8.3 Work Breakdown Structure (WBS)

14. What is a Work Breakdown Structure (WBS) and how is it used?

Model answer: WBS (Work Breakdown Structure) is the process of dividing a complex project into simpler, manageable tasks. Larger tasks are broken into manageable chunks of work that can be easily supervised and estimated. The first main deliverable is identified, then higher-level tasks are broken into smaller chunks to the level of detail suited to the project. WBS can be displayed as a tree structure, a list or tables. It is developed before dependencies are identified and activity durations are estimated, and is used to identify the tasks before constructing the Gantt chart and the CPM/PERT networks.
From §8.3 WBS subsection.
Source: Book §8.3
📖 §8.3 Gantt chart

15. What is a Gantt chart and what does it show?

Model answer: A Gantt chart is a horizontal bar graph used for scheduling tasks and tracking the progress of (energy management) projects; it was developed by Henry Gantt during the era of scientific management. Each task is drawn as a bar whose length shows how long the task is expected to take, so the duration is easily shown. The horizontal axis is a time scale (absolute or relative to the project start), typically in weeks or months; rows of bars show the start and end dates of individual tasks. Bars may overlap where tasks run in parallel.
From §8.3 Gantt chart subsection.
Source: Book §8.3
📖 §8.3 Limitation of Gantt chart

16. State the main limitation of a Gantt chart and how a network diagram compares.

Model answer: The main limitation of a Gantt chart is that it does not normally show the logical interdependencies between predecessor and successor activities very well. Such requirements are best served by the network diagram, which shows the logic clearly but does not have a time-scale axis like the Gantt chart. So the two are complementary: the network diagram shows dependency logic, while the Gantt chart shows timing/duration on a time axis.
From §8.3 'Limitation of Gantt chart'. Also basis of Objective Q2 (network has no time scale).
Source: Book §8.3 (Objective Q2 basis)
📖 §8.3 Gantt Chart Enhancement

17. In what ways can a basic Gantt chart be enhanced to communicate more information?

Model answer: A basic Gantt chart can be enhanced by: (1) a vertical marker line through a given day to mark the present point in time and show the activities scheduled simultaneously (revealing the need for simultaneous resources); (2) shading each bar in proportion to progress so each activity's status is seen at a glance; (3) depicting dependencies using link lines or colour codes; (4) specifying resource allocation for each task; and (5) showing milestones.
From §8.3 'Gantt Chart Enhancement' subsection.
Source: Book §8.3
📖 §8.3 Project networking — AOA vs AON

18. Distinguish between Activity-On-Arrow (AOA) and Activity-On-Node (AON) representations of a project network.

Model answer: In Activity-On-Arrow (AOA), activities are represented by arrows that are connected at events (nodes shown as circles) to show the preceding/succeeding dependencies. In Activity-On-Node (AON), the activities are represented as the nodes, and the events that signify the beginning or ending of activities are depicted as arcs or lines between the nodes; AON is the method used in project management software packages.
From §8.3 - two ways of representing activities (AOA and AON).
Source: Book §8.3
📖 §8.3 Dummy activity

19. What is a dummy activity and why is it used?

Model answer: A dummy activity is required when two or more activities have identical starting and ending events. It is shown as a dotted line and has zero duration and consumes no resources; it exists only to show a logical dependency. For example, a dummy can be inserted to ensure that activity C starts only after both activity B and activity D are completed. It enforces correct logic in the network without representing real work.
From §8.3 - dummy activity definition. (Notes confirm dotted, zero duration, no resources, logic only.)
Source: Book §8.3
📖 §8.3 Critical Path Method (CPM) — origin & benefits

20. What is the Critical Path Method (CPM), and what benefits does it provide?

Model answer: CPM (Critical Path Method) is a deterministic project-management technique developed by DuPont in 1957 to manage the shutdown and restart of chemical plants for maintenance. It uses one fixed time per activity (it does not account for variation in completion time). Benefits: it provides a graphical view of the project; it predicts the time required to complete the project; and it shows which activities are critical to maintaining the schedule and which are not.
From §8.3 CPM subsection - origin (DuPont 1957) and the three stated benefits.
Source: Book §8.3
📖 §8.3 Steps in CPM project planning

21. List the six steps in CPM project planning.

Model answer: (1) Specify the individual activities (from the WBS; single start and single finish). (2) Determine the sequence of the activities (decide which precede and which follow). (3) Draw the network diagram. (4) Estimate the completion time for each activity (CPM uses one fixed deterministic time). (5) Identify the critical path (the longest path through the network). (6) Update the CPM diagram as the project progresses (a new critical path may emerge).
From §8.3 'Steps in CPM Project Planning'.
Source: Book §8.3
📖 §8.3 CPM — ES, EF, LF, LS parameters

22. Define the four time parameters ES, EF, LF and LS used to identify the critical path, with their formulas.

Model answer: ES (Earliest Start) = the earliest time an activity can start, given its predecessors must be completed first. EF (Earliest Finish) = ES + t (earliest start plus activity duration). LF (Latest Finish) = the latest time an activity can be completed without delaying the project. LS (Latest Start) = LF - t (latest finish minus activity duration). ES/EF are found in the forward pass and LF/LS in the backward pass; t is the activity duration.
From §8.3 'Identify the Critical Path' - the four parameters defined verbatim.
Source: Book §8.3
📖 §8.3 Forward pass (earliest start rule) & backward pass (latest finish rule)

23. State the earliest-start-time rule (forward pass) and the latest-finish-time rule (backward pass).

Model answer: Earliest start time rule (forward pass): the earliest start time for an activity leaving a particular node is equal to the largest of the earliest finish times (EF) for all activities entering that node. Latest finish time rule (backward pass): the latest finish time for an activity entering a particular node is equal to the smallest of the latest start times (LS) for all activities leaving that node. Forward pass uses EF = ES + t starting at ES = 0; backward pass uses LS = LF - t starting from the project duration.
From §8.3 - the two rules stated verbatim in the book.
Source: Book §8.3
📖 §8.3 Float or Slack — definition & formula (float = LS−ES = LF−EF)

24. Define float (slack) for an activity and give the formula. (Objective Q9/Q5 basis)

Model answer: The total float (slack/free time) of an activity is the time between its earliest and latest start time, or between its earliest and latest finish time. It is the amount of time an activity can be delayed past its earliest start (or earliest finish) without delaying the project. Formula: Float = LS - ES = LF - EF. Example from the book: slack for activity C = LS(8) - ES(5) = LF(12) - EF(9) = 3 weeks.
From §8.3 - float definition and worked C=3 weeks example. Matches Objective Q5 and Q9.
Source: Book §8.3
📖 §8.3 Critical path & critical activities

25. Define the critical path and critical activities, and state their significance.

Model answer: The critical path is the sequence (longest-duration path) of activities from start to finish in which none of the activities have slack - i.e. ES = LS and EF = LF for every activity on it. Critical activities are the activities lying on the critical path; they have zero float and cannot be delayed without delaying the whole project. The critical path identifies the minimum time to complete the project: if any activity on it is shortened or extended, the project time shortens or extends accordingly.
From §8.3. Critical path = zero-slack longest path = minimum project time. Matches Objective Q7 and Q4.
Source: Book §8.3
📖 §8.3 Project crashing & use of float

26. What is project crashing, and what is the rule for adding or saving resources on a network?

Model answer: If resources are spent to speed up activities, it should be done only for critical activities; shortening the critical path by adding resources to critical activities is called project crashing. Conversely, if resources are to be saved by lengthening some activities, this should be done only for non-critical activities, and only up to the limit of their float. The book also notes the CPM diagram should be updated as the project progresses, since a new critical path may emerge.
From §8.3 Float/Slack subsection - definition of crashing and the resource-saving rule.
Source: Book §8.3
📖 §8.3 PERT — definition, origin & assumptions

27. What is PERT, what does it assume, and where was it developed?

Model answer: PERT (Program Evaluation and Review Technique) is a probabilistic network model that allows for randomness in activity completion times. It was developed in the late 1950s for the U.S. Navy's Polaris project, which had thousands of contractors, and it has the potential to reduce both the time and cost of a project. Unlike CPM (one certain time), PERT uses three time estimates - most likely (Tm), optimistic (To) and pessimistic (Tp) - and assumes a beta probability distribution (approximated as a normal/bell-shaped distribution) for the activity times.
From §8.3 PERT subsection - definition, beta distribution assumption, Polaris origin.
Source: Book §8.3
📖 §8.3 PERT — T_E, standard deviation & variance formulae

28. State the PERT formulas for expected time, standard deviation and variance of an activity.

Model answer: Expected (mean) time: TE = (To + 4*Tm + Tp) / 6, where To = optimistic time, Tm = most likely time, Tp = pessimistic time (note the most-likely term is weighted by 4). Standard deviation: sigma = (Tp - To) / 6. Variance: V = sigma^2 = ((Tp - To)/6)^2. The project variance is found by summing the variances of the activities on the critical path only; this allows the probability of completing the project by a given date to be estimated.
From §8.3 PERT subsection. Units match the activity time units (e.g. weeks; variance in weeks^2). Basis of Objective Q1 (10,15,20 -> 15) and Q10 (15,18,27 -> 19).
Source: Book §8.3 (Objective Q1/Q10 basis)
📖 §8.3 Benefits of PERT

29. What information/benefits does PERT provide?

Model answer: PERT provides: the expected project completion time; the probability of completion before a specified date; the critical-path activities that directly impact the completion time; the activities that have slack time and can lend resources to critical-path activities; and the start and end dates of activities. The variance in project completion time is obtained by summing the variances of the critical-path activities, enabling probabilistic completion estimates.
From §8.3 'Benefits of PERT' subsection.
Source: Book §8.3
📖 §8.5 Planning Budget — three categories of project cost

30. Into which three categories do project costs fall, and what does each include?

Model answer: Project costs fall into three categories: (1) Human Resource Costs - costs associated with project personnel, including salaries and the cost of benefits (e.g. vacation time, health insurance). (2) Administrative Costs - everyday costs that support the project work but are not tied to a specific task, e.g. local telephone expense, copier paper, heating expenses, support personnel. (3) Resource Costs - things specific to the project such as materials for specific tasks, equipment leases, long-distance telephone, and travel expenses.
From §8.5 Planning Budget - the three cost categories.
Source: Book §8.5
📖 §8.5 Planning Budget — direct vs indirect costs

31. Differentiate between direct costs and indirect costs in a project budget.

Model answer: Direct costs are costs that can be directly attributed to the project work, such as team-member salaries, equipment rentals and training for team members. Indirect costs are not specific to the project - they are shared costs, for example the lease cost of a building where the project team works alongside other (non-project) employees, or administrative staff/managers/functional members who assist but are not assigned project tasks. Each organisation has its own procedures for accounting for indirect costs, with guidance from Finance/Accounting.
From §8.5 'Direct Costs versus Indirect Costs'.
Source: Book §8.5
📖 §8.5 Budget Estimation — analogous vs bottom-up estimating

32. Distinguish between analogous estimating and bottom-up estimating.

Model answer: Analogous estimating establishes an estimate for the current project based on the actual costs of previous projects that are similar in size and scope (a top-down approach using historical data). Bottom-up estimating establishes individual estimates for each task and adds them all together to determine the total estimate for the project. The WBS is used to identify, estimate and assign costs: the work-package level shows individual task costs, while higher WBS levels show rolled-up total costs.
From §8.5 Budget Estimation subsection.
Source: Book §8.5
📖 §8.4 Implementation Plan for Top Management

33. What should top management identify before implementing a project, and what role does monitoring play at the implementation stage?

Model answer: Before implementation, top management should identify their strengths and weaknesses (internal forces) and opportunities and threats (external forces) - i.e. a SWOT view. Strengths and opportunities are positive forces to be exploited; weaknesses and threats are hindrances for which means of overcoming must be devised. Monitoring is important at the implementation phase to ensure the project runs to schedule; it is a continuous process put in place before implementation begins, must appear on the work plan, and should involve all stakeholders so problems can be identified and corrected.
From §8.4 Implementation Plan for Top Management - SWOT and monitoring.
Source: Book §8.4
📖 §8.7 Construction / commissioning process

34. What is project construction (commissioning), and to what does it apply?

Model answer: Project construction is the process of assuring that all systems and components of an industrial installation are designed, installed, tested, operated and maintained according to the operational requirements of the user. The commissioning process applies not only to new projects but also to existing units and systems undergoing expansion, renovation or revamping. In practice it is the integrated application of engineering techniques and procedures to check, inspect and test every operational component - from individual instruments and equipment up to modules, subsystems and complete systems - across all phases up to final handover to the owner.
From §8.7 Construction subsection.
Source: Book §8.7
📖 §8.8 Measurement and Verification (M&V)

35. What is Measurement and Verification (M&V) and where is it primarily applied?

Model answer: Measurement and Verification (M&V), sometimes called monitoring and verification, is a process used to determine energy and demand savings. Its primary application is in energy efficiency projects where the return on the capital investment is tied to the projected energy savings. M&V becomes a central part of a contract when the contract payments or performance guarantee depend on the magnitude of the energy savings achieved. It is primarily focused on the risks that affect the measurement/determination of savings, which are defined in the terms of the contract between the participants.
From §8.8 Measurement and Verification.
Source: Book §8.8
📖 §8.8 M&V — energy savings equation & adjustments

36. State the general M&V equation for energy savings and explain the 'Adjustments' term.

Model answer: Energy or demand savings are found by comparing measured energy use before and after the program: Energy Savings = Base-year Energy Use - Post-Retrofit Energy Use ± Adjustments. The 'Adjustments' term brings the energy use in the two periods to the same set of conditions; conditions commonly affecting energy use are weather, occupancy, plant throughput and equipment operation. Adjustments are derived from identifiable physical facts, may be positive or negative, and are commonly made to restate base-year energy use under post-retrofit conditions (made routinely, e.g. for weather, or as necessary, e.g. when a second shift is added).
From §8.8 - the savings equation and meaning of Adjustments.
Source: Book §8.8
📖 §8.6 Procurement Procedures

37. What is the role of the procurement department in a project?

Model answer: Many organisations have a procurement department that helps purchase the items needed for the project and may also help create the budget. The project manager should always check with the procurement department to determine whether there are procedures to follow when preparing the budget, and should work with the procurement department when purchasing services from vendors or contractors.
From §8.6 Procurement Procedures.
Source: Book §8.6
📖 §8.2 Technical Design — elements of the feasibility study

38. What elements must a sound technical feasibility study (detailed project report) identify for a project to be a viable investment?

Model answer: A sound technical feasibility study must identify in detail: the proposed new technologies, process modifications, equipment replacements and other measures in the project; the product/technology/material supply chain (locally available or imported, reliability of supply); any special technical difficulties (installation, maintenance, repair) and the associated skills required; preliminary designs and schematics for all major equipment, with design requirements, manufacturer's name/contact and capital cost estimate; and an organisational and management plan for implementation, including timetable, personnel requirements, staff training, project engineering and other logistical issues.
From §8.2 Technical Design subsection - the elements of the feasibility study / detailed project report.
Source: Book §8.2 Technical Design
📖 §8.3 CPM — worked Example 8.1 (Tables 8.2 & 8.3)

39. In the textbook CPM example (activities A-I), the total of all durations is 51 weeks yet the project finishes in 26 weeks. Explain why, and name the critical path.

Model answer: The total of all activity durations (51 weeks) assumes activities are done one after another, but the network shows several activities can be carried out simultaneously (e.g. A and B run in parallel). So the project duration is not the sum of all durations but the length of the longest path through the network. For this example the critical path is A-E-F-G-I (all zero float) and the project duration is 26 weeks; non-critical floats are B=6, C=3, D=2, H=3 weeks.
From Example 8.1 and Table 8.3. Demonstrates critical path = longest path = min duration despite parallelism.
Source: Book §8.3 (worked Example 8.1)
📖 §8.3 PERT — expected time numerical (Book EOC Objective Q1)

40. An activity has optimistic time 10 days, most likely 15 days and pessimistic 20 days. Calculate its expected completion time.

Model answer: Expected time TE = (To + 4*Tm + Tp) / 6 = (10 + 4*15 + 20) / 6 = (10 + 60 + 20) / 6 = 90 / 6 = 15 days. Note the answer is NOT the simple average; the most-likely value is weighted by 4. (For the related Objective Q10: To=15, Tm=18, Tp=27 gives TE = (15 + 72 + 27)/6 = 114/6 = 19 days.)
From Objective Q1 (answer 15 days) and Q10 (answer 19 days). Worked using the PERT TE formula.
Source: Book §8.3 (Objective Q1)
📖 §8.3 PERT — end-of-chapter Solved Example (T_E, variance, critical path)

41. For the book's solved PERT example, show how TE and variance are computed for activity B (To=2, Tm=8, Tp=20), and state the resulting critical path and duration.

Model answer: Expected time TE = (To + 4*Tm + Tp)/6 = (2 + 4*8 + 20)/6 = (2 + 32 + 20)/6 = 54/6 = 9 weeks. Variance V = ((Tp - To)/6)^2 = ((20 - 2)/6)^2 = (18/6)^2 = 3^2 = 9.00 weeks^2 (standard deviation sigma = 3 weeks). In that solved example the paths are A-B-D-E-H-I = 38, A-C-E-H-I = 39 and A-C-F-G-H-I = 34 weeks; the critical path is A-C-E-H-I with project duration 39 weeks.
From the solved example before the Questions section. Demonstrates TE, variance and path comparison.
Source: Book §8.3 (solved end example)
📖 Book-1 §8.3 PERT — formulas, verified against the chapter solved example (activity E)

42. For a PERT activity with optimistic time 6, most likely 8 and pessimistic 22 weeks, calculate the expected time, standard deviation and variance.

Model answer: Expected time TE = (To + 4*Tm + Tp)/6 = (6 + 4*8 + 22)/6 = (6 + 32 + 22)/6 = 60/6 = 10 weeks. Standard deviation sigma = (Tp - To)/6 = (22 - 6)/6 = 16/6 = 2.67 weeks. Variance V = sigma^2 = (16/6)^2 = 256/36 = 7.11 weeks^2. (This matches activity E of the textbook solved example.)
Confirmed vs Book-1 §8.3 — these figures reproduce activity E of the chapter’s solved example (6/8/22 → Tₑ = 10 weeks, V = 7.11), so the working can be checked line by line against the printed table. Formulas used are the book’s: Tₑ = (Tₒ + 4Tₘ + Tₚ)/6 and V = ((Tₚ − Tₒ)/6)². Remember only critical-path variances are summed to get the project variance.
Source: AI practice
📖 Book-1 §8.3 Steps in CPM Project Planning + Example 8.1 (forward/backward pass rules)

43. List the six steps in CPM project planning given in the guidebook, and state the forward-pass and backward-pass rules used to find the critical path.

Model answer: THE SIX STEPS IN CPM PROJECT PLANNING (guidebook §8.3): 1. Specify the individual activities — taken from the Work Breakdown Structure; the project must have a single start activity and a single finish activity. 2. Determine the sequence of those activities — decide which activities must precede and which must follow others. 3. Draw the network diagram — originally activity-on-node (AON), though some planners use activity-on-arrow (AOA); insert a dummy (dotted, zero duration, no resources) where two or more activities share identical starting and ending events. 4. Estimate the completion time for each activity — from past experience or knowledgeable persons. CPM is deterministic, so ONE fixed time is used per activity (PERT instead uses Tₑ = (Tₒ + 4Tₘ + Tₚ)/6). 5. Identify the critical path — the longest-duration path through the network. 6. Update the CPM diagram as the project progresses — actual times replace estimates and a new critical path may emerge. FORWARD PASS (gives ES and EF): start at the origin node with ES = 0 and use EF = ES + t. Earliest start time rule — the earliest start time for an activity leaving a particular node equals the LARGEST of the earliest finish times of all activities entering that node. BACKWARD PASS (gives LF and LS): start at the completion node with LF = the project duration and use LS = LF − t. Latest finish time rule — the latest finish time for an activity entering a particular node equals the SMALLEST of the latest start times of all activities leaving that node. FLOAT AND CRITICAL PATH: Float (slack) = LS − ES = LF − EF. The critical path is the sequence of activities from start to finish with ZERO slack (ES = LS and EF = LF); its length is the minimum time to complete the project. Shortening a critical activity by adding resources is called crashing; resources may be saved only on non-critical activities, up to the limit of their float.
Corrected — Book-1 §8.3: the model answer previously gave a generic six-step solving routine instead of the guidebook’s own "Steps in CPM Project Planning" (specify activities → sequence → draw network → estimate times → identify critical path → update diagram). It now quotes those six steps and then adds the book’s earliest-start-time rule (largest entering EF) and latest-finish-time rule (smallest leaving LS) from Example 8.1. Float = LS − ES = LF − EF, and zero-float activities form the critical path.
Source: AI practice
📖 Book-1 §8.3 CPM — Example 8.1 / chapter solved example (path-enumeration method)

44. A small project has activities: A (Start, 3 wks), B (after A, 4), C (after A, 1), D (after C, 3), E (Start, 2), F (after B, 2); Finish needs D, E, F. Find the critical path and project duration.

Model answer: List all start-to-finish paths and add durations: A-B-F = 3+4+2 = 9 weeks; A-C-D = 3+1+3 = 7 weeks; E = 2 weeks. The longest path is A-B-F = 9 weeks, so the critical path is A-B-F and the project duration is 9 weeks. Floats: A, B, F have zero float (critical); for the A-C-D chain float = 9-7 = 2 weeks (C and D each have 2 weeks slack); E has 9-2 = 7 weeks slack.
Confirmed vs Book-1 §8.3 — the method is the book’s own: enumerate every start-to-finish path, sum the durations and take the longest as the critical path (as done on p.209 for A-C-E-H-I = 39 weeks). Working here: A-B-F = 3+4+2 = 9; A-C-D = 3+1+3 = 7; E = 2 → critical path A-B-F, project duration 9 weeks. Floats: A, B, F = 0 (critical); C and D = 2 weeks each; E = 7 weeks. Durations are drill figures; the method is book-grounded.
Source: AI practice
📖 §8.3 CPM benefits & essential difference between CPM and PERT

45. What are the benefits of the Critical Path Method (CPM)? Explain how PERT differs from CPM.

Model answer: CPM benefits: it provides a graphical view of the project, predicts the time required to complete the project, and shows which activities are critical to the schedule and which are not (it also identifies float/slack and the minimum project duration). CPM uses deterministic single-time estimates and considers time and cost (origin: DuPont 1957). PERT differs in that it is probabilistic: it uses three time estimates (optimistic, most likely, pessimistic) combined as TE = (To + 4Tm + Tp)/6 to account for uncertainty, and is suited to research/novel projects with uncertain durations (origin: US Navy Polaris).
Reported in the Sep 2025 paper. Answer assembled from the book's CPM benefits and CPM-vs-PERT material.
Source: 25th Exam Sep 2025

Long questions (10 marks) — 16

📖 §8.3 CPM — Example 8.1 (Tables 8.2 & 8.3, pp.198-200)

1. Example 8.1 (Illustration of CPM). The Work Breakdown Structure of a project is given with single deterministic activity times. Activity (Immediate Predecessor, Completion time in weeks): A(-,5), B(-,6), C(A,4), D(A,3), E(A,1), F(E,4), G(D&F,14), H(B&C,12), I(G&H,2). (a) Draw the network. (b) Do the forward and backward pass and tabulate ES, EF, LS, LF. (c) Find the float/slack of every activity. (d) List every path with its duration, and identify the critical path and the total project duration. (10 marks)

Model answer: CPM uses one fixed (deterministic) time per activity. The sum of all durations is 51 weeks, but many activities run in parallel so the real project time is the length of the longest path. FORWARD PASS (EF = ES + t; when several arrows enter a node ES = the LARGEST entering EF): A: ES 0, EF 5 | B: ES 0, EF 6 | C: ES 5, EF 9 | D: ES 5, EF 8 | E: ES 5, EF 6 | F: ES 6, EF 10 | G: ES max(EF_D 8, EF_F 10)=10, EF 24 | H: ES max(EF_B 6, EF_C 9)=9, EF 21 | I: ES max(EF_G 24, EF_H 21)=24, EF 26. Project duration = 26 weeks. BACKWARD PASS (LS = LF - t; when several arrows leave a node LF = the SMALLEST leaving LS), set LF_I = 26: I: LF 26, LS 24 | G: LF 24, LS 10 | H: LF 24, LS 12 | F: LF 10, LS 6 | D: LF 10, LS 7 | E: LF 6, LS 5 | C: LF 12, LS 8 | B: LF 12, LS 6 | A: LF 5, LS 0. SUMMARY TABLE (Activity Dur ES EF LS LF Float): A 5 0 5 0 5 0 (critical) B 6 0 6 6 12 6 C 4 5 9 8 12 3 D 3 5 8 7 10 2 E 1 5 6 5 6 0 (critical) F 4 6 10 6 10 0 (critical) G 14 10 24 10 24 0 (critical) H 12 9 21 12 24 3 I 2 24 26 24 26 0 (critical) ALL PATHS & DURATIONS: A-C-H-I = 5+4+12+2 = 23 | A-D-G-I = 5+3+14+2 = 24 | B-H-I = 6+12+2 = 20 | A-E-F-G-I = 5+1+4+14+2 = 26 (longest). CRITICAL PATH = A-E-F-G-I, project duration = 26 weeks. Critical activities (zero float) are A, E, F, G, I. Non-critical floats: B=6, C=3, D=2, H=3 weeks - these can be delayed by that much without delaying the project.
Textbook worked Example 8.1 (single deterministic CPM times). ES/EF/LS/LF, floats and critical path A-E-F-G-I match Table 8.3 in the OCR guidebook exactly.
Source: unknown
📖 §8.3 PERT — Example 8.2 (Tables 8.4 & 8.5, pp.202-204)

2. Example 8.2 (Illustration of PERT). Prepare a PERT chart for the project below. Activity (Immediate Predecessor; optimistic To, most-likely Tm, pessimistic Tp in days): a(-,10,22,22), b(-,20,20,20), c(-,4,10,16), d(a,2,14,32), e(b&c,8,8,20), f(b&c,8,14,20), g(b&c,4,4,4), h(c,2,12,16), i(g&h,6,16,38), j(d&e,2,8,14). (a) Compute expected time Te and variance V for each activity. (b) Do the forward/backward pass for ES, EF, LS, LF and slack. (c) Identify the critical path, the project duration and the project variance. (10 marks)

Model answer: PERT is probabilistic and uses three time estimates. Te = (To + 4Tm + Tp)/6; standard deviation sigma = (Tp - To)/6; variance V = ((Tp - To)/6)^2. EXPECTED TIME & VARIANCE: a: Te=(10+88+22)/6=20, sigma=2, V=4 | b: Te=20, sigma=0, V=0 | c: Te=(4+40+16)/6=10, sigma=2, V=4 | d: Te=(2+56+32)/6=15, sigma=5, V=25 | e: Te=(8+32+20)/6=10, sigma=2, V=4 | f: Te=(8+56+20)/6=14, sigma=2, V=4 | g: Te=4, sigma=0, V=0 | h: Te=(2+48+16)/6=11, sigma=2.33, V=5.4 | i: Te=(6+64+38)/6=18, sigma=5.33, V=28.4 | j: Te=(2+32+14)/6=8, sigma=2, V=4. FORWARD/BACKWARD PASS (Activity Te ES EF LS LF Slack): a 20 0 20 0 20 0 (critical) b 20 0 20 1 21 1 c 10 0 10 4 14 4 d 15 20 35 20 35 0 (critical) e 10 20 30 25 35 5 f 14 20 34 29 43 9 g 4 20 24 21 25 1 h 11 10 21 14 25 4 i 18 24 42 25 43 1 j 8 35 43 35 43 0 (critical) CRITICAL PATH = a-d-j, project duration = 20+15+8 = 43 days. PROJECT VARIANCE = sum of variances on the critical path = V(a)+V(d)+V(j) = 4+25+4 = 33; project standard deviation = sqrt(33) = 5.74 days. This variance lets you compute the probability of finishing by any target date using the normal distribution.
Textbook worked Example 8.2 (PERT). Te, variances, ES/EF/LS/LF, slack and critical path a-d-j = 43 days match Table 8.5 in the OCR guidebook (note: the printed EF of d is a scan typo 25; 20+15 = 35).
Source: unknown
📖 §8.3 Solved Example, end of chapter (pp.208-209)

3. Solved Example. For the tasks, durations and predecessor relationships in the activity table below: (a) draw the network, (b) calculate the expected time for all tasks, (c) calculate the variance for all tasks, (d) determine all possible paths and their estimated durations, (e) identify the critical path. Activity (Immediate Predecessor; To, Tm, Tp in weeks): A(-,4,7,10), B(A,2,8,20), C(A,8,12,16), D(B,1,2,3), E(D&C,6,8,22), F(C,2,3,4), G(F,2,2,2), H(E&G,4,8,12), I(H,1,2,3). A dummy activity is needed so that E starts only after both C and D are complete. (10 marks)

Model answer: Te = (To + 4Tm + Tp)/6 ; V = ((Tp - To)/6)^2. EXPECTED TIME & VARIANCE (Activity Te V): A: (4+28+10)/6 = 7, V = ((10-4)/6)^2 = 1.00 B: (2+32+20)/6 = 9, V = 9.00 C: (8+48+16)/6 = 12, V = 1.78 D: (1+8+3)/6 = 2, V = 0.11 E: (6+32+22)/6 = 10, V = 7.11 F: (2+12+4)/6 = 3, V = 0.11 G: (2+8+2)/6 = 2, V = 0.00 H: (4+32+12)/6 = 8, V = 1.78 I: (1+8+3)/6 = 2, V = 0.11 ALL PATHS & DURATIONS (add the Te values): A-B-D-E-H-I = 7+9+2+10+8+2 = 38 A-C-E-H-I = 7+12+10+8+2 = 39 (longest) A-C-F-G-H-I = 7+12+3+2+8+2 = 34 CRITICAL PATH = A-C-E-H-I, project duration = 39 weeks. Project variance = sum of critical-path variances = V(A)+V(C)+V(E)+V(H)+V(I) = 1.00+1.78+7.11+1.78+0.11 = 11.78; project standard deviation = sqrt(11.78) = 3.43 weeks.
Textbook end-of-chapter Solved Example (9-activity PERT). Te values, variances, all three paths (38/39/34) and critical path A-C-E-H-I = 39 weeks match the OCR guidebook.
Source: unknown
📖 §8.3 Book end-of-chapter Long Question L-1 (p.211)

4. L-1. Construct a PERT/network diagram for the following project and find the critical path. Activity (Duration in weeks, Precedent): A(7, Start), B(3, A), C(1, B), D(8, A), E(2, D&C), F(1, D&C), G(1, D&C), H(3, F), I(2, H), J(1, E&G&I). (i) Draw the network. (ii) What is the critical path? (iii) What is the total duration to complete the project? (iv) What is the available slack in each activity? (10 marks)

Model answer: FORWARD PASS (EF = ES + t; ES = largest entering EF): A: 0-7 | B: 7-10 | C: 10-11 | D: 7-15 | E: ES max(D 15, C 11)=15, EF 17 | F: ES 15, EF 16 | G: ES 15, EF 16 | H: 16-19 | I: 19-21 | J: ES max(E 17, G 16, I 21)=21, EF 22. Project duration = 22 weeks. BACKWARD PASS (LF_J = 22; LS = LF - t; LF = smallest leaving LS): J: LF 22 LS 21 | I: LF 21 LS 19 | H: LF 19 LS 16 | F: LF 16 LS 15 | G: LF 21 LS 20 | E: LF 21 LS 19 | D: LF min(LS_E 19, LS_F 15, LS_G 20)=15, LS 7 | C: LF 15 LS 14 | B: LF 14 LS 11 | A: LF min(LS_B 11, LS_D 7)=7, LS 0. SLACK (LS - ES) for each activity: A=0, B=4, C=4, D=0, E=4, F=0, G=5, H=0, I=0, J=0. PATHS: A-B-C-E-J = 7+3+1+2+1 = 14 ; A-D-E-J = 7+8+2+1 = 18 ; A-D-G-J = 7+8+1+1 = 17 ; A-D-F-H-I-J = 7+8+1+3+2+1 = 22 (longest). (ii) CRITICAL PATH = A-D-F-H-I-J. (iii) Project duration = 22 weeks. (iv) Critical activities A, D, F, H, I, J have zero slack; non-critical slacks are B=4, C=4, E=4, G=5 weeks.
OCR end-of-chapter Long Question L-1. Critical path A-D-F-H-I-J = 22 weeks; full forward/backward pass and slacks computed and verified.
Source: unknown
📖 §8.3 Book end-of-chapter Long Question L-2 (p.212)

5. L-2. (i) Construct a PERT/network diagram for the data below. (ii) Identify the critical path and also compute the earliest start, earliest finish, latest start and latest finish of all activities. Activity (Precedent, Time in weeks): A(Start,3), B(A,4), C(A,1), D(C,3), E(Start,2), F(B,2), and the Finish depends on D, E and F. (10 marks)

Model answer: FORWARD PASS (EF = ES + t): A: 0-3 | E: 0-2 | B: ES 3, EF 7 | C: ES 3, EF 4 | D: ES 4, EF 7 | F: ES 7, EF 9. Project completion = max(EF_D 7, EF_E 2, EF_F 9) = 9 weeks. BACKWARD PASS (project finish = 9; LS = LF - t): F: LF 9 LS 7 | B: LF 7 LS 3 | D: LF 9 LS 6 | C: LF 6 LS 5 | E: LF 9 LS 7 | A: LF min(LS_B 3, LS_C 5)=3, LS 0. SUMMARY (Activity Dur ES EF LS LF Float): A 3 0 3 0 3 0 (critical) B 4 3 7 3 7 0 (critical) C 1 3 4 5 6 2 D 3 4 7 6 9 2 E 2 0 2 7 9 7 F 2 7 9 7 9 0 (critical) PATHS: A-B-F = 3+4+2 = 9 (longest) ; A-C-D = 3+1+3 = 7 ; E = 2. CRITICAL PATH = A-B-F, total project duration = 9 weeks. Floats: C=2, D=2, E=7 weeks.
OCR end-of-chapter Long Question L-2. Critical path A-B-F = 9 weeks; full ES/EF/LS/LF pass computed and verified.
Source: unknown
📖 §8.3 CPM network numerical — method per Example 8.1 (re-solved)

6. Construct a CPM network for the data below and solve it fully. Activity (Predecessor, Duration in weeks): A(Start,4), B(A,5), C(A,2), D(C,5), E(B,3), F(D&E,4). (a) Draw the network. (b) List every path with its duration and identify the critical path. (c) Compute ES, EF, LS, LF and float of all activities and the total project duration. (10 marks)

Model answer: FORWARD PASS (EF = ES + t): A: 0-4 | B: 4-9 | C: 4-6 | D: 6-11 | E: 9-12 | F: ES max(EF_D 11, EF_E 12)=12, EF 16. Project duration = 16 weeks. BACKWARD PASS (LF_F = 16; LS = LF - t): F: LF 16 LS 12 | E: LF 12 LS 9 | D: LF 12 LS 7 | B: LF 9 LS 4 | C: LF 7 LS 5 | A: LF min(LS_B 4, LS_C 5)=4, LS 0. SUMMARY (Activity Dur ES EF LS LF Float): A 4 0 4 0 4 0 (critical) B 5 4 9 4 9 0 (critical) C 2 4 6 5 7 1 D 5 6 11 7 12 1 E 3 9 12 9 12 0 (critical) F 4 12 16 12 16 0 (critical) PATHS: A-B-E-F = 4+5+3+4 = 16 (longest) ; A-C-D-F = 4+2+5+4 = 15. CRITICAL PATH = A-B-E-F, project duration = 16 weeks. Activities C and D each carry 1 week of float.
Re-solved and corrected: an earlier compilation gave 13 weeks / path A-C-D-F, which is arithmetically impossible (A-C-D-F only totals 15 and A-B-E-F totals 16). Correct answer verified by full forward/backward pass. Not in the OCR guidebook.
Source: unknown
📖 §8.3 Network numerical with project crashing — method per §8.3 (re-solved)

7. Construct a PERT/network diagram and solve. Activity (Duration in days, Predecessor): A(2, Start), B(3, A), C(5, A), D(4, B), E(6, B), F(5, C), G(7, D), H(3, E), I(1, F&G&H). (i) Identify the critical path. (ii) Find the total project duration. (iii) Find the slack for activities C and E. (iv) If activity G is crashed by 2 days, what is the new critical path and duration? (10 marks)

Model answer: FORWARD PASS: A 0-2 | B 2-5 | C 2-7 | D 5-9 | E 5-11 | F 7-12 | G 9-16 | H 11-14 | I: ES max(F 12, G 16, H 14)=16, EF 17. Project duration = 17 days. PATHS: A-B-D-G-I = 2+3+4+7+1 = 17 (longest) ; A-B-E-H-I = 2+3+6+3+1 = 15 ; A-C-F-I = 2+5+5+1 = 13. (i) CRITICAL PATH = A-B-D-G-I. (ii) Project duration = 17 days. (iii) Slack from backward pass: C: ES 2, LS 6 -> slack = 4 days. E: ES 5, LS 7 -> slack = 2 days. (iv) CRASHING G by 2 days (G becomes 5): path A-B-D-G-I = 2+3+4+5+1 = 15 and path A-B-E-H-I = 2+3+6+3+1 = 15. Two critical paths now co-exist (A-B-D-G-I and A-B-E-H-I) and the new project duration = 15 days. Crashing G by 2 shortens the project by 2 days; crashing it further would not help because A-B-E-H-I then governs.
Real exam numerical (not in OCR guidebook). Forward/backward pass and crashing re-verified.
Source: unknown
📖 §8.3 PERT numerical, activity-on-arrow — method per Example 8.2 (re-solved)

8. An R&D project has the tasks below given in activity-on-arrow (i-j) form with three time estimates (days). Activity (i-j, name, To, Tm, Tp): 1-2 A 4,6,8 ; 1-3 B 2,3,10 ; 1-4 C 6,8,16 ; 2-4 D 1,2,3 ; 3-4 E 6,7,8 ; 3-5 F 6,7,14 ; 4-6 G 3,5,7 ; 4-7 H 4,11,12 ; 5-7 I 2,4,6 ; 6-7 J 2,9,10. Draw the project network, compute the expected times, list all paths and find the critical path and project duration. (10 marks)

Model answer: Expected time Te = (To + 4Tm + Tp)/6: A(1-2)=6, B(1-3)=4, C(1-4)=9, D(2-4)=2, E(3-4)=7, F(3-5)=8, G(4-6)=5, H(4-7)=10, I(5-7)=4, J(6-7)=8. ALL PATHS from node 1 to node 7 (using Te): 1-2-4-6-7 (A-D-G-J) = 6+2+5+8 = 21 1-2-4-7 (A-D-H) = 6+2+10 = 18 1-3-4-6-7 (B-E-G-J) = 4+7+5+8 = 24 (longest) 1-3-4-7 (B-E-H) = 4+7+10 = 21 1-3-5-7 (B-F-I) = 4+8+4 = 16 1-4-6-7 (C-G-J) = 9+5+8 = 22 1-4-7 (C-H) = 9+10 = 19 CRITICAL PATH = 1-3-4-6-7 i.e. B-E-G-J, project duration = 24 days.
Real exam numerical (23rd exam, Mar 2023; not in OCR guidebook). Expected times and all paths re-verified.
Source: unknown
📖 §8.3 PERT numerical — paths, critical path & activity float (re-solved)

9. For a network whose activity expected durations (from Te = (To + 4Tm + Tp)/6) work out to tA=7, tB=9, tC=12, tD=2, tE=9, tF=3, tG=4, tH=8, tI=2, with logic A(start); B and C after A; D after B; E after D&C; F after D; G after F; H after E&G; I after H: (a) list all the paths and their durations, (b) identify the critical path, (c) determine the earliest start, earliest finish, latest start and latest finish of Task F. (10 marks)

Model answer: ALL PATHS & DURATIONS: A-B-D-F-G-H-I = 7+9+2+3+4+8+2 = 35 A-C-E-H-I = 7+12+9+8+2 = 38 (longest) A-B-D-E-H-I = 7+9+2+9+8+2 = 37 (b) CRITICAL PATH = A-C-E-H-I, project duration = 38 weeks. (c) Task F times: F follows D (D follows B follows A), so ES_F = EF_D = 7+9+2 = 18, EF_F = 18+3 = 21. On the backward pass the F-G branch must reach H (ES_H = 28 on the critical path), so LS_G = 24, giving LF_F = 24 and LS_F = 24-3 = 21. Therefore Task F: ES = 18, EF = 21, LS = 21, LF = 24 (float = 3 weeks).
Real exam numerical (Jul 2022; not in OCR guidebook). Paths and Task-F float re-verified.
Source: unknown
📖 §8.3 Network numerical — critical path & project duration (re-solved)

10. Draw a PERT chart for the tasks below, find the critical path and calculate the expected project duration. Task (Predecessor, Duration in weeks): A(-,3), B(-,5), C(-,7), D(A,8), E(B,5), F(C,5), G(E,4), H(F,5), I(D,6), J(G&H,4). (10 marks)

Model answer: FORWARD PASS: A 0-3, B 0-5, C 0-7 | D 3-11, E 5-10, F 7-12 | G 10-14, H 12-17, I 11-17 | J: ES max(EF_G 14, EF_H 17)=17, EF 21. Project completion = max(EF_I 17, EF_J 21) = 21 weeks. ALL PATHS & DURATIONS: A-D-I = 3+8+6 = 17 B-E-G-J = 5+5+4+4 = 18 C-F-H-J = 7+5+5+4 = 21 (longest) CRITICAL PATH = C-F-H-J, expected project duration = 21 weeks. Branch A-D-I (17) and branch B-E-G-J (18) both finish earlier and carry float (4 and 3 weeks respectively against the 21-week critical path).
Real exam numerical (24th exam, Sep 2024; not in OCR guidebook). Forward pass and paths re-verified.
Source: unknown
📖 §8.3 Critical path & path float — longest path governs (re-solved)

11. For a given network diagram, identify the total number of paths with their durations, the critical path, and the float on each path. The paths through the network are: Start-A-B-C-End = 46 days; Start-D-E-F-End = 33 days; Start-D-B-C-End = 41 days; Start-G-H-I-End = 28 days; Start-G-E-F-End = 31 days. (10 marks)

Model answer: There are five paths through the network: 1. Start-A-B-C-End = 46 days 2. Start-D-E-F-End = 33 days 3. Start-D-B-C-End = 41 days 4. Start-G-H-I-End = 28 days 5. Start-G-E-F-End = 31 days The CRITICAL PATH is the longest one: Start-A-B-C-End = 46 days; it has ZERO float and fixes the project duration at 46 days. PATH FLOAT = (critical-path duration) - (path duration): Path 2: 46 - 33 = 13 days Path 3: 46 - 41 = 5 days Path 4: 46 - 28 = 18 days Path 5: 46 - 31 = 15 days The float tells how much the activities unique to each non-critical path can slip before that path itself becomes critical.
Compilation exam question illustrating the path-listing / float method (not in OCR guidebook). Internally consistent; floats re-verified.
Source: unknown
📖 §8.3 CPM and PERT — methods and essential difference (Book EOC S-5)

12. Explain the Critical Path Method (CPM) and the Programme Evaluation and Review Technique (PERT). What is the essential difference between them? (10 marks)

Model answer: CPM (Critical Path Method): 1. Developed by DuPont in 1957 to plan the shut-down and restart of chemical plants for maintenance. 2. It is a DETERMINISTIC model - a single fixed time is used for each activity; it does not allow for variation in completion time. 3. Best suited to well-known, repetitive work where durations can be estimated with confidence; it emphasises both time AND cost. 4. Benefits: gives a graphical view of the project, predicts the time to complete it, and shows which activities are critical. PERT (Programme Evaluation and Review Technique): 5. Developed in the late 1950s for the U.S. Navy's Polaris missile project, which had thousands of contractors. 6. It is a PROBABILISTIC model that allows for randomness/uncertainty in activity times. 7. It uses THREE time estimates - optimistic (To), most-likely (Tm) and pessimistic (Tp) - and a beta distribution, with expected time Te = (To + 4Tm + Tp)/6. 8. It can give the probability of completing the project before a specified date, using the variance of the critical-path activities. ESSENTIAL DIFFERENCE: 9. CPM uses a single deterministic time estimate per activity, whereas PERT uses three time estimates to account for uncertainty in the activity durations. 10. All other steps (list activities, sequence them, draw the network, forward/backward pass, find float and the critical path) are the same in both techniques - only the time estimation differs.
Directly from OCR guidebook (Section 8.3, CPM and PERT text; Short Question S-5).
Source: unknown
📖 §8.2 Project Development Cycle (PDC) — six steps in detail

13. Describe the various steps in the Project Development Cycle (PDC) of an energy management project. (10 marks)

Model answer: The Project Development Cycle (PDC) has six steps: 1. PROJECT IDENTIFICATION AND SCREENING - components of the project are identified internally (by the facility manager during day-to-day energy management or from audits) or externally (through systematic audits by an ESCO / consultant). Projects are ranked on economic feasibility (IRR, NPV, cash flow, payback), sustainability of the savings, ease of measuring/monitoring/verifying savings, availability and adaptability of the technology, and other environmental/social benefits (e.g. reduced GHG and SOx emissions). 2. TECHNICAL DESIGN - a sound technical feasibility study covering the proposed technologies and process/equipment changes, the supply chain, special technical difficulties and skills required, preliminary designs with capital-cost estimates, and an organisational/management plan (timetable, personnel, training). 3. FINANCING - funds are arranged from internal sources (company reserves, revenue budget if payback < 1 year, new share capital) or external sources (bank loans, leasing, payment by savings, energy-services contract, private finance initiative), after discussion with finance managers. 4. CONTRACTING - since much of the project is executed through contracts, contractor competence is ensured, realistic resource/time plans are insisted upon, penalties and incentives are set, and project authorities retain independence to off-load contracts if delays are anticipated. 5. IMPLEMENTATION - the plans are put into action; measurements to control and measure success are set up, key personnel are won over, and there is timely, frequent communication so changes can be absorbed without loss of control. 6. PERFORMANCE MONITORING - a monitoring system anticipates deviations, analyses emerging problems and triggers corrective action; it focuses on critical aspects and physical milestones (not financial targets) and is kept simple. A periodic PROJECT REVIEW then compares actual with projected performance, tests the original assumptions and provides a documented log for future decisions.
Directly from OCR guidebook (Section 8.2, Project Development Cycle text).
Source: unknown
📖 §8.2 Contracting — types of contract for installation & commissioning

14. When an energy management project is implemented by an outside contractor, several types of contract may be used. Explain the main types of contract used for installation and commissioning. (10 marks)

Model answer: Because a substantial part of a project is executed through contracts, proper contract management is critical - contractor competence must be ensured, discipline and realistic time/resource plans insisted upon, penalties set for failure and incentives for good performance, and the project authority must keep the freedom to off-load a contract if delays loom. The main contract types are: 1. TRADITIONAL CONTRACT (fixed-price / lump-sum) - all specifications are given to a contractor who buys and installs the equipment at a fixed price (or cost-plus-mark-up). The contractor takes the risk of unforeseen problems in exchange for a larger profit. Appropriate when dealing with unknown vendors or when the work is judged risky. 2. EXTENDED TECHNICAL GUARANTEE / SERVICE - the contract offers extended guarantees on the performance of selected equipment and/or service and maintenance agreements. 3. EXTENDED FINANCING TERMS - the contractor provides an extended lease or other financing vehicle in which the payment schedule can be based on the expected energy savings. 4. GUARANTEED SAVINGS PERFORMANCE CONTRACT - all or part of the savings is guaranteed by the contractor, and all or part of the cost of the equipment/services is paid down out of the savings as they are achieved. 5. SHARED SAVINGS PERFORMANCE CONTRACT - the contractor provides the financing and is paid an agreed fraction of the ACTUAL savings as they are achieved; this payment is used to pay down the debt cost of the equipment and/or services.
Directly from OCR guidebook (Section 8.2, Contracting text; Short Question S-3 covers items 4 and 5).
Source: unknown
📖 §8.3 Float/slack, critical path and project crashing

15. Explain the terms float (slack), critical path and project crashing in project network analysis. How are ES, EF, LS and LF used to find the float, and how is crashing applied? (10 marks)

Model answer: 1. The critical path is identified by four parameters for each activity: Earliest Start (ES), Earliest Finish (EF = ES + t), Latest Finish (LF) and Latest Start (LS = LF - t). 2. ES and EF come from a FORWARD PASS: starting at the origin with ES = 0, and where several activities enter a node the next ES equals the LARGEST entering EF. 3. LS and LF come from a BACKWARD PASS: starting at the end node with LF = project duration, and where several activities leave a node the LF equals the SMALLEST leaving LS. 4. FLOAT (or slack) is the time an activity can be delayed without delaying project completion: Float = LS - ES = LF - EF. 5. It shows how much allowance each activity has; e.g. if LS = 8 and ES = 5 (or LF = 12 and EF = 9) the activity has 3 weeks of float. 6. The CRITICAL PATH is the sequence of activities from start to finish with ZERO float, i.e. ES = LS and EF = LF for every activity on it. 7. The critical path is the LONGEST-duration path and it equals the MINIMUM time needed to complete the project; a delay to any critical activity delays the whole project. 8. PROJECT CRASHING is shortening the project by adding resources to speed up activities - and resources should be spent ONLY on CRITICAL activities, because shortening a non-critical activity does not reduce the project time. 9. Conversely, to SAVE resources you may deliberately lengthen only NON-CRITICAL activities, up to the limit of their float. 10. As the project progresses, the network is updated with actual times; a new critical path may then emerge, so the analysis is repeated.
Directly from OCR guidebook (Section 8.3, CPM steps 5-6, Float/Slack text).
Source: unknown
📖 §8.3 PERT — expected time, variance, benefits and limitations

16. Explain how PERT estimates activity time and project completion. Give the formulae for expected time, standard deviation and variance, explain how the project variance is obtained, and list the benefits and limitations of PERT. (10 marks)

Model answer: 1. PERT is a probabilistic network model that allows for randomness in activity completion times and uses THREE time estimates: optimistic (To), most-likely (Tm) and pessimistic (Tp). 2. It assumes a beta probability distribution for each activity, for which the EXPECTED (mean) TIME is the weighted average Te = (To + 4Tm + Tp)/6. 3. STANDARD DEVIATION of an activity: sigma = (Tp - To)/6. 4. VARIANCE of an activity: V = sigma^2 = ((Tp - To)/6)^2. 5. The expected times are placed on the network and the usual forward pass (ES, EF) and backward pass (LS, LF) are done; the slack of each activity = LF - EF = LS - ES, and the critical path is the path with zero slack. 6. The VARIANCE IN PROJECT COMPLETION TIME is obtained by SUMMING the variances of the activities on the CRITICAL PATH only; the project standard deviation is the square root of that sum. 7. Assuming a normal distribution for the critical path, this variance lets you compute the PROBABILITY that the project finishes by a specified date. 8. Benefits of PERT: it gives the expected completion time, the probability of finishing before a target date, the critical activities that directly affect completion, the activities with slack that can lend resources to critical ones, and each activity's start/end dates. 9. Limitations: the three time estimates are subjective/judgemental; the true distribution may not be beta. 10. Also, because other paths can become critical if their activities slip, PERT (which uses only the critical-path distribution) consistently UNDERESTIMATES the expected project completion time.
Directly from OCR guidebook (Section 8.3, PERT text, formulae, Benefits and Limitations of PERT).
Source: unknown