Lesson DE4-DE6
DE4-DE6 Refining models, Gantt charts and resource levelling Quiz: AQA Further Maths, Unit 5
20 questions
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Lesson DE4-DE6, Refining models, Gantt charts and resource levelling: 20 multiple choice questions for the AQA Further Maths (7367), Unit 5: Optional application 3: discrete mathematics, written with Revision Ninja.
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The 20 questions
-
What is a Gantt (cascade) diagram in critical path analysis?
- A histogram showing the number of resources used in each time period
- A chart that shows the total float of each activity as a single number
- A bar chart in which each activity is drawn as a bar from its start time to its finish time against a time axis
- A network diagram with nodes showing durations and arcs showing precedence only
-
What is a resource histogram in critical path analysis?
- A chart showing only the activities on the critical path
- A chart showing how much of a resource is needed at each point in time, based on a schedule
- A bar chart of the total duration of each activity
- A graph of earliest start times plotted against latest start times
-
What is resource levelling?
- Adjusting the timing of non-critical activities within their float to reduce peaks in resource use, usually without extending the project
- Starting every activity at its latest start time
- Increasing the number of resources until every activity becomes critical
- Removing all non-critical activities from the project
-
What is meant by a heuristic procedure, as used in resource levelling?
- An algorithm that proves the project cannot be completed
- A method that always gives the exact optimal schedule
- A rule requiring every activity to be on the critical path
- A practical method that gives a good schedule quickly, but does not guarantee the optimal schedule
-
A model of a project assumes unlimited resources, but the real site has only a few workers. Which refinement is most appropriate?
- Add resource limits to the model, so that activities cannot run together when the workers available would be exceeded
- Replace every activity duration with the average duration of all activities
- Ignore the critical path, since the resources are assumed to be unlimited
- Remove all precedence constraints so that activities can start at once
-
Activities: A (3, no predecessors), B (4, none), C (2, after A), D (5, after B), E (3, after C and D). Workers: A 2, B 1, C 2, D 2, E 1, all started at earliest start times. What is the earliest time at which activity E can start in the early start schedule?
- Time 5
- Time 9
- Time 4
- Time 12
-
In a Gantt chart with activities drawn from their earliest start times, how is the total float of a non-critical activity usually shown?
- As a marker placed at the start of the critical path
- As a separate bar showing its latest start time only
- As a bar twice as long as its duration
- As a line or extension after its bar, showing how far it can slip without delaying the project
-
Activities: A (3, no predecessors), B (4, none), C (2, after A), D (5, after B), E (3, after C and D). Workers: A 2, B 1, C 2, D 2, E 1, all started at earliest start times. What is the peak number of workers needed at any time in the early start schedule?
- 6
- 4
- 3
- 5
-
Activities: A (3, no predecessors), B (4, none), C (2, after A), D (5, after B), E (3, after C and D). Workers: A 2, B 1, C 2, D 2, E 1, all started at earliest start times. In which time interval does the peak demand of workers occur in the early start schedule?
- Between time 4 and time 5
- Between time 0 and time 3
- Between time 9 and time 12
- Between time 5 and time 9
-
Activities: A (3, no predecessors), B (4, none), C (2, after A), D (5, after B), E (3, after C and D). Workers: A 2, B 1, C 2, D 2, E 1, all started at earliest start times. What is the total number of worker-days needed for the whole project?
- 12
- 27
- 30
- 24
-
Activities: A (3, no predecessors), B (4, none), C (2, after A), D (5, after B), E (3, after C and D). Workers: A 2, B 1, C 2, D 2, E 1, all started at earliest start times. Activity C has float 4. Why might resource levelling move C later within its float?
- Because moving C reduces the total number of worker-days needed
- Because moving C within its float can reduce peak demand without delaying the project
- Because C has zero float and so cannot be moved at all
- Because C is critical and must be moved to finish earlier
-
Activities: A (3, no predecessors), B (4, none), C (2, after A), D (5, after B), E (3, after C and D). Workers: A 2, B 1, C 2, D 2, E 1, all started at earliest start times. If only 3 workers are available at any time, is the early start schedule feasible?
- No, because the project duration of 12 exceeds the number of workers
- Yes, because the peak demand is only 3 between time 0 and time 3
- Yes, because the total of 27 worker-days is below the limit
- No, because the demand of 4 workers between times 4 and 5 exceeds the limit of 3
-
If resource levelling delays a critical activity, what is the effect on the project?
- The critical path changes so that it includes only non-critical activities
- The project duration is extended, since critical activities have no float
- The project is unchanged, because levelling moves only non-critical activities
- The project finishes earlier, since levelling shortens activity durations
-
When an activity is moved during resource levelling, what must remain true?
- It must stay within its float, so its predecessors and successors remain satisfied and the project duration is unchanged
- It must be moved onto the critical path
- Its float must be exactly equal to its duration
- It must always start at time zero
-
Which statement about a Gantt chart showing float is correct?
- Critical activities are drawn with float lines longer than those of non-critical activities
- A Gantt chart cannot show critical activities
- Float lines show the duration of each activity
- Float lines show how far each non-critical activity can slip, while critical activities show no float
-
A model assumes that one worker can carry out two activities at the same time. Which refinement is most appropriate?
- Replace each activity with a single node of zero duration
- Delete all activities with positive float from the network
- Add a constraint that each worker can do only one activity at a time, and check the resource histogram against this limit
- Remove the resource constraint, since the model is already accurate
-
What is the main purpose of a resource histogram when refining a critical path model?
- To show whether the schedule needs more resources than are available at any time, so that the schedule can be adjusted
- To replace the whole network with a single activity
- To find the critical path directly from the resource numbers
- To calculate the float of every activity exactly
-
Activity D needs 2 workers for 5 days. How many worker-days does D require?
- 5
- 10
- 2
- 7
-
Two activities that run in parallel each need 3 workers, but only 4 workers are available. Which statement is correct?
- They can run together, because each needs fewer than four workers
- They can run together, because their total of six workers is small
- They cannot run together without exceeding the resources, so one must be delayed within its float or extra workers must be provided
- They must be run in sequence, which leaves the project duration unchanged
-
Which statement about heuristic resource levelling is correct?
- It may not find the best possible schedule, but gives a practical schedule that respects precedence and resource limits
- It guarantees the smallest total float across the network
- It always finds the minimum possible project duration
- It requires that every activity be on the critical path
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