Scheduling workers

Lower bound on workers

If the project must finish in the critical time T, the number of workers needed is at least total of all durations ÷ T, rounded up. This ignores the order of activities, so it may not be achievable: it is a bound, not a schedule.

Building a schedule

Assume each activity needs one worker and cannot be split. Work through time. When a worker becomes free, give them an activity whose predecessors are all finished. D1 does not set a fixed scheduling algorithm, but a good rule of thumb is: if there is a choice, choose the activity with the lowest latest finish time (the most urgent), and if two tie, choose the critical one (less float). Draw the schedule as a bar chart with one row per worker.

Checking

Check every precedence condition: no activity starts before all its predecessors finish. Check every activity appears exactly once. State the finishing time.

Typical questions

• Can the project be done in the critical time with k workers? Use the lower bound and the Gantt chart (count activities that must happen at the same time).
• Schedule with a fixed number of workers and state the shortest completion time.
• Explain why a lower bound cannot be met.

Worked example

Find a lower bound for the number of workers needed to complete project J in the critical time.

1. Total of durations = 39
2. Critical time = 19
3. 39 ÷ 19 = 2.053

Answer: At least 3 workers

Check yourself

1. Find a lower bound for the number of workers needed to complete project J in its critical time.
   Show answer

   3. 39 ÷ 19 = 2.053, which rounds up to 3.

2. What is the sum of all the activity durations in project J?
   Show answer

   39 days. 4 + 6 + 3 + 5 + 4 + 7 + 2 + 5 + 3 = 39.

3. Project J is scheduled with only 2 workers, giving each free worker the available activity with the lowest latest finish time (if two tie, the one with less float). When does the project finish?
   Show answer

   20 days. Starts: A at 0 (worker 1), B at 0 (worker 2), C at 4 (worker 1), D at 6 (worker 2), F at 7 (worker 1), E at 11 (worker 2), H at 14 (worker 1), G at 15 (worker 2), I at 17 (worker 2). Last activity ends at 20.

4. Project J is scheduled with 3 workers in the same way. When does the project finish?
   Show answer

   19 days. With 3 workers every activity can start in time, finishing at 19, the critical time.

5. Two activities become available when a worker is free. Which should the worker be given?
   • The one with the longer duration
   • The one with the lower latest finish time
   • The one listed first alphabetically
   • The one with more float
   Show answer

   The one with the lower latest finish time. The lower latest finish time is the more urgent activity.

6. The lower bound for project J is 3 workers. Does this prove 3 workers are enough?
   • Yes: a lower bound is always achievable
   • Yes, if the critical path is unique
   • No: the bound is always too small by one
   • No: the bound ignores precedence, so a schedule must be checked
   Show answer

   No: the bound ignores precedence, so a schedule must be checked. The bound only compares total work with time. Here a schedule with 3 workers does finish at 19, but this had to be checked.

7. Unless told otherwise, what assumption is made about activities when scheduling?
   • They can be shared between workers
   • Each needs one worker and cannot be split once started
   • They can be done in any order
   • They all take one day
   Show answer

   Each needs one worker and cannot be split once started. D1 scheduling assumes one worker per activity and no interruption.

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Sources

• Pearson Edexcel International Advanced Level in Mathematics specification (issue 3)
• Pearson Edexcel IAL Decision Mathematics 1 scheme of work

Last updated 7 November 2026. Spotted a mistake? Tell us.