Scheduling Activities (Edexcel A Level Further Maths: Decision Maths 1)

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Scheduling Activities

What is meant by scheduling (activities)?

  • Scheduling activities is the process of assigning workers (resources) to activities within a project
  • The two types of problem that arise involve 
    • finding the minimum number of workers such that a project can be completed in its minimum project duration
      • The minimum project duration is also called the critical time of the project
    • finding the (new) minimum project duration given that their are constraints (restrictions) on the maximum number of workers available at any given time
  • In harder problems, certain workers may only be capable of carrying out particular activities

What assumptions are made in scheduling?

  • In scheduling activities the following assumptions are made
    • each activity requires only one worker
      • or one team of workers, i.e.  one resource
    • an activity is assigned the first available worker
    • if there is a choice of activities to assign a worker to choose the activity with the lower late end time
      • i.e.  the lower late event time at the activity end node
    • a worker can only work on one activity at a time
    • once a worker has started an activity, that activity needs to be completed

How do I schedule activities for a project that requires the minimum number of workers?

  • For this type of problem, the critical time of the project cannot change
    • the critical activities cannot be delayed
    • one worker (resource) will complete all the critical activities 
  • Using a Gantt chart and considering the non-critical activities
    • non-critical activities can be delayed, but only within their (total) float times
      • i.e.  early start times can be delayed but late end times cannot
    • visualise each activity as its bar being able to 'slide' (back and forth) within its box (solid and dotted)
      • aim for as few overlaps between activities as possible
    • activities can then be combined into as few rows as possible
      • by placing them 'back-to-back'
      • i.e.  where possible activities should start immediately after others end
    • each row on the (reduced) Gantt chart will then be completed by one worker
      • i.e.  the number of rows is the number of workers
  • Remember that the precedence of activities needs to be maintained
    • e.g.  Activity H, say, cannot move so it starts after activity I, as activity I depends on H being completed first
      • i.e.  H is an immediate predecessor of I

What is the lower bound for the (minimum) number of workers?

  • The lower bound for the number of workers needed such that a project is completed in its minimum project duration is the smallest integer that satisfies

Lower space bound space greater or equal than space fraction numerator Total space time space of space all space activities over denominator Minimum space project space duration end fraction

  • It is not always possible to schedule activities such that the lower bound can be met

How do I schedule activities for a project that has a maximum number of workers?

  • For this type of problem, there will be a maximum number of workers available at any point in time
    • this cannot be exceeded, even if it requires activities to be delayed and the project's critical time increased
  • Using a Gantt chart
    • find the minimum number of workers required to complete the project in its critical time
      • the Gantt chart would have already been largely reduced
      • do this using the process above
    • now consider how any activity (critical or non-critical) can be delayed such that
      •  at any time the Gantt chart uses no more rows than the (maximum) number of workers available
  • As in the first type of problem, the precedence of activities needs to be maintained
    • e.g.  Activity H, say, cannot move so it starts after activity I, as activity I depends on H being completed first
      • i.e.  H is an immediate predecessor of I

Exam Tip

  • Practice scheduling questions as exam preparation
    • there is not an algorithm as such and experience is the best way to become familiar with what to look for

Worked example

A precedence table and Gantt chart for a project are shown below.
Each activity requires one worker and times are given in days.

Activity Immediately preceding activities
A -
B -
C A
D B
E A, D
F B
G C
H C
I G, H
J E, F

reslev-sched-gantt-we-qu

a)
Find the lower bound for the minimum number of workers required to complete the project within its critical time.

table row cell fraction numerator Total space time space of space all space activities over denominator Minimum space project space duration end fraction end cell equals cell fraction numerator 23 plus 4 plus 3 plus 7 plus 6 plus 4 plus 4 over denominator 23 end fraction end cell row blank equals cell 51 over 23 end cell row blank equals cell 2.217 space... end cell end table

The lower bound is the smallest integer greater than or equal to 2.217 ...

The lower bound for the number of workers is 3

b)
Find the minimum number of workers required such that the project can be completed within its critical time.

Worker 1 ('row 1') will be assigned to all the critical activities ('back-to-back')
Worker 2 can start activity B at time zero, with activity D following immediately (at its early start time) and activity E immediately after that

sched-we-ans-b1

By 'sliding' activities F and H it can be seen that there will be (at least) one day where either E and F or F and H will need to happen simultaneously - therefore a third worker is needed
There is some flexibility in assigning worker 3 but activity H is dependent on activity C
Activity J is dependent on E and F but as long as F finishes by its late end time that will look after itself
The adjusted Gantt chart is

sched-gantt-we-ans-a

The (minimum) number of workers is the number of rows in the adjusted Gantt chart, of which there are 3

To complete the project within its critical time (23 days) a minimum of three workers will be required.

c)
Find the minimum project duration given that a maximum of two workers are available at any time.

Using the solution to part b) we can see that activities A, C, B, D and E efficiently occupy two workers
Delaying activity F so it starts immediately after E will delay the whole project by one day (24 days)
(Be careful describing this, activity F would be starting after day 14, so starts on day 15)

This leaves activity H - I is dependent on it (and G) and H itself is dependent on C
Assigning H to worker 2 would delay the project by longer than necessary (at least 28 days) but by assigning H to worker 1 (either between C and G or between G and I) means precedences are maintained

The adjusted Gantt chart is

sched-gantt-we-ans-b

Activity I is the last to finish at a time of 27 days

The minimum project time for a maximum of two workers is 27 days.

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Paul

Author: Paul

Paul has taught mathematics for 20 years and has been an examiner for Edexcel for over a decade. GCSE, A level, pure, mechanics, statistics, discrete – if it’s in a Maths exam, Paul will know about it. Paul is a passionate fan of clear and colourful notes with fascinating diagrams – one of the many reasons he is excited to be a member of the SME team.