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

Exam code: 9FM0

Paul

Written by: Paul

Reviewed by: Dan Finlay

Updated on

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 bound  Total time of all activitiesMinimum project duration

  • 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

Examiner Tips and Tricks

  • 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.

Total time of all activitiesMinimum project duration=23+4+3+7+6+4+423=5123=2.217 ...

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

Expertise: Maths Content Creator

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.

Dan Finlay

Reviewer: Dan Finlay

Expertise: Portfolio Lead

Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.