TD 608Project Management and Analysis

Part IProject Conception and Execution

Milind SohoniLecture 6

() February 7, 2010 1 / 14

The Scheduling Problem

Recall:

ID Duration After

VC1 1 mo -VC2 1 mo VC1AP 0.5 mo VC1SG 1 mo VC1SP 1 mo SG, APTP 0.5 mo SPF 2 mo APW 2 mo SPT 0.5 mo TP, WFP 0.5 mo T, VC2

W, F, SPTarget 10

Note: the target end-time is setto an abstract number, say 10.

Objectives:Compute feasible start andend times

I No task should startbefore its precedenceshave ended.

Identify the critical pathI Tasks for which delays will

impact the project

Compute slacksI maximum delays for

non-critical tasks

We could do this ad hoc forour small project.

For larger projects, it needs asystematic procedure.

CPM is one such scheme.

() February 7, 2010 2 / 14

The Scheduling Problem

Recall:

ID Duration After

VC1 1 mo -VC2 1 mo VC1AP 0.5 mo VC1SG 1 mo VC1SP 1 mo SG, APTP 0.5 mo SPF 2 mo APW 2 mo SPT 0.5 mo TP, WFP 0.5 mo T, VC2

W, F, SPTarget 10

VC2

SGAP

SP

F

T FP

W

TP

VC1

0.5 1

1 1

0.51

2 2

0.50.5

() February 7, 2010 3 / 14

The Activity Graph

VC2

SGAP

SP

F

T FP

W

TP

VC1

0.5 1

1 1

0.51

2 2

0.50.5

We will construct a graphwhere every node is anactivity and every directededge is a precedence :

Note that:

The relevant data about anactivity, i.e., its label and itsduration is stored at thenode.

If our modellling is correct,then the graph has no cycles

In this case, there is anactivity, viz. FP which hasno successor.

() February 7, 2010 4 / 14

Starting the Sequencing

VC2

SGAP

SP

F

T FP

W

TP

VC1

0.5 1

1 1

0.51

2 2

0.50.5

Next , we repeatedly

We remove a node which hasno successors

we remove all edges leadinginto this node

We maintain the sequence inwhich we remove these nodes

If the graph has no cycles, thenthis process terminates byexhausting all nodes!

Sequence: : FP, T, W

() February 7, 2010 5 / 14

Getting the sequence

VC2

SGAP

SP

F

T FP

W

TP

VC1

0.5 1

1 1

0.51

2 2

0.50.5

Next , we repeatedly

We remove a node which hasno successors

we remove all edges leadinginto this node

We maintain the sequence inwhich we remove these nodes

If the graph has no cycles, thenthis process terminates byexhausting all nodes!

Sequence : : FP, T, W, F, TP,SP, SG, AP, VC2, VC1

() February 7, 2010 6 / 14

Starting the assignment

VC2

SGAP

SP

F

T FP

W

TP

VC1

0.5 1

1 1

0.51

2 2

0.50.5

We go back to the graph and produce a table ofthe nodes in the reverse order of the sequence.

ID Duration After Start End

FP 0.5 mo T, VC2W, F, SP 9.5 10Target 10

We copy the target time as the end-timeof the last activity

The start time is endtime-duration=9.5

() February 7, 2010 7 / 14

The next few nodes

VC2

SGAP

SP

F

T FP

W

TP

VC1

0.5 1

1 1

0.51

2 2

0.50.5

Sequence : : FP, T, W, F, TP, SP, SG, AP,VC2, VC1

ID Duration After Start End

T 0.5 mo TP, W 9 9.5FP 0.5 mo T, VC2

W, F, SP 9.5 10

For the next task, we se all its successors

In this case, for T, the only successor is FP

We put the end-time as the earliest of allsuccesor start-times.

We put the start time asendtime-duration.

() February 7, 2010 8 / 14

Continuing like this ...

VC2

SGAP

SP

F

T FP

W

TP

VC1

0.5 1

1 1

0.51

2 2

0.50.5

Sequence : : FP, T, W, F, TP, SP, SG, AP,VC2, VC1

ID Duration After Start End

SP 1 mo SG, AP 6 7TP 0.5 mo SP 8.5 9F 2 mo AP 7.5 9.5W 2 mo SP 7 9T 0.5 mo TP, W 9 9.5FP 0.5 mo T, VC2

W, F, SP 9.5 10

Coming to SP, we see that:

Successors are TP and W, with start-times7 and 8.5.

Thus 7 is the minimum and the end-timeof SP.

The end-time is as before.

() February 7, 2010 9 / 14

Finally

VC2

SGAP

SP

F

T FP

W

TP

VC1

0.5 1

1 1

0.51

2 2

0.50.5

The earliest start-time isthe project start-time.

ID Duration After Start End

VC1 1 mo - 4 5VC2 1 mo VC1 8.5 9.5AP 0.5 mo VC1 5.5 6SG 1 mo VC1 5 6SP 1 mo SG, AP 6 7TP 0.5 mo SP 8.5 9F 2 mo AP 7.5 9.5W 2 mo SP 7 9T 0.5 mo TP, W 9 9.5FP 0.5 mo T, VC2

W, F, SP 9.5 10

We finally obtain the start and end timesof all activities.

Note that this is quite different from thead hoc schedule we had obtained earlier,e.g., VC2.

() February 7, 2010 10 / 14

Critical Path

VC2

SGAP

SP

F

T FP

W

TP

VC1

0.5 1

1 1

0.51

2 2

0.50.5

Crtical Path are allactivities whereindelays will affect thewhole project.

All activities with the samestart-time as the projectstart are initial activities.

All activities with the sameend-time as the project endare final activities.

All edges connectingactivities which do not havea gap are called criticaledges.

All activites which lie on anypath from initial to finaltasks along critical edges, arecritical activities

Here, the critical path has VC1,SG, SP, W, T, FP, same asbefore!

() February 7, 2010 11 / 14

Slacks

VC2

SGAP

SP

F

T FP

W

TP

VC1

0.5 1

1 1

0.51

2 2

0.50.5

Slack : Window in whichan activity needs to becompleted withoutaffecting project time.

For any activity locate all edges comingin. The latest end-time of these is thestart-time of the window.

For any activity locate all edges goingout. The earliest start-time of these isthe end-time of the window.

For every critical activity, the interval isthe alloted window, and there is noslack.

For a non-critical activity, the windowis larger. For example, for AP:

I VC1 comes in, and ends at 5.I SP, F do out and start at 6, 7.5.I window is [5,6]

Thus AP of duration 0.5 may startanytime between [5,0.5]

() February 7, 2010 12 / 14

Variations and other Mobilization Problems

We have seen how to schedulefrom the project-finish. A similaranalysis can be made from theproject-start end.

The basic points are:

Assumes a unique eventwithout a predecessor.

Peels of one event afteranother, without predecesors.This gives us the sequence.

Start-times of activitiesalloted as latest end-times ofpredecessors.

The general model is of course,Linear Programming, OR

PERTA variation of the CPMmethod which allows foruncertainty in the durations.

The Assignment ProblemEfficient allocation resourcesto activities.

Inventory control andware-housing

Planning the availability ofresources while minimizingstorage costs.

Sourcing and SupplyOptimum purchase strategyfrom various suppliers underdifferent cost regimes.

() February 7, 2010 13 / 14

A Problem and Variation

Conisder the following task list:

Id Dur. After Labour

A 3 - 3B 2 A 1C 2 A 4D 1 A 3E 6 B,C,D 7F 3 D 4G 3 E,F 7H 1 E,F 1I 5 F 6J 2 G,H,I 2

The last column indicates the unitsof labour which are needed to bedeployed for the duration of the task.

Compute a minimum timeduration schedule, assuminga start-time of 0.

What are the critical pathsin this schedule.

Based on this schedule,compute the total labourrequirement at any timemoment.

Is there any other scheduleof minimum time where thepeak labour requirement islower?

() February 7, 2010 14 / 14