Topics to cover in 2nd part(to p2)

Topics

Chapters

# of lectures

1. Project Management

8

2

2. Inventory

16

1.5

3. Decision Analysis

12

1.5

4. Queuing

13

1

Chapter 8 - Project Management

Chapter Topics (to p3)

Date

Topics

Mar 11

1. The Elements of Project Management

2. The Project Network Probabilistic Activity Times

Mar 25

1. Project Crashing and Time-Cost Trade-Off

2. Formulating the CPM/PERT Network as a Linear Programming Model

Project ManagementQuestions:

Why do we need to study Project Management?

How does a project management technique work?(to p4)(to p5)

ObjectiveThe main purpose is to govern the operations of a project such that all activities involved are well administrated and that we can also control its completion time(to p3)

Project management techniqueSteps to solve a project management problem:

to represent a project problem graphicallyto determine its completion timeto carry out sensitivity analysis, if any(to p6)(to p12)(to p29)

1. Represent a project problem graphicallySteps:

Gather all information and organize them in a table format that consists of: event, processing time, and precedent constraints as follows:

Draw a semantic network to represent them

Special case!(to p7)(to p9)(to p4)

Semantic network to represent themHere, we use three symbols:

node to represent stageline/branch to represent event

arrow to represent precedentconstraintExample(to p8)(to p6)

Example123A4CB203010Rule1: All nodes must starts from oneNode and ends with one node(to p7)

Special case!When two or events taken places in the same time interval(known an concurrent events)Consider the following example!

How to draw it?

(to p10)

Case 1123ABC357Wrong! Rule2: no node can havetwo outcomes and end with the same noteSolution (to p11)

Solutions for Rule 2Three ways to draw it:

12345ABCDummy 1=0Dummy 2 = 01234ABCDummy = 01234ABCDummy = 0Solution 1:Solution 2:Solution 3:What oneis better?A dummy activity showsa precedence relationshipReflects no processing time(to p6)

2. Determine its completion timeConsider the project network as shown in next slide

Question: Is it an easy way to find out the solution?

Answer: YES, it knows as Critical Path Method (CPM)(to p15)(to p13)

The Project Network All Possible Paths for Obtaining a Solution

Figure 8.3Expanded network for building a house showing concurrent activities.

Table 8.1Possible Paths to complete the House-Building Network

Then the completion time for paths A, B, C and D can be computed as (to p14)

The Project Network

Completion time for: path A: 12 3 4 6 7, 3 + 2 + 0 + 3 + 1 = 9 months (Critical Path) path B: 1 2 3 4 5 6 7, 3 + 2 + 0 + 1 + 1 + 1 = 8 months path C: 1 2 4 6 7, 3 + 1 + 3 + 1 = 8 months path D: 1 2 4 5 6 7, 3 + 1 + 1 + 1 + 1 = 7 monthsThe critical path is the longest path through the network; the minimum time the network can be completed.Figure 8.5Alternative paths in the network

This is theSolution!(to p12)

Critical Path Method (CPM)General concepts:For each branch of the project network, we firstly determine four values of ES, EF, LS and LFFor each branch, we compute their slack time,Slack time = (LS-ES) or (LF-EF)The critical path is located at branch that has slack time = 0(Do you know the reason why?)

How it works?(to p16)

How CPM works?Steps:Prepare the project networkConstruct a table as follows:Compute ES and EF Compute LS and LFCompute LS-ES or LF-EF

ESij = max (EFi) EFij = ESi + tij

with EF1=0Critical path when LS-ES=0(to p4)(to p17)(to p26)(to p22)

Compute ES and EFNote: When computing these values, the pattern is like moving zic-zac format by firstly computer ES12 and then adding it to EF12 and move to next branch by copying the max values of the branch 1-2 to say, 2-3 We compute them from top to bottom!Their relationship : Example 1:(to p18)(to p22)(to p19)

The starting point of ES and EFConsider:

ThenEF1 = 0

ES12 = max (EF1)EF12 = ES12 + t12= 0= 0 + t1212t12(to p17)

The overall computation is shown in next slide(to p20)

BranchesESij = max(EFi)EFij=ESij+tij1-22-32-43-44-54-65-66-7ES12= max(EF1)=ES23=max(EF2)=ES24=max(EF2)=ES34=max(EF3)=ES45=max(EF4)=ES46=max(EF4)=ES56=max(EF5)=ES67=max(EF6)=EF12=ES12+t12=EF23=ES23+t23=EF24=EF34=EF45=EF46=EF56=EF67=

- ES is the earliest time an activity can start.ESij = Maximum (EFi)- EF is the earliest start time plus the activity time.EFij = ESij + tij(note:you can compute these values and show in the network diagram as well)

Add all t to note 4 and take the longest timeMax (node 3+t34, node2+t24) max (5+0, 3+1)=max(5,4)=5 add all ti for note 2Max(node4+t46,node5+t56=max(5+3,5+1)=8Complete solution(to p4)(to p21)

Branch

ESij = max (EFi )

EFij = ESij + tij

1 -2

2-3

ES12 = max (EF1) = 0

ES23 = max (EF2) = 3

EF12 = ES12 + t12

= 0 + 3 =3

EF23=ES23+t23

= 3 + 2 = 5

2-4

3-4

4 -5

ES24 = max(EF2) = 3

ES34= max (EF3) = 5

ES45= max (ES4) = 5

EF24=ES24+t24

=3 + 1 = 4

EF34=ES34 + t34

= 5 + 0 = 5

EF45 = ES45 + t45

= 5 + 1 = 6

4 -6

5-6

6-7

ES46=max(EF4) = 5

ES56=max(EF5) = 6

ES67=max(EF6) =8

EF46=ES46+t46

=5 + 3 = 8

EF56=ES56 +t56

=6 + 1 = 7

EF67=ES67+t67

= 8+ 1 = 9

The Project Network Activity Scheduling- Earliest Times- ES is the earliest time an activity can start.ESij = Maximum (EFi)- EF is the earliest start time plus the activity time.EFij = ESij + tijFigure 8.6Earliest activity start and finish times

(to p20)

Compute LS and LFNote: We compute these values from the bottom to top, with assigning:

LSij = LFi -tij LFij = min LSj withthe end of LFij = EFij

Example: computing Figure 8.3

(to p23)

The overall computational is shown in next slide(to p24)

BranchesLSij = LFij-tijLFij=min(LSj)1-22-32-43-44-54-65-66-7LS12 = Li12-t12 =LS23 = LF23-t23 =LS24 = LF24-t24 =LS34 = LF34-t34 =

LS45 = LF45-t45 =LS46 = LF46-i46 =LS56 = LF56-t56 =LS67 = LF67-t67 =LF12=min(LS2)=LF23=min(LS3)=LF24=min(LS4)=LF34=min(LS4)=LF45=min(LS5)=LF46=min(LS6)=LF56=min(LS6)=LF67=min(LS7)=

- LS is the latest time an activity can start without delaying critical path time. LSij = LFij - tij- LF is the latest finish time LFij = Minimum (LSj)Start with the end node firstSame as EF67from the previous slideAgain, you can place these values onto the branches Min(node 6-t46,node5-t45)=Min(8-3,7-1)=Min(5,6)=5Min(node3-t23,node4-t24)=Min(5-2,5-1)=Min(3,4)=3Min(node 7-t67)=Min(9-1)=8(to p25)(to p22)

Branches

LSij=LFij-tij

LFij=min LSj

1-2

2-3

2-4

LS12=LF12-t12 = 3-3 =0

LS23=LF23-t23=5-2=3

LS24=LF24-t24=5-1=4

LF12 = Min(LS2) =3

LF23=Min(LS3) = 5

LF24=Min(LS4)=5

3-4

4-5

4-6

LS34=LF34-t34=5-0 = 5

LS45=LF45-t45 = 7-1=6

LS46=LF46-t46=8-3=5

LF34=Min(LS4) = 5

LF45=Min(LS5)=7

LF46=Min(LS6)=8

5-6

6-7

LS56=LF56-t56=8-1=7

LS67=LF67-t67=9-1=8

LF56=Min(LS6)=8

LF67=Min(LS67)=9

The Project Network Activity Scheduling - Latest Times- LS is the latest time an activity can start without delaying critical path time. LSij = LFij - tij- LF is the latest finish time LFij = Minimum (LSj)Figure 8.7Latest activity start and finish times

(to p24)

Compute LS-ES or LF-EFTwo ways you can achieve it:

by compiling slack, Sij by showing branches

(to p27)(to p28)(to p16)

The Project Network Calculating Activity Slack Time- Slack, Sij, computed as follows: Sij = LSij - ESij or Sij = LFij - EFij Table 8.2 Activity Slack

Figure 8.9Activity Slack

*What does it mean?(to p26)

The Project Network Activity Slack Slack is the amount of time an activity can be delayed without delaying the project. Slack time exists for those activities not on the critical path for which the earliest and latest start times are not equal. Shared slack is slack available for a sequence of activities.Figure 8.8Earliest activity start and finish times

(to p26)

Sensitivity AnalysisToday, we only consider one case

Probabilistic Activity Times

Refer to activity time estimates usually can not be made with certainty

PERT is known as the solution method(to p30)

PERTIn PERT, three different time estimations are applied: most likely time (m), the optimistic time (a) , andthe pessimistic time (b).

How do we make use of these three values?

(to p31)

Probabilistic Activity TimesWe used these values to estimate the mean and variance of a beta distribution:

mean (expected time):

variance:

How to use these values to solve a project network problem?(to p32)

PERTWe simply apply t values in CPM and determine the values of:ESEFLSLFS

and branches with slack = 0 still consider as critical paths

Example.(to p33)

Procedures for PERTStep 1: based on the values of a, b and m, determine the t and v values for each path Step 2: determine the critical path by using t values in the CPMStep 3: compute its corresponding means and standard deviations according.

Example Result implicationApplications(to p34)(to p38)(to p39)

PERT ExampleStep 1: computer t and v values

Step 2: determine the CPM

Step 3: determine v value(to p35)(to p36)(to p37)(to p33)

Step 1: computer t and v values Figure 8.11Network with mean activity times and variances

Table 8.3Activity Time Estimates for Figure 8.10

(to p34)

Step 2: determine the CPMFigure 8.12Earliest and latest activity times

Table 8.4Activity Earliest and Latest Times and Slack

(to p34)

Step 3: determine v value The expected project time is the sum of the expected times of the critical path activities. The project variance is the sum of the variances of the critical path activities. The expected project time is assumed to be normally distributed (based on central limit theorum). In example, expected project time (tp) and variance (vp) interpreted as the mean () and variance (2) of a normal distribution:

= 25 weeks 2 = 6.9 weeks

(to p34)

Critical Path Activity

Variance

1 ( 3

3 ( 5

5 ( 7

7( 9

1

1/9

16/9

4

total 62/9

Probability Analysis of the Project Network- Using normal distribution, probabilities are determined by computing number of standard deviations (Z) a value is from the mean.- Value is used to find corresponding probability in Table A.1, App. A.Figure 8.13Normal distribution of network duration

Critical value(to p33)

Consider when

x = 30

x = 22

Tutorial Assignment

(to p40)(to p41)(to p42)

Probability Analysis of the Project NetworkExample 12 = 6.9 = 2.63Z = (x-)/ = (30 -25)/2.63 = 1.90Z value of 1.90 corresponds to probability of .4713 in Appendix A of p715. Probability of completing project in 30 weeks or less : (.5000 + .4713) = .9713, or 97.13% (Why so high a probability rate?)Figure 8.14Probability the network will be completed in 30 weeks or less

(to p39)

Probability Analysis of the Project NetworkExample 2Z = (22 - 25)/2.63 = -1.14Z value of 1.14 (ignore negative) corresponds to probability of .3729 in Table A.1, appendix A.Probability that customer will be retained is .1271 (= 0.5- 0.3729) , or 12.71%(Again, why so low probability rate?)Figure 8.15Probability the network will be completed in 22 weeks or less

(to p39)

Tutorial AssignmentTry to use QM to solve CPM/PERT problems (see slide 19)

Exercises (Chapter 8)Old: 8, 10, 17New: 4, 6, 11(to p43)

Probability Analysis of the Project NetworkCPM/PERT Analysis with QM for WindowsExhibit 8.1(to p16)

The Project Network Activity Slack Slack is the amount of time an activity can be delayed without delaying the project. Slack time exists for those activities not on the critical path for which the earliest and latest start times are not equal. Shared slack is slack available for a sequence of activities.Figure 8.8Earliest activity start and finish times

The Project Network Calculating Activity Slack Time- Slack, Sij, computed as follows: Sij = LSij - ESij or Sij = LFij - EFij Table 8.2 Activity Slack

Figure 8.9Activity Slack

*