Em week 11 time management

ENG209: ENGINEERING MANAGEMENT (Elective Course for EEE Students) (2016 Fall, Week 11) Prof. Dr. Mete Gündoğan Slides Preparation Res. Asst. Cihat Öztürk (IE PhD Candidate) Res. Asst. Melda K. Akgün (IE MSc Candidate) 1

ENG209 Week_11

SECTION 4 TIME MANAGEMENT Lesson Content:  Project Scheduling and Control Techniques  Gantt Charts  CPM (Critical Path Method)  Program Evaluation and Review Technique (PERT)

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ENG209 Week_11

GANTT CHART • Graph or bar chart with a bar for each project activity that shows

passage of time • Provides visual display of project schedule

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GANTT CHART • Each bar shows the inception time, finish time and duration of an • • • •

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activity.

It is preferable because of simplicity and userfriendliness.

It doesn’t necessitate qualified employee Every line shows a different workbench, workpiece, product or order.

Columns show dates and horizontal size shows time (duration).

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GANTT CHART  Advantages  It is easy to understand  You can follow progress easily  Creation is easy  It is one of the most preferable methods to tell project to the

customers or managers.

 Disadvatages  It could be superficial.  Seeing precedence relationship could be impossible.

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Steps to Creating a Gantt Chart: Determine Project start date and deadline. ii. Gather all information surrounding the list of activities within a project – the Work Breakdown Structure may be useful for this. iii. Determine how long each activity will take iv. Evaluate what activities are dependant on others v. Create Graph shell including the timeline and list of activities. vi. Using either Forward Scheduling or Backward Scheduling, Begin to add bars ensuring to include dependencies and the full duration for each activity. i.

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Example: Activity

Duration

Dependant On

1. Read Literature 2. Conduct Literature Review 3. Arrange Client Visits 4. Prepare Surveys 5. Conduct Surveys 6. Analyse Surveys 7. Write Up

21 days 14 days 7 days 5 days 14 days 10 days 30 days

N/A 1 N/A 1 and 3 4 5 1,2,3,4,5 and 6

Project Start Date = 1st September Deadline = 20th Decemeber 7

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Example:

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Activity

Duration

Dependant On

1.read

21 days

N/A

2. Conduct literature

14 days

1

3. arrange

7 days

N/A

4. prepare

5 days

1 and 3

5. Conduct surveys

14 days

4

6. analyse

10 days

5

7. Write up

30 days

1,2,3,4,5 and 6

Gantt Chart - Example

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CPM (CRITICAL PATH METHOD) • CPM is a network diagramming technique used to predict total

project duration.

• CPM is an analysis technique with three main purposes:  To calculate the project’s finish date  To indentify to what extent each activity in the schedule can slip(float) without delaying the project  To identify the activities with highest risk that cannot slip without changing the project finish date

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What is CPM?  CPM calculates  The longest path of planned activities to the end of the project  The earliest and latest that each activity can start and finish without

making the project longer  Determines “critical” activities (on the longest path)  Prioritize activities for the effective management and to shorten

the planned critical path of a project by:  Pruning critical path activities  “Fast tracking" (performing more activities in parallel)  “Crashing the critical path" (shortening the durations of critical path

activities by adding resources) 11

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Benefits of CPM  Useful at many stages of project management  Mathematically simple  Give critical path and slack time  Provide project documentation  Useful in monitoring costs

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Limitations to CPM  Clearly defined, independent and stable activities  Specified precedence relationships  Over emphasis on critical paths  Deterministic CPM model  Activity time estimates are subjective and depend on judgment  PERT assumes a beta distribution for these time estimates, but the

actual distribution may be different  PERT consistently underestimates the expected project completion time due to alternate paths becoming critical 13

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The CPM Approach

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CPM Dictionary Activity:  A task which must be performed to complete the project i.e. : assamble, inspect and test  Represented by an arrow  Every activity needs a spesific durtion, cost or both of them.  Every activity has a start time and finish time.

ACTIVITY i

i DURATION (ARROW DIAGRAM)

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ACTIVITY A

FLOW CHART

ACTIVITY B

(PRECEDENCE DIAGRAM)

CPM Dictionary ï‚— Event: Signals the beginning or ending of an activity

Designates a point in time Represented by a circle (node) EVENT 1

ACTIVITY TIME

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EVENT 2

CPM Dictionary  Duration(D) :The duration is needed for actualising an    

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activity Early Start (ES) : the earliest date a task can start Early Finish (EF):the earliest date a task can be completed Late Start (LS) :the latest date a task can start without delaying the project and date Late Finish (LF) :the latest date a task can finish without delaying the project end date

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CPM Dictionary  Float (slack) :

amount of time that a task can be delayed without causing a delay to:  subsequent tasks (free float)  project completion date (total float)

 Path :

A connected sequence of activities leading from the starting event to the ending event  Critical path : the sequence of activities which add up to the longest overall duration. It is the shortest time possible to complete the project.  Critical activity : activity with zero float 18

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CPM Dictionary  Network :

Shows the sequential relationships among activities using nodes and arrows  Activity-on-node (AON) : Nodes represent activities, and arrows show precedence relationships  Activity-on-arrow (AOA) : Arrows represent activities and nodes are events for points in time

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DEFINITIONS A must finish before B

( Act. on Arrow )

A

( Act. On Node )

B

A

A A&B must finish before C

A

C

C

A&B start simultaneously

B

B

A must finish before C

A

C

A&B must finish before D B

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B

D

A

C

B

D

A

A&B must finish before C. A&B must finish before D

C

B

A must finish before D

A

C

B

D

A

D

B

E

C

F

D

A

D

A &B&C must finish before E B

C must finish before F

E

C

F

A must finish before B A must finish before C C&B start simultaneously B&C must finish before D 21

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B

C A

B

D

A

D C

Identify dummy  It is used when two or more activity have the same starting and

ending event.

Boil the water Fill water and milk to the cup

1

2

3

WRONG!!!!!

Worm the milk

 There could be just one activity between two event.

Boil the water

4

DUMMY Fill water and milk to the cup

1

2 Worm the milk

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3

CORRECT√√√

Logic dummy  If there is a comman event between two independent series activity, there could be

logic error.  i.e: We got an offer for a new job and operator training is needed. As soon as operator start to working and machine is installed , operator training will begin.

We wont spend extra time for machine investigation. It will start as soon as machine is installed.

Machine setup

Machine investigation

WRONG!!!!!

Hiring operator

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Operator training

Logic dummy  If there is a comman event between two independent series activity, there could be

logic error.  i.e: We got an offer for a new job and operator training is needed. As soon as operator start to working and machine is installed , operator training will begin.

We wont spend extra time for machine investigation. It will start as soon as machine is installed.

Machine setup

Machine investigation

DUMMY

Hiring operator

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Operator training

CORRECT√√√

NETWORK ï‚— Shows the sequential relationships among activities using nodes

and arrows D B

H

A C

E

FLOW DIAGRAM

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B

F

D

F

A

H

G C

E

G

PRECEDENCE DIAGRAM

NETWORK Questions to prepare activity network;     

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Is this a Start Activity?

Is this a Finish Activity?

What Activity Precedes this?

What Activity Follows this?

What Activity is Concurrent with this?

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Example: Laddering A 10 km

B 10 km

C 10 km

 Pipes will be lied along install 30 km.  Assumption: We have one kit for each activity;  Excavate (EX)  Lie pip (LP)  Back-fill land (BF) 27

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Example: Laddering 1

EX.

LP

2

3

BF

4

EX.

5

Part A 7

LP

6

BF

7

Part B EX.

8

LP

9

BF

10

Part C

****If the activities order serial, only one of the kits will be working when the others are waiting.******

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Example: Laddering EX Part A

Part B

Part C

1

4

7

2

EX

EX

5

LP

LP

3

6

LP 8

9

BF

BF

10

BF

****If we order activity like in the shape, it is not possible to complete the works. Because there is only a kit for every work.*** 29

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Example: Laddering EX A 1

2

LP A

3

BF A

DUMMY

EX B

LP B 4

DUMMY

5

6

7 DUMMY

BF B EX C

8

LP C

9

BF C

10

****We can complete the project as soon as possible and using sources efficiently with help of dummy variables.*** 30

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CPM Rules 1. Two nodes can be connected by at most one arc.

DUMMY

WRONG!!!!!

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CORRECT√√√

CPM Rules 2. Every preceeding activity of an event must be completed before succeding event start.

A

D

B E C

***D and E can not start before A,B and C was completed.*** 32

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CPM Rules 3. In a network time flow always actualises from right to the left.

LOOP

****You must avoid from vicious circle**** 33

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CPM Rules 4. Arrows doesn’t protract depending on duration.

5. Every event has a spesific number.

6. There is always a starting and an ending event.

7. You must avoid from unnecesary dummy activities.

8. The node representing the completion of an activity always has a larger number than the node representing the beginning of an activity.

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Example 2: Draw the CPM network according to given information.  Network beginning event is starting event for A.  When A finish, C and D will start.  When B finish,F and E will start.  When C finish, G will start.  When D,F and G finish, project will be completed.

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 Network beginning event is starting event for A.  When A finish, C and D will start.

Example 2:

 When B finish,F and E will start.  When C finish, G will start.  When D,F and G finish, project will be completed.

D

C

A

G

E

B

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F

Example 3:

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ACTIVITY

START EVENT

FINISH EVENT

A

1

2

B

1

3

C

1

5

D

2

4

E

2

6

F

3

4

DUMMY

3

5

G

3

6

H

4

6

I

5

6

ï‚— Draw the CPM network according to given information.

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ACTIVITY

START EVENT

FINISH EVENT

A

1

2

B

1

3

C

1

5

D

2

4

E

2

6

F

3

4

DUMMY

3

5

G

3

6

H

4

6

I

5

6

Example 3:

E

D

2

A

4

H

F

1

B

G

3

6 I

C 38

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5

Example 4:

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WORK

Duration (days)

Precedence

A

4

-

B

7

A

C

5

A

D

4

B

E

8

-

F

8

B, C, E

G

16

-

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Example 4: Work

Duration (Days)

Precedence

A

4

-

B

7

A

C

5

A

D

4

B

E

8

-

B

F

8

B, C, E

7

G

16

-

1

A 4

Activity on Arrow view

5

E 8 G 40

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16

4

3 C

2

D

4

F 8

5

Example 4: A

START

B

D

C

F

E

G 41

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Activity on Node view

Work

Days

Precedence

A

4

-

B

7

A

C

5

A

D

4

B

E

8

-

F

8

B, C, E

G

16

-

FİNİSH

CPM Calculations • How does the Critical Path Method calculate the project’s finish

date?

• Forward pass calculation  calculates Early Start and Early Finish date

• Backward pass calculation  calculates Late Start and Late Finish dates

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CPM Calculations • By the forward pass and backward pass we can find; • Critical / Uncritical events • Critical path • Float and total float • P.S: Backward pass calculations cannot be done before forward

pass calculations.

Finish Time ?

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FORWARD PASS • It is done from starting to the ending.

• Early Start (ES) and Early Finish ( EF ) times are calculated for every event.

• The early finis time of precedence events determine the early start time of all

succeding events.

0

2

5

A 2 0

6

11

B 1

15 E 4 15

0

Finish 15

Start 0

5 D 5

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Forward Pass ENG209 Week_11

5

11 C 6

11

13 F 2

FORWARD PASS ES=3

ES=10

ES

EF D Activity A

ES=9

 Early Start (ES) = Latest related early date of all immediate predecessors

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ES= immediate Predecessors’ ES max  Early finish (EF) = Early Start + Duration EF= ES + D ENG209 Week_11

BACKWARD PASS  It is done from ending to the starting.  Late Start ( LS ) and Late Finish ( LF ) times are calculated for every event.  The Late start time of succeding events determine the late start time of

preceeding events.

0

2

5

A 2 0

0

8

6

11

B 1 10

10

15 E 4

11

15

11

Start 0

15

Finish

0

0

5

5

D 5 0 46

15

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11 C 6

5

5

11 13

11

13

15

F 2 15 Backward Pass

15

BACKWARD PASS LF=3 D

LF=10

Activity A LS

LF

LF=9

 Late Finish (LF) = Earliest related late date of all immediate successors

LF= immediate Successors’ LF min  Late Start (LS) = Late Finish – Duration LS= LF-D 47

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BACKWARD PASS ï‚— The purpose of backward pass is to find the float.

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Float (Slack)  Project Float (slack) : Amount of time project can be delayed    

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without delaying externally imposed project completion date.

Total Float (slack) : The amount of time an activity can be delayed without delaying project end date.

Critical path activities have zero float.

Knowing float helps in better allocation of resources.

If project is behind an imposed external date, then project will have negative float.

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Float (Slack)

Backward Pass

ES

EF

LS

LF

ES : Early Start LS : Late Start EF : Early Finish LF : Late Finish

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Float = LS-ES

Forward Pass

OR

Float = LF-EF

Float (Slack) 0

2

5

A 2 0

0

8

6

11

B 1 10

10

15 E 4

11

15

11

Start 0

15

Finish

0

0

5

5

D 5 0

11

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11

C 6 5

5

13 F 2

11

ï‚— What is the total Float for activity A? 51

15

13

15

15

15

Float (Slack) 0

2

5

A 2

6 B 1

8

10

10

ES

ES

EF

LS

LF

11

EF 0

LS

2

8

A

 Total float = LF-ES-D = 10 – 0 – 2 = 8 ENG209 Week_11

10

Float Float

52

LF

A

Critical Path • Longest duration path in a Network Diagram • It is the shortest time to complete the project • It helps to determine the focus for project management efforts to

capacity control, cost control and time optimization.

• Near critical path is close in duration to critical path • Closer the near critical and critical paths are, the more risk project has

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Activities on Arrow - Forward Pass 1 ESx

X

2

tx

Y

3

ty

EFx

11

EFy

B

ESy

1 0

7

A

C

2

4

5

4

8 G ENG209 Week_11

4

F

4

8

11 E

54

D

3

16

5

19

Activities on Arrow -Backward Pass 1

X

2

tx

Y

3

ty

LFx

LSx

11 11

LFy

B

LSy

1 0

0

7

A

C

2

4

4

5

4

8 G ENG209 Week_11

4

F

4

8

11 11

E

55

D

3

16

5

19 19

Activities on Arrow – Critical Path 11 11

Total Float = Tyg – ( Txe + t ) TB=0B

Slack = Tye – ( Txe + t )

7

SB=0

TB=4D

3

4

TB=0

SB=4

SB=0 1 0

TB=0A t=4

SB=0

0

Txe Txg

2 4 Tye

TB=2C 5

SB=2

4

TB=0F 8

SB=0 11 11

5

19 19

Tyg

TB=3 SB=3

E 8 G 16

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4

TB=3 SB=3

56 Critical path activities have zero float. There colud be Critical path more than one.

Example 3: For a construction project the relationship of works and duration of each work is given.  Draw the CPM network according to given information.  Find the critical path.

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Example 3:

58

Works

Duration (months)

1-2

5

1-3

7

1-4

9

2-4

6

2-6

12

3-5

6

4-5

5

4-6

8

4-7

9

5-8

13

6-7

4

7-8

6

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Works Months

Example 3: 5

19 19

5 2

TF=0 FF=0 5

12

5

TF=3 FF=0 3 6

6

5 7 9 6 12 6 5 8 9 13 4 6

TF=0 FF=0 29 29

TF=0 FF=0 13 TF=0 FF=0 5

7 10 TF=3 FF=3 16 16 ENG209 Week_11

7

8

4

11 11

7

59

4

9 TF=3 FF=3 9 TF=2 FF=2

0

6

23 23

TF=0 8 FF=0

TF=0 6 FF=0

1 0

TF=0 FF=0

TF=2 FF=2

1-2 1-3 1-4 2-4 2-6 3-5 4-5 4-6 4-7 5-8 6-7 7-8

CP1= (1-2), (2-4), (4-6), (6-7), (7-8) CP2= (1-2), (2-4), (4-5), (5-8)

Example 4: Activity

Precedence

A

---

16

B

----

20

C

---

30

D

B

E

B

10

F

A

15

G H I

Duration (days)

b) c)

15

D D

a)

3 16

E,F,G

12

d) e)

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Create network diagram.

Find early start time for every activity.

Find late start time for every activity if the project is completed in the minumum time.

Find critical path.

Find the slacks.

Activity (days)

Example 4: a) Flow diagram

Precedence

A

---

16

B

----

20

C

---

D

B

E

B

10

F

A

15 D

3

D

16

I

E,F,G

F

2

30 15

G H

12

5

A 1

B

E 3

D C

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G

Duration

I

4 H 6

b) Find early start time for every activity Event

Precedence

ES

Time

EF

1

-

-

-

0

2

1

0

16

16

3

1

0

20

20

4

3

20

15

35

5

2

16

10

38

3

20

15

4

35

3

1

0

30

4

35

16

5

38

12

6

62

51

The early finish time of the project is 51 days.

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b) Find late start time for every activity Event

Succeeding events

LF

6

-

51

5

6

51

12

39

4

5

39

3

35

6

51

16

4

35

15

5

39

10

2

5

39

15

24

1

2

24

16

0

3

20

20

6

51

30

3

63

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Time

LS 51

20

b) Find critical path.

64

Events

EF

LS

1

0

0

2

16

24

3

20

20

4

35

35

5

38

39

6

51

51

In the table the events which early finish time equals late start time create critical path.

Critical path: 1 3 4 6 ENG209 Week_11

b) Find the slacks.

Activity

ES

LS

Time

Slack

1-2

0

24

16

8

1-3

0

20

20

0

1-6

0

51

30

21

2-5

16

39

15

8

3-4

20

35

15

0

3-5

20

39

10

9

4-5

35

39

3

1

4-6

34

51

16

0

5-6

38

51

12

1

Activity 2-5 can delay 8 days without extend the project time.

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Example 5: Company is planning to product a new mp3 player. Managers want to know; When the project will be completed? Which activities are critical? Starting and ending time of every activity.

66

Code

Activity

Succeding

Time (days)

A

Ä°dentification of product features

-

30

B

feasibility study

A

8

C

prototype production

B

8

D

raw material procurement

B

5

E

Prepare final design

H

2

F

Production

D,E

25

G

Personnel training

A

10

H

Personnel offers about prototype

C,G

3

I

Sales training

E

5

J

Ä°dentifying marketing mix

A

5

K

promotion campaign

I,J

15

Finish the project

F,K

2

L

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Example 5: D 5 B 8 C 8 A 30

G 10

J 5

H 3

E 2

F 25

L 2

I 5

K 15

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Example 5: 30;38 30;38 0;30 0;30

A 30

D 5

B 8 G 10 J 5

30;40 36;46

30;35 56;61

C 8

38;43 46;51

38;46 38;46

46;49 46;49

H 3

E 2 49;51 49;51

51;56 56;61

51;76 51;76

76;78 76;78

F 25

L 2

I 5

56;71 61;76

K 15

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Example 5: • If activity A delays 1 day, when the project will be completed?

• A is an critical activity. So all acitities succeding A will be

postponed 1 day. And the project time will lenght out 1 day.

• If activity D delays 7 day, when the project will be completed?

• D is not a critical activity. So we must think the slack of this

acitivty. Because the slack of this activity is 8 days, 7 days delay wont effect the project completion time.

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Example 5: • If activity D delays 7 day and activity J delays 25 days when the project will

be completed?

• Both of the activiy are not critical and delay time is smaller than slack time.

So project completion time.wont be effected.

• If activity K delays 6 day, when the project will be completed?

• K is not a critical activity. So we must think the slack of this acitivty. Because

delay time is bigger than slack project completition time will be effected.

• Project time will length out 1 day (6-5=1) • Activity K is a critical activity from nowç And critical path will change.

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