CSc 233 Fall 2010

Improvements on Beating a Dead HorseThe tribal wisdom of the Dakota Indians, passed on

from generation to generation, says that "When you discover that you are riding a dead horse,

the best strategy is to dismount."

However, in many organizations a range of far more advanced strategies are often employed.

SUCH AS…

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CSc 233 Fall 2010

• Buying a stronger whip • Changing riders • Saying things like, "This is the way we have always ridden this

horse." • Appointing a committee to study the horse • Arranging to visit other sites to see how others ride dead

horses • Lowering the standards so that dead horses can be included • Appointing a tiger team to revive the dead horse • Sending riders to training sessions to improve riding ability • Comparing the state of dead horses in today's environment • Changing the requirements to declare that "This horse is not

dead." • Declaring that "No horse is too dead to beat."

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CSc 233 Fall 2010

• Reclassifying the dead horse as "living impaired" • Doing a Cost Analysis study to see if contractors can ride the

horse more cheaply. • Purchasing a product that claims to make dead horses run

faster • Hiring an outside contractor to ride the dead horse • Harnessing several dead horses together to increase the speed • Providing additional funding and/or training to increase the

dead horse's performance • Forming a quality circle to find uses for dead horses • Saying "This horse was procured with cost as an independent

variable." • Doing a productivity study to see if lighter riders would

improve the dead horse's performance 3

CSc 233 Fall 2010

• Declaring that the horse is "better, faster and cheaper" dead (i.e. as the dead horse does not have to be fed, it is less costly, carries lower overhead, and therefore contributes substantially more to the mission of the organization than do some other horses)

• Rewriting the expected performance requirements for all horses

• Promoting the dead horse to a supervisory position...

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CSc 233 Fall 2010

PERT (Program Evaluation & Review Technique)CPM ( Critical Path Method)

• Used to plan, schedule, and control a wide variety of projects

• Schedule and coordinate the various tasks so that the project is completed on time

• Managers want to know:What is the expected project completion dateWhat is the scheduled start & completion date for each specific taskWhich tasks are “critical” and must be completed exactly as scheduled

to keep the project on scheduleHow long can “noncritical” tasks be delayed before they cause a delay in

the total project

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CSc 233 Fall 2010How to Draw a Pert Chart• Make a list of all the tasks that are required to complete the project.• Put the tasks in order by dependency.

For example, two tasks are design and coding. You can't design and code until the requirements are specified. Requirements comes before design and coding.

• Number (or letter) your sorted list, starting with assigning the lowest number (or letter) to the first task and the highest number to the last task.

• Draw the graph, with earlier tasks on the left and later tasks moving right. Tasks that can be done at the same time should be vertically aligned.

• Draw arrows from prerequisite tasks to the tasks that depend on them.

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CSc 233 Fall 2010

The Network?

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IMMEDIATETASK PREDECESSORS

A -B -C AD AE AF CG DH B, EI HJ F, G, I

CSc 233 Fall 2010

The Network

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IMMEDIATETASK PREDECESSORS Left to Right

A -B - C FC A AD A D GE A JF C EG DH B, E B H II HJ F, G, I

CSc 233 Fall 2010

Critical Path Calculation

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Left to Right

C FA

D GJ

E

B H I

ES EFTask ID

DurationLS LF

CSc 233 Fall 2010

Each Task

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Begin with the first task (ES = 0)Duration = t

ES Earliest start timeEF Earliest finish time (ES + t)

Earliest start: largest value of EF times for all predecessor tasksBegin after you have the EF for the last task (set LF = EF)

LF Latest finish timeLS Latest start time (LF – t)

Latest finish Rule: smallest value of LS times for all successor tasks

ES EFTask ID

DurationLS LF

CSc 233 Fall 2010

Assignment

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Task Predecessor DurationA - 10B - 5C A 20D B 5E C, D 20F D 15G D 5H E 10I F 5J G, H, I 10

1

A

B

CSc 233 Fall 2010

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Task Predecessor DurationA - 10B - 5C A 20D B 5E C, D 20F D 15G D 5H E 10I F 5J G, H, I 10

1 2

A C

B D

CSc 233 Fall 2010

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Task Predecessor DurationA - 10B - 5C A 20D B 5E C, D 20F D 15G D 5H E 10I F 5J G, H, I 10

1 2 3

A C E

FB D

G

CSc 233 Fall 2010

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Task Predecessor DurationA - 10B - 5C A 20D B 5E C, D 20F D 15G D 5H E 10I F 5J G, H, I 10

1 2 3 4

A C E H

F IB D

G

CSc 233 Fall 2010

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Task Predecessor DurationA - 10B - 5C A 20D B 5E C, D 20F D 15G D 5H E 10I F 5J G, H, I 10

1 2 3 4 5

A C E H

F IB D J

G

CSc 233 Fall 2010

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1 2 3 4 5

A C E H

F IB D J

G

Task ES + DUR = EFA 0 10 10B 0 5 5C 10 20 30D 5 5 10E 30 MAX(30,10) 20 50F 5 15 20G 10 5 15H 50 10 60I 20 5 25J 60 MAX(15,60,25) 10 70

CSc 233 Fall 2010

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1 2 3 4 5

A C E H

F IB D J

G

Task ES + DUR = EF A 0 10 10B 0 5 5C 10 20 30D 5 5 10E 30 MAX(30,10) 20 50F 5 15 20G 10 5 15H 50 10 60I 20 5 25J 60 MAX(15,60,25) 10 70

CSc 233 Fall 2010

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1 2 3 4 5

A C E H

F IB D J

G

Task ES + DUR = EF Task LF - DUR = LSA 0 10 10 A 10 10 0B 0 5 5 B 10 5 5C 10 20 30 C 30 20 10D 5 5 10 D 30 MIN(30,35,55) 5 25E 30 MAX(30,10) 20 50 E 50 20 30F 5 15 20 F 50 15 35G 10 5 15 G 60 5 55H 50 10 60 H 60 10 50I 20 5 25 I 55 5 50J 60 MAX(15,60,25) 10 70 J 70 10 60

CSc 233 Fall 2010

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Task ES + DUR = EF Task LF - DUR = LSA 0 10 10 A 10 10 0B 0 5 5 B 10 5 5C 10 20 30 C 30 20 10D 5 5 10 D 30 MIN(30,35,55) 5 25E 30 MAX(30,10) 20 50 E 50 20 30F 5 15 20 F 50 15 35G 10 5 15 G 60 5 55H 50 10 60 H 60 10 50I 20 5 25 I 55 5 50J 60 MAX(15,60,25) 10 70 J 70 10 60

Task LS ES LS - ES LF - EF LF EFA 0 0 0 0 10 10B 5 0 5 5 10 5C 10 10 0 0 30 30D 25 5 20 20 30 10E 30 30 0 0 50 50F 35 5 30 30 50 20G 55 10 45 45 60 15H 50 50 0 0 60 60I 50 20 30 30 55 25J 60 60 0 0 70 70

CRITICAL PATH: A C E H J

SLACK

CSc 233 Fall 2010

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6

)6

(

64

:each taskFor

22

ab

ab

bmaEt

−=

−=

++=

σ

σ ∑

∑=

=

taskCPEach

22 taskCPEach

σσ

tEE

CSc 233 Fall 2010

Project Tasks - example

21A E H I J

Task OptimisticMost

Probable PessimisticExpected

time Variancea m b

A 4 5 12 6 1.78B 1 1.5 5 2 0.44C 2 3 4 3 0.11D 3 4 11 5 1.78E 2 3 4 3 0.11F 1.5 2 2.5 2 0.03G 1.5 3 4.5 3 0.25H 2.5 3.5 7.5 4 0.69I 1.5 2 2.5 2 0.03J 1 2 3 2 0.11

TOTAL 32 5.33

Time Estimates

CSc 233 Fall 2010

PERT

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0.9656scheduleweek 20 a meeting ofy Probabilit

)(17 Time CompletionProject Expected

Time Completion Suggested""

82.165.132

65.133.52

ttTP

Et

tEtz

t

T

=≤

=

=−

=−

=

===

σ

σσ

CSc 233 Fall 2010

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CSc 233 Fall 2010

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CSc 233 Fall 2010

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ExpectedProject

CompletionTime

Probability Project is

completed earlier

Probability Project is completed later

CSc 233 Fall 2010

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CSc 233 Fall 2010

Draw the Network

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CSc 233 Fall 2010

Network

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ES EFTask ID

DurationLS LF

K LB F

I J

A D G HEND

C E

CSc 233 Fall 2010

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Homework Assignment