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5.1 Dr. Henry Deng Dr. Henry Deng Assistant Professor Assistant Professor MIS Department MIS Department UNLV UNLV IS 488 Information Technology Project Management

Dr. Henry Deng

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IS 488 Information Technology Project Management. Dr. Henry Deng. Assistant Professor MIS Department UNLV. Today. Course schedule and dates Questions from PERT lecture 1 In class exercise for PERT lecture 1 Review some of your ‘exam’ questions Activity time example Lecture 2 on PERT - PowerPoint PPT Presentation

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Page 1: Dr. Henry Deng

5.1

Dr. Henry DengDr. Henry Deng

Assistant ProfessorAssistant Professor

MIS DepartmentMIS Department

UNLVUNLV

IS 488 Information Technology Project

Management

Page 2: Dr. Henry Deng

5.2

TodayToday

• Course schedule and datesCourse schedule and dates• Questions from PERT lecture 1Questions from PERT lecture 1• In class exercise for PERT lecture 1In class exercise for PERT lecture 1• Review some of your ‘exam’ Review some of your ‘exam’

questionsquestions• Activity time exampleActivity time example• Lecture 2 on PERTLecture 2 on PERT• Exercise 2 on PERTExercise 2 on PERT• Project team and topicProject team and topic

Page 3: Dr. Henry Deng

5.3

Let’s try this exercise before Lecture 2 -Let’s try this exercise before Lecture 2 -Calculate: ES,EF,LS,LF, Slacks, and CP Calculate: ES,EF,LS,LF, Slacks, and CP

1

2

3

4 5

68

7

B2

A2

C1 D

3

E5 F1

G2

H1

I2

Page 4: Dr. Henry Deng

5.4

Solution: ES,EF,LS,LF, Slacks, and CPSolution: ES,EF,LS,LF, Slacks, and CP

1

2

3

4 5

68

7

B[0,2

]2[

0,2]

A[0,2]

2[1,3]

C[2,3]1[2,3]D

[3,6

]3[

3,6]

E[6,11]5[6,11] F[11,12]

1[11,12]

G[12,14]

2[12,14]

H[0,1]1[13,14]

I[14,

16]

2[14

,16]

CP: B,C,D,E,F,G, and I

Page 5: Dr. Henry Deng

5.5

PERT - Estimating activity timePERT - Estimating activity time

• Consider following questionConsider following question– What is the average waiting time in line

at the registrar’s office?• How would you go about calculating an How would you go about calculating an

average score?average score?

Page 6: Dr. Henry Deng

5.6

PERT - Estimating activity timePERT - Estimating activity time

• Consider following questionConsider following question– How long does it take to test codes for an

accounts receivable program?• How would you get an average score? How would you get an average score?

Different from previous question? Different from previous question?

Page 7: Dr. Henry Deng

5.7

PERT - Estimating activity timePERT - Estimating activity time

• Consider following questionConsider following question– How long does it take to get sufficient

responses to a RFP?• How would you estimate that?How would you estimate that?

Page 8: Dr. Henry Deng

5.8

PERT - Estimating activity timePERT - Estimating activity time

• Calculating the duration of the entire project Calculating the duration of the entire project and the scheduling of the specific activities and the scheduling of the specific activities depends on how we calculate time for each depends on how we calculate time for each activity.activity.

• Obtaining estimates for projects that are Obtaining estimates for projects that are repeat or projects that we have experience repeat or projects that we have experience with is relatively easy. Estimating activity time with is relatively easy. Estimating activity time for new and unique projects is significantly for new and unique projects is significantly more difficult. more difficult.

• To factor uncertainly into the network To factor uncertainly into the network analysis, often three estimates are used: analysis, often three estimates are used: Optimistic timeOptimistic time (a), (a), most probable timemost probable time (m), and (m), and pessimistic timepessimistic time (b). (b).

Page 9: Dr. Henry Deng

5.9

Estimating uncertain activity timeEstimating uncertain activity time

• The three estimates (a, m, b) enable the The three estimates (a, m, b) enable the systems analyst to develop the most likely systems analyst to develop the most likely activity time that ranges from the best activity time that ranges from the best possible (optimistic) time to the worst possible (optimistic) time to the worst possible (pessimistic) time. possible (pessimistic) time.

• The expected time (The expected time (tt) can be calculated using ) can be calculated using the following formula: the following formula:

tt = (a+4m+b)/6 = (a+4m+b)/6• To measure the dispersion or variation in the To measure the dispersion or variation in the

activity time values, the common statistical activity time values, the common statistical measure of the variance can be used: measure of the variance can be used:

2 2 == [(b-a)/6][(b-a)/6]22

((This formula assumes that a standard deviation is This formula assumes that a standard deviation is approximately 1/6 of the difference between the extreme values approximately 1/6 of the difference between the extreme values of the distribution: (b-a)/6. The variance is simply the square of of the distribution: (b-a)/6. The variance is simply the square of the standard deviationthe standard deviation).).

Page 10: Dr. Henry Deng

5.10

Example of estimating activity timeExample of estimating activity time

Consider the optimistic, most probable, and Consider the optimistic, most probable, and pessimistic time estimates forpessimistic time estimates for a project that involves a project that involves the following activities:the following activities:

ActivityActivity OptimisticOptimistic Most probableMost probable PessimisticPessimistic ((aa)) ( (mm)) ( (bb) )

--------------------------------------------------------------------------------------------------------------------------------------------------AA 44 55 1212BB 11 1.51.5 5 5CC 22 33 4 4DD 33 44 1111EE 22 33 4 4FF 1.51.5 22 2.5 2.5GG 1.51.5 33 4.5 4.5HH 2.52.5 3.53.5 7.5 7.5II 1.51.5 22 2.5 2.5JJ 11 22 3 3

Page 11: Dr. Henry Deng

5.11

Estimating time for activity AEstimating time for activity A

Using the expected time (Using the expected time (tt) formula ) formula tt = (a + 4m + b)/6 = (a + 4m + b)/6

we have an estimated average or we have an estimated average or expected completion time of expected completion time of

ttAA = [4 + 4(5) + 12]/6 = 36/6 = 6 weeks = [4 + 4(5) + 12]/6 = 36/6 = 6 weeksand using the variance formulaand using the variance formula

2 2 == [(b - a)/6][(b - a)/6]22

we can determine the measure of we can determine the measure of uncertainty or the variance for activity uncertainty or the variance for activity A:A:

22A A == [(12 - 4)/6][(12 - 4)/6]2 2 = (8/6) = (8/6)22 = 1.78 = 1.78

Page 12: Dr. Henry Deng

5.12

Estimating time for all activitiesEstimating time for all activities

ActivityActivity Expected timeExpected time VarianceVariance (in weeks)(in weeks)

--------------------------------------------------------------------------------------------------------------------------------------------------A [A [4 + 4(5) + 12]/6 4 + 4(5) + 12]/6 6 6 [(12 - 4)/6][(12 - 4)/6]22 1.781.78B B 22 0.440.44C [2C [2 + 4(3) + 4]/6 + 4(3) + 4]/6 3 3 [(4 - 2)/6][(4 - 2)/6]22 0.110.11DD 55 1.781.78E E 33 0.110.11FF 22 0.030.03GG 33 0.250.25H [2.5H [2.5 + 4(3.5) + 7.5]/6 + 4(3.5) + 7.5]/6 4 4 [(7.5 – 2.5)/6][(7.5 – 2.5)/6]22 0.690.69II 22 0.030.03JJ 22 0.110.11

Total Total 32 32

Once expected activity times are calculated, we can proceed Once expected activity times are calculated, we can proceed with the critical path calculations to determine the expected with the critical path calculations to determine the expected project completion time and a detailed activity schedule.project completion time and a detailed activity schedule.

Page 13: Dr. Henry Deng

5.13

Network with expected activity timesNetwork with expected activity times

1

2

3

5

4

6

7 8

A 6

B 2

E3

C3

G3

H4

J2

I2

F2

D5

Page 14: Dr. Henry Deng

5.14

Network with ES & EFNetwork with ES & EF

1

2

3

5

4

6

7 8

A [0,6

] 6

B [0,2]

2

E [6,9]

3

C [6,9]3

G [11,14]3

H [9,13]4

J [15,17]2

I [13

,15]

2

F [9,11]

2

D [6,11]

5

Page 15: Dr. Henry Deng

5.15

Network with ES, EF, LS & LFNetwork with ES, EF, LS & LF

1

2

3

5

4

6

7 8

A [0,6

]6

[0,6

]

B [0,2]

2 [7,9]

E [6,9]

3 [6,9]

C [6,9]3 [10,13]

G [11,14]3 [12,15]

H [9,13]4 [9,13]

J [15,17]2 [15,17]

I [13

,15]

2 [1

3,15

]

F [9,11]

2 [13,15]

D [6,11]

5 [7,12]

LatestFinishTime

LatestStartTime

EarliestStartTime

EarliestFinishTime

Page 16: Dr. Henry Deng

5.16

Activity schedule (in weeks)Activity schedule (in weeks)

Earliest Latest Earliest Latest Earliest Latest Earliest Latest Start StartStart Start FinishFinish Finish Slack Finish Slack CriticalCritical

Activity (ES) (LS) (EF) (LF) (LS - ES) Activity (ES) (LS) (EF) (LF) (LS - ES) Path?Path?

--------------------------------------------------------------------------------------------------------------------------------------------------------------------AA 0 0 0 0 6 6 6 6 00 YesYesBB 0 0 7 7 2 2 9 9 77CC 6 6 1010 9 9 13 13 44DD 6 6 7 7 1111 12 12 11EE 6 6 6 6 9 9 9 9 00 YesYesFF 9 9 1313 1111 15 15 44GG 11 11 1212 1414 15 15 11 HH 9 9 9 9 1313 13 13 00 YesYesII 13 13 1313 1515 15 15 00 YesYesJJ 15 15 1515 1717 17 17 00 YesYes

Critical path - A, E, H, I, and JCritical path - A, E, H, I, and J Project duration - 17 Project duration - 17 weeksweeks

Page 17: Dr. Henry Deng

5.17

Variance in critical path activitiesVariance in critical path activities

• Variation in critical path activities can cause Variation in critical path activities can cause variation in the project completion date.variation in the project completion date.

• If a non-critical activity is delayed beyond its If a non-critical activity is delayed beyond its slack time, then that activity would become slack time, then that activity would become part of the new critical path, and further delays part of the new critical path, and further delays would affect the project completion date. would affect the project completion date.

• Variation in critical path activities resulting in Variation in critical path activities resulting in shorter critical path will result in an earlier than shorter critical path will result in an earlier than expected completion date.expected completion date.

• The variance in the project duration is the The variance in the project duration is the same as the sum of the variance of the critical same as the sum of the variance of the critical path activities.path activities.

Page 18: Dr. Henry Deng

5.18

Probability of meeting deadline Probability of meeting deadline

• The expected (The expected (EE) project time () project time (TT) for the ) for the previous example is previous example is

EE((T)T) = = ttAA + + ttEE + + ttHH + + ttII + + ttJJ

= 6 + 3 + 4 + 2 + 2 = 17 weeks= 6 + 3 + 4 + 2 + 2 = 17 weeks

• The variance (The variance (22) for that example is) for that example is

Var (Var (TT) = ) = 2 2 = = 22AA + + 22

EE + + 22HH + + 22

II + + 22JJ

• Since standard deviation is the square Since standard deviation is the square root of the variance, then root of the variance, then

= = 22 = = 2.72 = 1.65 2.72 = 1.65

Page 19: Dr. Henry Deng

5.19

Estimating time for all activitiesEstimating time for all activities

ActivityActivity Expected timeExpected time VarianceVariance VarianceVariance (in weeks)(in weeks) 22 (for critical (for critical

path) path) --------------------------------------------------------------------------------------------------------------------------------------------------

AA (CP)(CP) 66 1.781.78 1.781.78B B 22 0.440.44CC 33 0.110.11DD 55 1.781.78EE ((CPCP)) 33 0.110.11 0.110.11FF 22 0.030.03GG 33 0.250.25HH ((CPCP)) 44 0.690.69 0.690.69II ((CPCP) ) 22 0.030.03 0.030.03J J ((CPCP) ) 2 2 0.110.11 0.110.11Total Total 32 32 2.72 Var (2.72 Var (TT) )

Standard deviation for critical path activities: Standard deviation for critical path activities: = = 22

= = 2.72 = 1.65 2.72 = 1.65

Page 20: Dr. Henry Deng

5.20

Probability of meeting deadlineProbability of meeting deadline

• Assuming a normal (bell-shaped) distribution Assuming a normal (bell-shaped) distribution of the project completion time allows us to of the project completion time allows us to compute the probability of meeting a specified compute the probability of meeting a specified project completion date. project completion date.

• Suppose the management has allowed 20 Suppose the management has allowed 20 weeks for the previous project. What is the weeks for the previous project. What is the probability that we will meet the 20-week probability that we will meet the 20-week deadline?deadline?

• We are looking for the probability of We are looking for the probability of T T <=20.<=20.• The z value for the normal distribution of T = 20 The z value for the normal distribution of T = 20

isisz = (20 - 17)/1.65 = 1.82z = (20 - 17)/1.65 = 1.82

• We need to use the normal distribution table.We need to use the normal distribution table.

Page 21: Dr. Henry Deng

5.21

Normal distribution of project timeNormal distribution of project time

---------------------------------------------------------------------------------------------------------------- 17 2017 20

Time (weeks)Time (weeks)

Page 22: Dr. Henry Deng

5.22

Page 23: Dr. Henry Deng

5.23

Normal distribution of project timeNormal distribution of project time

0.4656 + 0.50000.4656 + 0.5000 z = z =

= 0.9656= 0.9656 (20 -17)/1.65 (20 -17)/1.65

= 1.82= 1.82

pp((TT<= 20)<= 20)

---------------------------------------------------------------------------------------------------------------- 17 2017 20

Time (weeks)Time (weeks)

Page 24: Dr. Henry Deng

5.24

SummarySummary

• PERT procedure can be used to schedule PERT procedure can be used to schedule projects with uncertain activity times.projects with uncertain activity times.

• The three time estimates (optimistic, most The three time estimates (optimistic, most likely, pessimistic) help calculate an expected likely, pessimistic) help calculate an expected time and variance for each activity. time and variance for each activity.

• The time for critical path activities provides the The time for critical path activities provides the expected project completion time.expected project completion time.

• The sum of the variances of activities on the The sum of the variances of activities on the critical path provides the variance in the critical path provides the variance in the project completion time.project completion time.

• Normal probability distribution assumption Normal probability distribution assumption and procedures are used to compute the and procedures are used to compute the probability of the project being completed by a probability of the project being completed by a specific time.specific time.

Page 25: Dr. Henry Deng

5.25

Page 26: Dr. Henry Deng

5.26

Time-Cost Trade-off: CrashingTime-Cost Trade-off: Crashing

Video 6 Time CrashingVideo 6 Time Crashing

Page 27: Dr. Henry Deng

5.27

Resource LimitationsResource Limitations

critical path crashingcritical path crashing

(cost/time tradeoff)(cost/time tradeoff)

other methodsother methods

Page 28: Dr. Henry Deng

5.28

CrashingCrashing

• can shorten project completion time can shorten project completion time by adding extra resources (costs)by adding extra resources (costs)

• start off with start off with NORMAL TIME NORMAL TIME CPM CPM scheduleschedule

• get expected duration get expected duration TnTn, cost , cost CnCn• TnTn should be longest duration should be longest duration• CnCn should be most expensive in should be most expensive in

penalties, cheapest in crash costspenalties, cheapest in crash costs

Page 29: Dr. Henry Deng

5.29

Time ReductionTime Reduction

• to reduce activity time, pay for to reduce activity time, pay for more resourcesmore resources

• develop table of activities with develop table of activities with times and coststimes and costs

• for each activity, usually assume for each activity, usually assume linear relationship for relationship linear relationship for relationship between cost & timebetween cost & time

Page 30: Dr. Henry Deng

5.30

Crash ExampleCrash Example

Activity:Activity: programmingprogramming

Tn:Tn: 7 weeks7 weeks

Cn:Cn: $14,000$14,000 (7 weeks, 2 (7 weeks, 2 programmers)programmers)

if you add a third programmer, done in 6 if you add a third programmer, done in 6 weeksweeks

Tc:Tc: 6 weeks6 weeks

Cn:Cn: $15,000$15,000

cost slope = (15000-14000)/(6-7)=cost slope = (15000-14000)/(6-7)=-$1000/week-$1000/week

Page 31: Dr. Henry Deng

5.31

Example ProblemExample Problem

activityactivity Pred TnPred Tn CnCn TcTc CcCc slopeslopemaxmax

A requirements A requirements none 3none 3 can’t crashcan’t crashB programmingB programming A 7A 7 1400014000 66 1500015000 -1000-1000 1 week1 weekC get hardwareC get hardware A 1A 1 5000050000 .5.5 5100051000 -2000-2000 .5 week.5 weekD train users D train users B,C 3B,C 3 can’t crashcan’t crash

Crashing Algorithm:Crashing Algorithm:1 crash 1 crash only critical activitiesonly critical activities B only choiceB only choice

2 crash 2 crash cheapest currently criticalcheapest currently critical B is cheapestB is cheapest

3 after crashing one time period, recheck critical3 after crashing one time period, recheck critical

Page 32: Dr. Henry Deng

5.32

Crash ExampleCrash Example

Import critical software from Australia: Import critical software from Australia: late penaltylate penalty $500/d > 12 d $500/d > 12 d

A get import licenseA get import license 5 days5 days no predecessorno predecessor

B shipB ship 7 days7 days A is predecessorA is predecessor

C train usersC train users 11 days11 days no predecessorno predecessor

D train on systemD train on system 2 days2 days B,C predecessorsB,C predecessors

can crashcan crash C: $2000/day more than current for up to 3 daysC: $2000/day more than current for up to 3 days

B: faster boatB: faster boat 6 days6 days $300 more than current$300 more than current

bush planebush plane 5 days5 days $400 more than current$400 more than current

commercialcommercial 3 days3 days $500 more than current$500 more than current

Page 33: Dr. Henry Deng

5.33

Crash ExampleCrash Example

Original schedule: 14 days, $1,000 in penaltiesOriginal schedule: 14 days, $1,000 in penalties = = $1000$1000

crash B to 6 days:crash B to 6 days:13 days, $500 penalties, $300 cost 13 days, $500 penalties, $300 cost = = $800$800**

crash B to 5crash B to 5

C to 10:C to 10: 12 days, no penalties, $400+2000 cost12 days, no penalties, $400+2000 cost = = $2400$2400

to 11 days is worseto 11 days is worse

NOW A SELECTION DECISIONNOW A SELECTION DECISION

risk versus costrisk versus cost

Page 34: Dr. Henry Deng

5.34

Crashing LimitationsCrashing Limitations

• assumes assumes linear relationship linear relationship between between time and costtime and cost– not usually true (indirect costs don’t

change at same rate as direct costs)

• requires a lot of extra cost requires a lot of extra cost estimationestimation

• time consumingtime consuming

• ends with tradeoff decision ends with tradeoff decision

Page 35: Dr. Henry Deng

5.35

Resource ConstrainingResource Constraining

• CPM & PERT both assume CPM & PERT both assume unlimited resourcesunlimited resources

NOT TRUENOT TRUE– may have only a finite number of systems

analysts, programmers

• RESOURCE LEVELING RESOURCE LEVELING - balance the - balance the resource loadresource load

• RESOURCE CONSTRAINING RESOURCE CONSTRAINING - don’t - don’t exceed available resourcesexceed available resources