Week 9 Supplement Risk Management, Project Completion Times, Explaining the Statistics

Risk Management and Scheduling:

Estimating completion times

Week 9 Supplement: explaining the statistics09-01

Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall

Activity Completion Time Estimates Definition

Activity times follow a “Beta” probability distribution Optimistic and pessimistic times are lower/upper bounds Completion times will be clustered around “most likely” time

Activity completion time estimate Most likely (m): Most pessimistic: Time to complete < (b), probability = 0.99 Most optimistic: time to complete > (a), probability = 0.99

Activity completion time variance Completion times should fall within +/- 3 standard deviations

of average completion time Therefore “acceptable” standard deviation (s) equals 1/6 of

range of times The range is (b-a); therefore s =(b-a)/6 “Variance” is defined as s2

2

4 = TE

6

a m bActivity Duration

22 =

6

b aActivity Variance s

Asymmetrical (Beta) Distribution for Activity Duration Estimation

Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall 09-3

Optimistic timeMost likely time

Pessimistic time

Probability Distribution of Activity Completion Times

Activities complete in times less than the optimistic time with p=0.01

Activities complete in times greater than the pessimistic time with p=0.01

• Since the activity durations are skewed towards “pessimistic” durations… the activity duration Time Estimate (TE) will be slightly greater than m, the “most likely” activity completion time

• Activities will complete sometime between times a and b, with p = 0.98• Activities will complete in time greater than Activity TE with p = 0.5, and complete in time less than Activity TE with p = 0.5

TE

Activity TE


Project Completion Time Estimates

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Definition Project completion times follow a “Normal” probability

distribution Variations in activity completion times can affect project

completion time Completion times will be clustered around “most likely” time

Project completion time estimate (TE) Sum of the completion times of activities on the critical path Ignores activities not on the critical path: a simplifying

assumption

Project completion time variance Sum of the variances in completion times, for activities on the

critical path Project completion times should fall within +/- 3 standard

deviations of average completion time: 99.7% of completion times

Project Standard Deviation =


Probability Distribution of Project Completion Times Symmetrical (Normal) Distribution for Project Duration Estimation

Probabilities of project completion times – see chart on next page for derivations• 68.3% will be within +/- 1 standard deviation

of expected completion time• 95.6% within 2 std dev of expected• 99.7% within 3 std dev of expected

Z = Standard Deviation coefficient• for example, if Z= 2;

then Z * S = Z * 2 std dev

Since the activity Time Estimates (TEs) are skewed towards “pessimistic” durations, and because the project Time Estimate (TE) is the sum of the TEs for activities on the critical path, the project TE will be slightly greater than the “most likely” time for project completion. Since the individual project TE has already accounted for the skew in activity completion times, the distribution of project TEs is symmetrical.

TE; Z =0

Projects will complete in greater time than the predicted TE with p=0.5

Projects will complete in less time than the predicted TE with p=0.5

6http://www.six-sigma-material.com/Normal-Distribution.html

• The Pearson P 349 table is one-tailed; it can also be interpreted for two-tailed distributions

• For Z = 1; p=0.8413; the probability to the right of the mean is 0.8413 – 0.5 = 0.3413

• Therefore, the probability that a project will complete within +/- 1 std dev of TE, is 2 X 0.3414= 0.683

• For Z= 2; p = 0.9772• P (complete in +/- 2 std

dev of TE) is2 X (0.9772-0.5) = 0.956

• For Z= 3; p = 0.9987• P (complete in +/- 3 std

dev of TE) is2 X (0.9987-0.5) = 0.997


Example:• Determine the completion time for

which you are 95% confident …• Probability is 0.95 that project will

complete within this time• P= 0.05 that it will complete

outside• P (early) = 0.025• P (late) = 0.025

Probability Distribution of Project Completion Times Symmetrical (Normal) Distribution for Project Duration Estimation

• α = 1-0.95 = 0.05• 1- α/2 = 1- 0.05/2 =

0.025• From the Normal

Distribution table

Z 1-0.25 = Z 0.975 = 1.96

• P = 0.95 that completion time falls within the range of TE +/- Z times S • Since Z = 1.96, then completion time falls within the range of TE +/- 1.96 * S

95%2.5% 2.5%

Z 0.975 - Z 0.975

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• If p = 0.95 that project will complete within a certain time

• This is equivalent to p = 0.975 that project will complete by a certain time; orp (late) = 0.025

• For p =0.975

we find Z= 1.96

http://www.six-sigma-material.com/Normal-Distribution.html


Expected Project Completion time, and Variance in Project Completion Time

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Activity Opti-mistic (a)

Most Likely (m)

Pessi-mistic (b)

Expec-ted

Expected time of critical path activities

Activity time Variance [(b-a)/6]²

Variance of critical path activities only

A 3 4 11 5 ( = 1.78 1.78

B 2 5 8 5 1.00

C 3 6 9 6 6 1.00 1.00

D 8 12 20 12.7 13 4.00 4.00

E 3 5 12 5.8 2.25

F 2 4 7 4.2 4 0.69 0.69

G 6 9 14 9.3 1.78

H 1 2 4 2.2 2 0.25 0.25

Project 30 7.72

• Expected project completion time: 30 Sum of expected activity completion times = 30• Variance in project completion time: 7.72 ; and standard deviation = √ 7.72 = 2.78• Individual activities are independent of each other. Add the variances for activities on the Critical Path A-C-D-F-H = 7.72. • Variance in completion time for activities not on critical path, would not contribute to variances in project completion time


Z = 0.72Z=0

Probability of Completion by a specified time = p • Determine p based on “Z”

statistic

50% of projects will be early, 50% will be late

Probability that project completes within 32 weeks? • Example shows TE= 30 weeks, with s = 2.78 weeks• Need to compute Z, the standard deviation coefficient.

• Deadline = 32 weeks, or

2 weeks after expected completion

• Z as a proportion of S:

2 / 2.78 weeks = 0.72• Look up Z = 0.72 in Pearson

P. 349, to find p= 0.7642

• Therefore, 76% probability

project will complete

within 32 weeks;

24% after 32 weeks

Time in Weeks 30 32

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From Z= 0.72 We findp = 0.7642

http://www.six-sigma-material.com/Normal-Distribution.html


Probability of Project Completion (cont’d)Project Standard Deviation (σ) = Project Standard Deviation (σ) = Project Standard Deviation (σ) = 2.78 weeksZ = the number of standard deviations on the

target date = (32 – 30)/2.78 = 0.72 (Due date given of 32weeks)

Looking at Appendix A 0.72 indicates a probability 0.7642 (76.42%) of meeting the deadline.

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Probability of Project Completion (cont’d)To determine the due date to achieve a

certain probability of a project finishing by a certain date

Due Date = Expected date of completion + (Z x σ)To ensure a 95% likelihood of completing on

time Z = Due Date would have to be weeks.

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Z = 1.65Z=0

Probability of Completion by a specified time = p • Determine p based on “Z”

statistic

50% of projects will be early, 50% will be late

When will project complete, with probability of 0.95?• Example shows TE= 30 weeks, with s = 2.78 weeks• Need to compute Z, the standard deviation coefficient.

• To find Z, look up p = 0.95 in

Pearson P. 349; and Z = 1.65

• Completion time = TE + Z *S• = 30 weeks + 1.65 * 2.78 weeks

= 34.59 weeks

Time in Weeks 30 34.59

15http://www.six-sigma-material.com/Normal-Distribution.html

We find Z= 1.65

FromP = 0.95


Probability of Project Completion (cont’d)To determine the due date to achieve a

certain probability of a project finishing by a certain date

Due Date = Expected date of completion + (Z x σ)To ensure a 95% likelihood of completing on

time?Determine (from P 349 Pearson) that Z = 1.65 Due Date would have to be μ + 1.65* σ30 + 1.65*2.78 = 34.59 weeks.

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Week 9 Supplement Risk Management, Project Completion Times, Explaining the Statistics