Dr. Henry Deng

1
Dr. Henry Deng
IS 488 Information Technology Project Management

• Assistant Professor
• MIS Department
• UNLV

2
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
• Exercise 2 on PERT
• Project team and topic

3
Lets try this exercise before Lecture 2
-Calculate ES,EF,LS,LF, Slacks, and CP
E 5
4
5
D 3
2
F 1
C 1
B 2
8
3
6
A 2
I 2
1
G 2
H 1
7
4
Solution ES,EF,LS,LF, Slacks, and CP
E6,11 56,11
4
5
D3,6 33,6
2
F11,12 111,12
C2,3 12,3
B0,2 20,2
8
3
6
A0,2 21,3
I14,16 214,16
1
G12,14 212,14
H0,1 113,14
7
CP B,C,D,E,F,G, and I
5
PERT - Estimating activity time

• Consider following question
• What is the average waiting time in line at the
  registrars office?
• How would you go about calculating an average
  score?

6
PERT - Estimating activity time

• Consider following question
• How long does it take to test codes for an
  accounts receivable program?
• How would you get an average score? Different
  from previous question?

7
PERT - Estimating activity time

• Consider following question
• How long does it take to get sufficient responses
  to a RFP?
• How would you estimate that?

8
PERT - Estimating activity time

• Calculating the duration of the entire project
  and the scheduling of the specific activities
  depends on how we calculate time for each
  activity.
• Obtaining estimates for projects that are repeat
  or projects that we have experience with is
  relatively easy. Estimating activity time for new
  and unique projects is significantly more
  difficult.
• To factor uncertainly into the network analysis,
  often three estimates are used Optimistic time
  (a), most probable time (m), and pessimistic time
  (b).

9
Estimating uncertain activity time

• The three estimates (a, m, b) enable the systems
  analyst to develop the most likely activity time
  that ranges from the best possible (optimistic)
  time to the worst possible (pessimistic) time.
• The expected time (t) can be calculated using the
  following formula
• t (a4mb)/6
• To measure the dispersion or variation in the
  activity time values, the common statistical
  measure of the variance can be used
• ?2 (b-a)/62
• (This formula assumes that a standard deviation
  is approximately 1/6 of the difference between
  the extreme values of the distribution (b-a)/6.
  The variance is simply the square of the standard
  deviation).

10
Example of estimating activity time

• Consider the optimistic, most probable, and
  pessimistic time estimates for a project that
  involves the following activities
• Activity Optimistic Most probable Pessimistic
• (a) (m) (b)
• ------------------------------------------------
  -------------------------
• A 4 5 12
• B 1 1.5 5
• C 2 3 4
• D 3 4 11
• E 2 3 4
• F 1.5 2 2.5
• G 1.5 3 4.5
• H 2.5 3.5 7.5
• I 1.5 2 2.5
• J 1 2 3

11
Estimating time for activity A

• Using the expected time (t) formula
• t (a 4m b)/6
• we have an estimated average or expected
  completion time of
• tA 4 4(5) 12/6 36/6 6 weeks
• and using the variance formula
• ?2 (b - a)/62
• we can determine the measure of uncertainty or
  the variance for activity A
• ?2A (12 - 4)/62 (8/6)2 1.78

12
Estimating time for all activities

• Activity Expected time Variance
• (in weeks)
• -------------------------------------------------
  ------------------------
• A 4 4(5) 12/6 6 (12 - 4)/62
  1.78
• B 2 0.44
• C 2 4(3) 4/6 3 (4 - 2)/62
  0.11
• D 5 1.78
• E 3 0.11
• F 2 0.03
• G 3 0.25
• H 2.5 4(3.5) 7.5/6 4 (7.5
  2.5)/62 0.69
• I 2 0.03
• J 2 0.11
• Total 32
• Once expected activity times are calculated, we
  can proceed with the critical path calculations
  to determine the expected project completion time
  and a detailed activity schedule.

13
Network with expected activity times
C 3
2
5
F 2
D 5
A 6
E 3
G 3
J 2
1
4
7
8
B 2
I 2
H 4
3
6
14
Network with ES EF
C 6,9 3
2
5
F 9,11 2
D 6,11 5
A 0,6 6
J 15,17 2
G 11,14 3
E 6,9 3
1
4
7
8
B 0,2 2
I 13,15 2
H 9,13 4
3
6
15
Network with ES, EF, LS LF
C 6,9 3 10,13
Earliest Start Time
Earliest Finish Time
2
5
D 6,11 5 7,12
A 0,6 6 0,6
F 9,11 2 13,15
J 15,17 2 15,17
G 11,14 3 12,15
E 6,9 3 6,9
1
4
7
8
B 0,2 2 7,9
I 13,15 2 13,15
Latest Finish Time
Latest Start Time
H 9,13 4 9,13
3
6
16
Activity schedule (in weeks)

• Earliest Latest Earliest
  Latest
• Start Start Finish Finish
  Slack Critical
• Activity (ES) (LS) (EF)
  (LF) (LS - ES) Path?
• --------------------------------------------------
  --------------------------------
• A 0 0 6 6 0 Yes
• B 0 7 2 9 7
• C 6 10 9 13 4
• D 6 7 11 12 1
• E 6 6 9 9 0 Yes
• F 9 13 11 15 4
• G 11 12 14 15 1
• H 9 9 13 13 0 Yes
• I 13 13 15 15 0 Yes
• J 15 15 17 17 0 Yes
• Critical path - A, E, H, I, and J Project
  duration - 17 weeks

17
Variance in critical path activities

• Variation in critical path activities can cause
  variation in the project completion date.
• If a non-critical activity is delayed beyond its
  slack time, then that activity would become part
  of the new critical path, and further delays
  would affect the project completion date.
• Variation in critical path activities resulting
  in shorter critical path will result in an
  earlier than expected completion date.
• The variance in the project duration is the same
  as the sum of the variance of the critical path
  activities.

18
Probability of meeting deadline

• The expected (E) project time (T) for the
  previous example is
• E(T) tA tE tH tI tJ
• 6 3 4 2 2 17 weeks
• The variance (?2) for that example is
• Var (T) ?2 ?2A ?2E ?2H ?2I ?2J
• Since standard deviation is the square root of
  the variance, then
• ? ? ?2 ? 2.72 1.65

19
Estimating time for all activities

• Activity Expected time Variance Variance
• (in weeks) ?2 (for critical
  path) ------------------------------------------
  -------------------------------
• A (CP) 6 1.78 1.78
• B 2 0.44
• C 3 0.11
• D 5 1.78
• E (CP) 3 0.11 0.11
• F 2 0.03
• G 3 0.25
• H (CP) 4 0.69 0.69
• I (CP) 2 0.03 0.03
• J (CP) 2 0.11 0.11
• Total 32 2.72 Var
  (T)
• Standard deviation for critical path
  activities
• ? ? ?2 ? 2.72 1.65

20
Probability of meeting deadline

• Assuming a normal (bell-shaped) distribution of
  the project completion time allows us to compute
  the probability of meeting a specified project
  completion date.
• Suppose the management has allowed 20 weeks for
  the previous project. What is the probability
  that we will meet the 20-week deadline?
• We are looking for the probability of T lt20.
• The z value for the normal distribution of T 20
  is
• z (20 - 17)/1.65 1.82
• We need to use the normal distribution table.

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Normal distribution of project time

• -------------------------------------------------
  -------
• 17
  20
• Time (weeks)

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Normal distribution of project time

• 0.4656 0.5000 z
• 0.9656 (20 -17)/1.65
• 1.82
• p(Tlt 20)
• -------------------------------------------------
  -------
• 17
  20
• Time (weeks)

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Summary

• PERT procedure can be used to schedule projects
  with uncertain activity times.
• The three time estimates (optimistic, most
  likely, pessimistic) help calculate an expected
  time and variance for each activity.
• The time for critical path activities provides
  the expected project completion time.
• The sum of the variances of activities on the
  critical path provides the variance in the
  project completion time.
• Normal probability distribution assumption and
  procedures are used to compute the probability of
  the project being completed by a specific time.

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Time-Cost Trade-off Crashing

• Video 6 Time Crashing

27
Resource Limitations

• critical path crashing
• (cost/time tradeoff)
• other methods

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Crashing

• can shorten project completion time by adding
  extra resources (costs)
• start off with NORMAL TIME CPM schedule
• get expected duration Tn, cost Cn
• Tn should be longest duration
• Cn should be most expensive in penalties,
  cheapest in crash costs

29
Time Reduction

• to reduce activity time, pay for more resources
• develop table of activities with times and costs
• for each activity, usually assume linear
  relationship for relationship between cost time

30
Crash Example

• Activity programming
• Tn 7 weeks
• Cn 14,000 (7 weeks, 2 programmers)
• if you add a third programmer, done in 6 weeks
• Tc 6 weeks
• Cn 15,000
• cost slope (15000-14000)/(6-7)-1000/week

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Example Problem

• activity Pred Tn Cn Tc Cc slope max
• A requirements none 3 cant crash
• B programming A 7 14000 6 15000 -1000 1 week
• C get hardware A 1 50000 .5 51000 -2000 .5
  week
• D train users B,C 3 cant crash
• Crashing Algorithm
• 1 crash only critical activities B only choice
• 2 crash cheapest currently critical B is
  cheapest
• 3 after crashing one time period, recheck critical

32
Crash Example

• Import critical software from Australia late
  penalty 500/d gt 12 d
• A get import license 5 days no predecessor
• B ship 7 days A is predecessor
• C train users 11 days no predecessor
• D train on system 2 days B,C predecessors
• can crash C 2000/day more than current for up
  to 3 days
• B faster boat 6 days 300 more than current
• bush plane 5 days 400 more than current
• commercial 3 days 500 more than current

33
Crash Example

• Original schedule 14 days, 1,000 in
  penalties 1000
• crash B to 6 days13 days, 500 penalties, 300
  cost 800
• crash B to 5
• C to 10 12 days, no penalties, 4002000 cost
  2400
• to 11 days is worse
• NOW A SELECTION DECISION
• risk versus cost

34
Crashing Limitations

• assumes linear relationship between time and cost
• not usually true (indirect costs dont change at
  same rate as direct costs)
• requires a lot of extra cost estimation
• time consuming
• ends with tradeoff decision

35
Resource Constraining

• CPM PERT both assume unlimited resources
• NOT TRUE
• may have only a finite number of systems
  analysts, programmers
• RESOURCE LEVELING - balance the resource load
• RESOURCE CONSTRAINING - dont exceed available
  resources