IS 488 Information Technology
Project Management
Dr. Henry Deng
Assistant Professor
MIS Department
UNLV
5.1
Today
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•
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5.2
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
Let’s try this exercise before Lecture 2 Calculate: ES,EF,LS,LF, Slacks, and CP
4
E
5
5
2
8
3
6
1
7
5.3
Solution: ES,EF,LS,LF, Slacks, and CP
4
E[6,11]
5[6,11]
5
2
8
3
6
1
7
CP: B,C,D,E,F,G, and I
5.4
PERT - Estimating activity time
•
Consider following question
– What is the average waiting time in line
at the registrar’s office?
•
5.5
How would you go about calculating an
average score?
PERT - Estimating activity time
•
Consider following question
– How long does it take to test codes for an
accounts receivable program?
•
5.6
How would you get an average score?
Different from previous question?
PERT - Estimating activity time
•
Consider following question
– How long does it take to get sufficient
responses to a RFP?
•
5.7
How would you estimate that?
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).
5.8
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 = (a+4m+b)/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)/6]2
(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).
5.9
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
5.10
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)/6]2
we can determine the measure of
uncertainty or the variance for activity
A:
2A = [(12 - 4)/6]2 = (8/6)2 = 1.78
5.11
Estimating time for all activities
Activity
Expected time
Variance
(in weeks)
------------------------------------------------------------------------A
[4 + 4(5) + 12]/6 6 [(12 - 4)/6]2 1.78
B
2
0.44
C
[2 + 4(3) + 4]/6
3 [(4 - 2)/6]2 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)/6]2
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.
5.12
Network with expected activity times
C
3
2
5
A
6
1
3
5.13
4
G
3
H
4
6
7
J
2
8
Network with ES & EF
2
1
3
5.14
C [6,9]
3
5
4
G [11,14]
3
H [9,13]
4
6
7
J [15,17]
2
8
Network with ES, EF, LS & LF
2
1
4
3
5.15
C [6,9]
3 [10,13]
H [9,13]
4 [9,13]
5
G [11,14]
3 [12,15]
6
Earliest
Start
Time
7
Earliest
Finish
Time
J [15,17]
2 [15,17]
Latest
Start
Time
Latest
Finish
Time
8
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 JProject duration - 17 weeks
5.16
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.
5.17
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
5.18
Estimating time for all activities
--
Activity Expected time Variance
Variance
(in weeks) 2
(for critical path)
-----------------------------------------------------------------------
A
B
C
D
E
F
G
H
I
J
Total
(CP)
(CP)
(CP)
(CP)
(CP)
6
2
3
5
3
2
3
4
2
2
32
1.78
0.44
0.11
1.78
0.11
0.03
0.25
0.69
0.03
0.11
1.78
0.11
0.69
0.03
0.11
2.72 Var (T)
Standard deviation for critical path activities:
 =  2 =  2.72 = 1.65
5.19
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 <=20.
• 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.
5.20
Normal distribution of project time
-------------------------------------------------------17
20
Time (weeks)
5.21
5.22
Normal distribution of project time
0.4656 + 0.5000
= 0.9656
z=
(20 -17)/1.65
= 1.82
p(T<= 20)
-------------------------------------------------------17
20
Time (weeks)
5.23
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.
5.24
5.25
Time-Cost Trade-off: Crashing
Video 6 Time Crashing
5.26
Resource Limitations
critical path crashing
(cost/time tradeoff)
other methods
5.27
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
5.28
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
5.29
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
5.30
Example Problem
activity
Pred Tn
Cn
Tc
Cc
slope
max
A requirements
none 3 can’t crash
B programming
A 7 14000 6
15000 -1000 1
week
C get hardwareA 1 50000 .5
51000 -2000 .5 week
D train users B,C 3 can’t crash
Crashing Algorithm:
1 crash only critical activities
2 crash cheapest currently critical
cheapest
B only choice
B is
3 after crashing one time period, recheck critical
5.31
Crash Example
Import critical software from Australia: late penalty $500/d > 12 d
A get import license 5 days no predecessor
B ship
7 days A is predecessor
C train users
11 daysno 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
5.32
Crash Example
Original schedule: 14 days, $1,000 in penalties
=
$1000
crash B to 6 days:13 days, $500 penalties, $300 cost = $800*
crash B to 5
C to 10:
12 days, no penalties, $400+2000 cost = $2400
to 11 days is worse
NOW A SELECTION DECISION
risk versus cost
5.33
Crashing Limitations
• assumes linear relationship between
time and cost
– not usually true (indirect costs don’t
change at same rate as direct costs)
• requires a lot of extra cost
estimation
• time consuming
• ends with tradeoff decision
5.34
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 - don’t
exceed available resources
5.35