Project
Management
Dr. Ron Lembke
Operations Management
Can We Go Faster?
MacArthur Maze
• Key SF artery
– I-80 west to 880 south
– April 29, 2007, 8,600 gal gas
• Day 8 – CC Meyers gets job
• Day 25 – opened to traffic
• $876 k bid, $2.5m cost
– $200,000 per day incentive
– $5,000,000 bonus
China Coal
Plant
Official Deadline: June 29
Internal Goal: March 31
Delivered ON TIME
Lost Profits: $400 million US
1995: Pledged $100m to Princeton
Time-Cost Models
1. Identify the critical path
2. Find cost per day to expedite each node on
critical path.
3. For cheapest node to expedite, reduce it as
much as possible, or until critical path
changes.
4. Repeat 1-3 until no feasible savings exist.
Time-Cost Example
• ABC is critical path=30
Crash cost
per week
A
500
B
800
C
5,000
D
1,100
Crash
wks avail
2
3
2
2
D8
A 10
B 10
C 10
Cheapest way to gain 1
Week is to cut A
Time-Cost Example
• ABC is critical path=29
Crash cost
per week
A
500
B
800
C
5,000
D
1,100
Crash
wks avail
1
3
2
2
D8
A9
B 10
C 10
Wks
Incremental Total
Gained Crash $ Crash $
1
500
500
Cheapest way to gain 1 wk
Still is to cut A
Time-Cost Example
• ABC is critical path=28
Crash cost
per week
A
500
B
800
C
5,000
D
1,100
Crash
wks avail
0
3
2
2
D8
A8
B 10
C 10
Wks
Incremental Total
Gained Crash $ Crash $
1
500
500
2
500
1,000
Cheapest way to gain 1 wk
is to cut B
Time-Cost Example
• ABC is critical path=27
Crash cost
per week
A
500
B
800
C
5,000
D
1,100
Crash
wks avail
0
2
2
2
D8
A8
B9
C 10
Wks
Incremental Total
Gained Crash $ Crash $
1
500
500
2
500
1,000
3
800
1,800
Cheapest way to gain 1 wk
Still is to cut B
Time-Cost Example
• Critical paths=26 ADC & ABC
Crash cost
per week
A
500
B
800
C
5,000
D
1,100
Crash
wks avail
0
1
2
2
D8
A8
B8
C 10
Wks
Incremental Total
Gained Crash $ Crash $
1
500
500
2
500
1,000
3
800
1,800
4
800
2,600
To gain 1 wk, cut B and D,
Or cut C
Cut B&D = $1,900
Cut C = $5,000
So cut B&D
Time-Cost Example
• Critical paths=25 ADC & ABC
Crash cost
per week
A
500
B
800
C
5,000
D
1,100
Crash
wks avail
0
0
2
1
D7
A8
B7
C 10
Wks
Incremental Total
Gained Crash $ Crash $
1
500
500
2
500
1,000
3
800
1,800
4
800
2,600
5
1,900
4,500
Can’t cut B any more.
Only way is to cut C
Time-Cost Example
• Critical paths=24 ADC & ABC
Crash cost
per week
A
500
B
800
C
5,000
D
1,100
Crash
wks avail
0
0
1
1
D7
A8
B7
C9
Wks
Incremental Total
Gained Crash $ Crash $
1
500
500
2
500
1,000
3
800
1,800
4
800
2,600
5
1,900
4,500
6
5,000
9,500
Only way is to cut C
Time-Cost Example
• Critical paths=23 ADC & ABC
Crash cost
per week
A
500
B
800
C
5,000
D
1,100
Crash
wks avail
0
0
0
1
D7
A8
B7
C8
Wks
Incremental Total
Gained Crash $ Crash $
1
500
500
2
500
1,000
3
800
1,800
4
800
2,600
5
1,900
4,500
6
5,000
9,500
7
5,000
14,500
No remaining possibilities to
reduce project length
Time-Cost Example
• Now we know how much it
costs us to save any
number of weeks
• Customer says he will pay
$2,000 per week saved.
• Only reduce 5 weeks.
• We get $10,000 from
customer, but pay $4,500 in
expediting costs
• Increased profits = $5,500
D7
A8
B7
C8
Wks
Incremental Total
Gained Crash $ Crash $
1
500
500
2
500
1,000
3
800
1,800
4
800
2,600
5
1,900
4,500
6
5,000
9,500
7
5,000
14,500
No remaining possibilities to
reduce project length
What about
Uncertainty?
PERT Activity Times
• 3 time estimates
• Optimistic times (a)
• Most-likely time (m)
• Pessimistic time (b)


• Follow beta distribution
• Expected time: t = (a + 4m + b)/6
• Variance of times: v = (b - a)2/36
Project Times
• Expected project
time (T)
a  4m  b
ET 
6
• Sum of critical path
activity times, t
• Project variance (V)
• Sum of critical path
activity variances, v


b  a

2
2
36
Example
Activity
A
B
C
Project
a m
b
2 4
8
3 6.1 11.5
4 8 10
A
B
C
4.33
6.48
7.67
E[T] variance
4.33
1
6.48
2
7.67
1
18.5
4
a  4m  b 2  4 * 4  8
TA 

 4.33
6
6
2
2




b

a
8

2
2
A 

1
36
36
Sum of 3 Normal
Random Numbers
X  10
Average value of the sum is
equal to the sum of the averages
  10
2
Variance of the sum is equal to
the sum of the variances
X  20
 2  15
Notice curve of sum is more spread
out because it has large variance
X  30
 2  35
X  60
 2  60
10
20
30
40
50
60
Back to the Example:
Probability of <= 21 wks
Average time = 18.5, st. dev = 2
21 is how many standard deviations
above the mean?
21-18.5 = 2.5.
St. Dev = 2,
so 21 is 2.5/2 = 1.25 standard
deviations above the mean
Book Table (p. 498) says area between
Z=1.25 and –infinity is 0.8944
18.5
21
Probability <= 21 wks
= 0.8944 = 89.44%
Conclusion
• Determined most cost-effective way to
“crash” a project
– Cheapest way to crash a given number of
weeks
– Stop crashing when marginal cost exceeds
marginal benefit
• Computed project probabilities
– Use probabilities of each activity
– Can talk about likelihood of finishing
project in a given amount of time