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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.

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project management

Project Management

Dr. Ron Lembke

Operations Management

macarthur maze
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
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
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
Time-Cost Example

D 8

A 10

B 10

C 10

  • ABC is critical path=30

Crash cost Crash

per week wks avail

A 500 2

B 800 3

C 5,000 2

D 1,100 2

Cheapest way to gain 1

Week is to cut A

time cost example1
Time-Cost Example

D 8

A 9

B 10

C 10

  • ABC is critical path=29

Crash cost Crash

per week wks avail

A 500 1

B 800 3

C 5,000 2

D 1,100 2

Wks Incremental Total

Gained Crash $ Crash $

1 500 500

Cheapest way to gain 1 wk

Still is to cut A

time cost example2
Time-Cost Example

D 8

A 8

B 10

C 10

  • ABC is critical path=28

Crash cost Crash

per week wks avail

A 500 0

B 800 3

C 5,000 2

D 1,100 2

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 example3
Time-Cost Example

D 8

A 8

B 9

C 10

  • ABC is critical path=27

Crash cost Crash

per week wks avail

A 500 0

B 800 2

C 5,000 2

D 1,100 2

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 example4
Time-Cost Example

D 8

A 8

B 8

C 10

  • Critical paths=26 ADC & ABC

Crash cost Crash

per week wks avail

A 500 0

B 800 1

C 5,000 2

D 1,100 2

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 example5
Time-Cost Example

D 7

A 8

B 7

C 10

  • Critical paths=25 ADC & ABC

Crash cost Crash

per week wks avail

A 500 0

B 800 0

C 5,000 2

D 1,100 1

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 example6
Time-Cost Example

D 7

A 8

B 7

C 9

  • Critical paths=24 ADC & ABC

Crash cost Crash

per week wks avail

A 500 0

B 800 0

C 5,000 1

D 1,100 1

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 example7
Time-Cost Example

D 7

A 8

B 7

C 8

  • Critical paths=23 ADC & ABC

Crash cost Crash

per week wks avail

A 500 0

B 800 0

C 5,000 0

D 1,100 1

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 example8
Time-Cost Example

D 7

A 8

B 7

C 8

  • 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

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

pert activity times
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
Project Times
  • Expected project time (T)
    • Sumofcritical path activity times, t
  • Project variance (V)
    • Sum of critical path activity variances, v
example

A

B

C

6.48

4.33

7.67

Example

Activitya m bE[T]variance

A 2 4 8 4.33 1

B 3 6.1 11.5 6.48 2

C 4 8 10 7.67 1

Project 18.5 4

sum of 3 normal random numbers
Sum of 3 Normal Random Numbers

Average value of the sum is

equal to the sum of the averages

Variance of the sum is equal to

the sum of the variances

Notice curve of sum is more spread

out because it has large variance

10 20 30 40 50 60

back to the example probability of 21 wks
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

Probability <= 21 wks

= 0.8944 = 89.44%

18.5 21

conclusion
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