Project Management - PERT/CPM

What is project management?

Consider building a house:

Step A: Prepare site. (5 days)Step B: Build foundation. (8 days)Step C: Frame walls and roof. (15 days)Step D: Rough in Plumbing (12 days)Step E: Rough in Electrical (10 days)Step F: HVAC Venting (8 days)Step G: Drywall (11 days)Step H: Finish Electrical (5 days)Step I: Finish Plumbing (4 days) Step M: Paint (5 days)Step J: Finish HVAC (2 days) Step N: Landscape (5 days)Step K: Install Kitchen (8 days)Step L: Install Baths (14 days)

Project Management - PERT/CPM

Let each node represent a project event/milestone (node 1 isstart of project, node 11 is end of project).

Let each arc represent a project task/job.

Each arc is identified by a job letter and duration. Note thedummy jobs indicating precedence that jobs H and I must complete before K or L begins.

 

2A,5

1 3 4 5 6 9 10

7

8

11B,8 C,15

D,12

F,8

E,10G,11

H,5

I,4

J,2K,8

L,14

M,5

N,5

0

0

Project Management - PERT/CPM

What questions might project managers be interested in?

• How long will the project take?• Can I add manpower or tools to reduce the overall project length?• To which tasks should I add manpower?• What tasks are on the critical path?• Is the project on schedule?• When should materials and personnel be in place to begin a task?• Other?…

Project Management - Examples

• University Convocation Center• Windsor Engine Plant• Other major construction projects• Large defense contracts• NASA projects (space shuttle)• Maintenance planning of oil refineries, power plants, etc…• other…

Project Management – Minimum Completion Time

2A,3

1

3

4 5C,4

D,2B,1

E,5

0

LP Solution: Let ti be the time of event i.

Min Z = t5 – t1

s.t. t2 – t1 >= 3 t3 – t2 >= 0

t3 – t1 >= 1t4 – t2 >= 4t4 – t3 >= 2t5 – t4 >= 5 ti >= 0 for all i

Project Management – Critical Path

2A,3

1

3

4 5C,4

D,2B,1

E,5

0

LP Solution: insert Lindo Solution here

How do you find the critical path from the Lindo solution?

Project Management – Minimum Completion Time and Critical Path

2A,3

1

3

4 5C,4

D,2B,1

E,5

0

Solution by Network Analysis:

Let earliest time of node j, Uj, be the earliest time at which eventj can occur.

Set U1 = 0then U2 = U1 + t12 = 0 + 3 = 3

U3 = Max{U1 + t13 , U2 + t23} = Max{1,3} = 3U4 = Max{U3 + t34 , U2 + t24} = Max{5,7} = 7U5 = U4 + t45 = 12

Project Management – Minimum Completion Time and Critical Path

2A,3

1

3

4 5C,4

D,2B,1

E,5

0

Solution by Network Analysis:

Let latest time of node j, Vj, be the latest time at which eventj can occur while still completing project by minimum theminimum completion time, Um .

Set V5 = U5 = 12then V4 = V5 - t45 = 12 - 5 = 7

V3 = V4 - t34 = 7 – 2 = 5V2 = Min{V4 - t24 ,V3 - 0 } = 3

V1 = Min{V2 - t12 ,V3 – t13} = 0

Project Management – Minimum Completion Time and Critical Path

2A,3

1

3

4 5C,4

D,2B,1

E,5

0

Solution by Network Analysis:

To find the critical path, solve for slack time = Vj - Uj. All events with slack time equal to 0, and tasks connecting these events are on the critical path.

V5 - U5 = 12 – 12 = 0V4 - U4 = 7 – 7 = 0

V3 - U3 = 5 – 3 = 2V2 - U2 = 3 – 3 = 0

V1 - U1 = 0 – 0 = 0Critical Path: 1->2->4->5

CPM – Critical Path MethodCan normal task times be reduced?

Is there an increase in direct costs?• Additional manpower• Additional machines• Overtime, etc…

Can there be a reduction in indirect costs?• Less overhead costs• Less daily rental charges• Bonus for early completion• Avoid penalties for running late• Avoid cost of late startup

CPM addresses these cost trade-offs.

CPM – Critical Path MethodExample:

Job PredecessorsNormal Time

(days)Crash Time

(days)Cost of Crashing

per Day ($)A - 10 7 4B - 5 4 2C B 3 2 2D A,C 4 3 3E A,C 5 3 3F D 6 3 5G E 5 2 1H F,G 5 4 4

Overhead cost = $5/day

CPM – Critical Path MethodEnumerative Approach:

Reduce job H by 1 day: Total Cost improves by $5 - $4 = $1.

Reduce job A by 2 days: Total cost improves by $10 - $8 = $2.

Reduce job A by an additional day, and job B by a day? Total costimproves by $5 - $4 - $2 = -$1. Therefore do not take this action.

Reduce job A by an additional day, and job C by a day? Total costimproves by $5 - $4 - $2 = -$1. Therefore do not take this action.

Evaluate combinations of reducing path 3-4-6 and 3-5-6 by one day.D & E = $5 - $3 - $3 = -$1 F & E = $5 - $5 - $3 = -$3D & G = $5 - $3 - $1 = $1 F & G = $5 - $5 - $1 = -$1Therefore, reduce job D & G by 1 day: TC improves by $5 - $3 -$1 = $1.

Overall improvement: $1 + $2 + $1 = $4.

CPM – Critical Path MethodLP Approach:

Let tij – decision variable for time to complete task connecting events i and j. kij – normal completion time of task connecting events i and j. lij – minimum completion time of task connecting events i and j. Cij – incremental cost of reducing task connecting events i and j.

Model I: Given project must be complete by some time T, which tasksshould be reduced to minimize the total cost?

Min

s.t.

i j

ijijij tkCZ )(

01

i

n

ijijij

ijij

tTtt

ktl

ttt for all jobs (i,j)

for all jobs (i,j)

for all i

CPM – Critical Path MethodLP Approach:

Model II: Given an additional budget of $B for “crashing” tasks, what minimum project completion time can be obtained while staying within your budget?

Min

s.t.

1ttZ n

0

)(

i

iji j

ijij

ijijij

ijij

t

BtkC

ktl

ttt for all jobs (i,j)

for all jobs (i,j)

for all i