MGT400 4

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MGT400 – Project Management Chapter 9: Reducing Project Duration

(page 304-323)

Where We Are Now

Lesson content:

• Why reducing project duration?

• Options when accelerating a project: – Resource constraints

– No resource constraints.

• Review of project costs: – Direct Costs

– Indirect Costs

• Determining Cost of reducing time of project – Task/Activity Crashing

– Cost-Duration trade off using Project Network

• Practical considerations

Why Reducing Project Duration?

• Reasons for imposed project duration dates: • Time-to-market pressures

• Unforeseen delays

• Incentive contracts (bonuses for early completion)

• Imposed deadlines and contract commitments

• Overhead and public goodwill costs

• Pressure to move resources to other projects

• Time Is Money: Cost-Time Tradeoffs • Reducing the time of a critical activity usually increases additional

direct costs. • Solutions focus on reducing activities on the critical path to shorten overall

duration of the project.

• Reducing the time of project, reduces indirect costs.

Pg. 306

Options for Accelerating Project Completion

• Resources Not Constrained

• Adding resources

• Outsourcing project work

• Scheduling overtime

• Establishing a core project team

• Resources Constrained

• Fast-tracking

• Critical-chain

• Reducing project scope

• Compromise quality

Pg. 308-312

Reminder of Project Costs

• Project Indirect Costs

• Costs that cannot be associated with any particular work package or project activity. • Supervision, administration, renting offices, and interest

• Costs that vary (increase) with time. • Reducing project time directly reduces indirect costs.

• Project Direct Costs

• Normal costs that can be assigned directly to a specific work package or project activity. • Labor (overtime) , materials, equipment, and subcontractors

• Reducing activities duration increases direct costs.

Pg. 313

Reducing Project Duration: Determining the Cost

Compute total costs for specific durations and compare to benefits of reducing project time.

Search critical activities for lowest direct-cost activities to shorten project duration.

Identifying costs of reducing project time

Gather information about direct and indirect costs of specific project durations.

Pg. 313

Review: Project Cost–Duration Graph

FIGURE 9.1

Pg. 313

Constructing a Project Cost–Duration Graph • Find total direct costs for

selected project durations.

• Find total indirect costs for selected project durations.

• Sum direct and indirect costs for these selected project durations.

• Compare additional cost alternatives for benefits.

Constructing a Project Cost–Duration Graph

• Determining Activities to Shorten

• Review each activity.

• Determine what extra cost associated with reduction of to the ‘absolute’ minimum time.

• This is called Crashing the task.

• Shorten the activities with the smallest increase in cost per unit of time.

Pg. 314

Activity Graph

FIGURE 9.2

Pg. 315

Constructing a Project Cost–Duration Graph

• Assumptions:

• The cost relationship is linear.

• Normal time assumes low-cost, efficient methods to complete the activity.

• Crash time represents a limit—the greatest time reduction possible under realistic conditions.

• Slope represents a constant cost per unit of time.

• All accelerations must occur within the normal and crash times.

Pg. 315

Cost–Duration Trade-off Example

FIGURE 9.3

Step 1: Find out Max. Crash Time and Slope for each activity

Step 2 Max. Crash Time = Normal Time – Crash Time Slope = (Crash cost – Normal cost) / (Normal time – Crash time)

Pg. 316-317

Cost–Duration Trade-off Example

FIGURE 9.3

Step 1: Find out Max. Crash Time and Slope for each activity

Step 2 Max. Crash Time = Normal Time – Crash Time Slope = (Crash cost – Normal cost) / (Normal time – Crash time)

Activity A Max. Crash Time = Normal Time – Crash Time M. Crash time = 3 – 2 = 1

Slope = (Crash cost – Normal cost) / (Normal time – Crash time) Slope = ($70 - $50) / (3 -2) = $20

Pg. 316-317

Cost–Duration Trade-off Example

FIGURE 9.3

Step 1: Find out Max. Crash Time and Slope for each activity

Step 2 Max. Crash Time = Normal Time – Crash Time Slope = (Crash cost – Normal cost) / (Normal time – Crash time)

? ?

Activity B Max. Crash Time = Normal Time – Crash Time M. Crash time = ?

Slope = (Crash cost – Normal cost) / (Normal time – Crash time) Slope = ?

Pg. 316-317

Cost–Duration Trade-off Example

FIGURE 9.3

Step 1: Find out Max. Crash Time and Slope for each activity

Step 2 Max. Crash Time = Normal Time – Crash Time Slope = (Crash cost – Normal cost) / (Normal time – Crash time)

Activity B Max. Crash Time = Normal Time – Crash Time M. Crash time = 6 – 4 = 2

Slope = (Crash cost – Normal cost) / (Normal time – Crash time) Slope = ($160 - $80) / (6 -2) = $40

Pg. 316-317

Cost–Duration Trade-off Example

FIGURE 9.3

Step 1: Find out Max. Crash Time and Slope for each activity

Step 2 Max. Crash Time = Normal Time – Crash Time Slope = (Crash cost – Normal cost) / (Normal time – Crash time)

Activity C Max. Crash Time = Normal Time – Crash Time M. Crash time = 10 – 9 = 1

Slope = (Crash cost – Normal cost) / (Normal time – Crash time) Slope = ($90 - $60) / (10 - 9) = $30

Pg. 316-317

Cost–Duration Trade-off Example

FIGURE 9.3

Step 1: Find out Max. Crash Time and Slope for each activity

Step 2 Max. Crash Time = Normal Time – Crash Time Slope = (Crash cost – Normal cost) / (Normal time – Crash time)

Activity D, E, F, G Max. Crash Time = Normal Time – Crash Time M. Crash time = ?

Slope = (Crash cost – Normal cost) / (Normal time – Crash time) Slope = ?

? ?

Pg. 316-317

Cost–Duration Trade-off Example

FIGURE 9.3

Step 1: Find out Max. Crash Time and Slope for each activity

Step 2 Max. Crash Time = Normal Time – Crash Time Slope = (Crash cost – Normal cost) / (Normal time – Crash time)

Activity D, E, F, G Max. Crash Time = Normal Time – Crash Time M. Crash time = ?

Slope = (Crash cost – Normal cost) / (Normal time – Crash time) Slope = ?

Pg. 316-317

Cost–Duration Trade-off Example

FIGURE 9.3

Step 1: Find out Max. Crash Time and Slope for each activity

Step 2 Max. Crash Time = Normal Time – Crash Time Slope = (Crash cost – Normal cost) / (Normal time – Crash time)

Activity D, E, F, G Max. Crash Time = Normal Time – Crash Time M. Crash time = ?

Slope = (Crash cost – Normal cost) / (Normal time – Crash time) Slope = ?

Pg. 316-317

Cost–Duration Trade-off Example

We want to reduce 6 time unit.

Pg. 316-317

Cost–Duration Trade-off Example

FIGURE 9.3

Step 3: Reduce the Project Duration (CPM is given/found as A, D, F, G Activities) Project reducing required/asked as 6 time-units (hours, days, weeks, months etc.)

Step 4: Start reducing with the minimum slope, until to reach target reducing time units

Project crashing (reducing) is calculated as follow; Activity A => 1 x $20 = $20 Activity D => 4 x $25 = $100 Activity F => 1 x $30 = $30 Total reducing cost is = $150 (total of A, D, F)

Cumulative total cost (after crashing) is $450 + $150 = $600

Pg. 316-317

Total cost (before crashing) $450

Cost–Duration Trade-off Example (cont’d)

FIGURE 9.4

Pg. 316-317

Cost–Duration Trade-off Example (cont’d)

FIGURE 9.4 (cont’d)

Pg. 316-317

Summary Costs by Duration

FIGURE 9.5

22 775

Pg. 316-317

Project Cost–Duration Graph

FIGURE 9.6

Pg. 317

Practical Considerations

• Using the Project Cost–Duration Graph

• Crash Times

• Linearity Assumption

• Choice of Activities to Crash Revisited

• Time Reduction Decisions and Sensitivity

Pg. 318

What if Cost Not Time Is the Issue?

• Commonly Used Options for Cutting Costs

• Reduce project scope

• Have owner take on more responsibility

• Outsourcing project activities or even the entire project

• Brainstorming cost savings options

Pg. 322

Project Crashing Example (Reducing Duration)

Activity

Time (weeks) Cost (KWD) Max. Crash

Normal Crash Normal Crash Time Slope

A 2 1 2200 2400

B 5 2 3600 3900

C 3 1 2400 2700

D 6 2 4100 4500

E 4 2 4900 5200

F 5 3 1600 1850

Given data; CPM is A, C, E, F activities Target is 5 week crashing (reduce the duration)

Questions; 1.Calculate the Max. Crash time and Slope 2.Crash the project for 5 weeks (calculate crashing cost) 3.Calculate the total cost of the project (before and after crashing)

Project Crashing Example (Reducing Duration)

Activity

Time (weeks) Cost (KWD) Max. Crash Slope

Normal Crash Normal Crash Time

A 2 1 2200 2400 1 200

B 5 2 3600 3900 3 100

C 3 1 2400 2700 2 150

D 6 2 4100 4500 4 100

E 4 2 4900 5200 2 150

F 5 3 1600 1850 2 125

Step 1: Max. Crash Time = Normal Time – Crash Time and Slope = (Crash cost – Normal cost) / (Normal time – Crash time)

Project Crashing Example (Reducing Duration)

Activity

Time (weeks) Cost (KWD) Max. Crash Slope

Normal Crash Normal Crash Time

A 2 1 2200 2400 1 200

B 5 2 3600 3900 3 100

C 3 1 2400 2700 2 150

D 6 2 4100 4500 4 100

E 4 2 4900 5200 2 150

F 5 3 1600 1850 2 125

Step 2: CPM is A, C, E, F activities Target is 5 w. crashing (reduce the duration)

Project crashing (reducing) is required as 5 weeks, therefore the calculation is; Activity F => 2 x 125 KWD = 250 KWD Activity E => 2 x 150 KWD = 300 KWD Activity C => 1 x 150 KWD = 150 KWD Project reducing (crash) cost is = 700 KWD

Project Crashing Example (Reducing Duration)

Activity

Time (weeks) Cost (KWD) Max. Crash Slope

Normal Crash Normal Crash Time

A 2 1 2200 2400 1 200

B 5 2 3600 3900 3 100

C 3 1 2400 2700 2 150

D 6 2 4100 4500 4 100

E 4 2 4900 5200 2 150

F 5 3 1600 1850 2 125

Step 3: Calculate the total cost of the project (before and after crashing)

Total Cost = 18800 KWD (sum of Normal Cost) Before crashing…

Cumulative Total Cost = 18800 KWD + 700 KWD = 19500 KWD (sum of Normal Cost + crashing cost) After crashing…

Key Terms

Crashing

Crash point

Crash time

Direct costs

Fast-tracking

Indirect costs

Outsourcing

Project cost–duration graph