Time-Cost Tradeoffs

Time-Cost Tradeoff Example

TaskNormal

Duration Normal Cost

Marginal Cost to Crash One

Week

A 7 $60 $8B 6 $85 $5C 15 $55 $10D 10 $120 $4

A

B

C

D

Start End

Time-Cost Tradeoff Example (cont’d)

Project

Duration

(weeks) Critical Path(s) Task(s) Reduced

Total Direct

Cost

22 Start-A-C-End - $320

21 Start-A-C-End A $328

Start-A-B-End

20 Start-A-C-End C $338

Start-A-B-End

19 Start-A-C-End C $348

Start-A-B-End

18 Start-A-C-End A, B $361

Start-A-B-End

Linear Time-Cost TradeoffIn theory, the normal or expected duration of a task can be reduced by assigning additional resources to the task

Time

Cost

Crash Point

Normal Point

Slope (bj) = Increase in cost by reducing task by one time unit

Normal time =Crash time =

Normal cost =

Crash cost =

tjNtj

c

Cjc

CjN

Balancing Overhead & Direct Costs

Project Duration

Cost

Indirect (overhead)

Costs

Direct Costs

Total Cost

Crash Time

Normal TimeMinimum Cost Solution

Time-Cost Tradeoff (Direct Costs Only)Given Normal point with cost and time

and Crash point with cost and time

Assume constant marginal cost of crashing task j =

Decision Variables: Sj = Starting time of task j

END = End time of project

tj = Duration of task j

Minimize Total Direct Cost =

Sj ≥ Si + ti for all tasks i Pj

for all tasks in project

END = Tmax

tj, Sj ≥ 0

CjN

Cjc tj

c

tjN

bj = Cj

c - CjN

tjc - tj

N

bj tjj

tjc Š tj Š tj

N

General Time-Cost Tradeoffs

where

I = indirect (overhead) cost/time period

P = penalty cost/time period if END is delayed beyonddeadline Tmax

L = number of time periods project is delayed beyond deadline Tmax

Minimize Total Costs = + I (END) + P L

bj tjj

QUESTION: HOW TO DEFINE L?

Software Project Schedules

“Observe that for the programmer, as for the chef, the urgency of the patron may govern the scheduled completion of the task, but it cannot govern the actual completion. An omelet, promised in ten minutes, may appear to be progressing nicely. But when it has not set in ten minutes, the customer has two choices--wait or eat it raw. Software customers have the same choices.

The cook has another choice; he can turn up the heat. The result is often an omelet nothing can save--burned in one part, raw in another.”

F.P. Brooks, “The Mythical Man-Month”, Datamation, Vol 20, No 12 (Dec, 1974),

pp. 44-52.

Coordination Costs (Software Development Project)

n Assume you want to develop program that will require (approximately) 50,000 lines of PERL code

n A typical programmer can write approximately 1500 lines of code per weekn Coordination time is M (M-1)/2 weeks

No. of Programmers

No. of Weeks Coding

No. of Coordination

Weeks

Total Number of

Weeks

1 33.33 0 33.332 16.67 1 17.673 11.11 3 14.114 8.33 6 14.335 6.67 10 16.676 5.56 15 20.567 4.76 21 25.768 4.17 28 32.179 3.70 36 39.7010 3.33 45 48.3311 3.03 55 58.03

0

2

4

6

8

10

12

0 10 20 30 40 50 60 70

Total Number of Weeks

No. of

Pro

gra

mm

ers

(Cost)

Brook’s Law

“Adding manpower to a late software project makes it later.”

n

F.P. Brooks, “The Mythical Man-Month”, Datamation, Vol 20, No 12 (Dec, 1974), pp. 44-52.

Compressing New Product Development Projects

Traditional Method

Design follows a sequential pattern where information about the new product is slowly accumulated in consecutive stages

Stage 0 Stage 1 Stage N

New Product Development Process

Overlapped Product Design

Allows downstream design stages to start before preceding upstream stages have finalized their specifications….

Stage 0

Stage 1

Stage N

Issues and Tradeoffs

What are the tradeoffs when moving from a traditional sequential product design process to an overlapped product design process?

• Increased uncertainty (that leads to additional work)

• Can add additional resources to tasks to reduce duration--but costs are increased

Classic PERT Model Defined

• Since task durations are now random variables, time of any milestone (e.g., end of project) is now RV

• Assume all tasks are statistically independent

• Use values of j to identify expected critical path

• Since time of event (e.g., ESk) is now sum of independent RV’s, central limit theorem specifies that ESk is approximately normally distributed with mean E[ESk] and variance Var[ESk]

where there exists s paths to task k

Expected early start time of task k = EESk = maxs

jtasks j on path s

Classic PERT Model (cont’d)

Expect Project Duration = E[ESEND] = jtasks j on CP

Variance of Project Duration = Var[ESEND] = j2

tasks j on CP

Thus, expected project duration is defined as:

Using central limit theorem and standard normal distribution:

P ESEND Š Tmax = P z Š Tmax - E ESEND

Var ESEND

PERT Example #1

Duration Estimates ExpectedTask Description Predecessors Optimistic Pessimistic Likely Duration Variance

A Requirements Analysis none 2 14 6 6.67 4.00B Programming A 4 12 7 7.33 1.78C Hardware acquisition A 2 13 8 7.83 3.36D User training A 12 18 14 14.33 1.00E Implementation B, C 3 7 5 5.00 0.44F Testing E 3 7 4 4.33 0.44

END End of project D, F 0 0 0 0.00 0.00

StartTask A

Requirements Analysis

Task C Hardware

Acquisition

Task B Programming

Task F Testing

Task D User

Training

Task E Implementation

End

PERT Example #1 (cont’d)

ExpectedTask Path Early Start Variance Due Date Zi Pr(zi)

B,C,D Start-A 6.67 4.00 6 -0.33 0.37E Start-A-C 14.50 7.36 15 0.18 0.57F Start-A-C-E 19.50 7.81 20 0.18 0.57

End Start-A-C-E-F-End 23.83 8.25 25 0.41 0.66

PERT Expected Duration = 23.83 Expected CP = {Start, A, C, E, F, End}PERT Variance = 8.250

StartTask A

Requirements Analysis

Task C Hardware

Acquisition

Task B Programming

Task F Testing

Task D User

Training

Task E Implementation

End

PERT Example #2

Task B

B = 12

B2 = 4

Task D

D = 3

D2 = 1

Task A

A = 4

A2 = 2

Task C

C = 10

C2 = 5

ENDSTART

Example #3: Discrete Probabilities

Task A Task B Task C Task DValue Prob Value Prob Value Prob Value Prob

7 0.333 2 0.2 5 0.2 3 0.3

8 0.333 12 0.8 15 0.2 12 0.7

9 0.333 25 0.6

START END

Task A(8.0)

Task B(10.0)

Task C(19.0)

Task D(9.3)

Example #3 (cont’d) Task A Task B Task C Task D Critical Prob of Length PATHS

Combination Value Prob Value Prob Value Prob Value Prob Path CP of CP A,D B, D C1 7 0.333 2 0.2 5 0.2 3 0.3 A, D 0.004 10 0.004 0.000 0.0002 7 0.333 2 0.2 5 0.2 12 0.7 A, D 0.009 19 0.009 0.000 0.0003 7 0.333 2 0.2 15 0.2 3 0.3 C 0.004 15 0.000 0.000 0.0044 7 0.333 2 0.2 15 0.2 12 0.7 A, D 0.009 19 0.009 0.000 0.0005 7 0.333 2 0.2 25 0.6 3 0.3 C 0.012 25 0.000 0.000 0.0126 7 0.333 2 0.2 25 0.6 12 0.7 C 0.028 25 0.000 0.000 0.0287 7 0.333 12 0.8 5 0.2 3 0.3 B, D 0.016 15 0.000 0.016 0.0008 7 0.333 12 0.8 5 0.2 12 0.7 B, D 0.037 24 0.000 0.037 0.0009 7 0.333 12 0.8 15 0.2 3 0.3 B, D 0.016 15 0.000 0.016 0.000

10 7 0.333 12 0.8 15 0.2 12 0.7 B, D 0.037 24 0.000 0.037 0.00011 7 0.333 12 0.8 25 0.6 3 0.3 C 0.048 25 0.000 0.000 0.04812 7 0.333 12 0.8 25 0.6 12 0.7 C 0.112 25 0.000 0.000 0.11213 8 0.333 2 0.2 5 0.2 3 0.3 A, D 0.004 11 0.004 0.000 0.00014 8 0.333 2 0.2 5 0.2 12 0.7 A, D 0.009 20 0.009 0.000 0.00015 8 0.333 2 0.2 15 0.2 3 0.3 C 0.004 15 0.000 0.000 0.00416 8 0.333 2 0.2 15 0.2 12 0.7 A, D 0.009 20 0.009 0.000 0.00017 8 0.333 2 0.2 25 0.6 3 0.3 C 0.012 25 0.000 0.000 0.01218 8 0.333 2 0.2 25 0.6 12 0.7 C 0.028 25 0.000 0.000 0.02819 8 0.333 12 0.8 5 0.2 3 0.3 B, D 0.016 15 0.000 0.016 0.00020 8 0.333 12 0.8 5 0.2 12 0.7 B, D 0.037 24 0.000 0.037 0.00021 8 0.333 12 0.8 15 0.2 3 0.3 B, D 0.016 15 0.000 0.016 0.00022 8 0.333 12 0.8 15 0.2 12 0.7 B, D 0.037 24 0.000 0.037 0.00023 8 0.333 12 0.8 25 0.6 3 0.3 C 0.048 25 0.000 0.000 0.04824 8 0.333 12 0.8 25 0.6 12 0.7 C 0.112 25 0.000 0.000 0.11225 9 0.333 2 0.2 5 0.2 3 0.3 A, D 0.004 12 0.004 0.000 0.00026 9 0.333 2 0.2 5 0.2 12 0.7 A, D 0.009 21 0.009 0.000 0.00027 9 0.333 2 0.2 15 0.2 3 0.3 C 0.004 15 0.000 0.000 0.00428 9 0.333 2 0.2 15 0.2 12 0.7 A, D 0.009 21 0.009 0.000 0.00029 9 0.333 2 0.2 25 0.6 3 0.3 C 0.012 25 0.000 0.000 0.01230 9 0.333 2 0.2 25 0.6 12 0.7 C 0.028 25 0.000 0.000 0.02831 9 0.333 12 0.8 5 0.2 3 0.3 B, D 0.016 15 0.000 0.016 0.00032 9 0.333 12 0.8 5 0.2 12 0.7 B, D 0.037 24 0.000 0.037 0.00033 9 0.333 12 0.8 15 0.2 3 0.3 B, D 0.016 15 0.000 0.016 0.00034 9 0.333 12 0.8 15 0.2 12 0.7 B, D 0.037 24 0.000 0.037 0.00035 9 0.333 12 0.8 25 0.6 3 0.3 C 0.048 25 0.000 0.000 0.04836 9 0.333 12 0.8 25 0.6 12 0.7 C 0.112 25 0.000 0.000 0.112

6.8% 32.0% 61.1%

Example #3 (cont’d)

Length of Cumulative

CP's Prob Prob10 0.004 0.0011 0.004 0.0112 0.004 0.0115 0.108 0.1219 0.019 0.1420 0.019 0.1621 0.019 0.1824 0.224 0.4025 0.599 1.00

Task A Task B Task C Task D

6.8% 32.0% 61.1% 38.8%

Criticality Indices

Expected Project Duration = 23.22

Monte-Carlo Simulation (PERT Example 1)Task Duration Early Latest Total Expected

Task (Uniform Dist) Start Finish Slack Duration Variance

A 4.99 0 4.99 0.00 6.67 4.00B 4.75 4.99 9.74 0.00 7.33 1.78C 3.38 4.99 9.74 1.36 7.83 3.36D 12.20 4.99 21.02 3.83 14.33 1.00E 5.94 9.74 15.68 0.00 5.00 0.44F 5.34 15.68 21.02 0.00 4.33 0.44

END 0.00 21.02 21.02 0.00 0.00 0.00

Run Project Duration t(B) t(C) t(D) t(E) t(F)1 31.07 1 0 0 1 12 27.41 0 1 0 1 13 23.97 1 0 0 1 14 28.93 0 1 0 1 15 26.85 1 0 0 1 16 28.82 0 0 1 0 07 28.77 0 1 0 1 1

197 30.37 0 1 0 1 1198 29.78 1 0 0 1 1199 25.33 1 0 0 1 1200 29.70 0 1 0 1 1

Ave 27.13 48.5% 42.0% 9.5% 90.5% 90.5%Var 16.777

Project Makespan Lower Limit Upper Limit

95% Confidence interval 26.56 27.7299% Confidence interval 26.37 27.90

Calculating Confidence Intervals

For a confidence interval, we can use the sample mean and the estimated standard error of the mean

where s is the sample standard deviation and n is the number of trials

Using a normal approximation, a (1- ) two-sided confidence interval is given by

sX = sn

X

X -+ z/2 sX

New Product Development Projects

START

Lease Mfg/Office

Space

Identify/hire staff

Design of physical unit

Electronics design Software

Assemble prototype Beta test prototype

END

Beta test fails (with probability of 0.25)

and rework is needed

Beta test fails (with probability of 0.25)

and rework is needed

New Product Development Projects (cont’d)

START

Lease Mfg/Office

Space

Identify/hire staff

Design of physical unit

Electronics design Software

Assemble prototype Beta test prototype

END

Beta test fails and rework is needed

Prob = .25

Prob = .75

Critical Chain and the Theory of Constraints (TOC)

• Use deterministic CPM model with buffers to deal with any uncertainties,

• Place project buffer after last task to protect the customer’scompletion schedule,

• Exploit constraining resources (make certain that resources are fully utilized),

• Avoid wasting time slack time by encouraging early task completions,

• Carefully monitor the status of the buffer(s) and communicate this status to other project team members on a regular basis, and

• Make certain that the project team is 100 percent focused on critical chain tasks

Project “Goal” (according to Goldratt): Meet Project Due Date

Project Buffer Defined

• Project Buffer is placed at the end of the project to protect the customer’s promised due date

PERT Example #1 Revisited with Project Buffer

Start

Task BProgramming

User Task D

Usertraining

Task EImplementation

End Project

Buffer

Task Arequirements

analysis

Task CHardwareacquisition

Task F

Testing

Calculating Project Buffer Size

For tasks k on critical chain, we can calculate project buffer using following formula that project will be completed within worst-case duration estimates around 90 percent of the time:

For those “who want a scientific approach to sizing buffers....”

Buffer = tasks k on critical chain

tkp - k

2

Implications of Project Uncertainty

Assume that the duration of both tasks A and B are described by a normal distribution with a mean of 30 days

START END

Task A

Task B

What is the probability that the project will be completed within 30 days?

Uncertainty and Worker Behavior

Consider a project with two tasks that must be completed serially

The duration of each task is described by a RV with values Ti (i = 1, 2)

Values of T 1 Prob Values of T 2 Prob

7 0.3 14 0.58 0.4 18 0.59 0.3

8.0 16

Start Task 1 Task 2 End

Parkinson’s Law (Expanding Work)“Work expands so as to fill the time available for its completion”

Professor C.N. Parkinson (1957)

Set a deadline D = 24 days

So T(D) = project makespan (function of D) where

E[T(D)] = E(T1) + E(T2) + E[max(0, D - T1 - T2)]

Values of T 1 Prob Values of T 2 Prob

Project Makespan Prob

7 0.3 14 0.5 24 0.157 0.3 18 0.5 25 0.158 0.4 14 0.5 24 0.28 0.4 18 0.5 26 0.29 0.3 14 0.5 24 0.159 0.3 18 0.5 27 0.15

E[T(D)] = 25 days

Procrastinating Worker

Set a deadline D = 24 days

E’[T(D)] = E(T1) + E(T2) + E{max[0, D - T1 - E(T2)]}

Can show that E[T(D)] ≥ E’[T(D)] ≥ D

What are the implications for project managers?

Values of T 1 Prob

E[Delay] = max[0, D - T1 - E(T2)] E[Makespan]

7 0.3 1 248 0.4 0 249 0.3 0 25

8 0.3 24.30

Schoenberger’s Hypothesis

An increase in the variability of task durations will increase the expected project duration….

Schoenberger’s Hypothesis Illustrated

START END

Task A

Task B

Duration of Task A Probability

Duration of Task B Probability

12 0.1 10 0.514 0.8 15 0.516 0.1

14.0 12.5

Schoenberger’s Hypothesis Illustrated

RealizationTask A

DurationTask B

Duration Probability Max (A, B)

1 12 10 0.05 122 14 10 0.4 143 16 10 0.05 164 12 15 0.05 155 14 15 0.4 156 16 15 0.05 16

Duration of Task A Probability

Duration of Task B Probability

12 0.3 10 0.514 0.4 15 0.516 0.3

14.0 12.5

Increasing the variance of Task A:

Results in an increased expected duration = 14.65 days

Expected duration equals 14.55 days

Risk Management

• All projects involve some degree of risk

• Need to identify all possible risks and outcomes

• Need to identify person(s) responsible for managing project risks

• Identify actions to reduce likelihood that adverse events will occur

Risk Analysis

Risk Exposure (RE) or Risk Impact =

(Probability of unexpected loss) x (size of loss)

Example: Additional features required by clientLoss: 3 weeksProbability: 20 percentRisk Exposure = (.20) (3 weeks) = .6 week

How to Manage Project Risks?

Preventive Actions• Actions taken in anticipation of adverse events• May require action before project actually begins• Examples?

Contingency Planning• What will you do if an adverse event does occur?• “Trigger point” invokes contingency plan• Frequently requires additional costs

Risk and Contracts

High Low Low High

Degree of Risk

Contractor ClientFixed Price Contract Cost Plus Contract

Firm price

Elements can be

renegotiated IncentivesT&M

with limits

Cost Plus with

IncentivesTime &

materials

Tornado Diagram

Wage Rate

Direct Labor Hours

Material Units Needed

Early Completion Bonus

Material Unit Cost

Interest rates

Energy costs

Overhead

Project Cost ($000's)

$1290

$1265

$1260

$13101310$1350

$1350

$1380

$1400

$1700

$1720

$1680

$1690

$1640

$1620

$1625

$17601760

$1500 $1600 $1800$1700$1400$1300$1200

Sensitivity Chart

Wage Rate 0.85Direct Labor Hours 0.73

Material Units Needed 0.62Early Completion Bonus -0.45

Material Unit Cost 0.42Interest rates 0.28Energy costs 0.19

Overhead 0.10

0 0.5 1.0-0.5

Rank Order Correlation with Total Project Cost

Van Allen Company

Task

Immediate Predecessors Time Cost Time Cost

Start - 0 - 0 -A Start 3 $60 5 $40B Start 1 $50 5 $30C B 5 $70 10 $40D A 2 $60 7 $40E A 2 $50 6 $30F C, D 5 $90 11 $60G C, D 4 $60 6 $30H G 1 $40 5 $20I E, G 1 $50 4 $20

End F, H, I 0 - 0 -

Strike Expected(wks) Prob Duration

3 0.45 1.354 0.3 1.205 0.25 1.25

E[Strike Duration] 3.80