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More on Estimation

In general, effort estimation is based on severalparameters and the model ( E= a + b*S**c ): Personnel

Environment

Quality Size or Volume of work

Process

where S is the Size and a, b, and c are constants estimated with

other parameters

Some popular Effort Estimation methodologies:

Function Point

COCOMO (Constructive Cost Model)

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simplified Function Point

Proposed by Albrecht of IBM as an alternative metricto lines of code count for S, size of product.

Based on 5 major areas and a complexity table ofsimple, average and complex set of weights as follows:

input 3 4 6 output 4 5 7

inquiry 3 4 6

master files 7 10 15

interfaces 5 7 10

The Unadjusted Function Point is : UFP = w*Inp + w2*Out + w3*Inq + w4*MastF +w5*Intf

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Function Point (cont.)

14 technical complexity factors are included, each valued

between 0 and 5: data communications

distributed data

performance criteria

heavy hardware usage

high transaction rates

online data entry

online updating

complex computations

ease of installation ease of operation

portability

maintainability

end-user efficiency reusability

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Function Point (cont.)

The sum of 14 technical complexity factors can

have values of 0 through 70.

The the Total Complexity Factor(TCF) is

defined as:

TCF = .65 + (.01 * Sum of 14 technical complexityfactors)

TCF may have values of 0.65 through 1.35.

Finally, Function Point (FP) is defined as:

FP = UFP * TCF

For Cost and Productivity per FP, one may use

historical data.

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Simple Function Point Example

Consider the previous POWER function project and

use Simple weights from the table :

2 inputs, 1 output, 0 inquiry, 0 master file, and 0 interfaces

UFP = 3* 2 + 4*1 + 3*0 + 7*0 + 5*0 = 10

consider the 14 complexity factors : 0-data comm; 0-distrib

data; 0-perf criteria; 0-hardware usage; 0-transaction rate; 1-online data entry; 0-end user efficiency; 0-online update; 5-

complex computation;0-reusability; 0-ease of install; 0-ease of

operation; 0-portability; 1-maintainability:

TCF = .65 + (.01 * 7 ) = .72

FP = UFP * TCF FP = 10 * .72

FP = 7.2

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Function Point Example (cont.)

What does 7.2 function points mean in terms of

schedule and cost estimates ?

One can receive guidance from IFPUG

(International Function Point User Group) to

get some of the $/FP or person-days/FP data. With old IBM services division data of 20

function points per person month to perform

complete development, 7.2 FP translates to

approximately.36 person months or (22days *

.36 = 7.9 person days) of work.

Assume $7k/person-month, .36 person months

will cost about$2.5k.

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Some Function Points Drawbacks

Requires trained people to perform estimates ofwork volume or product size, especially the 14

technical complexity factors.

While IFPUG may furnish some broader data,Cost and Productivity figures are different from

organization to organization.

e.g. the IBM data takes into account of lots ofcorporate overhead cost

Some of the Complexity Factors are not that

important or complex with todays tools.

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COCOMO Estimating Technique

Developed by Barry Boehm in early 1980s who had a longhistory with TRW and government projects (initially, LOCbased )

Later modified into COCOMO II in the mid-1990s (FP

preferred but LOC is still used)

Assumed process activities : Product Design

Detailed Design

Code and Unit Test Integration and Test

Utilized by some but most of the software industry people stillrely on experience and/or own company proprietary data.

Note : No Requirements !

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COCOMO I Basic Form for Effort

Effort = A * B * (size ** C)

Effort =person months

A = scaling coefficient

B = coefficient based on 15 parameters

C = a scaling factor for process

Size = delivered source (K) lines of code

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COCOMO I Basic form for Time

Time = D * (Effort ** E)

Time = total number ofcalendar months

D = A constant scaling factor for schedule

E = a coefficient to describe the potential

parallelism in managing software development

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COCOMO I

Originally based on 56 projects

Reflecting 3 modes of projects

Organic : less complex and flexible process

Semidetached: average projectEmbedded: complex, real-time defense

projects

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COCOMO I

For the basic forms:

Effort = A * B *(size ** C)

Time = D * (Effort ** E)

Organic : A = 3.2 ; C = 1.05 ; D= 2.5; E = .38

Semidetached : A = 3.0 ; C= 1.12 ; D= 2.5; E = .35

Embedded : A = 2.8 ; C = 1.2 ; D= 2.5; E = .32

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Coefficient B

Coefficient B is an effort adjustment factor based on 15parameters which varied from very low, low,nominal,high,very high to extra high

B = Product of (15 parameters)

Product attributes: Required Software Reliability : .75 ; .88; 1.00; 1.15; 1.40;

Database Size : ; .94; 1.00; 1.08; 1.16;

Product Complexity : .70 ; .85; 1.00; 1.15; 1.30; 1.65

Computer Attributes

Execution Time Constraints : ; ; 1.00; 1.11; 1.30; 1.66

Main Storage Constraints : ; ; 1.00; 1.06; 1.21; 1.56

Virtual Machine Volatility : ; .87; 1.00; 1.15; 1.30;

Computer Turnaround time : ; .87; 1.00; 1.07; 1.15;

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Coefficient B (cont.)

Personnel attributes

Analyst Capabilities : 1.46 ; 1.19; 1.00; .86; .71;

Application Experience : 1.29; 1.13; 1.00; .91; .82;

Programmer Capability : 1.42; 1.17; 1.00; .86; .70;

Virtual Machine Experience : 1.21; 1.10; 1.00; .90; ; Programming lang. Exper. : 1.14; 1.07; 1.00; .95; ;

Project attributes

Use of Modern Practices : 1.24; 1.10; 1.00; .91; .82; Use of Software Tools : 1.24; 1.10; 1.00; .91; .83;

Required Develop schedule : 1.23; 1.08; 1.00; 1.04; 1.10;

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An example

Consider an average project of10Kloc:

Effort = 3.0 * B * (10** 1.12) = 3 * 1 * 13.2 = 39.6 pm

Where B = 1.0 (all nominal)

Time = 2.5 *( 39.6 **.35) = 2.5 * 3.6 = 9 months

This requires an additional 8% more effort and36% more

schedule time forproduct plan and requirements:

Effort = 39.6 + (39.6 * .o8) = 39.6 + 3.16 =42.76 pm

Time = 9 + (9 * .36) = 9 +3.24 = 12.34 months

I am cheating here!

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Try the POWER Function Example

POWER function was assumed to be 100 locof C++ code, fairly simple (or Organic), and

nominal for B factor:

Effort = 3.2 * 1 * ( .1 ** 1.05) = appr. 0.3 person- month

Time = 2.5 * ( .3 ** .38) = appr. 1.5 months

Note that while it takes only (.3 * 22 person days =6.7 person days), the total project duration will be

1.5 months.

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Some COCOMO I concerns

Is our initial loc estimate accurate enough ?

Are we interpreting each parameter the same

way ? Do we have a consistent way to assess the range

of values for each of the 15 parameters ?

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COCOMO II

Effort performed at USC with many industrialcorporations participating (still guided by Boehm)

Has a database of over 80 some projects

Early estimate, preferred to useFunction Point instead ofLOC for size; later estimate may use LOC for size.

Coefficient B based on 15 parameters for early estimate is

rolled up to 7 parameters, and for late estimates use 17parameters.

Scaling factor for Process has 6 categories ranging invalue from .00 to .05, in increments of .01

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Lets look at Our 3 Estimates for

POWER

Work Breakdown Structure and personal

experience:

Effort : 5 person days

Time : 3 calendars (done in parallel)

Function Point and using IBM service divisiondata:

Effort : 7.9 person days

Time : 7.9 calendar days (done with 1 person)

COCOMO I and using Organic, 100 loc, and

nominal for all B factors:

Effort : 6.7 person days

Time : 1.5 calendar months

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Estimation Models

There are many tools and models , butnone seem to have dominated the software

field.

Many practicing professionals still dependon personal and proprietary data.

Some other models :

Walston-Felix

SLIM (commercial-proprietary)