Specification From Examples

Julia JohnsonDept. of Math & Computer Science

Laurentian UniversitySudbury, Ontario

Canada

Problem

To describe system characteristics by providing examples of systems that exhibit those characteristics.

Outline1. Problem Statement2. Criticism of Existing Solutions3. Suggested Solution 3.1 Rough Sets 3.2 Strength of a Rule 3.3 Rough Mereology 3.4 RM System Specification4. Conclusions

ag

ag2 agmag1 …

ag11 ag12…

ag1n…

ag21 ag20agm1 agmp

…

M

BL

B1

L2

B3

L3

µB (B3,B1) =

B ≥ .25

µL (L3,L2) =

L ≥ .4

C1 C5µM (C5,C1) =

M ≥ .14

Rough Mereology

Mereology ≡ Theory of “Part of” relation, Lesniewski

Rough Mereology – Theory of Relation “Part of to a degree”, Polkowski & Skowron

Applications of Rough Mereology Control – Skowron & Polkowski 1994 Warsaw Politecnica Building – Poitr 1998-99 Polish Academy of Science Scheduling – Johnson 1998-99 University of Regina/University of

Waterloo

µ (x,y) is read “the degree in which x is a part of y”

-the rough inclusion function

For each construction of objects from sub-objects, we form a vector,

B L M

Where if M1 = O(B1L1) and M2 = O(B2L2)

Then B = µB(B1,B2) L = µL(L1,L2)

M = µM(M1,M2)

M2 is constructed from B2 and L2

The vector means

If µB(B1,B2) > B (B1 is part of B2 to degree at least B)

And µL(L1,L2) > L (L1 is part of L2 to degree at least L)

Then µM(M1,M2) > M (M1 is part of M2 to degree at least M)

B1

L2

B3

L2

µB (B3,B1) =

B ≥ .25

µL (L2,L2) =

L ≥ 1

Rough Mereology

C1 C4µM (C4,C1) =

M ≥ .28

(A) µ(x,y) Є [0,1]

(B) µ(x,x) = 1

(C) If µ(x,y) = 1 then µ(z,y) > µ(z,x) for each object z

A null object is any object n which satisfies

(D) µ(n,y) = 1 for every object y

Some Properties of µ

We wish to learn functions f from a set of vectors.

B

L

Mf

B1

B2

B3 . . .

Bn

L1

L2

L3 . . .

Ln

M1

M2

M3

.

.

. Mn

Back to the Problem at Hand

To describe system characteristics by providing examples of systems that exhibit those characteristics.

To determine system cost by providing examples of systems whose design, maintenance and overall costs are known.

Suppose we know the following:

•Maintenance requirements Maintk and Maintq similar to degree at least Maint , possibly k=q

•Cost1 and Cost2 , respectively, of the two systems O(Design1, Maint1) and O(Design2, Maint2) similar by at least Cost.

•Specs Designi and Designj similar to degree at least Design , i,j not necessarily distinct

We wish to learn function f from a set of vectors.

Maint

Design

Costf

Design1

Design2

Design3

. . .

Designn

Maint1

Maint2

Maint3

. . .

Maintn

Cost1

Cost2

Cost3

.

. . Costn

Table 1: Criteria Inferred from Application Data

Design Goal SYS1 SYS2 SYS3ResponseTimeThroughputMemoryRobustnessReliabilityAvailability

FaultToleranceSecuritySafetyUtilityUsability

slowlow

smallpoorlittlelowpoorpoorpoorpooreasy

fasthigh

mediumaveragemarginalmoderateaverageaverageaverageaverage

moderate

fasthighlargegoodgreathighgoodgoodgoodgood

difficultAcceptable? no yes no

(ResponseTime, slow) and (Throughput, low) (Acceptable, no),

(ResponseTime, fast) and (Memory, medium) (Acceptable, yes),

(Throughput, high) and (Memory, large) (Acceptable, no).

Uncertain (or possible) rules are:(ResponseTime, fast) and (Throughput, high)

(Acceptable, yes),(ResponseTime, fast) and (Throughput, high)

(Acceptable, no).

Table 3: Rough Inclusion for Table 1

System SYS1 SYS2 SYS3SYS1 1 0 0SYS2 0 1 .18SYS3 0 .18 1

Table 2: Criteria Dictated by the Customer

Cost SYS1 SYS2 SYS3DevelopmentDeploymentUpgrade

MaintenanceAdministration

lowlowhighhighlittle

moderatehighlowhighgreat

highhighlowlowgreat

Acceptable? yes yes no

Table 4: Rough Inclusion for Table 2

System SYS1 SYS2 SYS3SYS1 1 .2 0SYS2 .2 1 .6SYS3 0 .6 1

Table 5: Maintenance Criteria

Maintenance SYS1 SYS2 SYS3ExtensibilityModifiabilityAdaptabilityPortabilityReadabilityTraceability

easyeasyeasyeasyeasyeasy

difficultmoderate

easyeasy

difficultmoderate

difficultdifficultdifficultdifficultdifficultdifficult

Acceptable? yes yes no

Table 6: Rough Inclusion for Table 5

System SYS1 SYS2 SYS3SYS1 1 .33 0SYS2 .33 1 .33SYS3 0 .33 1

D1 D2 M1 M2 C1 C2

S1S1S1S2S2S3S1S1S1S2S2S3S1S1S1S2S2S3S1S1S1S2S2S3S1S1S1S2S2S3S1S1S1S2S2S3

S1S2S3S2S3S3S1S2S3S2S3S3S1S2S3S2S3S3S1S2S3S2S3S3S1S2S3S2S3S3S1S2S3S2S3S3

S1S1S1S1S1S1S1S1S1S1S1S1S1S1S1S1S1S1S2S2S2S2S2S2S2S2S2S2S2S2S3S3S3S3S3S3

S1S1S1S1S1S1S2S2S2S2S2S2S3S3S3S3S3S3S2S2S2S2S2S2S3S3S3S3S3S3S3S3S3S3S3S3

S1S1S1S1S1S1S1S1S1S1S1S1S1S1S1S1S1S1S1S1S1S1S1S1S1S1S1S1S1S1S1S1S1S1S1S1

S1S1S1S1S1S1S1S1S1S1S1S1S1S1S1S1S1S1S1S1S1S1S1S1S1S1S1S1S1S1S1S1S1S1S1S1

Table 7: Partial List of Arguments for µ

D M C D M C

1001

.6711001

.67111001

.67

11111.2.2.2.2.2.2.4.4.4.4.4.4.4

111111111111111111

1001

.6711001

.6711001

.671

111111.6.6.6.6.61111111

111111111111111111

Table 8: Partial List of Arguments for

RM – ‘acceptable degree’

What? How?

? µ (X, Y) threshold vectors composition of objects agents message passing

What? How?

µ (X, Y) threshold vectors composition of objects agents message passing

Simplicity user – satisfaction learnability ease of use comprehensibility user - friendliness

Summary & Conclusions1. Our objective is to describe system characteristics such as user

friendliness by providing examples of systems that exhibit such characteristics.

2. The computer recognizes a pattern and generates rules for what a user friendly system, for example, would be.

3. This is possible because computers are able to provide imprecise solutions to problems.

4. We have demonstrated the feasibility of applying rough sets/rough mereology to the problem of systems requirements systems.