specification from examples julia johnson dept. of math & computer science laurentian university...
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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.
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